An introduction to Lebesgue Measure

and Fourier Analysis

Laurent W. Marcoux

December 20, 2022

 i

Preface to the Third Edition - September 2022

We are now at the third instance of the notes. As always, there may be typos,
and it is crucial that you analyse what is written and not just accept things without
thinking.
Thanks to M. Esipova, M.E. Gusak, J. Vendryes, Y. Wu and D. Yang for catching
some of the remaining typos/errors.

Preface to the Second Edition - July 2019

Welcome to the second edition, also known as “Release Candidate 1 ”. Doubtless

there are typos/(hopefully small) errors, and you are welcome to let me know if you
find some. Oh, and make sure that you read the last sentence of the Preface to the
First Edition...
I would like to thank K. Santone and A. Tiwary for catching a great number of
typos/errors for me.

Preface to the First Edition - January 2018

These notes are a work in progress, and - this being the “first edition” - they
are replete with typos. As of April 10, 2018, I have had the opportunity to look
over Chapters 1 - 4. That is not to say that they are mistake-free. It is, instead, an
admission that I simply haven’t had the chance to look over the remaining Chapters.
A student should approach these notes with the same caution he or she would
approach buzz saws; they can be very useful, but you should be thinking the whole
time you have them in your hands. Enjoy.
ii

The reviews are in!

From the moment I picked your book up until I laid it down I was
convulsed with laughter. Someday I intend reading it.
Groucho Marx

This is not a novel to be tossed aside lightly. It should be thrown with

great force.
Dorothy Parker

The covers of this book are too far apart.

Ambrose Bierce

I read part of it all the way through.

Samuel Goldwyn

Reading this book is like waiting for the first shoe to drop.
Ralph Novak

Thank you for sending me a copy of your book. I’ll waste no time
reading it.
Moses Hadas

Sometimes you just have to stop writing. Even before you begin.
Stanislaw J. Lec
That’s not writing, that’s typing.
Truman Capote
 Contents

1. Riemann integration 1
2. Lebesgue outer measure 16
3. Lebesgue measure 27
4. Lebesgue measurable functions 42
5. Lebesgue integration 57
6. Lp Spaces 83
7. Hilbert spaces 109
8. Fourier analysis - an introduction 123
9. Convolution 134
10. The Dirichlet kernel 159
11. The Féjer kernel 171
12. Which sequences are sequences of Fourier coefficients? 191
Bibliography 197
Index 199

iii
 1. RIEMANN INTEGRATION 1

1. Riemann integration

What the world needs is more geniuses with humility, there are so
few of us left.
Oscar Levant

1.1. In first year, we study the Riemann integral for functions defined on inter-
vals, taking values in R. The same circle of ideas can be greatly extended, as we
shall now see.

Note. Much of the theory contained in these notes applies equally well to
the setting of real- or complex-valued functions and vector spaces. When writing a
statement which remains valid in either context, we shall simply write K to denote
the base field. In other words, we shall always have K ∈ {R, C}.

1.2. Definition. Let X be a vector space over the field K. A seminorm on X

is a function ν ∶ X → R satisfying
(i) ν(x) ≥ 0 for all x ∈ X.
(ii) ν(κx) = ∣κ∣ ν(x) for all κ ∈ K and x ∈ X.
(iii) ν(x + y) ≤ ν(x) + ν(y) for all x, y ∈ X.
If ν(x) = 0 implies that x = 0, we say that ν is a norm on X, and refer to the
ordered pair (X, ν) as a normed linear space (nls).

1.3. Remarks.
(a) Typically we shall denote seminorms by ν or µ. When the seminorm is
known to be a norm, we shall more often write ∥ ⋅ ∥.
(b) Let ν be a seminorm on a vector space X. Let z ∈ X denote the zero vector.
It follows from condition (ii) above that

ν(z) = ν(0z) = 0 ν(z) = 0.

In other words, if ∥ ⋅ ∥ is a norm on X and x ∈ X, then ∥x∥ = 0 if and only

if x = 0.
(c) We leave it as an exercise for the reader to prove that condition (c) implies
that ∣ν(x) − ν(y)∣ ≤ ν(x − y) for all x, y ∈ X.
(d) In a standard abuse of terminology, we often refer to the normed linear
space X, equipped with the norm ∥ ⋅ ∥.
2 L.W. Marcoux Introduction to Lebesgue measure

1.4. Example. Let X be a compact, Hausdorff topological space. Let X =

C(X, K) ∶= {f ∶ X → K∣f is continuous}. Given y ∈ X, define
νy ∶ C(X, K) → R
f ↦ ∣f (y)∣.
It is left as an exercise for the reader to verify that νy is a seminorm on C(X, K) for
all y ∈ K.
More generally, if Ω ⊆ X, the map νΩ ∶ C(X, K) → R defined by
νΩ (f ) ∶= sup ∣f (x)∣
x∈Ω
defines a seminorm on C(X, K).
We leave it to the reader to verify that νΩ is a norm if and only if Ω is dense
in X. In that case, νΩ = νX . This norm is typically denoted by ∥ ⋅ ∥∞ , and is
referred to as the supremum norm or “sup norm”on X. Having said that, in
this course we shall be dealing with certain Banach spaces of equivalence classes
of functions, equipped with an “essential supremum norm”, and this norm is also
typically denoted by ∥ ⋅ ∥∞ . In order to distinguish between the two, in these notes
we shall denote the supremum norm on C(X, K) by ∥ ⋅ ∥sup .

1.5. The norm ∥ ⋅ ∥ on a nls X gives rise to a metric via the formula:
d∶ X×X → R
(x, y) ↦ ∥x − y∥.
We refer to this as the metric induced by the norm. (That this is indeed a
metric is left as an exercise to the reader.) When referring to metric properties of
X, it is understood that we are referring to the metric induced by the norm, unless
it is explicitly stated otherwise.
1.6. Definition. A normed linear space (X, ∥ ⋅ ∥) is said to be a Banach space
if (X, d) is a complete metric space, where d is the metric on X induced by the norm.

1.7. Examples.
(a) The motivating example is X = K itself, where the norm is given by the
absolute value function. Since (K, ∣ ⋅ ∣) is complete, it is a Banach space.
Of course, C is a one-dimensional Banach space over C, and a two-
dimensional Banach space over R.
(b) Let N ≥ 1 be an integer. For x = (xn )N N
n=1 ∈ K , we define three functions:
● ∥x∥1 ∶= ∣x1 ∣ + ∣x2 ∣ + ⋯ + ∣xN ∣;
● ∥x∥∞ ∶= max(∣x1 ∣, ∣x2 ∣, . . . , ∣xN ∣); and
1
● ∥x∥2 ∶= (∑N 2 2
n=1 ∣xn ∣ ) .
It is a routine exercise that ∥ ⋅ ∥1 and ∥ ⋅ ∥∞ define norms on KN , and that
KN becomes a Banach space when equipped with either of these norms.
It is a slightly more interesting exercise (left to the reader) to show that
N
(K , ∥ ⋅ ∥2 ) is also a Banach space. The standard proof that ∥ ⋅ ∥2 satisfies
 1. RIEMANN INTEGRATION 3

the triangle inequality as a function on KN relies on the Cauchy-Schwarz

Inequality. See Chapter 7 for more details.
(c) More generally, if N ≥ 1 is an integer and 1 < p < ∞, we may define
∥ ⋅ ∥p ∶ KN → R via
1
N p
∥(xn )N p
n=1 ∥p = ( ∑ ∣xk ∣ )
k=1

to obtain a norm on KN . The proof of the triangle inequality is somewhat

more delicate than in the above cases. We shall return to this in Chapter 6,
and in the assignments.
It is an interesting fact (which the reader is encouraged to prove) that
for x ∈ Kn ,
lim ∥x∥p = ∥x∥∞ .
p→∞

(d) We saw in Example 1.4 that C([0, 1], K) is a nls when equipped with the
norm
∥f ∥sup = sup ∣f (x)∣ = max ∣f (x)∣.
x∈[0,1] x∈[0,1]

It is clear that convergence of a sequence (fn )n in this norm to a function

f ∈ C([0, 1], K) is simply uniform convergence of (fn )∞ n=1 to f , as studied
in first-year calculus.
Once again, we leave it as an exercise for the reader to show that
(C([0, 1], K), ∥ ⋅ ∥sup ) is complete, and is thus a Banach space.
(e) With X = C([0, 1], K) as above, define
1
∥f ∥1 ∶= ∫ ∣f (x)∣dx.
0

Then (C([0, 1], K), ∥ ⋅ ∥1 ) is a nls, but it is not a Banach space. The details
are yet again left to the reader.
(f) Let (X, ∥ ⋅ ∥X ) and (Y, ∥ ⋅ ∥Y ) be normed linear spaces over the field K. Let
T ∶ X → Y be a K-linear map. Consider

∥T ∥ ∶= sup{∥T x∥Y ∶ x ∈ X, ∥x∥X ≤ 1}.

Set
B(X, Y) ∶= {T ∶ X → Y ∣ T is linear and ∥T ∥ < ∞}.
As we shall see in the assignments, B(X, Y) is a vector space over K, and
∥ ⋅ ∥ ∶ B(X, Y) → R defines a norm on B(X, Y), referred to as the operator
norm on B(X, Y). One can show that (B(X, Y), ∥ ⋅ ∥) is complete if and
only if (Y, ∥ ⋅ ∥Y ) is complete.
4 L.W. Marcoux Introduction to Lebesgue measure

1.8. Definition. Let (X, ∥ ⋅ ∥) be a Banach space, a < b ∈ R, and f ∶ [a, b] → X

be a function.
A partition of [a, b] is a finite set
P ∶= {a = p0 < p1 < ⋯ < pN = b}
for some integer N ≥ 1. The set of all partitions of [a, b] is denoted by P[a, b].
Given P as above, a finite set P ∗ = {p∗k ∶ 1 ≤ k ≤ N } satisfying pk−1 ≤ p∗k ≤ pk ,
1 ≤ k ≤ N is said to be a set of test values for the partition. We then define the
corresponding Riemann sum
N
S(f, P, P ∗ ) ∶= ∑ f (p∗k )(pk − pk−1 ).
k=1

1.9. Remarks.
(a) When (X, ∥ ⋅ ∥) = (R, ∣ ⋅ ∣), then this is the usual Riemann sum that one
studies in first-year calculus. In particular, in the case where 0 ≤ f (x) for
all x ∈ [a, b], we see that S(f, P, P ∗ ) estimates the area under the curve
y = f (x), x ∈ [a, b].
(b) Observe that in general,
N
1
S(f, P, P ∗ ) = ∑ λk f (p∗k ),
b−a k=1
pk − pk−1
where λk ∶= , 1 ≤ k ≤ N . Clearly λk ≥ 0 for all k, while ∑N
k=1 λk = 1.
b−a
1
Thus S(f, P, P ∗ ) is a convex combination of the f (p∗k )’s, and as
b−a
such, “averages” f over [a, b].

1.10. Example. Let X = C([−π, π], C), equipped with the supremum norm
∥ ⋅ ∥sup . Let 1 ≤ n ∈ N be a fixed integer and consider the function f ∶ [0, 1] → X given
by
[f (x)](θ) = e2πx sin (nθ) + cos x cos (nθ), θ ∈ [−π, π].
1 1
If P = {0, 10 , 2 , 1}, and if P ∗ = { 50
1 1 4
, 3 , 5 }, then
1
S(f, P, P ∗ ) = (eπ/25 sin (nθ) + cos(1/50) cos (nθ))( − 0)
10
1 1
+ (e2π/3 sin (nθ) + cos(1/3) cos (nθ))( − )
2 10
1
+ (e8π/5 sin (nθ) + cos(4/5) cos (nθ))(1 − ).
2
 1. RIEMANN INTEGRATION 5

1.11. Definition. Let a < b be real numbers and (X, ∥ ⋅ ∥) be a Banach space.
We say that a function f ∶ [a, b] → X is Riemann integrable if there exists a vector
x0 ∈ X such that for any ε > 0, there exists a partition P ∈ P[a, b] with the property
that if Q is any refinement of P and Q∗ is any choice of test values for Q, then
∥x0 − S(f, Q, Q∗ )∥ < ε.

1.12. Remark. We note that if such a vector x0 as above exists, then it is

unique. Indeed, suppose that y0 ∈ X also satisfies the condition in Definition 1.11.
If y0 ≠ x0 , we let ε = ∥y0 − x0 ∥/2 > 0. Choose partitions P1 ∈ P[a, b] (resp. P2 ∈
P[a, b]) as in the definition of Riemann integrability corresponding to ε and x0 (resp.
corresponding to ε and y0 ). Let R = P1 ∪ P2 , so that R is a common refinement of
P1 and P2 .
If Q is any refinement of R and if Q∗ is any set of test values for Q, then –
noting that Q is again a common refinement of P1 and P2 – we see that
2ε = ∥y0 − x0 ∥
≤ ∥y0 − S(f, Q, Q∗ )∥ + ∥S(f, Q, Q∗ ) − x0 ∥
<ε+ε
= 2ε,
an obvious contradiction. Thus x0 = y0 , and the Riemann integral is unique.
When it exists, we refer to this unique vector x0 as the Riemann integral of
f over [a, b], and we write
b b
x0 = ∫ f =∫ f (s)ds.
a a

As is the case with the usual version of Riemann integration, the usefulness of
the Cauchy Criterion below is that it allows us to verify that a given function is
Riemann integrable without first having to know what its integral is.

1.13. Theorem. [The Cauchy Criterion for Riemann integrability.]

Let X be a Banach space, a < b be real numbers and f ∶ [a, b] → X be a function.
The following conditions are equivalent.
(a) f is Riemann integrable over [a, b].
(b) For all ε > 0 there exists a partition R ∈ P[a, b] with the property that if P
and Q are refinements of R, and if P ∗ and Q∗ are test values for P and Q
respectively, then
∥S(f, P, P ∗ ) − S(f, Q, Q∗ )∥ < ε.
Proof.
(a) implies (b). This is a standard argument which is left as an exercise for
the reader.
6 L.W. Marcoux Introduction to Lebesgue measure

(b) implies (a).

For each integer n ≥ 1, choose a partition Rn ∈ P[a, b] such that for all
refinements P, Q ⊇ Rn of Rn , and any associated choices of test values P ∗
and Q∗ we have
1
∥S(f, P, P ∗ ) − S(f, Q, Q∗ )∥ < .
n
∗
For each N ≥ 1, set WN ∶= ∪Nn=1 Rn and fix a choice WN of test values for
∗
WN . Define xn = S(f, Wn , Wn ), n ≥ 1.
If n2 ≥ n1 ≥ N ≥ 1, then
∥xn2 − xn1 ∥ = ∥S(f, Wn2 , Wn∗2 ) − S(f, Wn1 , Wn∗1 )∥
1 1
< ≤ ,
n1 N
as Wn1 and Wn2 are both refinements of RN .
From this it readily follows that (xn )∞ n=1 is a Cauchy sequence in X.
Since X is a Banach space and as such a complete metric space, we find
that
x ∶= lim xn
n→∞
b
exists in X. There remains to show that x = ∫a f .

To that end, let ε > 0 and choose N > 0 such that

1 ε
(i) < , and
N 2
ε
(ii) k ≥ N implies that ∥x − xk ∥ < .
2
If V is a refinement of WN , then it is also a refinement of RN , and hence
for any choice V ∗ of test values for V ,
∥x − S(f, V, V ∗ )∥ ≤ ∥x − xN ∥ + ∥xN − S(f, V, V ∗ )∥
1
< + ∥S(f, WN , WN∗ ) − S(f, V, V ∗ )∥
N
ε 1
< +
2 N
ε ε
< +
2 2
= ε.
b
By definition, x = ∫a f , and f is Riemann integrable over [a, b].
◻

1.14. Theorem. Let (X, ∥ ⋅ ∥) be a Banach space and a < b ∈ R. If f ∶ [a, b] → X

is continuous, then f is Riemann integrable over [a, b].
Proof. The proof is a routine adaptation of the fact that every continuous, real-
valued function on [a, b] is Riemann integrable.
 1. RIEMANN INTEGRATION 7

Since f is continuous on the compact set [a, b], and since X is a metric space,
we see that f is uniformly continuous on [a, b]. Let ε > 0 and choose δ > 0 such that
ε
x, y ∈ [a, b] and ∣x − y∣ < δ implies that ∥f (x) − f (y)∥ < .
2(b − a)
Let R = {a = r0 < r1 < r2 < ⋯ < rN = b} ∈ P[a, b] be a partition of [a, b] such that
rj − rj−1 < δ for all 1 ≤ j ≤ N , and let R∗ = {rj∗ }N
j=1 be a set of test values for R.
Suppose that P = {pi }i=0 is any refinement of R, and that P ∗ = {p∗i }M
M
i=1 is a set
of test values for P . Then we can find a sequence 0 = k0 < k1 < k2 < ⋯ < kN ∶= M
such that
pkj = rj , 1 ≤ j ≤ N.
In other words,
P = {a = r0 = p0 = pk0 < p1 < ⋯ < pk1 = r1 < pk1 +1 < ⋯ < pk2 = r2 < ⋯ < pkN = rN = b}.
Now S(f, R, R∗ ) = ∑N ∗ ∗ M ∗
j=1 f (rj )(rj − rj−1 ). while S(f, P, P ) = ∑i=1 f (pi )(pi − pi−1 ).
But then
N
S(f, R, R∗ ) = ∑ f (rj∗ )(rj − rj−1 )
j=1
N kj
∗
= ∑ f (rj ) ∑ (pi − pi−1 )
j=1 i=kj−1 +1
N kj
=∑ ∑ f (rj∗ )(pi − pi−1 ),
j=1 i=kj−1 +1

while
N kj
S(f, P, P ∗ ) = ∑ ∑ f (p∗i )(pi − pi−1 ).
j=1 i=kj−1 +1

But for kj−1 + 1 ≤ i ≤ kj , we have that rj−1 ≤ rj∗ , p∗i ≤ rj , and therefore ∣rj∗ − p∗i ∣ ≤
rj − rj−1 < δ. From our estimate above, we see that

N kj
∥S(f, R, R∗ ) − S(f, P, P ∗ )∥ = ∥ ∑ ∗ ∗
∑ (f (rj ) − f (pi ))(pi − pi−1 )∥
j=1 i=kj−1 +1

N kj
≤∑ ∑ ∥f (rj∗ ) − f (p∗i )∥(pi − pi−1 )
j=1 i=kj−1 +1

N kj
ε
<∑ ∑ (pi − pi−1 )
j=1 i=kj−1 +1 2(b − a)

N j k
ε
= ∑ ∑ (pi − pi−1 )
2(b − a) j=1 i=kj−1 +1
8 L.W. Marcoux Introduction to Lebesgue measure

M
ε
∑(pi − pi−1 )
=
2(b − a) i=1
ε
= (pM − p0 )
2(b − a)
ε
= .
2
As such, if Q is any other refinement of R and if Q∗ is any set of test values for
Q, then
ε
∥S(f, Q, Q∗ ) − S(f, R, R∗ )∥ < ,
2
whence
∥S(f, Q, Q∗ ) − S(f, P, P ∗ )∥ ≤ ∥S(f, Q, Q∗ ) − S(f, R, R∗ )∥+
∥S(f, R, R∗ ) − S(f, P, P ∗ )∥
ε ε
< +
2 2
= ε.
By the Cauchy Criterion 1.13, f is Riemann integrable.
◻
We provide a second proof of the above result in the Appendix to this section.

We now turn our attention to real-valued functions, with the intention of show-
ing what one of the failings of the Riemann integral is, and thereby (hopefully)
motivating the study of the Lebesgue integral in the forthcoming chapters.

1.15. Example. Given a subset E ⊆ R of the real numbers, we define the

characteristic function (or indicator function) of E to be
χE ∶ R → R
⎧
⎪1 if s ∈ E
⎪
s ↦ ⎨
⎪
⎪ 0 if s ∈/ E.
⎩
Consider E = Q ∩ [0, 1]. We claim that the Riemann integral
1
∫ χE (s)ds
0
does not exist.

Indeed, observe that E and [0, 1] ∖ E are each dense in [0, 1]. Let
P = {0 = p0 < p1 < p2 < ⋯ < pN = 1}
be any partition of [0, 1]. For 1 ≤ n ≤ N , choose p∗n ∈ Q ∩ [pn−1 , pn ], and choose
qn∗ ∈ Qc ∩ [pn−1 , pn ], so that P ∗ ∶= {p∗1 , p∗2 , . . . , p∗N } and Q∗ ∶= {q1∗ , q2∗ , . . . , qN
∗
} are
 1. RIEMANN INTEGRATION 9

both sets of test values for P . Then

N
S(χE , P, P ∗ ) = ∑ f (p∗n )(pn − pn−1 )
n=1
N
= ∑ 1(pn − pn−1 )
n=1
= pN − p0
=1−0
= 1,
while
N
S(χE , P, Q∗ ) = ∑ f (qn∗ )(pn − pn−1 )
n=1
N
= ∑ 0 (pn − pn−1 )
n=1
= 0.
It follows that for any choice of 0 < ε < 1, the Cauchy Criterion fails for χE , and as
such χE is not Riemann integrable over [0, 1].

1.16. Remark. Recall that Q and thus E ∶= Q ∩ [0, 1] is denumerable. Write

E = {qn }∞
n=1 , and define En ∶= {q1 , q2 , . . . , qn }, n ≥ 1. With fn ∶= χEn , n ≥ 1, we find
that
0 ≤ f1 ≤ f2 ≤ f3 ≤ ⋯ ≤ χE .
In fact, for each s ∈ [0, 1], we find that
χE (s) = lim fn (s).
n→∞

We say that the sequence (fn )∞

is increasing and that it converges pointwise
n=1
to χE .
Since each fn is continuous except at finitely many points in the interval [0, 1]
(in fact it is constantly equal to zero except at finitely many points), it is routine to
verify that each fn is Riemann integrable and that
1
∫ fn (s)ds = 0,
0
n ≥ 1. Nevertheless,
1 1 1
0 = lim ∫ fn (s)ds ≠ ∫ ( lim fn )(s)ds = ∫ χE (s)ds,
n→∞ 0 0 n→∞ 0
as the latter quantity does not exist.
We now seek to develop a more flexible and “forgiving” integral which will correct
such “pathological” behaviour.
10 L.W. Marcoux Introduction to Lebesgue measure

1.17. A heuristic approach. Whereas the Riemann integral partitions the

domain of a function f ∶ [a, b] → R into intervals, and associates to each such parti-
tion a step function, our new approach will partition the range of f into subintervals
[yk−1 , yk ], 1 ≤ k ≤ N . We then set Ek = {x ∈ [a, b] ∶ f (x) ∈ [yk−1 , yk ]}, 1 ≤ k ≤ N .
b
This allows us to estimate “∫a f ” by so-called simple functions
N
∑ yk mEk ,
k=1
where mEk denotes the “measure” (a generalization of length) of Ek . Since Ek need
not be a particularly nice set (one need only consider f = χQ∩[0,1] as above), our
first goal is to make sense of the “measure” or “length” of as many subsets of R as
possible.
 1. RIEMANN INTEGRATION 11

Appendix to Section 1.
1.18. There are more notions of integration in a Banach space than just Rie-
mann integration. Indeed, in general, there exist more notions of integration than
one can shake a stick at, even if one has strong arms, a very light stick, ample
dexterity and a solid and enviable history of stick-shaking.
This notion, however, will prove sufficient for our purposes. In Theorem 1.14,
we showed that every continuous function from a closed, bounded interval in R to a
Banach space is Riemann integrable over that closed interval. A minor modification
will show that every piecewise continuous, bounded function from a closed interval
to a Banach space is Riemann integrable in the sense of Definition 1.11.

1.19. Culture. The notion of integrating in a Banach space is not simply some
arcane and useless generalization of integration of real- or complex-valued functions.
Let 1 ∈ n be an integer, and let H ∶= Cn , equipped with the Euclidean norm ∥ ⋅ ∥2 .
Consider (B(Cn ), ∥ ⋅ ∥), where ∥ ⋅ ∥ denotes the operator norm on B(H) = B(Cn ).
We leave it as an exercise for the reader to show that every linear map T ∶ Cn →
C satisfies ∥T ∥ < ∞, and thus we may identify B(Cn ) with Mn (C), the space of
n

n × n matrices with entries in C with respect to a fixed, orthonormal basis {ek }nk=1
for Cn .
Let d1 , d2 , . . . , dn ∈ C be n distinct points, and let D ∈ B(Cn ) be the unique
linear operator whose corresponding matrix is the diagonal matrix
⎡d1 0 ... 0 ⎤⎥
⎢
⎢ d 0 . . . 0 ⎥⎥
⎢ 2
⎢ ⎥
⎢ ⋱ ⎥.
⎢ ⎥
⎢
⎢ d n−1 0 ⎥
⎥
⎢ d n⎦
⎥
⎣
Suppose that ∅ ≠ ∆ ⊆ {1, 2, . . . , n}.
Let Γ ⊆ C be a piecewise smooth curve in C for which
⎧
⎪1 if k ∈ ∆
⎪
ind(Γ, dk ) = ⎨
⎪
⎪ 0 otherwise.
⎩
It can be shown that
1 −1
P ∶= ∫ (sI − D) ds
2πi Γ
is the orthogonal projection P = diag(p1 , p2 , . . . , pn ), where pk = 1 if k ∈ ∆ and pk = 0
otherwise.
The astute reader will have observed the striking similarity of the integral above
to Cauchy’s Integral Formula. This is not a coincidence. Such integrals are studied
in much greater generality in the theory of Banach algebras.
12 L.W. Marcoux Introduction to Lebesgue measure

As promised, here is a second proof of Theorem 1.14.

1.20. Theorem. (Theorem 1.14 revisited). Let (X, ∥ ⋅ ∥) be a Banach space and
a < b ∈ R. If f ∶ [a, b] → X is continuous, then f is Riemann integrable over [a, b].
Proof. We shall in fact show that if PN ∈ P([a, b]) is a regular partition of [a, b]
b−a
into 2N subintervals of equal length N , and if PN∗ = PN ∖ {a} is the set of test
2
values for PN consisting of “right-hand endpoints” of the subintervals of PN , then
the sequence (S(f, PN , PN∗ ))∞ N =1 converges in X to
b
∫ f (s) ds.
a
We begin by showing that the sequence (S(f, PN , PN∗ ))∞
N =1 is Cauchy, and there-
fore converges to something in X, which we temporarily designate by y.

Since f is continuous on the compact set [a, b], and since X is a metric space,
we see that f is uniformly continuous on [a, b]. Let ε > 0 and choose δ > 0 such that
ε
x, y ∈ [a, b] and ∣x − y∣ < δ implies that ∥f (x) − f (y)∥ < .
b−a
b−a
For each N ≥ 1, let PN be as above, and choose M ≥ 1 such that M < δ. If
2
K ≥ L ≥ M , then PM ⊆ PL ⊆ PK ; indeed, writing
PL = {a = p0 < p1 < ⋯ < p2L = b},
and
PK = {a = q0 < q1 < ⋯ < q2K = b},
and setting p∗j
= pj , 1 ≤ j ≤ 2L and qs∗ = qs , 1 ≤ s ≤ 2K , we see that pj = qj 2K−L for all
0 ≤ j ≤ 2L , and that for 1 ≤ j ≤ 2L ,
ε
∥f (p∗j ) − f (qs∗ )∥ < , (j − 1) 2K−L < s ≤ j 2K−L .
b−a
Thus
X
X
X 2L j 2K−L X
X
X
∥S(f, PL , PL∗ ) − S(f, PK , PK ∗ X X
)∥ = X
X
X
X ∑ ∑ (f (p j ) − f (q s ))(q s − q s−1 X
) X
X
X
X
X X
X
Xj=1 s=(j−1) 2K−L +1
X X
X
2L j 2K−L
≤∑ ∑ ∥f (pj ) − f (qs )∥(qs − qs−1 )
j=1 s=(j−1) 2K−L +1

2L j 2K−L
ε
≤∑ ∑ (qs − qs−1 )
j=1 s=(j−1) 2K−L +1
b−a
K
ε 2
= ∑ qs − qs−1
b − a s=1
ε
= (b − a) = ε.
b−a
 1. RIEMANN INTEGRATION 13

So if we set yN = S(f, PN , PN∗ ), N ≥ 1, then the above argument shows that (yN )∞
N =1
is a Cauchy sequence in X, and as such, admits a limit y ∈ X.
b
There remains to show that y = ∫a f (s) ds. The proof is almost identical to that
above.
b−a
With ε > 0, choose T ≥ 1 such that T < δ and also such that
2
∥y − S(f, PT , PT∗ )∥ < ε.
Let R = {a = r0 < r1 < ⋯ < rJ = b} be any refinement of PT = {a = p0 < p1 < ⋯ < p2T =
b}. Thus there exists a sequence
0 = j0 < j1 < ⋯ < j2T = J
such that
rjk = pk , 0 ≤ k ≤ 2T .
Let R∗ be any set of test values for R. If jk−1 + 1 ≤ s ≤ jk , then ∣p∗k − rs∗ ∣ ≤ ∣pk − pk−1 ∣ =
b−a
< δ, 1 ≤ k ≤ 2T and so
2T
2T jk
∥S(f, PT , PT∗ ) − S(f, R, R∗ )∥ ≤ ∑ ∑ ∥f (p∗k ) − f (rs∗ )∥(rs − rs−1 )
k=1 s=jk−1 +1
T
jk
ε 2
< ∑ ∑ (rs − rs−1 )
b − a k=1 s=jk−1 +1
ε
= (b − a) = ε.
b−a
Thus
∥y − S(f, R, R∗ )∥ ≤ ∥y − S(f, PT , PT∗ )∥ + ∥S(f, PT , PT∗ ) − S(f, R, R∗ )∥ < ε + ε = 2ε.
This clearly shows that
b
y=∫ f (s) ds.
a
◻
14 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 1.

Exercise 1.1.
Let ∅ ≠ X be a compact, Hausdorff space. Prove that for each ∅ ≠ Ω ⊆ X, the
function
νΩ ∶ C(X, K) → R
f ↦ supx∈Ω ∣f (x)∣
defines a seminorm on C(X, K), and that it is a norm if and only if Ω is dense in X.

Exercise 1.2.
(a) Let ν be a seminorm on a vector space Y over K. Prove that
∣ν(x) − ν(y)∣ ≤ ν(x − y) for all x, y ∈ Y.
(b) Let (X, ∥ ⋅ ∥) be a nls. Prove that the map
d∶ X×X → R
(x, y) ↦ ∥x − y∥
defines a metric on X.

Exercise 1.3.
Let N ≥ 1 be an integer. Define three functions from KN to R as follows: for
x = (xn )N N
n=1 ∈ K , we set
(a) ∥x∥1 ∶= ∣x1 ∣ + ∣x2 ∣ + ⋯ + ∣xN ∣;
(b) ∥x∥∞ ∶= max(∣x1 ∣, ∣x2 ∣, . . . , ∣xN ∣); and
1
(c) ∥x∥2 ∶= (∑N 2 2
n=1 ∣xn ∣ ) .
Prove that each of these functions defines a norm on KN .

Exercise 1.4.
Let N ≥ 1 be an integer, and let x ∈ KN . Prove that
lim ∥x∥p = ∥x∥∞ .
p→∞

Exercise 1.5.
(a) Prove that the nls (C([0, 1], K), ∥⋅∥sup ) is complete, and that it is therefore
a Banach space.
1
(b) Recall that ∥f ∥1 ∶= ∫0 ∣f (x)∣dx defines a norm on C([0, 1], K). Prove that
(C([0, 1], K), ∥ ⋅ ∥1 ) is not complete.

Exercise 1.6.
Let X be a Banach space, a < b be real numbers and f ∶ [a, b] → X be a function.
Suppose that f is Riemann integrable over [a, b].
 1. RIEMANN INTEGRATION 15

Prove that for all ε > 0 there exists a partition R ∈ P[a, b] with the property
that if P and Q are refinements of R, and if P ∗ and Q∗ are test values for P and Q
respectively, then
∥S(f, P, P ∗ ) − S(f, Q, Q∗ )∥ < ε.

Exercise 1.7.
Let X be a Banach space, a < b be real numbers, and g ∶ [a, b] → X be a bounded,
piecewise-continuous function. Show that g is Riemann integrable over [a, b].

Exercise 1.8.
1
Prove the claim from Remark 1.16, namely: ∫0 fn (s) ds = 0 for all n ≥ 1,
∞
where Q ∩ [0, 1] = {qn }n=1 , and where fn is the characteristic function of En ∶=
{q1 , q2 , . . . , qn }, n ≥ 1.

Exercise 1.9. This question will be used in Chapter 9.

Let (X, ∥ ⋅ ∥) be a Banach space, and let a < b ∈ R. Let g ∶ [a, b] → K and
f ∶ [a, b] → X be continuous, and set ∥f ∥sup ∶= sup{∥f (x)∥ ∶ x ∈ [a, b]}. Observe that
∥f ∥sup is finite as x ↦ ∥f (x)∥ is continuous on the compact set [a, b].
Then
b b
∥∫ f (x)g(x)dx∥ ≤ ∥f ∥sup ∫ ∣g(x)∣dx.
a a
16 L.W. Marcoux Introduction to Lebesgue measure

2. Lebesgue outer measure

If you want to know what God thinks of money, just look at the people
he gave it to.
Dorothy Parker

2.1. Our goal in this section is to define a “measure of length” for as many
subsets of R as possible. We would like our new notion to agree with our intuition
in the cases we know; for example, it seems reasonable to ask that our generalized
notion of “length” of a finite interval (a, b) should be (b − a) when a < b in R. We
shall therefore use this as our starting point, and we shall use this intuition to extend
our notion of “length” to a greater variety of sets by approximation.

For a ≤ b ∈ R, we define the length of (a, b) to be b − a, and we write

ℓ((a, b)) ∶= b − a.
We also set ℓ(∅) = 0, and ℓ((−∞, b)) = ℓ((a, ∞)) = ℓ((−∞, ∞)) = ∞ for all a, b ∈ R.
In this way we have defined ℓ(I) whenever I is an open interval in R.

2.2. Definition. Let E ⊆ R. A countable collection {In }∞n=1 of open intervals

∞
is said to be a cover of E by open intervals if E ⊆ ∪n=1 In .
For each subset E of R, we define a quantity m∗ E ∈ [0, ∞] ∶= [0, ∞) ∪ {∞} as
follows:
∞
m∗ E ∶= inf{ ∑ ℓ(In ) ∶ {In }∞
n=1 a cover of E by open intervals}.
n=1

In order to help make the text more readable, and since these are the only covers
of sets we will consider in these notes, we shall abbreviate the expression “cover of E
by open intervals” to “cover of E”. For any set X, we denote by P(X) = {Y ∶ Y ⊆ X}
the power set of X.

2.3. Definition. Let ∅ ≠ X be a set. An outer measure µ on X is a function

µ ∶ P(X) → [0, ∞]
which satisfies
(a) µ∅ = 0;
(b) if E ⊆ F ⊆ X, then µE ≤ µF . We say that µ is monotone increasing;
and
(c) if Fn ⊆ X for all n ≥ 1, then
∞
µ(∪∞
n=1 Fn ) ≤ ∑ µ(Fn ).
n=1
 2. LEBESGUE OUTER MEASURE 17

It is worth noting that by virtue of (b), condition (c) is equivalent to condition:

(d) if E, F1 , F2 , F3 , . . . ⊆ X and if E ⊆ ∪∞
n=1 Fn , then
∞
µE ≤ ∑ µFn .
n=1
Condition (c) (or equivalently (d)) is generally referred to as the countable sub-
additivity or σ-subadditivity of µ.
2.4. Proposition. The function m∗ defined in Definition 2.2 is an outer mea-
sure on R.
Proof.
(a) Let E = ∅. With In = ∅, n ≥ 1, it is clear that {In }∞
n=1 is a cover of E, and
so
∞ ∞
0 ≤ m∗ ∅ ≤ ∑ ℓ(In ) = ∑ 0 = 0.
n=1 n=1
Thus m∗ ∅ = 0.
(b) Let E ⊆ F ⊆ R.
If {In }∞
n=1 is a cover of F , then it is also a cover of E. It follows
immediately from the definition that
m∗ E ≤ m∗ F.
(c) Suppose that {En }∞ n=1 is a countable collection of subsets of R. We wish
∞
to prove that m∗ (∪∞ ∗
n=1 En ) ≤ ∑n=1 m En .
∞
If ∑n=1 m∗ En = ∞, then we have nothing to prove. Thus we consider
the case where ∑∞ ∗ ∞
n=1 m En < ∞. Set E ∶= ∪n=1 En .
(n)
Let ε > 0 and for each n ≥ 1, choose a cover {Ik }∞ k=1 of En such that
∞
(n) ∗ ε
∑ ℓ(Ik ) < m En + .
k=1 2n
(n)
Then {Ik }∞k,n=1 is a cover of E, and so
∞ ∞
(n)
m∗ E ≤ ∑ ∑ ℓ(Ik )
n=1 k=1
∞
∗ ε
≤ ∑ (m En + )
n=1 2n
∞
= ( ∑ m∗ En ) + ε.
n=1
(Here we have used the fact that if 0 ≤ an ∈ R, n ≥ 1, and if σ ∶ N → N is
any permutation, then ∑∞ ∞
n=1 an = ∑n=1 aσ(n) .)
Since ε > 0 was arbitrary,
∞
m∗ E ≤ ∑ m∗ En .
n=1
18 L.W. Marcoux Introduction to Lebesgue measure

2.5. Corollary. Let E ⊆ R be a countable set. Then m∗ E = 0.

Proof. Suppose that E is denumerable, say E = {xn }∞ n=1 .
ε ε
Let ε > 0 and for each n ≥ 1, set In = (xn − 2n+1 , xn + 2n+1 ). Then {In }∞
n=1 is a
cover of E, and therefore
∞ ∞
ε
0 ≤ m∗ E ≤ ∑ ℓ(In ) = ∑ n
= ε.
n=1 n=1 2

Since ε > 0 was arbitrary, we have m∗ E = 0.

The case where E is finite is left as an exercise.
◻

2.6. Corollary. The outer measure of the rational numbers is 0; i.e. m∗ Q = 0.

We have defined outer measure m∗ E for any subset E of R, and we have done this
based upon an intuitive notion of what the length of an open interval (a, b) should
be, namely b − a. At first glance, it seems obvious that m∗ (a, b) = ℓ(b − a) = b − a.
But upon reflection, we see that this is not how m∗ (a, b) is defined. This leaves
us with an interesting problem: how does our notion of measure m∗ (a, b) of an
interval compare with this notion of length? On the one hand, it is clear that
m∗ (a, b) ≤ ℓ(a, b) = b − a, since I1 ∶= (a, b) and In = ∅, n ≥ 2 yields a cover of (a, b).
On the other hand, the notion of outer measure of (a, b) requires us to consider all
covers of (a, b) by intervals, not only the obvious cover by the interval (a, b) itself.
We now turn to this problem. It will prove useful to first consider the outer measure
of closed, bounded intervals [a, b], as these are compact. Because of this, we will be
able to replace general covers of [a, b] (by open intervals) with finite covers of [a, b]
(by open intervals).

2.7. Proposition. Let a < b ∈ R. Then

(a) m∗ [a, b] = b − a, and therefore
(b) m∗ (a, b] = m∗ [a, b) = m∗ (a, b) = b − a.
Proof.
(a) Let ε > 0 and note that with I1 = (a − 2ε , b + 2ε ) and In = ∅, n ≥ 2, the
collection {In }∞
n=1 is a cover of [a, b] by open intervals and thus
∞
m∗ [a, b] ≤ ∑ ℓ(In ) = ℓ(I1 ) = (b − a) + ε.
n=1

Since ε > 0 was arbitrary, m∗ [a, b] ≤ b − a.

We now turn the question of obtaining a lower bound for m∗ [a, b].
Suppose that {In }∞n=1 is an cover of [a, b] by open intervals. (Note:
without loss of generality, we may assume that In ≠ ∅, 1 ≤ n < ∞.) We
must show that ∑∞ n=1 ℓ(In ) ≥ b − a.
 2. LEBESGUE OUTER MEASURE 19

Since [a, b] is compact, we can find a finite subcover {I1 , I2 , . . . , IN } of

[a, b]. If ℓ(In ) = ∞ for some 1 ≤ n ≤ N , then the inequality
∞ N
∑ ℓ(In ) ≥ ∑ ℓ(In ) ≥ b − a
n=1 n=1
trivially holds. Thus we shall assume that ℓ(In ) < ∞ for all 1 ≤ n ≤ N , and
we may then write In = (an , bn ), 1 ≤ n ≤ N .
Since a ∈ [a, b] ⊆ ∪N n=1 In , we can find 1 ≤ n1 ≤ N such that a ∈ In1 =
(an1 , bn1 ). If bn1 > b, we stop.
Otherwise, a < bn1 ≤ b, so bn1 ∈ [a, b] ⊆ ∪N n=1 In , and we can find n2 ∈
{1, 2, . . . , N } ∖ {n1 } such that bn1 ∈ (an2 , bn2 ). If bn2 > b, we stop.
Otherwise, an1 < a < bn1 < bn2 ≤ b, so bn2 ∈ [a, b] ⊆ ∪N n=1 In , and we can
find n3 ∈ {1, 2, . . . , N } ∖ {n1 , n2 } such that bn2 ∈ (an3 , bn3 ). If bn3 > b, we
stop.
Eventually this process must end, since we have only N < ∞ intervals,
and each stage nk is chosen from among {1, 2, . . . , N } ∖ {n1 , n2 , . . . , nk−1 }.
Suppose therefore that 1 ≤ M ≤ N is the minimal integer such that bnM > b.
Then
[a, b] ⊆ ∪M
k=1 (ank , bnk ).

Figure 1. An example where M = 6.

Now,
∞ N
∑ ℓ(In ) ≥ ∑ ℓ(In )
n=1 n=1
M
≥ ∑ ℓ((ank , bnk ))
k=1
= (bn1 − an1 ) + (bn2 − an2 ) + ⋯ + (bnM − anM )
= bnM + (bnM −1 − anM ) + (bnM −2 − anM −1 ) + ⋯ + (bn1 − an2 ) − an1
> bnM − an1
> b − a.
Since {In }∞
n=1 was an arbitrary cover of [a, b], it follows that
m∗ [a, b] ≥ b − a.
Combining this with the reverse inequality above, we conclude that
m∗ [a, b] = b − a.
20 L.W. Marcoux Introduction to Lebesgue measure

(b) Consider the interval (a, b].

For all 0 < ε < b − a, we have that [a + ε, b] ⊆ (a, b] ⊆ [a, b], and thus
monotonicity of Lebesgue outer measure implies that
(b − a) − ε = m∗ [a + ε, b] ≤ m∗ (a, b] ≤ m∗ [a, b] = b − a.
Since ε was arbitrary (subject to 0 < ε < b − a), we conclude that m∗ (a, b] =
b − a.
The remaining cases are similar, and are left as an exercise for the
reader.
◻

2.8. Corollary. Let a, b ∈ R. Then

m∗ (−∞, b) = m∗ (−∞, b] = m∗ (a, ∞) = m∗ [a, ∞) = m∗ R = ∞.
Proof. Consider the interval (−∞, b).
By monotonicity of Lebesgue outer measure, for each n ≥ 1,
n = m∗ [b − n, b) ≤ m∗ (−∞, b),
and thus m∗ (−∞, b) = ∞.
The remaining cases are similar, and are left as an exercise for the reader.
◻

2.9. Corollary. Let a < b ∈ R. Then [a, b] is uncountable.

Proof. This follows immediately from Proposition 2.7 and Corollary 2.5.
◻

2.10. Definition. Let µ be an outer measure on R. We say that µ is transla-

tion invariant if for all E ⊆ R and all κ ∈ R, we have that
µ(E + κ) = µE,
where E + κ ∶= {x + κ ∶ x ∈ E}.

2.11. Proposition. Lebesgue outer measure is translation-invariant on R.

Proof. Observe that {In }∞ n=1 is a cover of E (by open intervals) if and only if
∞
{In + κ}n=1 is a cover of E + κ (by open intervals).
Moreover, for any interval I = (a, b), where a < b, we have that
ℓ(I) = b − a = (b + κ) − (a + κ) = ℓ(I + κ),
while if I is of the form (−∞, b), (a, ∞) or (−∞, ∞), then ℓ(I) = ∞ = ℓ(I + κ).
Thus
∞
m∗ E = inf{ ∑ ℓ(In ) ∶ {In }∞
n=1 a cover of E by open intervals}
n=1
∞
= inf{ ∑ ℓ(In + κ) ∶ {In + κ}∞
n=1 a cover of E + κ by open intervals}
n=1
= m∗ (E + κ).
 2. LEBESGUE OUTER MEASURE 21

2.12. Countable subaddivity of Lebesgue outer measure guarantees that if En

is a subset of R for all n ≥ 1 and if E = ∪∞
n=1 En , then
∞
m∗ E ≤ ∑ m∗ En .
n=1
Of course, we can not expect equality – for example, we might have En = [0, 1] for
all n ≥ 1, in which case E ∶= ∪∞
n=1 En = [0, 1] and
∞
m∗ E = m∗ [0, 1] = 1 < ∞ = ∑ m∗ En .
n=1
Clearly the issue in this example is the fact that the sets En are not disjoint. If
A = [0, 2] ∪ [7, 11], then our intuition tells us that if outer measure is going to be
a reasonable notion of “length”, then we should have m∗ A = 2 + 4 = 6. In fact, we
can prove directly that this is the case (we shall see a more general version of this
in the Assignments).
From this, we might be tempted to conjecture that given a disjoint collection
{En }∞
n=1 of subsets of R, we have that
∞
m∗ (⊍∞ ∗
n=1 En ) = ∑ m En .
n=1
As we shall now discover, this fails spectacularly. The failure of this equality leads
us to the notion of non-measurable sets, which will be the subject of the next section
of the notes.
2.13. Theorem. There does not exist a translation-invariant outer measure µ
on R satisfying the conditions:
(a) µ(R) > 0;
(b) µ[0, 1] < ∞; and
(c) µ is σ-additive: that is, if {En }∞
n=1 is a countable collection of disjoint
subsets of R, then with E ∶= ∪∞
n=1 n ,
E
∞
µE = ∑ µEn .
n=1
As a consequence, we see that Lebesgue outer measure m∗ is not σ-additive.
Proof. We shall argue by contradiction. Suppose, to the contrary, that there exists
such an outer measure µ.
Step 1. We consider the relation ∼ on R defined by x ∼ y if y − x ∈ Q. We leave it
as an exercise for the reader to show that ∼ is indeed an equivalence relation. For
each x ∈ R, denote the equivalence class of x under this relation by [x]. Clearly
[x] = {x + q ∶ q ∈ Q}.
Let F ∶= {[x] ∶ x ∈ R} = R/ ∼ denote the set of all equivalence classes of elements
in R. Observe that
(i) for any x ∈ R, [x] = Q + x is dense in R (since Q is dense in R), and
22 L.W. Marcoux Introduction to Lebesgue measure

(ii) given any x, y ∈ R, [x] = [y] if and only if y − x ∈ Q.

For each F ∶= [x] = Q + x ∈ F, we have that F is dense in R. This allows us to
choose a unique representative xF ∈ [x] ∩ [0, 1] of that equivalence class. Note that
to do this simultaneously for all F ∈ F requires the Axiom of Choice!
Of course, F = xF + Q.

Step 2. As is well-known, every equivalence relation on a set partitions that set

into disjoint equivalence classes, and so
R = ⊍{F = [xF ] ∶ F ∈ F}.
Let us define the set V ∶= {xF ∶ F ∈ F}. Then
R = ∪{V + q ∶ q ∈ Q}.
We shall refer to V as Vitali’s set, in honour of G. Vitali, who first “constructed”
it. As we shall see, it is interesting enough to merit this special notation.

Step 3. We claim that p ≠ q ∈ Q implies that V + p ∩ V + q = ∅, so that the sets

{V + q}q∈Q are in fact pairwise disjoint. Furthermore, we show that µV > 0. Indeed,
suppose that p ≠ q ∈ Q, and that z ∈ (V + p) ∩ (V + q). Then there exist F1 , F2 ∈ F
such that
z = xF1 + p = xF2 + q.
It follows that xF2 − xF1 = p − q ∈ Q, so that [xF1 ] = [xF2 ]. But V contained a unique
representative from each equivalence class, and therefore F1 = F2 , whence xF1 = xF2
and so p = q, a contradiction.
Thus
R = ⊍{V + q ∶ q ∈ Q}.
∞
Write Q = {qn }n=1 , which we may do as it is denumerable. Since µ is translation
invariant and σ-additive,
∞ ∞
0 < µ(R) = µ(⊍∞
n=1 (V + qn )) = ∑ µ(V + qn ) = ∑ µV.
n=1 n=1
Thus µV > 0, which in turn implies that µR = ∞.

Step 4. The set R ∶= Q ∩ [0, 1] is denumerable, and so we may write R = {rn }∞

n=1 .
Note that V ⊆ [0, 1] implies that V + rn ⊆ [0, 2] for all n ≥ 1.
Observe that [0, 2] = [0, 1] ∪ ([0, 1] + 1), and thus
µ[0, 2] ≤ µ[0, 1] + µ([0, 1] + 1) = 2µ[0, 1] < ∞.
Finally, our hypothesis that µ is σ-additive and translation invariant implies
that
∞ ∞
∞ = ∑ µV = ∑ µ(V + rn ) = µ(⊍∞
n=1 (V + rn )) ≤ µ[0, 2] < ∞,
n=1 n=1
a contradiction.
This contradiction proves that an outer measure µ on R satisfying all of the
above conditions can not exist.
◻
 2. LEBESGUE OUTER MEASURE 23

2.14. In light of the above Theorem, we are faced with a difficult choice. We
may either
(a) content ourselves with the notion of Lebesgue outer measure m∗ for all
subsets E of R, which agrees with our intuitive notion of length for in-
tervals, but in so doing we must sacrifice the highly-desirable property of
σ-additivity; or
(b) restrict the domain of our function m∗ to a more “tractable” family of
subsets of R, where we might in fact be able to prove that the restriction
of m∗ to this family is σ-additive.
The standard approach is the second, and it is the one we shall adopt. The
next section is devoted to describing our tractable family of sets, and to proving the
σ-additivity of the restriction of m∗ to this collection.
24 L.W. Marcoux Introduction to Lebesgue measure

Appendix to Section 2.
2.15. For those with a historical bent (and there are exciting new treatments
for that now), the “construction” of the set V defined in Step 2 of Theorem 2.13 is
due to the Italian mathematician Giuseppe Vitali [6]. Strictly speaking, this isn’t
an explicit construction. The Axiom of Choice was invoked to prove the existence
of such a set V.

In the next Chapter, we shall give a name to the “tractable” family of sets
described in Paragraph 2.14. Indeed, we shall refer to elements of this family as
“Lebesgue measurable sets”. The set V under discussion is an example of a non-
measurable set – see Exercise 2. More generally, however, a Vitali set is a set
B ⊆ [0, 1] for which x ≠ y ∈ B implies that that x − y ∈ R ∖ Q; i.e. B contains at most
one representative from each coset of Q in R/Q.

As we have just remarked, Vitali’s proof of the existence of a non-measurable

subset of R relied on the Axiom of Choice. In fact, it was shown by Robert Solovay [5]
that the Axiom of Choice is required to prove the existence of a non-measurable set,
in the sense that the assertion that every subset of R is measurable is consistent
with the Zermelo-Fraenkel axioms (ZF) of set theory. Rumour has it that his mother
never spoke to him again after that.

There are other examples of non-measurable sets, including Bernstein sets, of

which your humble author knows nothing. Those interested might wish to consult
John C. Oxtoby’s Measure and Category [4] for further details.
2.16. Theorem 2.13 raises an interesting question: is the issue with σ-additivity
due to the fact that we are considering infinitely many disjoint sets? In other
words, might it still be possible to find a translation-invariant outer measure µ on
R satisfying
(a) µ(R) > 0;
(b) µ[0, 1] < ∞; and
(c) µ is finitely-additive: that is, given {E1 , E2 , . . . , EN } disjoint subsets of
R, then with E ∶= ⊍Nn=1 En ,
N
µE = ∑ µEn ?
n=1

Suppose that such a measure µ exists. Let {En }∞ n=1 be a countable family of
pairwise disjoint sets. Since µ is an outer measure on R, it is countably subadditive,
and so
∞
µ(⊍∞
n=1 En ) ≤ ∑ µEn .
n=1
By monotonicity and finite-additivity of µ, we find that for all N ≥ 1,
N ∞
N ∞
∑ µEn = µ(⊍n=1 En ) ≤ µ(⊍n=1 En ) ≤ ∑ µEn .
n=1 n=1
 2. LEBESGUE OUTER MEASURE 25

Taking limits as N tends to infinity, we obtain:

∞ ∞
∞
∑ µEn ≤ µ(⊍n=1 En ) ≤ ∑ µEn ,
n=1 n=1
or equivalently, that
∞
µ(⊍∞
n=1 En ) = ∑ µEn .
n=1
In other words, such an outer measure µ would be countably additive, contradicting
Theorem 2.13.
26 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 2.

Exercise 2.1.
Let E ⊆ R be a finite set. Prove that m∗ E = 0.

Exercise 2.2.
Prove the remaining cases from Proposition 2.7 and Corollary 2.8. That is, prove
that if a < b ∈ R, then
m∗ [a, b) = m∗ (a, b) = b − a,
while
m∗ (−∞, b] = m∗ (a, ∞) = m∗ [a, ∞) = m∗ R = ∞.

Exercise 2.3.
Prove that
m∗ ([0, 2] ∪ [7, 11]) = 6.

Exercise 2.4.
Let ∼ be the relation on R defined by x ∼ y if y − x ∈ Q. Prove that ∼ is an
equivalence relation on R.

Exercise 2.5.
Let m∗ denote Lebesgue outer measure on R, and let V ⊆ [0, 1] denote Vitali’s
set from the proof of Theorem 2.13.
Prove that
m∗ V > 0.
 3. LEBESGUE MEASURE 27

3. Lebesgue measure

I can speak Esperanto like a native.

Spike Milligan

3.1. As mentioned at the end of the previous Chapter, our strategy will be to
restrict the domain of Lebesgue outer measure to a smaller collection of sets, where
Lebesgue outer measure will be σ-additive. We shall refer to this collection as the
collection of Lebesgue measurable sets.
We begin with Carathéodory’s definition of a Lebesgue measurable set, since it
is the most practical definition to use. Later we shall see that Lebesgue measurable
sets are “almost” countable intersections of open sets, in a way which we shall make
precise.
3.2. Definition. A set E is R is said to be Lebesgue measurable if, for all
X ⊆ R,
m∗ X = m∗ (X ∩ E) + m∗ (X ∖ E).
We denote by M(R) the collection of all Lebesgue measurable sets.

3.3. Remarks. Since our attention in this course is almost exclusively focused
upon Lebesgue measure, we shall allow ourselves to drop the adjective “Lebesgue”
and refer only to “measurable sets”.
Informally speaking, we see that a set E ⊆ R is measurable provided that it is
a “universal slicer ”, in the sense that it “slices” every other set X into two disjoint
sets, namely X ∩ E and X ∖ E, where Lebesgue outer measure is additive!
We also note that the inequality
m∗ X ≤ m∗ (X ∩ E) + m∗ (X ∖ E)
is free from the σ-subadditivity of Lebesgue outer measure. In checking to see
whether a given set is measurable or not, it therefore suffices to verify that the
reverse inequality holds for all sets X ⊆ R.

Before we proceed to the examples, which shall obtain a result which allows us
to show that the set M(R) of Lebesgue measurable sets itself has an interesting
structure.
3.4. Definition. Let Y be a non-empty set. A collection Ω ⊆ P(Y ) is said to
be an algebra of subsets of Y if
(a) Y ∈ Ω;
(b) E ∈ Ω implies that E c ∶= Y ∖ E ∈ Ω; and
(c) if N ≥ 1 and E1 , E2 , . . . , EN ∈ Ω, then
E ∶= ∪N
n=1 En ∈ Ω.
28 L.W. Marcoux Introduction to Lebesgue measure

We say that Ω is a σ-algebra of subsets of Y if Ω is an algebra of sets which

satisfies the additional property:
(d) if Fn ∈ Ω for all n ≥ 1, then
F ∶= ∪∞
n=1 Fn ∈ Ω.

Informally, we often say that Ω is a σ-algebra.

3.5. Theorem. The collection M(R) of Lebesgue measurable sets in R is a

σ-algebra of subsets of R.
Proof.
(a) Let us first verify that R ∈ M(R).
If X ⊆ R, then X ∩ R = X, while X ∩ Rc = X ∩ ∅ = ∅. Thus
m∗ (X ∩ R) + m∗ (X ∩ Rc ) = m∗ X + m∗ ∅ = m∗ X + 0 = m∗ X.
By definition, R ∈ M(R).
(b) The fact that M(R) is closed under complementation is clear, since the
definition of a measurable set is symmetric in E and R ∖ E.
(c) Let {En }∞ ∞
n=1 ⊆ M(R). We must show that E ∶= ∪n=1 En ∈ M(R).
Step 1. For each N ≥ 1, consider HN ∶= ∪N n=1 En . We shall argue by induction
that HN ∈ M(R) for all N ≥ 1.
For N = 1, this is trivially true, as H1 = E1 ∈ M(R) by hypothesis.
Next suppose that 1 ≤ M ∈ N and that HM ∈ M(R). Let X ⊆ R be
arbitrary. The induction hypothesis says that
m∗ X = m∗ (X ∩ HM ) + m∗ (X ∖ HM ).
But EM +1 ∈ M(R), and so
m∗ (X ∖ HM ) = m∗ ((X ∖ HM ) ∩ EM +1 ) + m∗ ((X ∖ HM ) ∖ EM +1 ).
By the subadditivity of m∗ ,
m∗ X = m∗ (X ∩ HM ) + m∗ (X ∖ HM )
= m∗ (X ∩ HM ) + m∗ ((X ∖ HM ) ∩ EM +1 )) + m∗ ((X ∖ HM ) ∖ EM +1 )
≥ m∗ (X ∩ (HM ∪ EM +1 )) + m∗ (X ∖ (HM ∪ EM +1 ))
= m∗ (X ∩ HM +1 ) + m∗ (X ∖ HM +1 )
Since the reverse inequality holds for any outer measure, we see that
m∗ X = m∗ (X ∩ HM +1 ) + m∗ (X ∖ HM +1 ),
and thus HM +1 ∈ M(R).
By induction, HN ∈ M(R) for all N ≥ 1.
Step 2. Next we shall write each HN as a disjoint union of sets in M(R).
Let F1 ∶= H1 , and for n ≥ 2, set Fn ∶= Hn ∖ Hn−1 . Clearly Fi ∩ Fj = ∅ for
1 ≤ i ≠ j, and HN = ⊍Nn=1 Fn for all N ≥ 1.
 3. LEBESGUE MEASURE 29

Let N ≥ 2 be an integer and note that HN −1 , HN ∈ M(R) implies that

c c c
HN −1 , HN ∈ M(R) by (b). By Step 1, (HN ∪ HN −1 ) ∈ M, and using (b)
once again,
c
FN = HN ∖ HN −1 = (HN ∪ HN −1 )c ∈ M(R).

Step 3. Now we claim that if X ⊆ R, then for each N ≥ 1,

N
m∗ (X ∩ (⊍N ∗
n=1 Fn )) = ∑ m (X ∩ Fn ).
n=1

The claim is trivially true when N = 1.

Suppose that 1 ≤ M ∈ N and that
M
m∗ (X ∩ (⊍M ∗
n=1 Fn )) = ∑ m (X ∩ Fn ).
n=1

By the measurability of FM +1 and the fact that all Fj ’s are disjoint,

m∗ (X ∩ (⊍M +1 ∗ M +1
n=1 Fn )) = m ((X ∩ (⊍n=1 Fn )) ∖ FM +1 )+

m∗ ((X ∩ (⊍M +1
n=1 Fn )) ∩ FM +1 )

= m∗ (X ∩ (⊍M ∗
n=1 Fn )) + m (X ∩ FM +1 )
M
= ∑ m∗ (X ∩ Fn ) + m∗ (X ∩ FM +1 )
n=1
by the induction hypothesis
M +1
= ∑ m∗ (X ∩ Fn ).
n=1

This completes the induction step and therefore proves our claim.
Step 4. Finally, observe that E = ∪∞ ∞ ∞
n=1 En = ∪n=1 Hn = ⊍n=1 Fn . We shall use
this to prove that E ∈ M(R).
Let X ⊆ R. For all N ≥ 1, HN ∈ M(R) and so
m∗ X = m∗ (X ∩ HN ) + m∗ (X ∖ HN )
= m∗ (X ∩ (⊍N ∗
n=1 Fn )) + m (X ∖ HN )

≥ m∗ (X ∩ (⊍N ∗
n=1 Fn )) + m (X ∖ E) as (X ∖ E) ⊆ (X ∖ HN )
N
= ∑ m∗ (X ∩ Fn ) + m∗ (X ∖ E) by Step 3.
n=1

Taking limits, we see that

∞
m∗ X ≥ ∑ m∗ (X ∩ Fn ) + m∗ (X ∖ E).
n=1
30 L.W. Marcoux Introduction to Lebesgue measure

Keeping in mind that m∗ is σ-subadditive, we note that

m∗ (X ∩ E) = m∗ (X ∩ (⊍∞
n=1 Fn ))
= m∗ (⊍∞
n=1 (X ∩ Fn ))
∞
≤ ∑ m∗ (X ∩ Fn ).
n=1
Combining these last two estimates, we conclude that
m∗ X ≥ m∗ (X ∩ E) + m∗ (X ∖ E),
and therefore that E ∈ M(R).
◻

It’s high time that we produce examples of Lebesgue measurable sets. Thanks
to the previous Theorem, given a subset S ⊆ M(R), the entire σ-algebra generated
by S (namely the smallest σ-algebra of subsets of R which contains S – why should
this exist?) is also contained in M(R).
3.6. Proposition.
(a) If E ⊆ R and m∗ E = 0, then E ∈ M(R).
(b) For all b ∈ R, E ∶= (−∞, b) ∈ M(R).
(c) Every open and every closed set is Lebesgue measurable.
Proof.
(a) Let E ⊆ R be a set with m∗ E = 0, and let X ⊆ R. By monotonicity of outer
measure,
m∗ (X ∩ E) ≤ m∗ E = 0,
and
m∗ (X ∩ E c ) ≤ m∗ X.
Thus
m∗ X = 0 + m∗ X ≥ m∗ (X ∩ E) + m∗ (X ∖ E).
As we have seen, this is the statement that E ∈ M(R).
(b) Fix b ∈ R and set E = (−∞, b). Let X ⊆ R be arbitrary. We must show that
m∗ X ≥ m∗ (X ∩ E) + m∗ (X ∖ E).
If m∗ X = ∞, then there is nothing to prove, and so we assume that
m X < ∞. Let ε > 0 and let {In }∞
∗
n=1 be a cover of X by open intervals such
that
∞
∗
∑ ℓ(In ) < m X + ε < ∞.
n=1
It follows that each interval In has finite length, and so we may write
In = (an , bn ), n ≥ 1. (As always, there is no harm in assuming that each
In ≠ ∅, otherwise we simply remove In from the cover.)
Set Jn = In ∩ E = (an , bn ) ∩ (−∞, b). Clearly each Jn , n ≥ 1 is an open
interval, possibly empty.
 3. LEBESGUE MEASURE 31

Let Kn = In ∖ E = (an , bn ) ∩ [b, ∞). Then

Kn ∈ {∅, [b, bn ), (an , bn )},

depending upon the values of an and bn . But then we can find cn < dn in
R such that
Kn ⊆ Ln ∶= (cn , dn )
and ℓ(Ln ) − m∗ Kn < ε
2n ,
n ≥ 1.
In particular, for each n ≥ 1, In ⊆ Jn ∪ Ln and
ε
(ℓ(Jn ) + ℓ(Ln )) − ℓ(In ) < .
2n
Now
∞
m∗ X > ( ∑ ℓ(In )) − ε
n=1
∞
ε
> ( ∑ (ℓ(Jn ) + ℓ(Ln ) − )) − ε
n=1 2n
∞ ∞ ∞
ε
= ∑ ℓ(Jn ) + ∑ ℓ(Ln ) − ∑ n
− ε.
n=1 n=1 n=1 2

But X ∩ E ⊆ ∪∞
n=1 Jn and X ∖ E ⊆ ∪∞
n=1 Ln , and so

m∗ X > m∗ (X ∩ E) + m∗ (X ∖ E) − 2ε.

Since ε > 0 was arbitrary,

m∗ X ≥ m∗ (X ∩ E) + m∗ (X ∖ E),

proving that E ∈ M(R).

(c) Fix b ∈ R. We have just seen that (−∞, b) ∈ M(R). Since M(R) is an
algebra of sets, E c = [b, ∞) ∈ M(R) as well. But then En ∶= [b + n1 , ∞) ∈
M(R) for all n ≥ 1, and since the latter is a σ-algebra,

(b, ∞) = ∪∞
n=1 En ∈ M(R) for all b ∈ R.

If a < b, then (a, b) = (−∞, b) ∩ (a, ∞) ∈ M(R). Since R ∈ M(R) by

Theorem 3.5, we see that we have shown that every open interval lies in
M(R).
We saw in the Assignments that every open set G ∈ G is a countable
(disjoint) union of open intervals. Since M is a σ-algebra, this means that
every open set G ∈ M. But M is also closed under complementation, and
so every closed set lies in M as well.
◻

Now that we know that M(R) ≠ ∅, the following definition makes sense.
32 L.W. Marcoux Introduction to Lebesgue measure

3.7. Definition. Let m∗ denote Lebesgue outer measure on R. We define

Lebesgue measure m on R to be the restriction of m∗ to M(R). That is, Lebesgue
measure is the function
m ∶ M(R) → [0, ∞]
E ↦ m∗ E.

Let us recall that our strategy was to try to restrict the domain of m∗ sufficiently
to allow the restriction of m∗ to this smaller collection of sets to be σ-additive. The
next result shows that M(R) may serve as such a domain.

3.8. Theorem. Lebesgue measure is σ-additive as a function on M(R). That

is, if En ∈ M(R) for all n ≥ 1 and if Ei ∩ Ej = ∅ for all 1 ≤ i ≠ j < ∞, then
∞
m(⊍∞
n=1 En ) = ∑ mEn .
n=1

Proof. We leave the proof of this to the Assignments.

◻

3.9. Corollary. There exist non-measurable sets.

Proof. Suppose otherwise. Then every subset of R is Lebesgue measurable, and
thus M(R) = P(R) and m = m∗ . But in Theorem 2.13, we saw that Lebesgue outer
measure m∗ is not σ-additive on P(R). This contradicts Theorem 3.8.
Thus M(R) ≠ P(R).
◻
More interestingly, we have the following result, which we leave to the exercises.
Recall Vitali’s set V from Theorem 2.13.

3.10. Proposition. Vitali’s set V is not Lebesgue measurable.

3.11. Definition. The σ-algebra of sets generated by the collection G = {G ⊆

R ∶ G is open} of all open subsets of R is called the σ-algebra of Borel subsets
of R, and is denoted by
Bor (R).
(That Bor (R) exists follows from Exercise 1 below.)
By Theorem 3.5 and Proposition 3.6 above, we see that
Bor (R) ⊆ M(R).
 3. LEBESGUE MEASURE 33

3.12. Remarks. Note that since Bor (R) is a σ-algebra which contains all
open subsets of R, it also contains all closed subsets of R. In fact, we could have
defined Bor (R) to be the σ-algebra of subsets of R generated by the collection
F ∶= {F ⊆ R ∶ F is closed} of closed subsets of R, and concluded that it would have
contained G.
Given a family A ⊆ P(R) with ∅, R ∈ A, set
Aσ ∶= {∪∞
n=1 An ∶ An ∈ A, n ≥ 1}
Aδ ∶= {∩∞
n=1 An ∶ An ∈ A, n ≥ 1}.
We refer to elements of Aσ as A-sigma sets and elements of Aδ as A-delta sets.
Observe that Gδ and Fσ are both subsets of Bor (R).

3.13. Admittedly, our definition of a Lebesgue measurable set is not the most
intuitive definition, and Carathéodory’s definition is quite different from Lebesgue’s
original definition, which we shall now investigate.

3.14. Theorem. Let E ⊆ R. The following statements are equivalent.

(a) E is Lebesgue measurable; i.e. E ∈ M(R).
(b) For every ε > 0 there exists an open set G ⊇ E such that
m∗ (G ∖ E) < ε.
(c) There exists a Gδ -set H such that E ⊆ H and
m∗ (H ∖ E) = 0.
In other words, up to a set of measure zero, every Lebesgue measurable set is
a Gδ -set. As we shall see in the assignments, up to a set of measure zero, every
Lebesgue measurable set is a Fσ -set as well.

Proof.
(a) implies (b).
Case 1. Suppose that mE < ∞.
Let ε > 0 and choose a cover {In }∞
n=1 of E by open intervals such that
∞
∑ ℓ(In ) < mE + ε.
n=1
Set G = ∪∞
n=1 In so that G is open and E ⊆ G. Then
∞ ∞
mG ≤ ∑ mIn = ∑ ℓ(In ) < mE + ε.
n=1 n=1
Also,
G = (G ∩ E) ∪ (G ∖ E) = E ⊍ (G ∖ E),
and so
mE + m(G ∖ E) = mG < mE + ε,
whence
m∗ (G ∖ E) = m(G ∖ E) < ε.
34 L.W. Marcoux Introduction to Lebesgue measure

Case 2. Suppose that mE = ∞.

Let ε > 0, and for each n ≥ 1, let En = E ∩[−n, n]. Note that En ∈ M(R),
as the latter is a σ-algebra of sets. Moreover, En ⊆ [−n, n] implies that
mEn ≤ m[−n, n] = 2n < ∞. From Case 1 above, we can find Gn and open
set such that En ⊆ Gn and m(Gn ∖ En ) < 2εn , n ≥ 1. Let G = ∪∞ n=1 Gn , so
that G is open, and
E = ∪∞ ∞
n=1 En ⊆ ∪n=1 Gn = G.
If x ∈ G ∖ E, then x ∈/ En for all n ≥ 1, and there exists N ≥ 1 such that
x ∈ GN . That is, x ∈ GN ∖EN . Thus G∖E ⊆ ∪∞ n=1 (Gn ∖En ) (in fact equality
is easily seen to hold), and so
∞ ∞
ε
m(G ∖ E) ≤ ∑ m(Gn ∖ En ) ≤ ∑ n
= ε.
n=1 n=1 2
(b) implies (c).
For each n ≥ 1, choose Gn ⊆ R open so that E ⊆ Gn , and m∗ (Gn ∖E) < n1 .
Set H ∶= ∩∞n=1 Gn , so that E ⊆ H, and H ∈ Gδ .
By monotonicity of outer measure, for each n ≥ 1 we have
1
m∗ (H ∖ E) ≤ m∗ (Gn ∖ E) < ,
n
and so
m∗ (H ∖ E) = 0.
(c) implies (a).
Suppose there exists H ∈ Gδ ⊆ M(R) such that E ⊆ H and m∗ (H ∖E) =
0. By Proposition 3.6, m∗ (H ∖ E) = 0 implies that H ∖ E ∈ M(R) with
m(H ∖ E) = 0. But M(R) is a σ-algebra of sets, and thus it is an algebra,
from which we deduce that
E = H ∖ (H ∖ E) ∈ M(R).
◻

3.15. Example. The Cantor middle thirds set. Recall from Corollary 2.5
that if E ⊆ R is countable, then m∗ E = 0. By Proposition 3.6, it follows that
E ∈ M(R), and thus mE = m∗ E = 0. In other words, every countable set is Lebesgue
measurable with Lebesgue measure zero.
We shall now construct an uncountable set C – in fact one whose cardinality is
c, the cardinality of the real line R – whose measure mC is equal to zero. Since
mR = ∞, we see that the Lebesgue measure of a set is not so much a reflection of
its cardinality, as much as a question of how the points in the set are distributed.
Having said that, when the set in question is countable, the above argument shows
that it is always “thinly distributed”, in this analogy.
The Cantor set is typically obtained as the intersection of a countable family
of sets, each iteratively constructed from the previous as follows:
We set C0 = [0, 1], and for n ≥ 1, we set Cn = 13 Cn−1 ∪ ( 32 + 13 Cn−1 ). Thus
 3. LEBESGUE MEASURE 35

● C1 = [0, 13 ] ∪ [ 32 , 1];
● C2 = [0, 91 ] ∪ [ 92 , 13 ] ∪ [ 96 , 79 ] ∪ [ 89 , 1],
1 2 3 6 7 8 9
● C3 = [0, 27 ]∪[ 27 , 27 ]∪[ 27 , 27 ]∪[ 27 , 27 ]∪[ 18 19 20 21 24 25 26
27 , 27 ]∪[ 27 , 27 ]∪[ 27 , 27 ]∪[ 27 , 1];
● ⋮
The figure below shows each of the sets Cn , 0 ≤ n ≤ 7.

Figure 2. an illustration from http://mathforum.org/mathimages/index.php/Cantor Set.

The Cantor middle thirds set is defined as the intersection of all of these
sets, i.e.
C ∶= ∩∞
n=0 Cn .

Alternatively, beginning with C0 = [0, 1], one can think of obtaining C1 from C0
by removing the (open) “middle third” interval ( 31 , 23 ), resulting in the two intervals
which comprise C1 above. To obtain C2 from C1 , one removes the (open) “middle
third” of each of the two intervals in C1 , and so on. This motivates the term middle
thirds in the above nomenclature.
It should be clear from the construction above that
(a) C0 ⊇ C1 ⊇ C2 ⊇ C3 ⊇ ⋯ ⊇ C. Furthermore, each set Cn is closed, n ≥ 0 and
mC0 = 1 < ∞. From our work in the assignments,
2n
m∗ C = lim mCn = lim = 0.
n→∞ n→∞ 3n

(b) Being closed and bounded, C is compact – hence measurable with mC =

0. Also, C0 is compact and the collection {Cn }∞
n=0 clearly has the Finite
Intersection Property. Thus C = ∩∞
n=0 C n ≠ ∅. (We shall in fact show that
C is uncountable!)

Now let us approach things from a different angle. Given x ∈ [0, 1], consider the
ternary expansion of x, namely
x = 0.x1 x2 x3 x4 . . .
where xk ∈ {0, 1, 2} for all k ≥ 1. As with decimal expansions, the expression above
is meant to express the fact that
∞
xk
x= ∑ k
.
k=1 3
36 L.W. Marcoux Introduction to Lebesgue measure

Non-uniqueness of this expansion is a problem here, as it is with decimal expansions.

For example,
1
= 0.0222222⋅ = 0.1000000⋯.
3
We leave it as an exercise for the reader to show that the expansion of x ∈ [0, 1)
is unique except when there exists N ≥ 1 such that
r
x= N for some 0 < r < 3N , where 3 ∤ r.
3
When this is the case, we have that
x = 0.x1 x2 x3 x4 ⋯xN ,
where xN ∈ {1, 2}.
If xN = 2, we shall use that expression.
If xN = 1, then
x = 0.x1 x2 x3 ⋯xN −2 xN −1 1000⋯
= 0.x1 x2 x3 ⋯xN −2 02222⋯,
and we shall agree to adopt the second expression.
Finally, we shall use the convention that
1 = 0.2222⋯.
With this convention, over x ∈ [0, 1] has a unique ternary expansion.
Now
● x ∈ C1 if and only if x1 ≠ 1;
● x ∈ C2 if and only if x ∈ C1 ∩ C2 , i.e. if and only if x1 ≠ 1 ≠ x2 .
● x ∈ C3 if and only if x ∈ C2 ∩ C3 , i.e., if and only if 1 ∈/ {x1 , x2 , x3 }.
More generally, for N ≥ 1, x ∈ CN if and only 1 ∈/ {x1 , x2 , . . . , xN }.
From this it follows that x ∈ C if and only if xn ≠ 1, n ≥ 1. In other words,
C = {x = 0.x1 x2 x3 x4 . . . ∶ xn ∈ {0, 2} for all n ≥ 1}.
As we shall see in the assignments, the map
φ∶ C → [0, 1]
x = 0.x1 x2 x3 x4 . . . ↦ y = ∑∞ yn
n=1 2n ,
where yn = x2n , n ≥ 1 is a surjection from C onto [0, 1]. (It is profitable to think of
y as the binary expansion of an arbitrary element of [0, 1].)
But then the cardinality ∣C∣ of C is greater than or equal to c = ∣[0, 1]∣, the
cardinality of the continuum. Since C ⊆ R, we also have that ∣C∣ ≤ c, and so the
Schröder-Bernstein Theorem (see Theorem 3.18 below) implies that
∣C∣ = c.
Thus C is an uncountable, measurable set whose Lebesgue measure is nonethe-
less equal to 0.
 3. LEBESGUE MEASURE 37

Appendix to Section 3.

3.16. With regards to Theorem 3.14 above, first note that the definition of
Lebesgue outer measure says that for all H ⊆ R,
∞
m∗ H = inf{ ∑ ℓ(In ) ∶ {In }∞
n=1 is a cover of H}.
n=1

Let ε > 0, H ⊆ R and choose a cover {In }∞ ∞

n=1 of H. Set G ∶= ∪n=1 In , so that G is
an open set in R. By the σ-subadditivity and monotonicity of m∗ ,
∞ ∞
m∗ H ≤ m∗ G ≤ ∑ m∗ (In ) = ∑ ℓ(In ) < m∗ H + ε.
n=1 n=1

It is important to realize that this does not say that

m∗ (G ∖ H) < ε.
For a trivial counterexample, one might take H = [0, ∞) and G = R. Then
m∗ H = m∗ G, and so m∗ H ≤ m∗ G + ε for all ε > 0, and yet
m∗ (G ∖ H) = m∗ (−∞, 0] = ∞.
A far more interesting example comes from Vitali’s set V. Recall that V ⊆ [0, 1]
is non-measurable, and we have seen (see Exercise 5) that it has positive, but finite
Lebesgue outer measure. Indeed, by monotonocity of Lebesgue outer measure,
0 < m∗ V ≤ 1.
The above construction shows that for each n ≥ 1, we can find Gn ⊆ R open such
that V ⊆ Gn and
1
0 ≤ m∗ Gn − m∗ V < .
n
The non-measurability of V, combined with Theorem 3.14, shows that there exists
ε0 > 0 such that if G ⊆ R is open and V ⊆ G, then
m∗ (G ∖ V) ≥ ε0 .
In particular,
m∗ (Gn ∖ V) ≥ ε0 for all n ≥ 1.
Oh my. Wicked. Very, very wicked indeed.

3.17. Carathéodory’s original definition of a Lebesgue measurable set E (see

Definition 3.2) asks that for every X ⊆ R, we must have
m∗ X = m∗ (X ∩ E) + m∗ (X ∖ E).
As it turns out (this will be an assignment question), if A ⊆ R is Lebesgue
measurable with mA < ∞, and if E ⊆ A, then the following are equivalent:
(a) E is Lebesgue measurable.
(b) m∗ A = m∗ (A ∩ E) + m∗ (A ∖ E).
38 L.W. Marcoux Introduction to Lebesgue measure

In other words, instead asking that m∗ be additive with respect to the decom-
position X = E ⊍ (X ∖ E) for every set X that contains E, it suffices to ask that
this condition holds in the single, solitary case where X = A!
In particular, this shows that if E is bounded (i.e. there exists M > 0 such that
E ⊆ [−M, M ]), then E is Lebesgue measurable if and only if
m∗ (E) + m∗ ([−M, M ] ∖ E) = 2M.
Suddenly Carathéodory’s definition of a Lebesgue measurable set doesn’t seem
so bad!

It might be worthwhile to remind the reader of the Schröder-Bernstein Theorem

and of its proof.

3.18. Theorem. The Schröder-Bernstein Theorem. Let A and B be

sets. If ∣A∣ ≤ ∣B∣ and ∣B∣ ≤ ∣A∣, then ∣A∣ = ∣B∣.
Proof.
Step 1. If Z is any set and φ ∶ P(Z) → P(Z) is increasing in the sense that
X ⊆ Y ⊆ Z implies that φ(X) ⊆ φ(Y ), then φ has a fixed point; that is, there
exists T ⊆ X such that φ(T ) = T .
Indeed, let T = ∪{X ⊆ Z ∶ X ⊆ φ(X)}. If X ⊆ Z and X ⊆ φ(X), then X ⊆ T and
so φ(X) ⊆ φ(T ). That is, X ⊆ Z and X ⊆ φ(X) implies X ⊆ φ(T ), and thus
T = ∪{X ⊆ Z ∶ X ⊆ φ(X)} ⊆ φ(T ).
But then φ(T ) ⊆ Z and φ(T ) ⊆ φ(φ(T )), so that φ(T ) is one of the sets appearing
in the definition of T - i.e. φ(T ) ⊆ T .
Together, these imply that φ(T ) = T . (We remark that it is entirely possible
that T = ∅.)

Step 2. Given sets A, B as above and injections κ ∶ A → B and λ ∶ B → A, define

φ ∶ P(A) → P(A)
X ↦ A / λ[B / κ(X)].
Suppose X ⊆ Y ⊆ A. Then κ(X) ⊆ κ(Y ).
Hence
B / κ(X) ⊇ B / κ(Y ), so
λ(B / κ(X)) ⊇ λ(B / κ(Y )), which implies
A / λ(B / κ(X)) ⊆ A / λ(B / κ(Y )), which in turn implies
φ(X) ⊆ φ(Y ).
Step 3. By Steps One and Two, there exists T ⊆ A such that T = φ(T ) =
A / λ[B / κ(T )].
 3. LEBESGUE MEASURE 39

Define
f∶ A → B
⎧
⎪
⎪κ(a) if a ∈ T ,
a ↦ ⎨ −1
⎪
⎪λ (a) if a ∈ A / T .
⎩
Observe that λ is a bijection between B / κ(T ) and A / T , and that κ is a
bijection between T and κ(T ), so that f is a bijection between A and B.
◻
40 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 3.

Exercise 3.1.
Let E ∈ M(R), so that E is a Lebesgue measurable set. Let κ ∈ R, and set
E + κ ∶= {x + κ ∶ x ∈ E}. Prove that E + κ ∈ M(R), and that mE = m(E + κ).
Thus Lebesgue measure m is translation-invariant.

Exercise 3.2.
Let V ⊆ [0, 1] be Vitali’s set from Theorem 2.13. Prove that V is not Lebesgue
measurable.

Exercise 3.3.
Let S ⊆ M(R).
Prove that there exists a σ-algebra N ⊆ M(R) of subsets which contains S and
with the property that if K is any σ-algebra of measurable sets which contains S,
then N ⊆ K. We say that N is the σ-algebra generated by S.

Exercise 3.4.
Let x ∈ [0, 1) and consider the ternary expansion of x given by
x = 0.x1 x2 x3 x4 . . . .
Show that the expansion is unique except when there exists N ≥ 1 such that
r
x= N for some 0 < r < 3N , where 3 ∤ r.
3

Exercise 3.5.
Prove that the cardinality ∣M(R)∣ of the collection of Lebesgue measurable sets
is equal to that of the collection P(R) ∖ M(R) of non-measurable sets.

Exercise 3.6. Assignment

Prove that every open subset G ⊆ R is a countable union of disjoint, open
intervals.

Exercise 3.7. Assignment

Prove that Lebesgue measure is σ-additive on M(R).

Exercise 3.8. Assignment

Let E ⊆ R. Prove that the following statements are equivalent.
(a) E is Lebesgue measurable.
(b) For every ε > 0 there exists a closed set F ⊆ E such that
m∗ (E ∖ F ) < ε.
(c) There exists an Fσ -set H such that H ⊆ E and
m∗ (E ∖ H) = 0.
 3. LEBESGUE MEASURE 41

Exercise 3.9. Assignment Question.

Suppose that {En }∞n=1 is an increasing sequence of Lebesgue measurable sets;
i.e.
E1 ⊆ E2 ⊆ E3 ⊆ ⋯.
∞
Let E = ∪n=1 En , so that E ∈ L(R), as the latter is a σ-algebra. Prove that
mE = lim mEn .
n→∞
42 L.W. Marcoux Introduction to Lebesgue measure

4. Lebesgue measurable functions

Don’t ever wrestle with a pig. You’ll both get dirty, but the pig will
enjoy it.
Cale Yarborough

4.1. Let H ⊆ R. We denote by M(H) the collection of all Lebesgue measurable

subsets of H. When E ∈ M(R), it follows that
M(E) = {F ∩ E ∶ F ∈ M(R)}.
We leave it as an exercise for the reader to prove that M(E) is a σ-algebra of subsets
of E.

4.2. Definition. Let (X, d) be a metric space and E ∈ M(R). A function

f ∶ E → X is said to be Lebesgue measurable if
f −1 (G) ∶= {x ∈ E ∶ f (x) ∈ G} ∈ M(E)
for every open set G ⊆ X.
We denote the set of Lebesgue measurable X-valued functions on E by L(E, X).

It is easily verified that this is equivalent to asking that f −1 (F ) ∈ M(E) for

every closed subset F of X.
In the definition above, we have insisted that the domain of the function be a
measurable set. Part of the reason for this is that (at the very least) we would want
the constant functions to be measurable, and this happens if and only if the domain
of our function is measurable.
As was the case with (Lebesgue) measurable sets, in light of the fact that we
are dealing almost exclusively with Lebesgue measure in these notes, we drop the
adjective “Lebesgue” and henceforth refer simply to measurable functions.

4.3. Proposition. Let E ⊆ R be a measurable set. Then every continuous

function f ∶ E → X is measurable.
Proof. To see this, note that the continuity of f implies that f −1 (G) is a (relatively)
open subset of E for all open sets G ⊆ R. But a set L ⊆ E is relatively open provided
that L = U ∩ E, where U ⊆ R is open. In our case, once we choose U0 ⊆ R open
such that f −1 (G) = U0 ∩ E, it follows from Theorem 3.5 and Proposition 3.6 that
U0 ∈ M(R), whence f −1 (G) ∈ M(E).
Thus f is measurable.
◻
 4. LEBESGUE MEASURABLE FUNCTIONS 43

4.4. Example. Let E ∈ M(R) and H ⊆ E. Consider the characteristic

function or indicator function of H, namely
χH ∶ E → R
⎧
⎪0
⎪ if x ∈ E ∖ H
x ↦ ⎨
⎪
⎪1 if x ∈ H.
⎩
If G ⊆ R is open, then
⎧
⎪∅ if G ∩ {0, 1} = ∅
⎪
⎪
⎪
⎪
⎪H
⎪ if G ∩ {0, 1} = {1}
χ−1
H (G) = ⎨
⎪
⎪
⎪E ∖ H if G ∩ {0, 1} = {0}
⎪
⎪
⎪
⎩E
⎪ if G ∩ {0, 1} = {0, 1}.
It follows that χH is Lebesgue measurable if and only if H is a Lebesgue mea-
surable set.

4.5. Proposition. Let E ∈ M(R), and suppose that (X, dX ) and (Y, dY ) are
metric spaces. Suppose furthermore that f ∶ E → X is measurable and that g ∶ X → Y
is continuous. Then g ○ f ∶ E → Y is measurable.
Proof. Let G ⊆ Y be open. Then g −1 (G) ⊆ X is open, as g is continuous. But f is
measurable, so
(g ○ f )−1 (G) = f −1 (g −1 (G)) ∈ M(E),
proving that g ○ f is measurable.
◻

4.6. Example. Let E ⊆ R be a measurable set and let f ∶ E → K be a measur-

able function. If g ∶ K → R is the function g(x) = ∣x∣, x ∈ K, then g is continuous and
so g ○ f = ∣f ∣ is measurable.
It is perhaps worth noting that the converse to this is false.

4.7. Proposition. Let E ∈ M(R) and f, g ∶ E → K be functions. The following

statements are equivalent.
(a) f and g are measurable.
(b) The map
h ∶ E → (K2 , ∥ ⋅ ∥2 )
x ↦ (f (x), g(x))
is measurable.
Proof.
(a) implies (b).
Let A, B ⊆ K be open sets. Then A × B = {(a, b) ∶ a ∈ A, b ∈ B} is open
in K2 , and furthermore, every open set G ⊆ C is a countable union of sets
of this form (see the exercises).
44 L.W. Marcoux Introduction to Lebesgue measure

Let G ⊆ K2 be open, and choose open sets An , Bn in K such that

G = ∪∞
n=1 An × Bn . Then

h−1 (G) = ∪∞ −1 ∞ −1 −1
n=1 h (An × Bn ) = ∪n=1 f (An ) ∩ g (Bn ).

But each f −1 (An ) ∩ g −1 (Bn ) is measurable, being the intersection of mea-

surable sets. Since M(E) is a σ-algebra, h−1 (G) ∈ M(E).
Thus h is measurable.
(b) implies (a).
Suppose next that h is measurable. The maps πi ∶ K2 → K, i = 1, 2 de-
fined by π1 (w, z) = w and π2 (w, z) = z are continuous. By Proposition 4.5,
f = π1 ○ h and g = π2 ○ h are measurable.
◻

4.8. Proposition. Let E ∈ M(R) and let f, g ∶ E → K be measurable. Then

(a) the constant functions are all measurable;
(b) f + g is measurable;
(c) f g is measurable; and
(d) if for all x ∈ E we have that g(x) ≠ 0, then f /g is measurable.
In particular, L(E, K) is an algebra.
Proof.
(a) Let κ ∈ K be fixed and suppose that φ ∶ E → K is the constant function
φ(x) = κ, x ∈ E. If G ⊆ K is open, then either κ ∈ G, in which case
φ−1 (G) = E or κ ∈/ G, in which case φ−1 (G) = ∅. Either way, φ−1 (G) is
measurable, whence φ is measurable.
Next we shall show that L(E, K) is closed under sums, products and quotients. Let
h ∶ E → K2 be the function h(x) = (f (x), g(x)). As we saw in Proposition 4.7, h is
measurable.
(b) Let σ ∶ K2 → K be the function σ(x, y) = x + y. Then σ is continuous and
so by Proposition 4.5,
f +g =σ○h
is measurable.
(c) Let µ ∶ K2 → K be the function µ(x, y) = xy. Then µ is continuous and so
by Proposition 4.5,
fg = µ ○ h
is measurable.
x
(d) Let δ ∶ K × (K ∖ {0}) → K be the function δ(x, y) = . Then δ is continuous
y
and so by Proposition 4.5,
f
=δ○h
g
is measurable.
 4. LEBESGUE MEASURABLE FUNCTIONS 45

Thus the set L(E, K) forms a subalgebra of the algebra KE of all functions from E
into K, and the constant function φ(x) = 1 for all x ∈ E clearly serves as the identity
of this algebra.
◻

4.9. Remark. Note that C is a metric space where d(w, z) ∶= ∣w − z∣, w, z ∈ C.

Moreover, the map
γ∶ C → R2
x + iy ↦ (x, y)
is a homemorphism.
Let E ∈ M(R) and suppose that f ∶ E → C is a function.
● If f is measurable, then γ ○ f = (Re f, Im f ) is measurable by Proposi-
tion 4.5. By Proposition 4.7, each of the functions g = Re f and h = Im f is
measurable.
● If g1 = Re f and g2 = Im f are both measurable, then by Proposition 4.7, so
is h ∶= (g1 , g2 ). Then f = γ −1 ○ h is measurable by Proposition 4.5
In other words, a complex-valued function is measurable if and only if its real and
imaginary parts are measurable. In light of this, we shall prove a number of results
for real-valued measurable functions (allowing us to bypass some routine technical
details), and leave it to the reader to formulate and prove the corresponding results
for complex-valued functions. But first, we stop for a result which will prove useful
in the second half of the notes. Recall that T ∶= {z ∈ C ∶ ∣z∣ = 1}.

4.10. Proposition. Let E ∈ M(R) and suppose that f ∶ E → C is measurable.

There exists a measurable function u ∶ E → T such that

f = u ⋅ ∣f ∣.

Proof. Since {0} ⊆ C is closed and f is measurable, K ∶= f −1 ({0}) ∈ M(E). From

Example 4.4, we have that χK is a measurable function, and therefore f + χK is
measurable as well. Note also that x ∈ E implies that (f + χK )(x) ≠ 0.
f + χK
Set u = . Then u is measurable by Proposition 4.8, and clearly f = u ⋅ ∣f ∣.
∣f + χK ∣
◻

At the moment, given a set E ∈ M(R), in order to verify that a function f ∶

E → R is measurable, we must check that f −1 (G) is measurable for all G ⊆ R
open, or equivalently that f −1 (F ) is measurable for all F ⊆ R closed. Given how
many different open (resp. closed) subsets of R there are – this threatens to be an
Herculean, so as not to say a Sisyphean task. The following result makes the process
much more manageable.
46 L.W. Marcoux Introduction to Lebesgue measure

4.11. Proposition. Let E ∈ M(R) and let f ∶ E → R be a function. The

following statements are equivalent.
(a) f is measurable.
(b) f −1 ((a, ∞)) ∈ M(E) for all a ∈ R.
(c) f −1 ((−∞, b]) ∈ M(E) for all b ∈ R.
(d) f −1 ((−∞, b)) ∈ M(E) for all b ∈ R.
(e) f −1 ([a, ∞)) ∈ M(E) for all a ∈ R.
Proof.
(a) implies (b). Since (a, ∞) is open for all values of a ∈ R, this is an
immediate consequence of the definition of measurability.
(b) implies (c). Let b ∈ R. By hypothesis, we have that f −1 ((b, ∞)) ∈ M(E).
But then

f −1 ((−∞, b]) = E ∖ f −1 ((b, ∞)) ∈ M(E),

since the latter is a σ-algebra of sets, by Exercise 1.

(c) implies (d). For each integer n ≥ 1, f −1 ((−∞, b − n1 ]) ∈ M(E) by hypoth-
esis, and thus
1
f −1 ((−∞, b)) = ∪∞ −1
n=1 f ((−∞, b − ]) ∈ M(E).
n
(d) implies (e). The proof is similar to that of (b) implies (c), and is left as
an exercise.
(e) implies (a). Observe first that if a, b ∈ R, then
1
f −1 ((a, ∞)) = ∪∞ −1
n=1 f ([a + , ∞)) ∈ M(E),
n
while
f −1 ((−∞, b)) = E ∖ f −1 ([b, ∞)) ∈ M(E).
If a < b, then

f −1 ((a, b)) = f −1 ((a, ∞)) ∩ f −1 ((−∞, b)) ∈ M(E).

Thus f −1 (I) is open for every open interval I ⊆ R.

Finally, if G ⊆ R is open, then there exists a countable family of open
intervals In such that G = ∪∞ n=1 In . (As seen in Exercise 6, we can choose
the In ’s to be disjoint, though this is not necessary here.) Since M(E) is
a σ-algebra,
f −1 (G) = ∪∞ −1
n=1 f (In ) ∈ M(E).

By definition, f is measurable.
◻
 4. LEBESGUE MEASURABLE FUNCTIONS 47

4.12. Corollary. Let E ∈ M(R) and f ∶ E → R be a function. The following

statements are equivalent.
(a) f is measurable.
(b) f −1 (B) ∈ M(E) for all B ∈ Bor (R).
Proof. This is left to the Assignments.
◻

4.13. Remark. Let E ∈ M(R) and suppose that f ∶ E → R is a function. We

define
f + (x) = max(f (x), 0) x∈E
−
f (x) = max(−f (x), 0) x ∈ E.
Note that f = f + − f − and that ∣f ∣ = f + + f − .
It follows from Examples 4.6 and Proposition 4.8 that if f is measurable, then
so are
∣f ∣ + f ∣f ∣ − f
f+ = and f− = .
2 2
Combining this with with Remark 4.9, we see that every complex-valued mea-
surable function is a linear combination of four non-negative, real-valued measurable
functions.

4.14. We are now going to examine a number of results that deal with pointwise
limits of sequences of measurable, real-valued functions. It will prove useful to
include the case where the limit at a given point exists as an extended real number;
that is, when the sequence diverges to ∞, or to −∞. While this is useful for treating
measure-theoretic and analytic properties of sequences of functions, there is a price
to pay: the extended real numbers have poor algebraic properties. In particular, we
can not add ∞ to −∞, and so the class of functions we shall examine will not form
a vector space.

4.15. Definition. We define the extended real numbers to be the set

R ∶= R ∪ {−∞, ∞},
also written R = [−∞, ∞].
By convention, we shall define
● α + ∞ = ∞ = ∞ + α for all α ∈ R ∪ {∞};
● α + −∞ = −∞ = −∞ + α for all α ∈ R ∪ {−∞};
● α ⋅ ∞ = ∞ ⋅ α = (−∞) ⋅ (−α) = (−α) ⋅ (−∞) = ∞ if 0 < α ∈ R or α = ∞;
● α ⋅ ∞ = ∞ ⋅ α = (−∞) ⋅ (−α) = (−α) ⋅ (−∞) = −∞ if α < 0 ∈ R or α = −∞;
● 0 = 0 ⋅ ∞ = ∞ ⋅ 0 = 0 ⋅ (−∞) = (−∞) ⋅ 0.
Observe that we define neither ∞ − ∞ nor −∞ + ∞.
48 L.W. Marcoux Introduction to Lebesgue measure

4.16. Definition. Given H ⊆ R, we refer to a function f ∶ H → R as an

extended real-valued function.
If E ∈ M(R), an extended real-valued function f ∶ E → R is said to be Lebesgue
measurable if
(a) f −1 (G) ∈ M(E) for all open sets G ⊆ R, and
(b) f −1 ({−∞}), f −1 ({∞}) ∈ M(E).
We denote the set of Lebesgue measurable extended real-valued functions on E
by
L(E, R) = {f ∶ E → R ∶ f is measurable}.
We shall often have occasion to refer to the non-negative elements of L(E, R),
and so we also define the notation
L(E, [0, ∞]) = {f ∈ L(E, R) ∶ 0 ≤ f (x) for all x ∈ E}.

Note. We remark that condition (a) above can be replaced with the condition that
f −1 (F ) ∈ M(E) for all closed sets F ⊆ R.

As was the case with real-valued measurable functions, in testing whether or not
a given extended real-valued function is measurable or not, it suffices to check that
the inverse images of certain intervals are measurable. In what follows, we write
(a, ∞] to mean (a, ∞) ∪ {∞} and [−∞, b) = (−∞, b) ∪ {−∞} for all a, b ∈ R.
4.17. Proposition. Let E ∈ M(R) and suppose that f ∶ E → R is a function.
The following statements are equivalent.
(a) f if Lebesgue measurable.
(b) For all a ∈ R, f −1 ((a, ∞]) ∈ M(E).
(c) For all b ∈ R, f −1 ([−∞, b)) ∈ M(E).
Proof. The proof of this Proposition is left as an exercise for the reader.
◻

4.18. Proposition. Let E ∈ M(R) and suppose that (fn )∞ n=1 is a sequence
of extended real-valued, measurable functions on E. The following (extended real-
valued) functions are also measurable.
(a) g1 ∶= supn≥1 fn ;
(b) g2 ∶= inf n≥1 fn ;
(c) g3 ∶= lim supn≥1 fn ; and
(d) g4 ∶= lim inf n≥1 fn .
Proof.
(a) By Proposition 4.17 above, it suffices to prove that g1−1 ((a, ∞]) ∈ M(E) for
all a ∈ R.
But
g1−1 ((a, ∞]) = ∪∞ −1
n=1 fn ((a, ∞]).
Since each fn−1 ((a, ∞]) ∈ M(E) (as each fn is measurable), we have that
g1−1 ((a, ∞]) ∈ M(E).
 4. LEBESGUE MEASURABLE FUNCTIONS 49

Hence g1 is measurable.
(b) The proof of (b) is similar to that of (a). For each b ∈ R,
g2−1 ([−∞, b)) = ∪n≥1 fn−1 ([−∞, b)) ∈ M(E).
Thus g2 is measurable.
(c) For each N ≥ 1, hN ∶= supn≥N fn is measurable by (a) above. Clearly
h1 ≥ h2 ≥ h3 ≥ ⋯.
But then lim supn≥1 fn = limn→∞ hn = inf n≥1 hn , and this is measurable by
(b) above.
(d) The proof of this is similar to that of (c), and is left as an exercise.
◻
The next result is an immediate corollary of the above Proposition.
4.19. Corollary. Let E ∈ M(R) and suppose that (fn )∞ n=1 is a sequence of
real-valued, measurable functions on E such that f (x) = limn→∞ fn (x) exists as an
extended real number for each x ∈ E.
Then f ∈ L(E, R); i.e. f is measurable.
4.20. Definition. Let E ∈ M(R) and φ ∶ E → R be a function. We say that
φ is simple if ran φ is finite. Suppose that ran φ = {α1 < α2 < ⋯ < αN }, and set
En ∶= φ−1 ({αn }), 1 ≤ n ≤ N . We shall say that
N
φ = ∑ αn χEn
n=1
is the standard form of φ.

4.21. Proposition. Let E ∈ M(R) and suppose that φ ∶ E → R is a simple

function with range ran φ = {α1 < α2 < ⋯ < αN }. The following statements are
equivalent.
(a) φ is measurable.
(b) If φ = ∑Nn=1 αn χEn is the standard form of φ, then En ∈ M(E), 1 ≤ n ≤ N .
Proof.
(a) implies (b).
Suppose that φ is measurable. Let 1 ≤ n ≤ N . If αn ∈ R, then {αn }
is a closed set, and thus En = φ−1 ({αn }) is measurable. If α1 = −∞, then
φ−1 ({α1 }) ∈ M(E) by definition of measurability, and similarly if αN = ∞,
then φ−1 ({αN }) ∈ M(E) by definition of measurability.
(b) implies (a).
Suppose that En ∈ M(E) for all 1 ≤ n ≤ N . Then χEn is measurable
for each n, by Example 4.4. For any a ∈ R,
φ−1 ((a, ∞]) = ∪{En ∶ a < αn }.
Thus φ−1 ((a, ∞]) is a finite (possibly empty) union of measurable sets, and
as such it is measurable.
50 L.W. Marcoux Introduction to Lebesgue measure

By Proposition 4.18, φ is measurable.

◻

4.22. Example. The standard form is not the only way of expressing a simple
function as a linear combination of characteristic functions.
Consider φ ∶ R → R defined as φ = χQ + 9χ[2,6] . Then ran φ = {0, 1, 9, 10}.
Set
E1 = φ−1 ({0}) = R ∖ (Q ∪ [2, 6])
E2 = φ−1 ({1}) = Q ∖ [2, 6]
E3 = φ−1 ({9}) = [2, 6] ∖ Q
E4 = φ−1 ({10}) = Q ∩ [2, 6].
Then
φ = 0χE1 + 1χE2 + 9χE3 + 10χE4
is the standard from of φ.

4.23. Definition. Let V be a vector space over K, where K = R or K = C. A

subset C ⊆ V is said to form a (real) cone if
(a) C ∩ −C = {0}, where −C = {−w ∶ w ∈ C} and
(b) y, z ∈ C and 0 ≤ κ ∈ R imply that
κy + z ∈ C.

4.24. Example.
(a) Let V = R3 , and let C = {(x, y, z) ∈ V ∶ 0 ≤ x, y, z}. Then C is a cone.
(b) Let V = C and let
π 2π
C = {w ∈ C ∶ w = reiθ , ≤ θ < , 0 ≤ r < ∞}.
6 6
Then C is a cone. We mention in passing that C is not closed in C.
(c) Let V = C([0, 1], C) ∶= {f ∶ [0, 1] → C ∶ f is continuous}. Let
C ∶= {f ∈ C([0, 1], C) ∶ f (x) ≥ 0 for all x ∈ [0, 1]}.
Then C is a cone.

4.25. Remark. Let E ∈ M(R) be a measurable set. We shall denote by

Simp(E, R)
the set of all simple, real-valued, measurable functions on E. We leave it as an
exercise for the reader to show that Simp(E, R) is an algebra, and thus a vector
space over R.
It will also be useful to adopt the following notation:
Simp(E, [0, ∞)) ∶= {φ ∈ Simp(E, R) ∶ 0 ≤ φ(x) for all x ∈ E}.
Observe that this is a real cone in Simp(E, R).
 4. LEBESGUE MEASURABLE FUNCTIONS 51

We denote by
Simp(E, R)
the set of all simple, extended, measurable real-valued functions on E, and we set
Simp(E, [0, ∞]) ∶= {φ ∈ Simp(E, R) ∶ 0 ≤ φ(x) for all x ∈ E}.
Alas, Simp(E, R) is not a vector space over R, since if φ ∈ Simp(E, R) and
φ−1 ({−∞, ∞}) ≠ ∅, then φ does not admit an additive inverse (recall that we have
not defined −∞ + ∞ in R).

The next result will be the key to our definition of Lebesgue integrability of
functions in the next section.
4.26. Proposition. Let E ∈ M(R) and f ∶ E → [0, ∞] be a measurable function.
Then there exists an increasing sequence
φ1 ≤ φ2 ≤ φ3 ≤ ⋯ ≤ f
of simple, real-valued functions φn such that
f (x) = lim φn (x) for all x ∈ E.
n→∞
Proof. The proof of this Proposition is left as an Assignment question.
◻
52 L.W. Marcoux Introduction to Lebesgue measure

Appendix to Section 4.

4.27. Let us now examine the extended real numbers from a somewhat different
point of view. You may consider the discussion below “culture”.

Definition. Let (X, τX ) be a topological space. A compact topological space

(Y, τY ) is said to be a compactification of X if there exists a dense subset Z ⊆ Y
and a homeomorphism
ρ ∶ X → Z.

In general, there are a great many compactifications of topological space. Some

(including the Stone-Čech compactification) are more important than others.
A full treatment of these, however, is beyond the scope of these notes. Instead, we
refer the interested reader to the excellent monograph of Stephen Willard [8].

4.28. There are, nevertheless, two particular compactifications which are of

interest to us here, and these are simple enough to briefly describe.
In fact, we have already seen one. Consider the following topology τ2 on the set
R of extended real numbers. A subset G ⊆ R belongs to τ2 if an only if:
(a) G∩(−∞, ∞) is open in R (with its usual topology inherited from the metric
d(x, y) = ∣x − y∣, x, y ∈ R);
(b) ∞ ∈ G implies that there exists a ∈ R such that (a, ∞] ⊆ G; and
(c) −∞ ∈ G implies that there exists b ∈ R such that [−∞, b) ⊆ G.
That this is indeed a topology on R is left as an exercise for the reader. It is
worth observing that τ2 is the topology on R generated by the sets {(a, ∞] ∶ a ∈ R},
in the sense that this collection forms a subbase for τ2 . A different subbase for τ2
is the collection {[−∞, b) ∶ b ∈ R}. Note that these are precisely the families of “test
sets” which appear in Proposition 4.17.

The proof of the following result is left as yet another exercise for the reader.
We emphasize that the topology on [0, 1] ⊆ R is the usual (relative) topology that
it inherits as a subset of R, itself equipped with the usual topology.

4.29. Theorem. (R, τ2 ) is homeomorphic to the interval [0, 1], under a home-
omorphism that sends R ⊆ R to the dense subset (0, 1) ⊆ [0, 1].

It follows that (R, τ2 ) (or equivalently [0, 1]) is a compactification of R. This

particular compactification is often referred to as the two-point compactification
of R. (This motivates the subscript “2” which appears in τ2 .)
 4. LEBESGUE MEASURABLE FUNCTIONS 53

4.30. The second interesting and useful compactification of R we wish to con-

sider is the so-called one-point compactification of R, also known as the Alek-
sandrov compactification of R.
Here, deferring to standard notation, we set α[R] ∶= R ∪ {∞}. We define a
topology τ1 on α[R] as follows: a subset G ⊆ α[R] belongs to τ1 if and only if:
(a) G∩(−∞, ∞) is open in R (with its usual topology inherited from the metric
d(x, y) = ∣x − y∣, x, y ∈ R); and
(b) ∞ ∈ G implies that there exists 0 < a ∈ R such that (a, ∞) ∪ (∞, −a) ⊆ G.
Condition (b) may be replaced by the condition that the neighbourhoods of ∞
must be of the form {∞} ∪ (R ∖ K), where K ⊆ R is compact (in the usual topology
on R).

Recall our notation: T = {z ∈ C ∶ ∣z∣ = 1}. The topology on T which we are

considering below is the usual (relative) topology that it inherits as a subset of C,
equipped with its usual topology.
4.31. Theorem. (α[R], τ1 ) is homeomorphic to T, under a homeomorphism
that sends R ⊆ α[R] to the dense subset T ∖ {−1} of T.

It follows that α[R] is a compactification of R. Since it is homeomorphic to the

familiar set T, we often just think of the one-point compactification of R as T itself.
4.32. We have described these compactifications of R merely to place the results
of this Chapter and of later Chapters that deal with Fourier Series in context. People
have lived fruitful and productive lives without knowing diddly-squat about either
of these compactifications. Well, if you can call that living.
4.33. Unlike the notions of continuity, of piecewise-continuity and even of Rie-
mann integrability, measurability of functions behaves unbelievably well under point-
wise limits of functions, as Proposition 4.18 shows. The fact that in many ways,
Lebesgue integration respects pointwise limits is what makes it such a cogent and
powerful tool. This is what we shall turn to next.
54 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 4.

Exercise 4.1.
Let E ∈ M(R). Prove that M(E) is a σ-algebra of sets.

Exercise 4.2.
Let (X, d) be a metric space.
(a) Let E ∈ M(R) be a measurable set. Verify that a function f ∶ E → X is
measurable if and only if f −1 (F ) ∈ M(R) for every closed subset F of X.
(b) Let H ⊆ R be a set. Verify that a constant function f ∶ H → X is measurable
if and only if H is measurable.

Exercise 4.3.
Let f ∶ R → K be a function and g ∶ K → R be the absolute value function
g(x) = ∣x∣, x ∈ K. Suppose that g ○ f = ∣f ∣ is measurable.
Give an example to show that f need not be measurable.

Exercise 4.4.
(a) Let G ⊆ K2 be an open set. Prove that there exists open sets An , Bn ⊆ K,
n ≥ 1, such that
G = ∪∞n=1 An × Bn .
(b) Suppose that G ⊆ R2 is a non-empty open set. Prove that there exist
countably many rectangles Rn = (an , bn ) × (cn , dn ) ⊆ R2 such that G =
∪∞
n=1 Rn .

Exercise 4.5.
Prove that the functions σ, µ and δ from Proposition 4.8 are all continuous.

Exercise 4.6.
Formulate and prove an analogue of Proposition 4.11 for complex-valued func-
tions.

Exercise 4.7.
Complete the proof of (d) implies (e) in Proposition 4.11.

Exercise 4.8. Assignment

Prove Corollary 4.12.

Exercise 4.9.
Let E ∈ M(R). Show that an extended real-valued function f ∶ E → R is
measurable if and only the following two conditions hold.
(a) f −1 (F ) ∈ M(E) for all closed sets F ⊆ R, and
(b) f −1 ({−∞}) and f −1 ({∞}) ∈ M(E).
 4. LEBESGUE MEASURABLE FUNCTIONS 55

Exercise 4.10.
Let E ∈ M(R) and suppose that f ∶ E → R is a function. Prove that the following
statements are equivalent.
(a) f if Lebesgue measurable.
(b) For all α ∈ R, f −1 ((α, ∞]) ∈ M(E).
(c) For all β ∈ R, f −1 ([−∞, β)) ∈ M(E).

Exercise 4.11.
Let E ∈ M(R) and suppose that (fn )∞n=1 is a sequence of extended real-valued,
measurable functions on E. Complete the proof of Proposition 4.18 by showing that
g ∶= lim inf fn
n≥1
is measurable.

Exercise 4.12.
Let E ∈ M(R). Prove that Simp(E, R) is an algebra over R.

Exercise 4.13. Assignment

Let E ∈ M(R) and f ∶ E → [0, ∞] be a measurable function. Prove that there
exists an increasing sequence
0 ≤ φ1 ≤ φ2 ≤ φ3 ≤ ⋯ ≤ f
of simple, real-valued functions φn such that
f (x) = lim φn (x) for all x ∈ E.
n→∞

Exercise 4.14.
Let E and F be measurable sets in R and suppose that f ∶ E → R and g ∶ F → R
are functions.
(a) Define the function
f̂ ∶ R → R
⎧
⎪f (x) if x ∈ E
⎪
x ↦ ⎨ .
⎪
⎪ 0 if x ∈/ E
⎩
Prove that f is measurable if and only if f̂ is measurable.
(b) Suppose that E ∩ F = ∅. Prove that the function h ∶ E ∪ F → R defined by
⎧
⎪f (x) if x ∈ E
⎪
h(x) = ⎨
⎪
⎪g(x) if x ∈ F
⎩
is measurable if and only if both f and g are measurable.
(c) Does the conclusion from (b) hold if E ∩ F ≠ ∅? Either prove that it does,
or find a counterexample to show that it doesn’t.
56 L.W. Marcoux Introduction to Lebesgue measure

Exercise 4.15. Assignment Question.

(a) Let f ∶ R → [0, ∞] be a measurable function. Show that there exists an
increasing sequence of measurable, simple functions φn ∶ R → [0, ∞) so that
f (x) = lim φn (x) for all x ∈ R.
n→∞
(b) Let E ∈ M(R) and let g ∶ E → [0, ∞] be a measurable function. Show
that there exists an increasing sequence of measurable, simple functions
ψn ∶ E → [0, ∞) so that
g(x) = lim ψn (x) for all x ∈ E.
n→∞

Hint for (a): For each n ≥ 1, partition the interval [0, n) into n2n equal subintervals
−1
Ek,n = [ 2kn , k+1 n
2n ), 0 ≤ k < (n2 ) − 1. Set En2 ,n = [n, ∞]. Use the sets f (Ek,n ),
n
n
0 ≤ k ≤ n2 to build φn .
Hint for (b): This should be very short, otherwise you are doing something wrong.

Exercise 4.16.
Let E ⊆ R be a set of measure zero, and let f ∶ E → R be any function whatsoever.
Prove that f is measurable.
 5. LEBESGUE INTEGRATION 57

5. Lebesgue integration

I know that there are people who do not love their fellow man, and I
hate people like that.
Tom Lehrer

5.1. Our approach to defining the Lebesgue integral of a measurable function f

on a measurable set E ⊆ R will be an iterative one. We shall begin this by defining
the integral of a simple, non-negative, extended real-valued function. We will then
use this definition to define the integral of f when f ∶ E → [0, ∞] is measurable, and
derive a number of consequences of our definition.
Following this, we shall design our notion of Lebesgue integration so that it is
linear; this will require us to impose certain conditions on the range of the functions
involved.

5.2. Definition. Let E ∈ M(R) and φ ∈ Simp(E, [0, ∞]). Let

N
φ = ∑ αn χEn
n=1

denote the standard form of φ. (Since φ is measurable, so is En , 1 ≤ n ≤ N.)

We define
N
∫ φ ∶= ∑ αn mEn ,
E n=1
and observe that ∫E φ ∈ [0, ∞].
If F ⊆ E is measurable, we define
N
∫ φ = ∫ φ ⋅ χF = ∑ αn m(F ∩ En ).
F E n=1

We remind the reader that by convention, we have defined 0 ⋅ ∞ = 0. Thus if

n = 1 and αn = 0 and mEn = ∞, or conversely if n = N and αn = ∞ and mEn = 0,
then αn mEn = 0, whence ∫En αn χEn = 0.

5.3. Example.
(a) Let φ = 0χ[4,∞) + 17χQ∩[0,4) + 29χ[2,4)∖Q . Then

∫ φ = 0 m[4, ∞) + 17 m(Q ∩ [0, 4)) + 29 m([2, 4) ∖ Q)

[0,∞)
= 0 ⋅ ∞ + 17 ⋅ 0 + 29 ⋅ 2
= 58.
58 L.W. Marcoux Introduction to Lebesgue measure

(b) Let C ⊆ [0, 1] be the Cantor set from Example 3.15, and consider
φ = 1 χC + 2 χ[5,9] . Then

∫ φ = 1 m(C ∩ [0, 6]) + 2 m([5, 9] ∩ [0, 6])

[0,6]
= 1 ⋅ 0 + 2 ⋅ m([5, 6])
= 2.

Our definition of the integral of a simple, nonnegative measurable function cur-

rently requires us to express the function in standard form. Let us now relax this
condition.

5.4. Definition. Let E ∈ M(R) and let φ ∶ E → R be a simple, measurable

function. Suppose that
N
φ = ∑ αn χHn ,
n=1

where Hn ⊆ E is measurable and αn ∈ R, 1 ≤ n ≤ N . Observe that we are not

requiring that the αn ’s be distinct, nor that they be written in any particular order,
nor that E = ∪Nn=1 Hn .
We shall say that the above decomposition of φ is a disjoint representation
of φ if
Hi ∩ Hj = ∅, 1 ≤ i ≠ j ≤ N.

We emphasize that the measurability of the sets Hn , 1 ≤ n ≤ N is part of the

definition of a disjoint representation of φ.

5.5. Lemma. Let E ∈ M(R) and suppose that φ, ψ ∶ E → R are simple, real-
valued, measurable functions. Then there exist
(i) N ∈ N,
(ii) α1 , α2 , . . . , αN , β1 , β2 , . . . , βN ∈ R, and
(iii) H1 , H2 , . . . , HN ∈ M(E)
such that Hi ∩ Hj = ∅ if 1 ≤ i ≠ j ≤ N ,
N N
φ = ∑ αn χHn and ψ = ∑ βn χHn .
n=1 n=1

Remark. The key things to notice here are that the Hn ’s appearing in the decom-
positions of φ and ψ are the same, and the representations are disjoint.
Proof. Let φ = ∑M 1 M2
j=1 aj χEj and ψ = ∑k=1 bk χFk , where Ej , Fk are measurable subsets
of E for all 1 ≤ j ≤ M1 , 1 ≤ k ≤ M2 , the Ej ’s are pairwise disjoint, and the Fk ’s are
pairwise disjoint. (That such a decomposition exists is clear, as we may simply write
φ and ψ in standard form.)
 5. LEBESGUE INTEGRATION 59

Then {Ej ∩ Fk ∶ 1 ≤ j ≤ M1 , 1 ≤ k ≤ M2 } are disjoint, measurable sets and

M1 M2 M1 M2
φ = ∑ aj ( ∑ χEj ∩Fk ) = ∑ ∑ aj χEj ∩Fk ,
j=1 k=1 j=1 k=1

and similarly
M1 M2
ψ = ∑ ∑ bk χEj ∩Fk .
j=1 k=1
Relabel {Ej ∩ Fk ∶ 1 ≤ j ≤ M1 , 1 ≤ k ≤ M2 } as {Hn ∶ 1 ≤ n ≤ N } to complete the proof.
(The αn ’s and βn ’s are clearly just relabelings of the aj ’s and the bk ’s respectively.)
◻

5.6. Lemma. Let E ∈ M(R) and suppose that φ ∈ Simp(E, [0, ∞]). If φ =
∑N
n=1 αn χHnis any disjoint representation of φ, then
N
∫ φ = ∑ αn mHn .
E n=1

Proof. If ∪N ≠ E, then we can set HN +1 = E ∖ (∪N

n=1 Hn n=1 Hn ) and αN +1 = 0. As
such, we assume without loss of generality that E = ∪Nn=1 Hn .
Observe that since the Hn ’s are mutually disjoint,
ran φ ⊆ {αn }N
n=1 .
Of course, we might have αi = αj with 1 ≤ i ≠ j ≤ N , so allow us to write
ran φ = {β1 < β2 < ⋯ < βM },
for some 1 ≤ M ≤ N , and set Em = φ−1 ({βm }), 1 ≤ m ≤ M , so that each Em ∈ M(E).
Now, for 1 ≤ m ≤ M , we have that Em = ⊍{Hn ∶ αn = βm }. By definition,
M M
∫ φ = ∫ ∑ βk χEk = ∑ βk mEk
E E k=1 k=1
Finally, m(Ek ) = ∑{mHn ∶ αn = βk }, and thus
M M N
∫ φ = ∑ βk ( ∑ mHj ) = ∑ ( ∑ αj mHj ) = ∑ αn mHn .
E k=1 αj =βk k=1 αj =βk n=1
◻

5.7. Proposition. Let E ∈ M(R). If φ, ψ ∈ Simp(E, [0, ∞]) and 0 ≤ κ ∈ R,

then
(a) ∫E κφ + ψ = κ ∫E φ + ∫E ψ.
(b) If φ ≤ ψ, then ∫E φ ≤ ∫E ψ.
Proof. By Lemma 5.5, we can find common disjoint representations of φ and ψ,
say
N N
φ = ∑ α n χH n and ψ = ∑ βn χHn .
n=1 n=1
60 L.W. Marcoux Introduction to Lebesgue measure

(a) Then κφ + ψ = ∑N
n=1 (καn + βn )χHn is a disjoint representation of κφ + ψ,
and so by Lemma 5.6,
N
∫ (κφ + β) = ∑ (καn + βn )mHn
E n=1
N N
= κ( ∑ αn mHn ) + ∑ βn mHn
n=1 n=1

= κ ∫ φ + ∫ ψ.
E E

(b) Suppose that φ ≤ ψ. Then αn ≤ βn for all 1 ≤ n ≤ N , and so

N N
∫ φ = ∑ αn mHn ≤ ∑ βn mHn = ∫ ψ.
E n=1 n=1 E

Does part (a) of the above result hold if we consider κ = ∞?

5.8. Definition. Recall that for E ∈ M(R), we defined

L(E, [0, ∞]) = {f ∶ E → [0, ∞] ∶ f is measurable}.
For f ∈ L(E, [0, ∞]), we define the Lebesgue integral of f to be
new
∫ f = sup{∫ φ ∶ φ ∈ Simp(E, [0, ∞)), 0 ≤ φ ≤ f }.
E E

5.9. Remarks.
(a) We leave it as an exercise for the reader to show that the above definition
is equivalent to defining
new
∫ f = sup{∫ φ ∶ φ ∈ Simp(E, [0, ∞]), 0 ≤ φ ≤ f }.
E E

(The difference being that we now allow the simple functions to be extended
real-valued and non-negative, instead of just real-valued and non-negative.)
(b) The reason for putting the superscript “new” in the above integral is the
following. Observe that if φ ∈ Simp(E, [0, ∞]), we now have two definitions
for the integral of φ. That is, writing φ = ∑N n=1 αn χHn in standard form,
we have our original definition (Definition 5.2)
N
∫ φ = ∑ αn mHn ,
E n=1

while from Definition 5.8, our new definition of the integral of φ becomes
new
∫ φ = sup{∫ ψ ∶ ψ ∈ Simp(E, [0, ∞)), 0 ≤ ψ ≤ φ}.
E E
 5. LEBESGUE INTEGRATION 61

new
(c) It is entirely possible that ∫E f = ∞. For example, if φ = ∞ ⋅ χ[0,1] , then
new
∫[0,1] φ = ∞.
new
Alternatively, if f (x) = x, x ∈ [0, ∞), then ∫[0,∞) f = ∞. The proof of
this is left as an exercise for the reader.

5.10. Let us reconcile these two definitions. Once this is done, we will no longer
need to distinguish between the original and the new integral for non-negative,
simple, measurable functions, and so we shall drop the superscript “new” for the
integrals of non-negative, measurable extended-real valued functions altogether.
On the one hand, note that φ ∈ {ψ ∈ Simp(E, [0, ∞)), 0 ≤ ψ ≤ φ}, and so by
new
definition of ∫E φ, we have that
new
∫ φ≤∫ φ.
E E
On the other hand, if ψ ∈ Simp(E, [0, ∞)) and 0 ≤ ψ ≤ φ, then by Proposition 5.7
(b),
∫ ψ ≤ ∫ φ,
E E
and thus
new
∫ φ = sup{∫ ψ ∶ ψ ∈ Simp(E, [0, ∞)), 0 ≤ ψ ≤ φ} ≤ ∫ φ.
E E E

This proves that

new
∫ φ=∫ φ
E E
whenever φ is a non-negative, extended real-valued simple function.
Now that we have this, we shall drop the superscript new and simply write

∫ f
E
for the Lebesgue integral of an element f ∈ L(E, [0, ∞]).

5.11. Remark. We shall see in the Assignments that even when f is a rela-
tively innocuous-looking function (for example f (x) = x on [0, 1]), calculating the
Lebesgue integral of f directly from the definition is an arduous task. Fortunately,
Theorem 5.24 below will provide us with an alternate means of calculating the inte-
grals of a large family of (Riemann integrable) functions, by showing that in many
cases, the Lebesgue integral coincides with the Riemann integral. Of course, when
the function is sufficiently nice, we may apply the Fundamental Theorem of Calculus
to calculate the latter.

Sets of measure zero will play a central role in the theory that follows. The
reason for this lies partly in the fact that the Lebesgue integral “ignores” these sets,
in a sense which we shall now make precise.
62 L.W. Marcoux Introduction to Lebesgue measure

5.12. Definition. Let E ∈ M(R). We say that a property (P ) holds almost

everywhere (a.e.) on E if the set
B ∶= {x ∈ E ∶ (P ) does not hold}
has Lebesgue measure zero.

5.13. Example. Let E ∈ M(R). Given f, g ∈ L(E, R), we say that f = g almost
everywhere if
B ∶= {x ∈ E ∶ f (x) ≠ g(x)}
has measure zero.
More specifically, therefore, χQ = 0 = χC a.e. on R, where C is the Cantor set
from Example 3.15.

5.14. Lemma. Let E ∈ M(R) and let f, g and h ∶ E → [0, ∞] be functions.

Suppose that g and h are measurable.
(a) Suppose furthermore that E = X ⊍ Y , where X and Y are measurable. Set
f1 ∶= f ∣X and f2 ∶= f ∣Y . Then f is measurable if and only if both f1 and f2
are measurable. When such is the case,

∫ f = ∫ f1 + ∫ f2 .
E X Y
(b) If g ≤ h, then ∫E g ≤ ∫E h.
(c) If H ⊆ E is a measurable set, then

∫ g = ∫ g ⋅ χH ≤ ∫ g.
H E E
Proof. The proof of this lemma is left as a worthwhile exercise for the reader.
◻

5.15. Proposition. Let E ∈ M(R), and f, g ∈ L(E, [0, ∞]).

(a) If mE = 0, then ∫E f = 0.
(b) If f = g a.e. on E, then ∫E f = ∫E g.
Proof.
(a) Let φ ∈ Simp(E, [0, ∞)) with φ ≤ f , and let φ = ∑N
n=1 αn χHn denote the
standard representation of φ, where Hn ∈ M(E), 1 ≤ n ≤ N .
Then
N
∫ φ = ∑ αn mHn
E n=1
N
≤ ∑ αn mE
n=1
N
= ∑ αn 0
n=1
= 0.
 5. LEBESGUE INTEGRATION 63

Thus

∫ f = sup{∫ φ ∶ φ ∈ Simp(E, [0, ∞)) ∶ φ ≤ f } = 0.

E E
(b) Let B = {x ∈ E ∶ f (x) ≠ g(x)}, so that mB = 0. Then, using Lemma 5.14(a)
as well as part (a) above, we find that

∫ f =∫ f +∫ f
E E∖B B

=∫ g+0
E∖B

=∫ g+∫ g
E∖B B

= ∫ g.
E
◻

We now come to one of the major results in this course. In dealing with prop-
erties that hold almost everywhere on a measurable set E ∈ M(R) (as will be the
case in the Monotone Convergence Theorem below), we often have recourse to the
following line of argument: we isolate the “bad” set K of measure zero where the
property under consideration fails to hold, and the deal with the “good” set E ∖ K,
where the property holds everywhere. Using Proposition 5.15 and Lemma 5.14, we
can often “glue” these results together. This is a lot of quotation marks, which are
the written equivalent of randomly flailing arms. Let us see the strategy in action,
where it might make more sense.

5.16. Theorem. The Monotone Convergence Theorem.

Let E ∈ M(R). Let (fn )∞
n=1 be a sequence in L(E, [0, ∞]) and suppose that for
each n ≥ 1,
fn ≤ fn+1 a.e. on E.
Suppose furthermore that f ∶ E → [0, ∞] is a function and that
f (x) = lim fn (x) a.e. on E.
n→∞

Then f is measurable and

∫ f = n→∞
lim ∫ fn .
E E
Proof. Step One.
First we shall show that f is measurable. Let
E0 = {x ∈ E ∶ f (x) ≠ lim fn (x)},
n→∞

so that mE0 = 0 (and in particular E0 is measurable).

By Lemma 5.14, fn ∣E∖E0 is measurable for all n ≥ 1. Since f ∣E∖E0 is the pointwise
limit of the measurable functions fn ∣E∖E0 by hypothesis, it follows that f ∣E∖E0 is
measurable by Corollary 4.19.
64 L.W. Marcoux Introduction to Lebesgue measure

But f ∣E0 is also measurable, since for any β ∈ R, f ∣−1

E0 ((β, ∞]) ⊆ E0 implies
that 0 ≤ m(f ∣E0 ((β, ∞])) ≤ mE0 = 0, which in turn implies that f ∣−1
−1
E0 ((β, ∞]) is
measurable. (See Exercise 4.16 as well.)
By Lemma 5.14, f is measurable.

Step Two.
Next, for each n ≥ 1, set
En ∶= {x ∈ E ∶ fn (x) > fn+1 (x)},
so that mEn = 0, and thus En is measurable. Let B = E0 ∪ (∪∞
n=1 En ). Then
∞
m∗ B ≤ ∑ mEn = 0,
n=0

and so B ∈ M(E) and mB = 0. Define H = E ∖ B. Let gn = fn ∣H for all n ≥ 1, and

set g = f ∣H . Arguing as in Step One, each gn is measurable, as is g.
For x ∈ H, we have that
g1 (x) ≤ g2 (x) ≤ g3 (x) ≤ ⋯ ≤ g(x),
and in fact g(x) = limn→∞ gn (x).

Step Three. The first two steps were only to reduce the problem to the case where
the interesting properties hold everywhere. Now the real argument begins.
Since gn ≤ gn+1 ≤ g for all n ≥ 1, by Lemma 5.14, we have that

∫ gn ≤ ∫ gn+1 ≤ ∫ g
H H H
for all n ≥ 1, and thus
sup ∫ gn = lim ∫ gn ≤ ∫ g.
n≥1 H n→∞ H H

Conversely, suppose that φ ∈ Simp(H, [0, ∞]) and that φ ≤ g. Let 0 < ρ < 1. We
shall prove that
∫ ρ φ = ρ ∫ φ ≤ sup ∫ gn .
H H n≥1 H

Let x ∈ H. Either ρ φ(x) < g(x), or ρ φ(x) = 0. Setting

Hn ∶= {x ∈ H ∶ ρ φ(x) ≤ gn (x)} = (ρ φ − gn )−1 ([−∞, 0]),
we see that H1 ⊆ H2 ⊆ H3 ⊆ ⋯ is an increasing sequence of Lebesgue measurable
subsets of H with H = ∪∞
n=1 Hn .

Step Four. We claim that

lim ∫ φ = ∫ φ.
n→∞ Hn H
 5. LEBESGUE INTEGRATION 65

To see this, express φ = ∑N

k=1 αk χJk in standard form. Then
N
∫ φ = ∫ φ ⋅ χHn = ∑ αk m(Jk ∩ Hn ).
Hn H k=1
But for each 1 ≤ k ≤ N , we also see that
Jk ∩ H1 ⊆ Jk ∩ H2 ⊆ Jk ∩ H3 ⊆ ⋯
is an increasing sequence of measurable sets and
∪∞ ∞
n=1 (Jk ∩ Hn ) = Jk ∩ (∪n=1 Hn ) = Jk ∩ H = Jk .

By the Continuity of Lebesgue Measure (see Exercise 3.9),

lim m(Jk ∩ Hn ) = mJk .
n→∞
It follows that
N N
lim ∫ φ = lim ∑ αk m(Jk ∩ Hn ) = ∑ αk mJk = ∫ φ.
n→∞ Hn n→∞ H
k=1 k=1

Step Five. Hence

ρ ∫ φ = ρ ( lim ∫ φ) ≤ ρ ( lim ∫ gn ) ≤ 1 ⋅ ( lim ∫ gn ) = sup ∫ gn .
H n→∞ Hn n→∞ Hn n→∞ H n≥1 H

Since φ ∈ Simp([0, ∞]) was arbitrary (subject to the condition that 0 ≤ φ ≤ g,

we conclude that
ρ ∫ g ≤ sup ∫ gn .
H n≥1 H
But then 0 < ρ < 1 was also arbitrary, and so

∫ g ≤ sup ∫ gn .
H n≥1 H

Combining this with the reverse inequality from Step Three shows that

∫ g = sup ∫ gn = n→∞
lim ∫ gn .
H n≥1 H H

Step Six. There remains only to “glue” the above results together to get the desired
statement.
By Lemma 5.14,

∫ f = ∫ f + ∫ f = 0 + ∫ g = n→∞
lim ∫ gn = lim ∫ fn + ∫ fn = lim ∫ fn .
n→∞ n→∞
E B H H H B H E
◻
Steps Three to Five of the above proof provide a proof of the Monotone Con-
vergence Theorem in the case where the sequence (fn )∞
n=1 is everywhere increasing
and where the sequence tends to f everywhere.
66 L.W. Marcoux Introduction to Lebesgue measure

5.17. Let us remind ourselves of a “pathological” sequence of Riemann integral

functions we constructed in Remark 1.16. In that remark, we enumerated the set
E ∶= Q ∩ [0, 1] = {qn }∞
n=1 , and set f = χEn , where En ∶= {q1 , q2 , . . . , qn }, n ≥ 1. We
observed that
0 ≤ f1 ≤ f2 ≤ f3 ≤ ⋯ ≤ χE ,
1
and that each fn is Riemann integrable with ∫0 fn (x)dx = 0. Moreover,
χE (x) = lim fn (x) for all x ∈ [0, 1],
n→∞

and yet χE is not Riemann integrable.

We leave it as an exercise for the reader to show that each fn is Lebesgue
integrable with ∫[0,1] fn = 0, n ≥ 1. By the Monotone Convergence Theorem 5.16,
we find that χE ∈ L([0, 1], R) and

∫ χE = lim ∫ fn = lim 0 = 0.
[0,1] n→∞ [0,1] n→∞

This agrees with the fact that

0≤∫ f =∫ χE = mE ≤ mQ = 0.
[0,1] [0,1]

The point is that the limit function χE is Lebesgue integrable, even though it is not
Riemann integrable.

The first half the of the following Corollary extends results from Proposition 5.7.
There, we dealt with nonnegative, simple, measurable functions. We remove the
requirement that the functions be simple.

5.18. Corollary. Let E ∈ M(R).

(a) If f, g ∈ L(E, [0, ∞]) and κ ≥ 0, then

∫ κf + g = κ ∫ f + ∫ g.
E E E

(b) If (hn )∞ N
n=1 is a sequence in L(E, [0, ∞]) and if h(x) ∶= limN →∞ ∑n=1 hn (x)
for all x ∈ E, then h is measurable and
∞
∫ h = ∑ ∫ hn .
E n=1 E

(c) Let f ∈ L(E, [0, ∞]). If (Hn )∞

n=1 is a sequence in M(E) with Hi ∩ Hj = ∅
when 1 ≤ i ≠ j < ∞ and H = ⊍∞ n=1 Hn , then
∞
∫ f = ∑∫ f.
H n=1 Hn

Proof.
 5. LEBESGUE INTEGRATION 67

(a) From our work in the Assignments (see Exercise 4.15), we may choose
increasing sequences (φn )∞ ∞
n=1 and (ψn )n=1 in Simp(E, [0, ∞]) such that
f (x) = limn→∞ φn (x) and g(x) = limn→∞ ψn (x) for all x ∈ E.
By the Monotone Convergence Theorem,

∫ f = n→∞
lim ∫ φn
E E
and
∫ g = n→∞
lim ∫ ψn .
E E
Given 0 ≤ κ ∈ R, it follows that (κφn + ψn )∞
n=1 is again an increasing se-
quence of non-negative, simple, measurable, functions converging pointwise
to the function κf + g.
Applying the Monotone Convergence Theorem 5.16 once more, we see
that
∫ (κf + g) = n→∞
lim ∫ (κφn + ψn ) = lim κ ∫ φn + ∫ ψn = κ ∫ f + ∫ g.
n→∞
E E E E E E
(b) For N ≥ 1, set gN ∶= ∑N
n=1 hn .
Then each gN is measurable (exercise), and
N
g =
∫E N ∑n=1 ∫E n h by part (a). Furthermore,
0 ≤ g1 ≤ g2 ≤ ⋯.
Now h(x) = limN →∞ gN (x) for all x ∈ E, and so h is measurable by Corol-
lary 4.19.
By the Monotone Convergence Theorem,
N ∞
∫ h = Nlim
→∞
∫ gN = Nlim
→∞
∑ ∫ hn = ∑ ∫ hn .
E E n=1 E n=1 E

(c) For each n ≥ 1, set fn = f ⋅ χHn . Then fn is measurable for each n ≥ 1

(exercise) and
N
f ⋅ χH (x) = lim ∑ fn (x) for all x ∈ E.
N →∞ n=1

By part (b),

∫ f = ∫ f ⋅ χH
H E
∞
= ∫ ∑ fn
E n=1
∞
= ∑ ∫ fn
n=1 E
∞
= ∑ ∫ f ⋅ χHn
n=1 E
∞
= ∑∫ f.
n=1 Hn
◻
68 L.W. Marcoux Introduction to Lebesgue measure

5.19. Definition. The Lebesgue Integral

Let E ∈ M(R) and f ∈ L(E, R). We say that f is Lebesgue integrable on E if
+ −
∫ f <∞ and ∫ f < ∞,
E E
in which case we set
+ −
∫ f ∶= ∫ f − ∫ f .
E E E
We denote by L1 (E, R) the set of all extended real-valued Lebesgue integrable
functions on E, and by L1 (E, R) all real-valued Lebesgue integrable functions on E.

In Chapter 9 and later, we shall need to specify the variable with respect to
which we are integrating. In analogy to the usual notation for Riemann integration,
we shall write
∫ f = ∫ f (s)dm(s)
E E
to denote the Lebesgue integral of f with respect to the variable s.

5.20. Remarks. Let E ∈ M(R).

(a) By definition, every Lebesgue integrable function on E is Lebesgue mea-
surable.
(b) A measurable function f is Lebesgue integrable if and only if ∣f ∣ is Lebesgue
integrable. This fails in general if f is not assumed to be measurable.
(Consider f = χH − χ[0,1]∖H , where H ⊆ [0, 1] is any non-measurable set.
Clearly ∣f ∣ is Lebesgue integrable over [0, 1], while f is not.)
Note that this is also a distinguishing feature of Lebesgue integration
versus improper Riemann integrals. For example, the function
sin x
f (x) = , x≥1
x
∞
admits an (improper) Riemann integral ∫1 f (x)dx, but it is not Lebesgue
integrable over [1, ∞).
(c) If f ∈ L1 (E, R), then
m(f −1 ({−∞})) = 0 = m(f −1 ({∞})).
The proof of this is left as an exercise for the reader.
(d) From (c), it follows that for any f ∈ L1 (E, R), we can find an element
g ∈ L1 (E, R) such that f = g a.e. and g(x) ∈ R for all x ∈ E. Indeed, just
let H = {x ∈ E ∶ f (x) ∈ {−∞, ∞}}. By (c), mH = 0, and we may simply
choose g = f ⋅ χE∖H .
Note that this in turn implies that

∫ f = ∫ g.
E E
This will prove to be more useful than it might first appear to be. One
huge problem with L1 (E, R) is that it is not a vector space!!! One problem
 5. LEBESGUE INTEGRATION 69

lies in the fact that if f, g ∈ L1 (E, R), x ∈ E and f (x) = ∞, g(x) = −∞,
then what should (f + g)(x) be?
This is not an issue insofar as L1 (E, R) is concerned. As we shall see
in Chapter 6, we may establish an equivalence relation on L1 (E, R) by set-
ting f ∼ g if f = g a.e. on E. We can then turn the equivalence classes of
elements of L1 (E, R) into a vector space in a natural way. In most texts,
these equivalence classes are denoted by the same notation used to denote
functions, and indeed, they are often referred to as “functions”, although
technically speaking they are not. Being absolute sticklers for detail, and
inspired by our French heritage, we shall exercise as much caution as possi-
ble in the use of language, and will try to be as precise as humanly possible
in keeping the notation and terminology straight.
(e) Suppose that g ∶ E → C is a measurable function. Let us write
g = (g1 − g2 ) + i(g3 − g4 ),
where g1 = (Re g)+ , g2 = (Re g)− , g3 = (Im g)+ , and g4 = (Im g)− .
We shall say that g is Lebesgue integrable if each of g1 , g2 , g3 and g4
are, in which case we define

∫ g = (∫ g1 − ∫ g2 ) + i(∫ g3 − ∫ g4 ).
E E E E E
Of course, this is equivalent to requiring that Re g and Im g be Lebesgue
integrable, in which case we define

∫ g = ∫ (Re g) + i ∫ (Im g).

E E E
We denote by L1 (E, C) the set of all complex-valued Lebesgue inte-
grable functions on E.

The perspicacious reader will observe that we have carefully avoided all
notions of “extended” complex-valued functions.

5.21. Proposition. Let E ∈ M(R). Suppose that f, g ∈ L1 (E, R), and κ ∈ R.

Then
(a) κf ∈ L1 (E, R) and ∫E κf = κ ∫E f .
(b) f + g ∈ L1 (E, R) and ∫E (f + g) = ∫E f + ∫E g.
(c) Finally,
∣∫ f ∣ ≤ ∫ ∣f ∣.
E E
Proof. Recall from Corollary 5.18 that we have already shown that (a) and (b)
hold in the case where 0 ≤ f, g and κ ≥ 0.
(a) Let us write f = f + − f − .
Case 1. κ = 0.
Then ∫E κf = ∫E 0 = 0 = κ ∫E f .
Case 2. κ > 0.
70 L.W. Marcoux Introduction to Lebesgue measure

Then (κf )+ = κf + and (κf )− = κf − , so that

+ −
∫ κf ∶= ∫ (κf ) − ∫ (κf )
E E E

= ∫ κf − ∫ κf −
+
E E

= κ ∫ f − κ ∫ f− +
E E

= κ ∫ f.
E
Case 3. κ < 0.
Then (κf )+ = −κf − and (κf )− = −κf + , so that
+ −
∫ κf ∶= ∫ (κf ) − ∫ (κf )
E E E

= ∫ (−κ)f − ∫ (−κ)f +−
E E

= (−κ) ∫ f − (−κ) ∫ f + −
E E

= (−κ)(− ∫ f )
E

= κ ∫ f.
E
(b) Let h = f + g, so that h ∈ L(E, R), as the latter is a vector space. Write
h = h+ − h− .
Then
h+ , h− ≤ h+ + h− = ∣h∣ ≤ ∣f ∣ + ∣g∣ = f + + f − + g + + g − ,
so that
+ + − + −
∫ h ≤ ∫ f + ∫ f + ∫ g + ∫ g < ∞,
E E E E E
and
− + − + −
∫ h ≤ ∫ f + ∫ f + ∫ g + ∫ g < ∞.
E E E E E
It follows that h ∈ L1 (E, R).
Furthermore, h = f + g implies that h+ + f − + g − = h− + f + + g + , whence
+ − − − + +
∫ h +∫ f +∫ g =∫ h +∫ f +∫ g .
E E E E E E
From this it easily follows that
+ −
∫ h = ∫ h − ∫ h = ∫ f + ∫ g.
E E E E E
+ −
(c) Note that ∣f ∣ = f + f is measurable, and
+ −
∫ ∣f ∣ = ∫ f + ∫ f < ∞,
E E E
proving that ∣f ∣ ∈ L1 (E, R).
 5. LEBESGUE INTEGRATION 71

Finally,

∣∫ f ∣ = ∣∫ f + − ∫ f − ∣
E E E

≤ ∣∫ f + ∣ + ∣∫ f − ∣
E E

= ∫ f + ∫ f−
+
E E

= ∫ ∣f ∣.
E

5.22. At the moment, we have a number of results concerning Lebesgue inte-

grals, but we have not explicitly calculated the Lebesgue integrals of any functions,
other than simple functions. In the Assignments, you will be asked to compute the
Lebesgue integral of the function f (x) = x, x ∈ [0, 1] by hand. This will lead to an
unusual increase in swearing on your part, and a raising of the ole’ blood pressure.
Surely, you will tell yourself, there must be a better way!
We shall now demonstrate that in the case of bounded functions on closed,
bounded intervals, Riemann integrability implies Lebesgue integrability, and indeed
for a given Riemann integrable function, the Riemann and Lebesgue integrals coin-
cide. Since we have a number of tools to calculate Riemann integrals (for example,
the Fundamental Theorem of Calculus), this will prove to be the better way in
a large number of cases.
b
Just a quick remark about notation: we shall continue to use ∫a f to denote the
Riemann integral of Riemann integrable function f ∶ [a, b] → R and ∫[a,b] f to denote
its Lebesgue integral.
We start with a simple but useful Lemma.

5.23. Lemma. Let a < b ∈ R. If φ ∶ [a, b] → R is a step function, then φ is both

Riemann integrable and Lebesgue integrable, and

b
∫ φ=∫ φ.
[a,b] a

Proof. Let P = {a = p0 < p1 < p2 < ⋯ < pN = b} ∈ P([a, b]), and

N
φ = ∑ αn χ[pn−1 ,pn ) .
n=1
72 L.W. Marcoux Introduction to Lebesgue measure

Then
N
∫ φ = ∑ αn m[pn−1 , pn )
[a,b] n=1
N
= ∑ αn (pn − pn−1 )
n=1
N pn
= ∑∫ αn
n=1 pn−1

b N
=∫ ∑ αn χ[pn−1 ,pn )
a n=1
b
=∫ φ.
a
◻

5.24. Theorem. Let a < b ∈ R and f ∶ [a, b] → R be a bounded, Riemann-

integrable function. Then f ∈ L1 ([a, b], R) and
b
∫ f =∫ f.
[a,b] a

That is, the Lebesgue and Riemann integrals of f over [a, b] coincide.
Proof. Suppose that ∣f ∣ is bounded above on [a, b] by 0 < M ∈ R. Let g = M χ[a,b] .
Clearly 12 (f + g) is Riemann integrable and
1 b 1 b M (b − a)
∫ (f + g) = ∫ f + .
a 2 2 a 2
If we prove that 12 (f +g) ∈ L1 ([a, b], R), then it is readily seen that f ∈ L1 ([a, b], R)
and – in light of Lemma 5.23 –
1 1 M (b − a)
∫ (f + g) = ∫ f+ .
[a,b] 2 2 [a,b] 2
1
We have shown that by replacing f by 2 (f + g) if necessary, we may assume
from the outset that 0 ≤ f ≤ M on [a, b].

Cast your mind back to the halcyon days when you studied Chapter 1, and more
specifically to the Cauchy Criterion, Theorem 1.13. Recall that it asserts that for
each n ≥ 1 there exists a partition Rn ∈ P([a, b]) such that for all refinements X and
Y of Rn , and for all choices of test values X ∗ and Y ∗ for X and Y respectively,
1
∣S(f, X, X ∗ ) − S(f, Y, Y ∗ )∣ < .
n
N
Let QN ∶= ∪n=1 Rn , 1 ≤ N ∈ N, so that QN is a common refinement of R1 , R2 , . . . , RN .
Write
QN = {a = q0,N < q1,N < q2,N < ⋯ < qmN ,N = b}.
Set Hk,N = [qk−1,N , qk,N ), 1 ≤ k ≤ mN − 1 and HmN ,N = [qmN −1,N , qmN ,N ].
 5. LEBESGUE INTEGRATION 73

Define
αk,n = inf{f (t) ∶ t ∈ Hk,n }, 1 ≤ k ≤ mN ,
βk,n = sup{f (t) ∶ t ∈ Hk,n }, 1 ≤ k ≤ mN ,
mN mN
and set φN = ∑N αk,n χHk,n and ψN =
n=1 ∑k=1 βk,n χHk,n . ∑N
n=1 ∑k=1
Since each QN is a refinement of QN −1 , it is not hard to see that
φ1 ≤ φ2 ≤ φ3 ≤ ⋯ ≤ f ≤ ⋯ ≤ ψ3 ≤ ψ2 ≤ ψ1 .
Moreover, using Lemma 5.23 above, we obtain
b b
∫ φN = ∫ φN = inf{S(f, QN , Q∗N ) ∶ Q∗N test values for QN } ≤ ∫ f,
[a,b] a a
and similarly,
b b
∫ ψN = ∫ ψN = sup{S(f, QN , Q∗∗ ∗∗
N ) ∶ QN test values for QN } ≥ ∫ f.
[a,b] a a
But QN is a refinement of RN , and thus
1
∣S(f, QN , Q∗N ) − S(f, QN , Q∗∗
N )∣ <
N
for all choices of test values Q∗N and Q∗∗ N for QN .
b
It follows that ∫[a,b] φN ≤ ∫a f ≤ ∫[a,b] ψN and
1
∣∫ φN − ∫ ψN ∣ ≤ , N ≥ 1.
[a,b] [a,b] N
Let φ(x) = supn≥1 φn (x) = limn→∞ φn (x) ≤ f (x), and ψ(x) = inf n≥1 ψn (x) =
limn→∞ ψn (x) ≥ f (x), x ∈ [a, b]. Since each φn , ψn is measurable, so are φ and ψ,
and by the Monotone Convergence Theorem and Lemma 5.23,

∫ φ = lim ∫ φN
[a,b] n→∞ [a,b]
b
= lim ∫ φN
n→∞ a
b
=∫ f
a
b
= lim ∫ ψN
n→∞ a

= lim ∫ ψN
n→∞ [a,b]

=∫ ψ.
[a,b]

Thus ∫[a,b] ψ − φ = 0. But φ ≤ ψ, and so we must have φ = ψ a.e. on [a, b].

Finally, note that φ ≤ f ≤ ψ, which in turn implies that f = φ = ψ a.e. on [a, b].
Since φ is measurable, so is f , and
b
∫ f =∫ φ=∫ f < ∞.
[a,b] [a,b] a
74 L.W. Marcoux Introduction to Lebesgue measure

5.25. Corollary. Let a < b ∈ R and f ∶ [a, b] → C be a bounded, Riemann-

integrable function. Then f ∈ L1 ([a, b], C) and
b
∫ f =∫ f.
[a,b] a

Proof. Observe that f is bounded and Riemann-integrable if and only if its real and
imaginary parts are bounded and Riemann-integrable. The result now immediately
follows by applying Theorem 5.24 to each of these.
◻

Theorem 5.24 required some effort. Let us see that it was worth it.

5.26. Example. Let f (x) = x, x ∈ [0, 1]. In the Assignments, you computed
the Lebesgue integral of f over [0, 1] to be

1
∫ x= .
[0,1] 2

This was anything but easy.

Equipped with Theorem 5.24, it is child’s play. The function f is clearly bounded
and continuous (hence Riemann-integrable) over [0, 1], and so by that Theorem,
x=1
1 x2 1
∫ x=∫ x dx = ] = .
[0,1] 0 2 x=0 2

Yes, Theorem 5.24 is worth the effort.

1
5.27. Example. Let f (x) = , x ∈ E ∶= [1, ∞). We wish to determine ∫[1,∞) f .
x2
For each n ≥ 1, set fn ∶= f ⋅ χ[1,n] . Then f is measurable (because it is continuous
except at one point of E), and

0 ≤ f1 ≤ f2 ≤ ⋯,

with f (x) = limn→∞ fn (x) for all x ≥ 1. By Theorem 5.24,

n
∫ fn = ∫ fn for all n ≥ 1.
[1,n] 1
 5. LEBESGUE INTEGRATION 75

By the Monotone Convergence Theorem,

∫ f = lim ∫ fn
[1,∞) n→∞ [1,∞)
1
= lim ∫
n→∞ [1,n] x2
n 1
= lim ∫
n→∞ 1 x2
1 x=n
= lim − ]
n→∞ x x=1
1
= lim (− − (−1))
n→∞ n
= 1.

In this example, the Lebesgue integral of f returns the value of the improper
Riemann integral of f over [1, ∞). Two things are worth noting:

● first, it is possible for the improper Riemann integral of a measurable func-

tion f ∶ [1, ∞) → R to exist, even though the Lebesgue integral ∫[1,∞) f
does not exist! We shall see an example of this in the Assignments.
● Second, we don’t have the notion of an improper Lebesgue integral. The
domain of f , [1, ∞), is just another measurable set.

5.28. Example. The Monotone Convergence Theorem 5.16 states that if (fn )∞
n=1
is an (almost everywhere) increasing sequence of measurable functions on a set
E ∈ M(R), then f = limn→∞ fn is measurable and

∫ lim fn = lim ∫ fn .
E n→∞ n→∞ E

In the absence of the adjective “increasing”, we can not expect this result to hold.
For example, consider the sequence (fn )∞n=1 given by

fn ∶ [1, ∞) → R
⎧
⎪1
⎪ if 1 ≤ x ≤ en
x ↦ ⎨ nx .
⎪
⎪0 if x > en
⎩

It is easy to verify (exercise) that the sequence (fn )∞

n=1 converges uniformly to f = 0
on [1, ∞).
76 L.W. Marcoux Introduction to Lebesgue measure

Nevertheless, for all n ≥ 1, fn is easily seen to be Riemann-integrable and

bounded on [1, en ], and thus
1
∫ fn = ∫
[1,∞) [1,e ] nx
n

en 1
=∫
1 nx
n
log x x=e
= ]
n x=1
=1−0
= 1.
Hence limn→∞ ∫[1,∞) fn = 1 ≠ 0 = ∫[1,∞) f .
I have done a bit (but not a great deal) of research to try to determine why
the next result is referred to as Fatou’s Lemma instead of Fatou’s Theorem. One
possible explanation is that it can be used to prove a number of other useful results
very quickly, and as such, is a “facilitator”, to employ the jolly discourse of psycho-
babble. A second possibility (which I have not read anywhere) is that it is petty
jealousy on the part of his peers. In any case, there is a different result from complex
analysis known as Fatou’s Theorem. In order to state it, we first require the notion
of Lp -spaces, and so we defer its statement to the Appendix of Chapter 6.

5.29. Theorem. Fatou’s Lemma.

Let E ∈ M(R) and (fn )∞
n=1 be a sequence in L(E, [0, ∞]). Then

∫ limninf fn ≤ limninf ∫ fn .
E E
Proof. For each N ≥ 1, set gN = inf{fn ∶ n ≥ N }. By Proposition 4.18, gN is
measurable for all N and
g1 ≤ g2 ≤ g3 ≤ ⋯.
By the Monotone Convergence Theorem 5.16,

∫ limninf fn = ∫ lim gN = lim ∫ gN .

E E N →∞ N →∞ E
Now gN ≤ fn for all n ≥ N , and so

∫ gN ≤ ∫ fn for all n ≥ N,
E E
whence
∫ gN ≤ limninf ∫ fn .
E E
But this holds for any N ≥ 1, and so by taking limits and using the above equality,
we find that
∫ limninf fn = Nlim ∫ gN ≤ limninf ∫ fn .
→∞ E
E E
◻
 5. LEBESGUE INTEGRATION 77

5.30. Example. The inequality in Fatou’s Lemma can be strict.

For example, let fn = nχ(0, 1 ] , n ≥ 1. It is clear that for any 0 ≤ x ≤ 1,
n
limn→∞ fn (x) = 0. Thus

∫ lim inf fn = ∫ 0 = 0.
[0,1] n [0,1]

On the other hand,

1
∫ fn = n m((0, ]) = 1 for all n ≥ 1,
[0,1] n
and so lim inf n ∫[0,1] fn = 1.

5.31. Example. Let E = [0, ∞) ∈ M(R), and for each n ≥ 1, let fn = − n1 χ[n,2n] .
Then each fn is measurable and (fn )n converges uniformly to f ≡ 0 on E. A fortiori,
(fn )n converges pointwise to f .
Neverthess,

∫ limninf fn = ∫ 0 = 0 > −1 = limninf ∫ fn .

E E E
This shows that we cannot simply drop the assumption of non-negativity of
the functions (fn )n in the hypotheses of Fatou’s Lemma and hope for the same
conclusion.

5.32. Suppose that E ∈ M(R) and that f, g ∶ E → R are measurable. Suppose

furthermore that 0 ≤ ∣f ∣ ≤ g a.e. on E, and that ∫E g < ∞, i.e. that g ∈ L1 (E, R).
We claim that f ∈ L1 (E, R).

Indeed, let B = {x ∈ E ∶ ∣f (x)∣ > g(x)}, so that B is measurable with mB = 0.

For x ∈ E ∖ B, we have that
f + (x) ≤ ∣f (x)∣ ≤ g(x) and f − (x) ≤ ∣f (x)∣ ≤ g(x).
From Lemma 5.14 and Proposition 5.15 we deduce that
+
∫ f =∫ f+ ≤ ∫ g = ∫ g < ∞,
E E∖B E∖B E

and similarly,
−
∫ f =∫ f− ≤ ∫ g = ∫ g < ∞.
E E∖B E∖B E
+ −
Hence f = f − f ∈ L1 (E, R).

The following result is also one of the major results of measure theory.
78 L.W. Marcoux Introduction to Lebesgue measure

5.33. Theorem. The Dominated Convergence Theorem.

Let E ∈ M(R) and (fn )∞ n=1 is a sequence in L(E, R). Suppose that there exists
g ∈ L1 (E, R) such that ∣fn ∣ ≤ g a.e. on E, n ≥ 1. Suppose furthermore that f ∶ E → R
is a function and that
f (x) = lim fn (x) a.e. on E.
n→∞

Then f ∈ L1 (E, R) and

∫ f = n→∞
lim ∫ fn .
E E
Proof.
Step One. As with the Monotone Convergence Theorem, our first goal is to isolate
the “bad” set of points where the convergence of the sequence fails, or where the
sequence of functions is not bounded above by g.
For each n ≥ 1, set En = {x ∈ E ∶ ∣fn (x)∣ > g(x)} and set E0 ∶= {x ∈ E ∶ f (x) ≠
limn→∞ fn (x)}. Finally, set E∞ ∶= {x ∈ E ∶ g(x) = ∞}. By hypothesis, mEn = 0,
0 ≤ n < ∞, while mE∞ = 0 as g ∈ L1 (E, R). Thus, if we set B ∶= E∞ ∪ (∪∞ n=0 En ),
then B is measurable and
∞ ∞
0 ≤ mB ≤ mE∞ + ∑ mEn = 0 + ∑ 0 = 0,
n=0 n=0
or in other words, mB = 0. From this it follows that for all n ≥ 1, ∫B fn = 0.
Moreover, f ∣B is measurable, and ∫B f = 0.
Note that for x ∈ H ∶= E ∖ B, we have that f (x) = limn→∞ fn (x) and we also
have ∣fn (x)∣ ≤ g(x). Given that mB = 0, it follows from Lemma 5.14 that f ∣H is
measurable if and only if f is. Moreover, mB = 0 also implies that

∫ f = n→∞
lim ∫ fn
H H
if and only if
lim ∫ fn .
∫ f = n→∞
E E
In other words, by replacing E with H if necessary, we may assume without loss
of generality that ∣fn (x)∣ ≤ g(x) < ∞ and that f (x) = limn→∞ fn (x) for all x ∈ E.
We shall assume that we have done this.
Step Two.
Note that g − fn ≥ 0 on E, and thus by Fatou’s Lemma 5.29,

∫ g − ∫ lim sup fn = ∫ limninf g − fn

E E n E

≤ lim inf ∫ g − fn
n E

= ∫ g − lim sup ∫ fn ,
E n E

which – given that f (x) = limn fn (x) for all x ∈ E – is equivalent to

lim sup ∫ fn ≤ ∫ f.
n E E
 5. LEBESGUE INTEGRATION 79

But g + fn ≥ 0 on E as well, and hence a second application of Fatou’s Lemma

yields

∫ g + ∫ limninf fn = ∫ limninf (g + fn )
E E E

≤ lim inf ∫ (g + fn )
n E

= ∫ g + lim inf ∫ fn ,
E n E
or equivalently
∫ f ≤ limninf ∫ fn .
E E
Putting these two inequalities together shows that

∫ f = n→∞
lim ∫ fn .
E E
◻
80 L.W. Marcoux Introduction to Lebesgue measure

Appendix to Section 5.
5.34. Let E ∈ M(R) and g ∈ L1 (E, R). We have observed that the set B ∶=
{x ∈ E ∶ g(x) ∈ {−∞, ∞}} has measure zero. When necessary, as it was in the proof
of the Lebesgue Dominated Convergence Theorem, we were able to simply “excise”
this set from the domain and concentrate our attention to the set E ∖ B. So why
introduce the extended real-numbers at all?
Convenience. Given an increasing sequence (fn )∞ n=1 in L1 (E, [0, ∞)), the point-
wise limit f (x) ∶= limn→∞ fn (x) need not be real-valued. By introducing the ex-
tended real numbers, we are able to treat the limit function f as simply another
measurable function.
When we define the Lp -spaces in Chapter 6, we shall define each Lp (E) as equiv-
alence classes of (extended real-valued) functions. However, each such equivalence
class will always admit a representative which is real-valued function. Truly, fortune
smiles upon us.
 5. LEBESGUE INTEGRATION 81

Exercises for Section 5.

Exercise 5.1. Let f (x) = x, x ∈ [0, ∞). Prove that ∫[0,∞) f = ∞.

Exercise 5.2. Assignment Question.

Let f (x) = x, x ∈ R. Calculate the Lebesgue integral ∫[0,1] f directly from the
definition – that is, do not appeal to Theorem 5.24.

Exercise 5.3.
Prove Lemma 5.14; that is, let E ∈ M(R) and let f, g and h ∶ E → [0, ∞] be
functions. Suppose that g and h are measurable.
(a) Suppose furthermore that E = X ⊍ Y , where X and Y are measurable. Set
f1 ∶= f ∣X and f2 ∶= f ∣Y . Then f is measurable if and only if both f1 and f2
are measurable. When such is the case,

∫ f = ∫ f1 + ∫ f2 .
E X Y
(b) If g ≤ h, then ∫E g ≤ ∫E h.
(c) If H ⊆ E is a measurable set, then

∫ g = ∫ g ⋅ χH ≤ ∫ g.
H E E

Exercise 5.4.
Prove that if f ∈ L1 (E, R), then
m(f −1 ({−∞})) = 0 = m(f −1 ({∞})).

Exercise 5.5.
sin x
Let f (x) = , x ≥ 1.
x
∞
(a) Prove that the improper Riemann integral ∫1 f (x)dx exists.
(b) Prove that the Lebesgue integral ∫[1,∞) f = ∞ does not exist.

Exercise 5.6. For n ≥ 1, define the function

fn ∶ [1, ∞) → R
⎧ 1
⎪
⎪ if 1 ≤ x ≤ en
x ↦ ⎨ nx .
⎪
⎪0 if x > en
⎩
Verify that the sequence (fn )∞
n=1 converges uniformly to f = 0 on [1, ∞).

Exercise 5.7.
Let E ∈ M(R). Suppose that f, g ∈ L1 (E, C), and that κ ∈ C. Prove that
(a) the function κf + g ∈ L1 (E, C) and

∫ κf + g = κ ∫ f + ∫ g.
E E E
82 L.W. Marcoux Introduction to Lebesgue measure

(b) Prove that ∣f ∣ ∈ L1 (E, C) and

∣∫ f ∣ ≤ ∫ ∣f ∣.
E E

Exercise 5.8.
Let E ∈ M(R). Show that a measurable function f ∶ E → C lies in L1 (E, C) if
and only if ∣f ∣ ∈ L1 (E, C).
Show that this fails if we do not assume that f is measurable.

Exercise 5.9.
The following special case of the Dominated Convergence Theorem is easily
derived from the Monotone Convergence Theorem.
Let E ∈ M(R) and suppose that (fn )∞
n=1 is a decreasing sequence in L(E, [0, ∞])
with f1 ∈ L1 (E, [0, ∞]). Thus
f1 ≥ f2 ≥ f3 ≥ ⋯ ≥ 0.
Define f ∶ E → [0, ∞] by f (x) = limn→∞ fn (x), x ∈ E.
Prove that f is measurable and that

∫ f = n→∞
lim ∫ fn .
E E

Exercise 5.10.
Let E ∈ M(R). If φ, ψ ∈ Simp(E, [0, ∞]) and κ = ∞, prove or disprove that

∫ κφ + ψ = κ ∫ φ + ∫ ψ.
E E E
 6. Lp SPACES 83

6. Lp Spaces

I’ve been on food stamps and welfare. Anybody help me out? No!
Craig T. Nelson

6.1. Functional analysis is the study of normed linear spaces and the continuous
linear maps between them. Amongst the most important examples of Banach spaces
are the so-called Lp -spaces, and it is to these that we now turn our attention. The
reader may wish to refresh his/her memory as to the definition of a seminorm on a
vector space X over K (Definition 1.2).
6.2. Example. Let E ⊆ R be a Lebesgue measurable set, and suppose that
mE > 0. Recall that
L1 (E, K) = {f ∶ E → K ∶ f is measurable and ∫ ∣f ∣ < ∞}.
E

Define the map

ν1 ∶ L1 (E, K) → R
f ↦ ∫E ∣f ∣.
Observe that
● ν1 (f ) ≥ 0 for all f ∈ L1 (E, K).
● ν1 (0) = ∫E ∣0∣ = 0.
● If k ∈ K, then ν1 (kf ) = ∫E ∣kf ∣ = ∣k∣ ∫E ∣f ∣ = ∣k∣ ν1 (f ).
● If f, g ∈ L1 (E, K), then
ν1 (f + g) = ∫ ∣f + g∣ ≤ ∫ ∣f ∣ + ∣g∣ = ∫ ∣f ∣ + ∫ ∣g∣ = ν1 (f ) + ν1 (g).
E E E E
It follows that ν1 defines a seminorm on L1 (E, K). But if ∅ ≠ F ⊆ E is a set of
measure zero (for example, F = {x0 } for some point x0 ∈ E), then χF =/ 0 and yet
ν1 (χF ) = ∫ ∣χF ∣ = mF = 0.
E
In other words, ν1 (⋅) does not define a norm on L1 (E, K).

6.3. Proposition. Let W be a vector space over the field K, and suppose that ν
is a seminorm on W. Let N ∶= {w ∈ W ∶ ν(w) = 0}. Then N is a linear manifold in
W and so W/N is a vector space over K, whose elements we denote by [x] ∶= x + N .
Furthermore, the map
∥ ⋅ ∥ ∶ W/N → R
[x] ↦ ν(x)
defines a norm on W/N .
Proof. Clearly 0 ∈ N and thus N ≠ ∅. Suppose that v, w ∈ N , and k ∈ K. Then
ν(kv + w) ≤ ν(kv) + ν(w) = ∣k∣ν(v) + ν(w) = 0,
84 L.W. Marcoux Introduction to Lebesgue measure

and so kv + w ∈ N , proving that N is a linear manifold in W.

From elementary linear algebra theory, we know that W/N is a vector space
under the operations [x] + [y] ∶= [x + y] and k[x] = [kx] for all x, y ∈ W, k ∈ K. We
normally refer to this as the quotient space of W by N .
To see that ∥ ⋅ ∥ defines a norm on W/N , we first check that this function is
well-defined. Indeed, suppose that [v] = [w] in W/N . Then v − w ∈ N , and so
ν(v − w) = 0. But then ∣ν(v) − ν(w)∣ ≤ ν(v − w) = 0 implies that ν(v) = ν(w), and so
∥ ⋅ ∥ is well-defined, as claimed.
Now
● ∥[x]∥ = ν(x) ≥ 0 for all [x] ∈ W/N , and ∥[0]∥ = ν(0) = 0.
● If ∥[x]∥ = 0, then ν(x) = 0, so x ∈ N and therefore [x] = [0].
● If [x] ∈ W/N and k ∈ K, then ∥k[x]∥ = ∥[kx]∥ = ν(kx) = ∣k∣ν(x) = ∣k∣ ∥[x]∥;
and finally
● If [x], [y] ∈ W/N , then
∥[x] + [y]∥ = ∥[x + y]∥ = ν(x + y) ≤ ν(x) + ν(y) = ∥[x]∥ + ∥[y]∥.
Thus ∥ ⋅ ∥ is a norm on W/N , which completes the proof.
◻

6.4. Let us return to Example 6.2, where we determined that ν1 (⋅) defines a
seminorm on L1 (E, K).
Suppose that g ∈ N1 (E, K) ∶= {f ∈ L1 (E, K) ∶ ν1 (f ) = 0}. Then ∫E ∣g∣ = 0, and
therefore g = 0 a.e. on E. Conversely, if g = 0 a.e. on E, then ∫E ∣g∣ = 0 and therefore
g ∈ N1 (E, K). In other words,
N1 (E, K) = {g ∈ L1 (E, K) ∶ g = 0 a. e. on E}.
Thus [g] = [h] in L1 (E, K)/N1 (E, K) if and only if g − h ∈ N1 (E, K), which is to
say that g = h a.e. on E. By Proposition 6.3, the map ∥[f ]∥ ∶= ν1 (f ) defines a norm
on L1 (E, K) ∶= L1 (E, K)/N1 (E, K).
6.5. Definition. The space L1 (E, K) = L1 (E, K)/N1 (E, K) defined above is
referred to as “L1 of E”, and it is a normed linear space.
It is crucial to remember that the elements of L1 (E, K) are cosets of L1 (E, K);
that is to say, they are equivalence classes of functions which are equal a.e. on E.
Given an element [f ] of L1 (E, K), one can not speak of the value of the function
at a point in E, since we are not dealing with functions!
Our next goal is to perform a similar construction on a family of spaces indexed
by positive real numbers 1 < p < ∞.

6.6. Definition. Let E ∈ M(R), so that E is Lebesgue measurable. Let 1 < p <
∞ be a real number, and set
Lp (E, K) ∶= {f ∈ L(E, K) ∶ ∫ ∣f ∣p < ∞} = {f ∈ L(E, K) ∶ ∣f ∣p ∈ L1 (E, K)}.
E
 6. Lp SPACES 85

6.7. We would like to verify that Lp (E, K) is a vector space for all 1 < p < ∞,
1/p
and that νp (f ) ∶= (∫E ∣f ∣p ) defines a seminorm on Lp (E, K). Then we can once
again appeal to Proposition 6.3 to obtain a normed linear space as a quotient of
Lp (E, K).
The proof of this is, however, somewhat technical, and will require a couple of
auxiliary results.

6.8. Definition. Let 1 ≤ p ≤ ∞. We associate to p the number 1 ≤ q ≤ ∞ as

follows:
● If p = 1, we set q = ∞;
−1
● if 1 < p < ∞, then we set q = (1 − p1 ) ; and
● if p = ∞, we set q = 1.
We say that q is the Lebesgue conjugate of p. With the convention that
1/∞ ∶= 0, we see that in all cases,
1 1
+ = 1.
p q

When 1 < p < ∞, we see that the above equation is equivalent to each of the
equations
● p(q − 1) = q and
● (p − 1)q = p.
While these are trivial algebraic manipulations, it will prove useful to keep them in
mind in the proofs below.

6.9. Lemma. Young’s Inequality. Let 1 < p < ∞ and denote by q the
Lebesgue conjugate of p. Let 0 < a, b ∈ R.
(a) Then
ap bq
ab ≤ + .
p q
(b) Equality holds in the above expression if and only if ap = bq .
1 1
Proof. Consider the function g ∶ (0, ∞) → R given by g(x) = xp + − x. Clearly g
p q
is differentiable on (0, ∞) with g ′ (x) = xp−1 − 1. Thus g is clearly strictly decreasing
on (0, 1) and it is strictly increasing on (1, ∞). Furthermore,
1 1
g(1) = + − 1 = 0.
p q
In particular, g(x) ≥ 0 for all x ∈ (0, ∞), and g(x) = 0 if and only if x = 1.
a
(a) Set x0 ∶= q−1 > 0. Then
b
ap 1 a
0 ≤ g(x0 ) = (q−1)p + − q−1 ,
pb q b
86 L.W. Marcoux Introduction to Lebesgue measure

so that
a ap 1
≤ + .
bq−1 pbq q
That is,
ap bq
ab ≤ + .
p q
(b) From above, equality holds if and only if g(x0 ) = 0, which happens if and
only if x0 = 1. But this is clearly equivalent to the condition that a = bq−1 ;
i.e. ap = bp(q−1) = bq .
◻
Recall from Proposition 4.10 that if E ∈ M(R) and f ∶ E → K is a measurable
function, then there exists a measurable function u ∶ E → T such that f = u ⋅ ∣f ∣.
Clearly, if K = R, then the range of u is contained in {−1, 1}. Let us denote by u
the function u(x) = u(x) for all x ∈ E.

6.10. Theorem. Hölder’s Inequality. Let E ∈ M(R) and 1 < p < ∞ be a

real number, and let q denote the Lebesgue conjugate of p.
(a) If f ∈ Lp (E, K) and g ∈ Lq (E, K), then f g ∈ L1 (E, K) and
ν1 (f g) ≤ νp (f ) νq (g),
1/p 1/q
where νp (f ) = (∫E ∣f ∣p ) and νq (g) = (∫E ∣g∣q ) .
(b) Suppose that H ∶= {x ∈ E ∶ f (x) ≠ 0} has positive measure. If
f ∗ ∶= νp (f )1−p u ∣f ∣p−1 ,
then f ∗ ∈ Lq (E, K), νq (f ∗ ) = 1, and

ν1 (f f ∗ ) = ∫ f f ∗ = νp (f ).
E
Remark. In an unfortunate coincidence of terminology, we shall also refer to f ∗ as
the Lebesgue conjugate function of f .

Proof.
We first observe that by Proposition 4.8, f g is measurable.
(a) Note that if f = 0 a.e or g = 0 a.e. on E, then f g = 0 a.e. on E and there is
nothing to prove. It is easy to verify that given 0 < α, β ∈ K, αf ∈ Lp (E, K)
and βg ∈ Lq (E, K). Suppose that we can find α0 , β0 ≠ 0 such that

∫ ∣(α0 f )(β0 g)∣ ≤ νp (α0 f )νq (β0 g).

E
By dividing both sides of this inequality by ∣α0 ∣ ∣β0 ∣, we clearly have that

∫ ∣f g∣ ≤ νp (f )νq (g).
E
This shows that, by choosing α0 = νp (f )−1 and β0 = νq (g)−1 , we may assume
without loss of generality that νp (f ) = 1 = νq (g).
 6. Lp SPACES 87

Now, by Young’s Inequality, Lemma 6.9, we have that

∣f ∣p ∣g∣q
∣f g∣ ≤ + ,
p q
and so
ν1 (f g) = ∫ ∣f g∣
E
1 1
≤ ∫ ∣f ∣p + ∫ ∣g∣q
p E q E
1 1
= νp (f )p + νq (g)q
p q
1 1
= +
p q
=1
= νp (f ) νq (g).
This completes the first half of the proof.
(b) Now let f ∗ be defined as above. Then f ∗ is measurable, being the product
of measurable functions, and recalling that (p − 1)q = p, we have

νq (f ∗ )q = ∫ ∣f ∗ ∣q
E
q
= ∫ (νp (f )1−p ∣f ∣p−1 )
E

= νp (f )(1−p)q ∫ ∣f ∣(p−1)q
E
−p p
= νp (f ) ∫ ∣f ∣
E
= νp (f )−p νp (f )p
= 1.
Finally,

ν1 (f f ∗ ) = ∫ ∣f f ∗ ∣ = ∫ f f ∗ = νp (f )1−p νp (f )p = νp (f ).
E E
◻

6.11. Theorem. Minkowski’s Inequality.

Let E ∈ M(R) be a measurable set and 1 < p < ∞. If f, g ∈ Lp (E, K), then
f + g ∈ Lp (E, K) and
νp (f + g) ≤ νp (f ) + νp (g).
Proof. That f + g is measurable is clear, as each of f and g is. Observe that for
any 0 ≤ a, b we have that
(a + b)p ≤ (2 max(a, b))p ≤ 2p (ap + bp ).
88 L.W. Marcoux Introduction to Lebesgue measure

Thus
∣f + g∣p ≤ (∣f ∣ + ∣g∣)p ≤ 2p (∣f ∣p + ∣g∣p ).
It follows that
νp (f + g)p = ∫ ∣f + g∣p ≤ 2p (νp (f )p + νp (g)p ) < ∞,
E
and therefore f + g ∈ Lp (E, K). Set h = f + g, and let h∗ denote the conjugate
function of h. Then h∗ ∈ Lq (E, K), νq (h) = 1 and ν1 (h ⋅ h∗ ) = νp (h).
From this and Hölder’s Inequality we deduce that
νp (f + g) = νp (h)
= ν1 (h ⋅ h∗ )
= ν1 ((f + g)h∗ )
≤ ν1 (f ⋅ h∗ ) + ν1 (g ⋅ h∗ )
≤ νp (f )νq (h∗ ) + νp (g)νq (h∗ )
= νp (f ) + νp (g).
◻

6.12. Corollary. Let E ∈ M(R), and 1 < p < ∞. Then Lp (E, K) is a vector
space, and νp (⋅) defines a seminorm on Lp (E, K).
Proof. Clearly Lp (E, K) ⊆ L(E, K) by definition, and since the latter is a vector
space, it suffices that we prove that Lp (E, K) ≠ ∅, and that f, g ∈ Lp (E, K) and
k ∈ K implies that kf + g ∈ Lp (E, K).
Let ζ ∶ E → K be the zero function ζ(x) = 0 for all x ∈ E. Clearly ζ ∈ Lp (E, K)
and hence Lp (E, K) ≠ ∅.
If f ∈ Lp (E, K) and k ∈ K, then kf is measurable by Proposition 4.8 and
p p p
∫ ∣kf ∣ = ∣k∣ ∫ ∣f ∣ < ∞,
E E
so that kf ∈ Lp (E, K). If g ∈ Lp (E, K) as well, then by Minkowski’s inequality,
kf + g ∈ Lp (E, K). Thus Lp (E, K) is a vector space.
With f, g ∈ Lp (E, K) and k ∈ K as above,
● νp (f ) = ∫E ∣f ∣p ≥ 0, and νp (ζ) = ∫E ζ p = ∫E 0 = 0.
●
1/p 1/p
νp (kf ) = (∫ ∣kf ∣p ) = (∣k∣p ∫ ∣f ∣p ) = ∣k∣νp (f ).
E E
● νp (f + g) ≤ νp (f ) + νp (g) by Minkowski’s inequality.
Thus νp (⋅) defines a seminorm on Lp (E, K). As was the case with L1 (E, K), if
∅ ≠ F ⊆ E is a subset of Lebsegue measure zero, then 0 ≠ χF is measurable and
1/p
νp (χF ) = (∫ ∣χF ∣p ) = ∫ χF = m(F ∩ E) = 0.
E E
Thus νp (⋅) is not a norm.
◻
 6. Lp SPACES 89

Once again, we shall appeal to Proposition 6.3 to obtain a normed linear space
from a semi-normed linear space.

6.13. Definition. Let E ∈ M(R) and 1 < p < ∞. We define the Lp -space
Lp (E, K) ∶= Lp (E, K)/Np (E, K),
where Np (E, K) = {f ∈ Lp (E, K) ∶ νp (f ) = 0}. The Lp -norm on Lp (E, K) is the
norm defined by
∥ ⋅ ∥p ∶ Lp (E, K) → R
.
[f ] ↦ νp (f )

6.14. Remark. Consider f ∈ Np (E, K), so that f is measurable and ∫E ∣f ∣p = 0.

It follows that ∣f ∣p = 0 a.e. on E, and hence that f = 0 a.e. on E. Conversely, if f is
measurable and f = 0 a.e., so that ∣f ∣p = 0 a.e. on E as well. But then f ∈ Np (E, K).
In other words, Np (E, K) = {f ∶ E → K ∶ f = 0 a. e. on E} for all 1 < p < ∞, and
[f ] = [g] in Lp (E, K) if and only if f, g ∈ Lp (E, K) and f = g a.e. on E.
For the sake of completeness, and in keeping with the literature, let us restate
the Hölder and Minkowski Inequalities for Lp (E, K).

6.15. Theorem. Hölder’s Inequality.

Let E ∈ M(R) and 1 < p < ∞. Let q denote the Lebesgue conjugate of p.
(a) If [f ] ∈ Lp (E, K) and [g] ∈ Lq (E, K), then [f ][g] ∶= [f g] ∈ L1 (E, K) is
well-defined and
∥[f g]∥1 ≤ ∥[f ]∥p ∥[g]∥q .
(b) If 0 ≠ [f ] ∈ Lp (E, K) and f ∗ is the conjugate function of f , then [f ∗ ] ∈
Lq (E, K), ∥[f ∗ ]∥q = 1, and
∥[f ][f ∗ ]∥1 = ∥[f ]∥p .
Proof. The only part that does not follow immediately from Theorem 6.10 is the
well-definedness of the operation [f ][g] = [f g], and this is left as an exercise for the
reader.
◻

6.16. Theorem. Minkowski’s Inequality.

Let E ∈ M(R) be a measurable set and 1 < p < ∞. If [f ], [g] ∈ Lp (E, K), then
[f + g] ∈ Lp (E, K) and
∥[f ] + [g]∥p ≤ ∥[f ]∥p + ∥[g]∥p .
We are now in position to prove the completeness of Lp (E, K) for all 1 ≤ p < ∞.
90 L.W. Marcoux Introduction to Lebesgue measure

6.17. Theorem. Let E ∈ M(R) and 1 ≤ p < ∞. Then Lp (E, K) is a Banach

space.
Proof.
As we have already noted, that Lp (E, K) is a normed linear space follows from
Proposition 6.3. There remains to show that it is complete. Recall from the As-
signments that a normed linear space is complete if and only if every absolutely
summable series is summable. Our proof will appeal to this result.

Suppose that ([fn ])∞ n=1 is a sequence in Lp (E, K) and suppose furthermore that
∞
γ ∶= ∑n=1 ∥[fn ]∥p < ∞. Our strategy will be to use the representatives fn , n ≥ 1 of
the elements [fn ] ∈ Lp (E, K) to produce a measurable function h ∈ Lp (E, K) such
that h(x) = ∑∞ n=1 fn (x) almost everywhere on E. Then we shall show that in fact,
[h] = ∑∞n=1 n in the sense of norm convergence in Lp (E, K).
[f ]

Step One. First we must show that ∑∞ n=1 fn (x) converges almost everywhere on E.
To that end, for each N ≥ 1, we set gN ∶= ∑Nn=1 ∣fn ∣, and observe that gN ∈ Lp (E, R)
by Corollary 6.12. Observe also that

0 ≤ g1 ≤ g2 ≤ g3 ≤ ⋯.
p
Set g∞ = supN ≥1 gN , so that 0 ≤ g∞ ∈ L(E, [0, ∞]). Then g∞ = supN ≥1 gNp , and so by
the Monotone Convergence Theorem 5.16,
p
∫ g∞ = Nlim ∫ gNp
E →∞ E

= lim ∫ (∣f1 ∣ + ∣f2 ∣ + ⋯ + ∣fN ∣)p

N →∞ E
= lim (νp (∣f1 ∣ + ∣f2 ∣ + ⋯ + ∣fN ∣))p
N →∞
≤ lim (νp (f1 ) + νp (f2 ) + ⋯ + νp (fN ))p
N →∞
= lim (∥[f1 ]∥p + ∥[f2 ]∥p + ⋯ + ∥[fN ]∥p )p
N →∞
≤ γ p < ∞.

From this it follows that B ∶= {x ∈ E ∶ g∞ (x) = ∞} has measure zero. Let H ∶= E ∖ B

and g ∶= χH ⋅ g∞ . Then g ∈ L(E, [0, ∞)), and g = g∞ a.e. on E. But then
p p p
∫ g = ∫ g∞ ≤ γ ,
E E

and so g ∈ Lp (E, K) – i.e., [g] ∈ Lp (E, K) – and ∥[g]∥p ≤ γ. Next, note that for
x ∈ H,
∞ ∞
∣ ∑ fn (x)∣ ≤ ∑ ∣fn (x)∣ = g∞ (x) = g(x) < ∞,
n=1 n=1

and so ∑∞
n=1 fn (x) ∈ K exists, by the completeness of K.
 6. Lp SPACES 91

Step Two. For each N ≥ 1, set hN ∶= χH ⋅ (∑N n=1 fn ), so that hN ∈ Lp (E, K) ⊆

N
L(E, K), and [hN ] = ∑n=1 [fn ]. Furthermore, for x ∈ H,

N
∣hN (x)∣ ≤ ∑ ∣fn (x)∣ ≤ g(x),
n=1

while for x ∈ B, ∣hN (x)∣ = 0 = g(x). Thus ∣hN ∣ ≤ g on E, and as a trivial consequence,
∣hN ∣p ≤ g p on E. From this we conclude that for each N ≥ 1,

p p p
∫ ∣hN ∣ ≤ ∫ g ≤ γ .
E E

Define h(x) ∶= limN →∞ hN (x) ∈ K, x ∈ E. Note that h is measurable, being the

limit of measurable functions. From above, ∣h∣ ≤ g on E, and thus

p p p
∫ ∣h∣ ≤ ∫ g ≤ γ < ∞.
E E

It follows that h ∈ Lp (E, K); i.e. [h] ∈ Lp (E, K).

Step Three. Recall that we are trying to prove that ∑∞ n=1 [fn ] converges in
Lp (E, K). We chose a specific set of representatives, namely the fn ’s themselves,
and we showed that there exists an element h ∈ Lp (E, K) such that almost every-
where on E, namely on H ⊆ E, h = ∑∞ n=1 fn as a pointwise limit of functions. If we
can show that

N
lim ∥[h] − [hN ]∥p = lim ∥[h] − ∑ [fn ]∥p = 0,
N →∞ N →∞ n=1

then obviously we are done.

Considering that ∣hM − hN ∣p ≤ (∣hM ∣ + ∣hN ∣)p ≤ (g + g)p for all M and N and that
p
∫E (2∣g∣) < ∞ by virtue of 0 ≤ g being in Lp (E, K), an application of the Lebesgue
92 L.W. Marcoux Introduction to Lebesgue measure

Dominated Convergence Theorem 5.33 shows that

∥[h] − [hN ]∥p = νp (h − hN )
1/p
= (∫ ∣h − hN ∣p )
E
1/p
= (∫ lim ∣hM − hN ∣p )
E M →∞
1/p
= ( lim ∫ ∣hM − hN ∣p )
M →∞ E
1/p
= lim (∫ ∣hM − hN ∣p )
M →∞ E
= lim ∥[hM ] − [hN ]∥p
M →∞
M
= lim ∥ ∑ [fn ]∥
M →∞ n=N +1
p
M
≤ lim ∑ ∥[fn ]∥p
M →∞ n=N +1
∞
= ∑ ∥[fn ]∥p .
n=N +1
Since ∑∞
n=1 ∥[fn ]∥p = γ < ∞ by hypothesis, it is clear that
∞
lim ∥[h] − [hN ]∥p ≤ lim ∑ ∥[fn ]∥p = 0.
N →∞ N →∞ n=N +1

As noted above, this completes the proof.

◻
The case of p = ∞.
6.18. Definition. Let E ∈ M(R) and suppose that f ∈ L(E, K). We define the
essential supremum of f to be
ν∞ (f ) = ess sup(f ) ∶= inf{γ > 0 ∶ m{x ∈ E ∶ ∣f (x)∣ > γ} = 0}.
Set L∞ (E, K) = {f ∈ L(E, K) ∶ ν∞ (f ) < ∞}.

6.19. Examples.
(a) Let E = R and f = χQ be the characteristic function of the rationals. For
any γ > 0, m{x ∈ R ∶ ∣χQ (x)∣ > γ} ≤ mQ = 0, and so ν∞ (χQ ) = 0.
Clearly there was nothing special about Q in this example, other than
the fact that this set has Lebesgue measure zero.
(b) Suppose that a < b ∈ R and that f ∈ C([a, b], K).
We claim that f ∈ L∞ ([a, b], K) and that ν∞ (f ) = ∥f ∥sup . Indeed,
every continuous function is measurable, so f ∈ L([a, b], K). Also, [a, b]
being a compact set, there exists x0 ∈ [a, b] such that ∣f (x0 )∣ = ∥f ∥sup . By
 6. Lp SPACES 93

continuity of f , given ε > 0, there exists δ > 0 such that x ∈ [a, b] and
∣x − x0 ∣ < δ implies ∣f (x)∣ > ∣f (x0 )∣ − ε. Whether or not x0 is one of the
endpoints of the interval, it follows that there exist c < d ∈ (a, b) such that
x ∈ [c, d] implies that ∣f (x)∣ > ∣f (x0 )∣ − ε. But then
m{x ∈ [a, b] ∶ ∣f (x)∣ > ∣f (x0 )∣ − ε} ≥ d − c > 0,
and so ν∞ (f ) ≥ ∥f ∥sup − ε. Since ε > 0 was arbitrary, ν∞ (f ) ≥ ∥f ∥sup .
If γ > ∥f ∥sup , then m{x ∈ E ∶ ∣f (x)∣ > γ} = m∅ = 0, and thus ν∞ (f ) ≤ γ.
Hence ν∞ (f ) ≤ ∥f ∥sup . Combining these inequalities,
ν∞ (f ) = ∥f ∥sup .
In particular, f ∈ L∞ ([a, b], K).

6.20. Proposition. Let E ∈ M(R). Then L∞ (E, K) is a vector space over K,

and that ν∞ (⋅) is a seminorm on L∞ (E, K).
Proof. Since L∞ (E, K) ⊆ L(E, K), and since the latter is a vector space, we need
only apply the Subspace Test from linear algebra to conclude that L∞ (E, K) is a
vector space.
It is clear from the definition that ν∞ (f ) ≥ 0 for all f ∈ L∞ (E, K), and clearly
the constant function ζ(x) = 0, x ∈ E lies in L∞ (E, K), with ν∞ (ζ) = 0. Thus
L∞ (E, K) ≠ ∅.
Suppose next that f, g ∈ L∞ (E, K), and that 0 ≠ k ∈ K. Then kf ∈ L(E, K), and
a moment’s thought should convince the reader that this implies that
ν∞ (kf ) = inf{γ > 0 ∶ m{x ∈ E ∶ ∣kf (x)∣ > γ} = 0}
= inf{∣k∣ δ > 0 ∶ m{x ∈ E ∶ ∣kf (x)∣ > ∣k∣ δ} = 0}
= ∣k∣ inf{δ > 0 ∶ m{x ∈ E ∶ ∣f (x)∣ > δ} = 0}
= ∣k∣ ν∞ (f ) < ∞.
Hence kf ∈ L∞ (E, K) for all 0 ≠ k ∈ K. Since 0 ν∞ (f ) = 0 = ν∞ (ζ) = ν∞ (0 ⋅ f ) for
all f ∈ L∞ (E, K), we see that kf ∈ L∞ (E, K) and ν∞ (kf ) = ∣k∣ν∞ (f ) for all k ∈ K.
Finally, suppose that f, g ∈ L∞ (E, K) and let α > ν∞ (f ), β > ν∞ (g) be arbitrary.
Let Ef ∶= {x ∈ E ∶ ∣f (x)∣ > α} and Eg ∶= {x ∈ E ∶ ∣g(x)∣ > β}, so that mEf = 0 = mEg .
If x ∈ H ∶= E ∖ (Ef ∪ Eg ), then ∣(f + g)(x)∣ ≤ ∣f (x)∣ + ∣g(x)∣ ≤ α + β, and so
{x ∈ E ∶ ∣(f + g)(x)∣ > α + β} ⊆ Ef ∪ Eg .
From this it immediately follows that
m{x ∈ E ∶ ∣(f + g)(x)∣ > α + β} ≤ mEf + mEg = 0,
whence ν∞ (f + g) ≤ α + β. Since α > ν∞ (f ) and β > ν∞ (g) were arbitrary,
ν∞ (f + g) ≤ ν∞ (f ) + ν∞ (g).
This completes the proof.
◻
94 L.W. Marcoux Introduction to Lebesgue measure

6.21. Definition. As we did with the previous Lp -spaces, 1 ≤ p < ∞, we define

the subspace
N∞ (E, K) ∶= {f ∈ L∞ (E, K) ∶ ν∞ (f ) = 0}
of L∞ (E, K), and we denote by [f ] the equivalence class in
L∞ (E, K) ∶= L∞ (E, K)/N∞ (E, K)
of f ∈ L∞ (E, K).

Yet again we may appeal to Proposition 6.3 to conclude the following:

6.22. Theorem. Let E ∈ M(R). Then L∞ (E, K) is a normed-linear space,
where for [f ] ∈ L∞ (E, K), we set
∥[f ]∥∞ ∶= ν∞ (f ).
1
6.23. Let f ∈ L∞ (E, K). If we define Fn ∶= {x ∈ E ∶ ∣f (x)∣ > ν∞ (f ) + }, then
n
mFn = 0 for all n ≥ 1, and F ∶= ∪∞
n=1 Fn = {x ∈ E ∶ ∣f (x)∣ > ν∞ (f )} satisfies
∞ ∞
mF ≤ ∑ mFn = ∑ 0 = 0.
n=1 n=1
From this it follows that given [f ] ∈ L∞ (E, K), we may always choose a represen-
tative g ∈ [f ] in such a way as to guarantee that ∣g(x)∣ ≤ ∥[f ]∥∞ for all x ∈ E. Indeed,
given f ∈ L∞ (E, K), then as noted above, the set F = {x ∈ E ∶ ∣f (x)∣ > ν∞ (f )} has
measure zero. As such, the function g ∶= χE∖F ⋅ f is measurable, and differs from
f only on F , whence [g] = [f ], and clearly ∣g(x)∣ ≤ ν∞ (f ) = ν∞ (g) = ∥[g]∥∞ for all
x ∈ E.
Moreover, it is readily seen that ν∞ (f ) = 0 if and only if f = 0 almost everywhere
on E, and thus
N∞ (E, K) = {f ∈ L(E, K) ∶ f = 0 a.e. on E}.

There remains to show that L∞ (E, K) is complete.

6.24. Theorem. Let E ∈ M(R). Then L∞ (E, K) is a Banach space.
Proof. Let ([fn ])∞ n=1 be an absolutely summable sequence in L∞ (E, K), and set
γ ∶= ∑∞n=1 ∥[fn ]∥∞ < ∞. By our work in the Assignments, we see that it suffices to
prove that limN →∞ ∑N n=1 [fn ] ∈ L∞ (E, K) exists.
By the argument of the above paragraph, we may assume without loss of gen-
erality that for each x ∈ E and each n ≥ 1, ∣fn (x)∣ ≤ ν∞ (fn ) = ∥[fn ]∥∞ , and thus
∞ ∞
∑ ∣fn (x)∣ ≤ ∑ ∥[fn ]∥∞ = γ.
n=1 n=1
∞
As such, for each x ∈ E, the series ∑n=1 fn (x)
is absolutely summable, and hence
summable by the completeness of K. Define a function f ∶ E → K via
∞
f (x) ∶= ∑ fn (x), x ∈ E.
n=1
 6. Lp SPACES 95

For each N ≥ 1, set hN = ∑N n=1 fN ∈ L∞ (E, K). Since f is the pointwise limit of the
measurable functions hN , we see that f ∈ L(E, K). Moreover, the above estimate
shows that ν∞ (f ) ≤ γ < ∞, and thus f ∈ L∞ (E, K).
Let ε > 0 and choose N0 > 0 such that ∑∞ n=N0 +1 ∥[fn ]∥∞ < ε. For all N > N0 , we
have that
∥[f ] − [hN ]∥∞ = ν∞ (f − hN ).
But for x ∈ E,
∞
∣f (x) − hN (x)∣ = ∣ ∑ fn (x)∣
n=N +1
∞
≤ ∑ ∣fn (x)∣
n=N +1
∞
≤ ∑ ∥[fn ]∥∞
n=N +1
< ε.
Thus ∥[f ] − [hN ]∥∞ < ε, N ≥ N0 .
This shows that [f ] = limN →∞ [hN ] = limN →∞ ∑N
n=1 [fn ], as required.
◻

6.25. Recall that if E ∈ M(R), 1 < p < ∞, f ∈ Lp (E, K) and g ∈ Lq (E, K), where
q is the Lebesgue conjugate of p, then Hölder’s Inequality (Theorem 6.10) states
that f g ∈ L1 (E, K) and that
ν1 (f g) ≤ νp (f ) νq (g).
Let us obtain a version of this inequality for p = 1, which will prove especially
useful when we examine Fourier series in later chapters.

6.26. Theorem. Hölder’s Inequality. Let E ∈ M(R) with mE > 0.

(a) If f ∈ L1 (E, K) and g ∈ L∞ (E, K), then f g ∈ L1 (E, K) and
ν1 (f g) ≤ ν1 (f ) ν∞ (g).
(b) There exists a function f ∗ ∈ L∞ (E, K) such that ν∞ (f ∗ ) = 1 and
ν1 (f ⋅ f ∗ ) = ∫ f ⋅ f ∗ = ν1 (f ).
E
Proof.
(a) By the comments of Paragraph 6.23, we know that we may find g0 ∈
L∞ (E, K) such that g0 = g a.e. on E and ∣g0 (x)∣ ≤ ν∞ (g) = ν∞ (g0 ) for
all x ∈ E. Since g = g0 a.e. on E implies that ∣f g∣ = ∣f g0 ∣ a.e. on E, we find
that
∫ ∣f g∣ = ∫ ∣f g0 ∣.
E E
Because of this, we may assume without loss of generality (by replacing g
by g0 if necessary) that ∣g(x)∣ ≤ ν∞ (g) for all x ∈ E.
96 L.W. Marcoux Introduction to Lebesgue measure

Note that by Proposition 4.8, f g is measurable. But then

ν1 (f g) = ∫ ∣f g∣ ≤ ∫ ∣f ∣ ν∞ (g) = ν∞ (g) ∫ ∣f ∣ = ν1 (f ) ν∞ (g).
E E E
(b) Arguing as in Proposition 4.10, we may find a measurable function u ∶ E →
T such that
f = u ⋅ ∣f ∣.
But then with f ∗ ∶= u (i.e. f ∗ (x) = u(x) for all x ∈ E), ν∞ (f ∗ ) = 1,
∣f ∣ = f ⋅ f ∗ , and so
ν1 (f f ∗ ) = ∫ ∣f ⋅ f ∗ ∣ = ∫ f ⋅ f ∗ = ∫ ∣f ∣ = ν1 (f ).
E E E
◻
∗
Once again, we shall refer to f as the Lebesgue conjugate function of f .
As before, we obtain an immediate corollary.
6.27. Corollary. Let E ∈ M(R). If [f ] ∈ L1 (E, K) and [g] ∈ L∞ (E, K), then
[f ][g] ∶= [f g] ∈ L1 (E, K) is well-defined and
∥[f g]∥1 ≤ ∥[f ]∥1 ∥[g]∥∞ .
6.28. Corollary. Suppose that a < b ∈ R. Consider h ∈ C([a, b], K) and f ∈
L1 ([a, b], K).
Then h ⋅ f ∈ L1 ([a, b], K) and
ν1 (h ⋅ f ) ≤ ν1 (f ) ν∞ (h) = ν1 (f ) ∥h∥sup .
Proof. Clearly h is measurable and h ∈ L∞ ([a, b], K) with ∥h∥sup = ν∞ (h), by
Example 6.19.
The result now follows from Hölder’s Inequality, Theorem 6.26 above.
◻

6.29. It is interesting to consider the relationship between Lp -spaces for differing

values of p. In the section below, we consider the case where the underlying set E
has finite measure; the case where the measure of E is infinite is left to the exercises.

Let E ∈ M(R) and suppose that mE < ∞. Let [f ] ∈ L∞ (E, K). Then f ∈
L(E, K) and ∣f (x)∣ ≤ ∥[f ]∥∞ almost everywhere on E. For 1 ≤ p < ∞,
p p p
∫ ∣f ∣ ≤ ∫ ∥[f ]∥∞ = ∥[f ]∥∞ mE < ∞,
E E
proving that [f ] ∈ Lp (E, K), with
∥[f ]∥p ≤ ∥[f ]∥∞ (mE)1/p .
Thus L∞ (E, K) ⊆ Lp (E, K), 1 ≤ p < ∞ when mE < ∞.

Next, suppose that 1 ≤ p < r < ∞, and that [g] ∈ Lr (E, K). Again, g ∈ L(E, K)
and
p r p/r r r r
∫ ∣g∣ = ∫ (∣g∣ ) ≤ ∫ max(1, ∣g∣ ) ≤ ∫ (1 + ∣g∣ ) = mE + ∥[g]∥r < ∞.
E E E E
 6. Lp SPACES 97

It follows that [g] ∈ Lp (E, K); i.e. Lr (E, K) ⊆ Lp (E, K).

6.30. Next, suppose that a < b ∈ R. It follows from Example 6.19 that
[C([a, b], K)] ∶= {[f ] ∶ f ∈ C([a, b], K)} ⊆ L∞ ([a, b]).
Recall that
R∞ ([a, b], K) = {f ∶ [a, b] → K ∶ f is Riemann-integrable and bounded}.
As we saw in Corollary 5.25, f ∈ L([a, b], K) and thus [f ] ∈ L∞ ([a, b], K), by virtue
of its being bounded.

Our next major goal is to prove that the space [C([a, b], K)] is dense in each
of the spaces Lp ([a, b), K), 1 ≤ p < ∞. We shall accomplish this through a series of
approximations.
6.31. Before proving our next result, we introduce a bit of notation. Given
E ∈ M(R) and 1 ≤ p ≤ ∞, we set
Simpp (E, K) = Simp(E, K) ∩ Lp (E, K).
Since Simp(E, K) and Lp (E, K) are vector spaces over K, so is Simpp (E, K).
We leave it to the exercises for the reader to prove that if mE < ∞ or if p = ∞,
then Simpp (E, K) = Simp(E, K).

6.32. Proposition. Let E ∈ M(R) be a Lebesgue measurable set and 1 ≤ p < ∞.

Then [Simpp (E, K)] ∶= {[φ] ∶ φ ∈ Simpp (E, K)} is dense in (Lp (E, K), ∥ ⋅ ∥p ).
Proof.
Step One. Suppose that f ∈ Lp (E, [0, ∞)), and let ε > 0. From our Assignment
Questions, we know that we may find functions φn ∈ Simp(E, [0, ∞)), n ≥ 1 such
that
0 ≤ φ1 ≤ φ2 ≤ ⋯ ≤ f,
and f (x) = limn→∞ φn (x), x ∈ E.
Since
p p
∫ ∣φn ∣ ≤ ∫ ∣f ∣ < ∞,
E E
we see that φn ∈ Simpp (E, [0, ∞)), n ≥ 1.
Moreover,
∣f − φn ∣p ≤ ∣f ∣p , n ≥ 1,
and so by the Lebesgue Dominated Convergence Theorem 5.33,
lim νp (f − φn )p = lim ∫ ∣f − φn ∣p = 0.
n→∞ n→∞ E
From this we clearly deduce that given ε > 0, we can find φ ∈ Simpp (E, [0, ∞))
such that νp (f − φ) < ε.
Step Two.
Now let [g] ∈ Lp (E, K) be arbitrary, and note that g ∈ Lp (E, K). Recall that we
may then write
g = (g1 − g2 ) + i(g3 − g4 ),
98 L.W. Marcoux Introduction to Lebesgue measure

where gk ∈ Lp (E, [0, ∞)), 1 ≤ k ≤ 4. (If g is real-valued, then g3 = g4 = 0.)

Given ε > 0, by Step One, we can find φk ∈ Simpp (E, [0, ∞)) such that
ε
νp (gk − φk ) < , 1 ≤ k ≤ 4.
4
Let φ ∶= (φ1 − φ2 ) + i(φ3 − φ4 ). Then φ ∈ Simpp (E, K) (since, as noted above,
Simpp (E, K) is a vector space over K), and
∥[f ] − [φ]∥p = νp (f − φ)
= νp (((f1 − f2 ) + i(f3 − f4 )) − ((φ1 − φ2 ) + i(φ3 − φ4 )))
= νp ((f1 − φ1 ) − (f2 − φ2 ) + i(f3 − φ3 ) − i(f4 − φ4 ))
4
≤ ∑ νp (fk − φk )
k=1
4
ε
<∑
k=1 4
= ε.
In particular, [Simpp (E, K)] is dense in (Lp (E, K), ∥ ⋅ ∥p ), 1 ≤ p < ∞.
◻

6.33. Proposition. Let E ∈ M(R) be a Lebesgue measurable set. Then

[Simp(E, K)] ∶= {[φ] ∶ φ ∈ Simp(E, K)}
is dense in (L∞ E, K), ∥ ⋅ ∥∞ ).
Proof.
Step One. Suppose first that f ∈ L∞ (E, [0, ∞)), and that there exists M > 0
such that f (x) ≤ M for all x ∈ E.
Let ε > 0 and choose N > 0 such that N1 < ε. Decompose the interval [0, M ]
into M N intervals of equal length N1 , namely Ik ∶= [ k−1 k
N , N ), 1 ≤ k ≤ M N − 1, and
IM N ∶= [M − N1 , M ].
Set Hk ∶= f −1 (Ik ), 1 ≤ k ≤ M N , so that Hk is measurable by the measurability
of f , and set
MN
k−1
φ ∶= ∑ ( )χHk .
k=1 N
Clearly φ ∈ Simp(E, [0, ∞)), and
1
∣f (x) − φ(x)∣ ≤ <ε for all x ∈ E.
N
In particular,
ν∞ (f − φ) < ε.
Step Two. Now suppose that [g] ∈ L∞ (E, K) is arbitrary, so that g ∈
L∞ (E, K). As we have seen in paragraph 6.23, we may assume without loss of
generality that ∣g(x)∣ ≤ ν∞ (g) for all x ∈ E.
 6. Lp SPACES 99

Moreover, as was the case in Proposition 6.32, we may write

g = (g1 − g2 ) + i(g3 − g4 ),
where gk ∈ L∞ (E, [0, ∞)), and ∣gk (x)∣ ≤ ν∞ (g) for all x ∈ E and 1 ≤ k ≤ 4. Given
ε > 0, by Step One, we can find φk ∈ L∞ (E, [0, ∞)) such that ν∞ (gk − φk ) < 4ε ,
1 ≤ k ≤ 4. The remainder of the proof is similar to that of the case where 1 ≤ p < ∞,
and we leave it as an exercise for the reader.
We conclude that [Simp(E, R)] is dense in (L∞ (E, K), ∥ ⋅ ∥∞ ).
◻
We next consider a simple, but useful, result which we shall have occasion to
apply twice. Recall that a linear manifold in a normed linear space X is a vector
subspace which may or may not be closed in the norm-topology.

6.34. Lemma. Let (X, ∥⋅∥) be a Banach space, and suppose that B ⊆ X satisfies
span B = X. Suppose also that L ⊆ X is a linear manifold and that B ⊆ L. Then
X = L.
Proof. The key is to observe that since L is a linear manifold, L is a closed subspace
of X. Indeed, since X is a vector space, we may apply the Subspace Test from first-
year linear algebra. That is, it suffices to prove that L ≠ ∅, and that if x, y ∈ L and
κ ∈ K, then kx + y ∈ L.

To that end, note that 0 ∈ L implies that 0 ∈ L ≠ ∅. Let κ ∈ K, x, y ∈ L, and

choose sequences (xn )n , (yn )n ∈ L such that x = limn xn and y = limn yn .
Then κxn + yn ∈ L for all n ≥ 1 since L is a linear manifold, and
lim ∥(κx + y) − (κxn + yn )∥ = lim ∥κ(x − xn ) + (y − yn )∥
n n
≤ lim ∣κ∣ ∥x − xn ∥ + ∥y − yn ∥
n
= ∣κ∣ lim ∥x − xn ∥ + lim ∥y − yn ∥
n n
= ∣κ∣0 + 0
= 0,

whence κx + y = limn (κxn + yn ) ∈ L. Thus L is a closed subspace of X.

Since B ⊆ L and the latter is a closed subspace of X, it follows that

X = span B ⊆ L.
But L is obviously contained in X, and thus the two sets are equal.
◻
100 L.W. Marcoux Introduction to Lebesgue measure

6.35. Proposition. Let a < b ∈ R. If 1 ≤ p < ∞, then [Step([a, b], K)] is dense
in (Lp ([a, b], K), ∥ ⋅ ∥p ).
Proof. Let B ∶= {[χH ] ∶ H ⊆ E measurable}. Then span B = [Simp([a, b], K)], and
by Proposition 6.32, this is dense in Lp ([a, b], K). By Lemma 6.34, it suffices to
show that every [χH ] can be approximated arbitrarily well in the ∥ ⋅ ∥p -norm by
elements of [Step([a, b], K)].
Let H ⊆ [a, b] be a measurable set and ε > 0. Recall from Theorem 3.14 that
we can find an open set G ⊆ R such that H ⊆ G, and m(G ∖ H) < 2ε . Write G
as a disjoint union of open intervals G = ∪∞ n=1 (an , bn ). (Note that each interval
is finite, since mH ≤ m[a, b] < ∞, and m(G ∖ H) < ∞), implying that m(G) =
m(H) + m(G ∖ H) < ∞.)
Thus m(G) = ∑∞ n
n=1 (bn − an ) < ∞. Set Gn = ∪k=1 (ak , bk ), n ≥ 1, and choose N ≥ 1
such that
∞
ε
m(G ∖ GN ) = ∑ (bn − an ) < .
n=N +1 2
Set ψ = χGN ∩[a,b] and observe that ψ ∈ Step([a, b], R).
Moreover,
⎧
⎪1 = ∣1 − 0∣ if x ∈ H ∖ GN
⎪
⎪
⎪
⎪
⎪1 = ∣0 − 1∣
⎪ if x ∈ (GN ∩ [a, b]) ∖ H
∣χH (x) − ψ(x)∣ = ⎨
⎪
⎪
⎪0 = ∣0 − 0∣ if x ∈/ (GN ∪ H)
⎪
⎪
⎪
⎩0 = ∣1 − 1∣
⎪ if x ∈ (GN ∩ H).
It follows that
νp (χH − ψ) = ∫ ∣χH − ψ∣p
E

= ∫ ∣χH − ψ∣
E
= m(H ∖ GN ) + m((GN ∩ [a, b]) ∖ H)
≤ m(H ∖ GN ) + m(GN ∖ H)
≤ m(G ∖ GN ) + m(G ∖ H)
ε ε
< +
2 2
= ε.
It follows that ∥[χH ] − [ψ]∥p < ε, thus showing that [Step([a, b], R)] is dense in
(Lp ([a, b], R), ∥ ⋅ ∥p ).
◻
Observe that Lemma 6.34 greatly simplified the proof; instead of approximating
an arbitrary element in Lp ([a, b], K), or even an arbitrary element of [Simp(E, K)]
by (equivalence classes of) step functions, we reduced the problem to that of ap-
proximating characteristic functions of measurable sets. In the same way, in proving
that [C([a, b], K)] is dense in (Lp ([a, b], K), ∥⋅∥p ), 1 ≤ p < ∞, Lemma 6.34 will reduce
 6. Lp SPACES 101

the problem to that of approximating the characteristic function of an interval by a

continuous function in the νp (⋅)-seminorm.

6.36. Theorem. Let a < b ∈ R. If 1 ≤ p < ∞, then [C([a, b], K)] is dense in
(Lp ([a, b], K), ∥ ⋅ ∥p ).
Proof. Let B ∶= {[χ[r,s] ∶ a ≤ r < s ≤ b}. Then span B = [Step([a, b], K)], which is
dense in Lp ([a, b], K) by Proposition 6.35 above.
We may therefore appeal once again to Lemma 6.34, which implies that we
need only show that every such [χ[r,s] ] can be approximated arbitrarily well in the
∥ ⋅ ∥p -norm by elements of [C([a, b], K)].
To begin, choose a ≤ r < s ≤ b, and to dispense with a technicality, choose M > 0
2
such that r + M < s. For n ≥ M , define
⎧
⎪0 if a ≤ x ≤ r
⎪
⎪
⎪
⎪
⎪
⎪n(x − r) if r < x ≤ r + n1
⎪
⎪
⎪
fn (x) = ⎨1 if r + n1 < x ≤ s − n1
⎪
⎪
⎪
⎪
⎪
⎪n(s − x) if s − n1 < x ≤ s
⎪
⎪
⎪
⎩0
⎪ if s < x ≤ b.
(If it not entirely clear from the outset why we picked such a sequence of func-
tions, the reader would be well-advised to graph them. How could you approximate
a step function by a continuous function in a simpler way? The choice of only
defining fn for n ≥ M is to ensure that fn ≤ χ[r,s] for all such n.)
Observe that fn is continuous for each n ≥ M , being piecewise linear. Also,
x ∈/ [r, r + n1 ] ∪ [s − n1 , s] implies that
∣fn (x) − χ[r,s] (x)∣ = 0,
and for all x ∈ [a, b], ∣fn (x) − χ[r,s] (x)∣ ≤ 1. It follows that for all n ≥ M ,
1
p
νp (fn − χ[r,s] ) = (∫ ∣fn (x) − χ[r,s] (x)∣p )
[a,b]
1
p
≤ (∫ 1)
[r,r+ n
1
]∪[s− n
1
,s]
1
2 p
=( ) .
n
Thus 1
2 p
0 ≤ lim ∥[fn ] − [χ[r,s] ]∥p ≤ lim ( ) = 0,
n→∞ n→∞ n
and hence
lim [fn ] = [χ[r,s] ]
n→∞
in (Lp ([a, b], R), ∥ ⋅ ∥p ), completing the proof.
◻
102 L.W. Marcoux Introduction to Lebesgue measure

6.37. This leads to the following interesting result. Recall first that a topological
space is said to be separable if it admits a countable dense subset.
Secondly, recall that if (X, d) is a separable metric space, δ > 0 is a positive real
number and {xλ ∶ λ ∈ Λ} ⊆ X satisfies d(xα , xβ ) ≥ δ for all α ≠ β ∈ Λ, then Λ is
countable. The proof is left to the exercises.
6.38. Corollary. Let a < b ∈ R.
(a) If 1 ≤ p < ∞, then (Lp ([a, b], K), ∥ ⋅ ∥p ) is separable.
(b) The space (L∞ ([a, b], K), ∥ ⋅ ∥∞ ) is not separable.
Proof.
(a) First fix 1 ≤ p < ∞.
Recall from Section 6.29 that L∞ ([a, b], K) ⊆ Lp ([a, b], K). Moreover,
the proof of this assertion showed that if [f ], [g] ∈ L∞ ([a, b], K), then
∥[f ] − [g]∥p = ∥[f − g]∥p ≤ ∥[f ] − [g]∥∞ (b − a)1/p .
Let ε > 0, and let [h] ∈ Lp ([a, b], K). We know from Theorem 6.36 that
[C([a, b], K)] is dense in Lp ([a, b], K) with respect to the p-norm.
Thus we can find g ∈ C([a, b], K) such that
ε
∥[h] − [g]∥p < .
3
Of course, by the Weierstraß Approximation Theorem and Example 6.19
above, we know that we can find a polynomial p(x) = p0 + p1 x + ⋯ + pm xm
such that
ε
∥[g] − [p]∥∞ = ∥g − p∥sup < .
3(b − a)1/p
By Exercise 6.9 below, we can find a polynomial q(x) = q0 + q1 x + ⋯ + qm xm
such that qk ∈ Q + iQ for all 0 ≤ k ≤ m and
ε
∥[p] − [q]∥∞ = ∥p − q∥sup < .
3(b − a)1/p
Finally,
∥[h] − [q]∥p ≤ ∥[h] − [g]∥p + ∥[g] − [p]∥p + ∥[p] − [q]∥p
≤ ∥[h] − [g]∥p + ∥[g] − [p]∥∞ (b − a)1/p + ∥[p] − [q]∥∞ (b − a)1/p
ε ε ε
≤ +[ 1/p
] (b − a)1/p + [ ] (b − a)1/p
3 3(b − a) 3(b − a)1/p
ε ε ε
= + +
3 3 3
= ε.
Thus the family (Q + iQ)[x] of all polynomials with (complex) rational
coefficients, when viewed as continuous functions on [a, b], has the property
that [(Q + iQ)[x]] is dense in (Lp ([a, b], K), ∥ ⋅ ∥p ). Since there are at most
countably many elements in (Q + iQ)[x] and therefore in [(Q + iQ)[x]], we
see that Lp ([a, b], K) is separable, 1 ≤ p < ∞.
 6. Lp SPACES 103

(b) Now consider p = ∞, and suppose that a ≤ r1 < s1 ≤ b and a ≤ r2 < s2 ≤ b.

Suppose furthermore that either r1 ≠ r2 or that s1 ≠ s2 . By Exercise 6.10
below, the symmetric difference
[r1 , s1 ]∆[r2 , s2 ]
of the intervals [r1 , s1 ] and [r2 , s2 ] contains an interval, say [u, v] ⊆ [a, b],
with u < v.
For all x ∈ [u, v], ∣χ[r1 ,s1 ] (x) − χ[r2 ,s2 ] (x)∣ = 1, and so
∥[χ[r1 ,s1 ] ] − [χ[r2 ,s2 ] ]∥∞ = ∥[χ[r1 ,s1 ]∆[r2 ,s2 ] ]∥∞ = 1.
Let Λ ∶= {(r, s) ∈ R2 ∶ a ≤ r < s ≤ b}. Then Λ is uncountable. For
(r1 , s2 ) ≠ (r2 , s2 ) ∈ Λ, the above estimate shows that
∥χ[r1 ,s1 ] − χ[r2 ,s2 ] ∥∞ = 1.
By the comment preceding this Corollary (see Exercise 6.8), L∞ ([a, b], K)
is not separable.
◻
104 L.W. Marcoux Introduction to Lebesgue measure

Appendix to Section 6.

6.39. If the reader consults almost any other text on measure theory (and the
reader will absolutely fall in this author’s estimation if they do not), they will
almost assuredly observe that we have been incredibly pedantic in our approach
to these notes. Most texts will use the same notation, namely “f ”, to refer to
both a measurable function, as well as to its equivalence class in “Lp (E, K)”, where
E ∈ M(R) is a measurable set. It is left to the reader to keep track of when
they are dealing with a function, and when they are dealing with its image in
Lp (E, K). Statements such as “f (x) = x a.e. on [0, 1]” are used to hint that we
are talking about an equivalence class rather than a function. Truth be told (and
what’s the point of not telling the truth in a set of course notes?), we have always
felt ambivalent about this approach. Without having taken any courses in math
pedagogy, experience has taught us that people inevitably make the mistake of
treating an element of Lp (E, K) as a function – for example, referring to f (x), for
some x ∈ E, when this concept is no longer valid (for the equivalence class of f ).
For this reason, we have attempted to consistently denote functions by simple
letters, e.g. f and g, while denoting their equivalence classes using the bracket
notation [f ] and [g]. While this is more cumbersome, it has the advantage of
being more precise, and our hope is that for the person learning about measure
theory for the first time, it will help to keep the two concepts separate. Once
sufficient mathematical maturity is acquired (oh, maybe a month from now), the
reader should be able to consult other texts with sufficient sophistication to handle
any notation thrown at them.

6.40. Which brings us to another point. In order to deal with pointwise lim-
its of increasing functions for the Monotone Convergence Theorem 5.16, we in-
troduced the notion of extended real numbers, and then of measurable functions
f ∶ E → R (where again, E ∈ M(R) is a measurable set). In many textbooks deal-
ing with Lebesgue measure, the equivalence classes of functions in “Lp (E)” consist
of extended real-valued measurable functions. (We have restricted our attention to
equivalence classes of real-valued measurable functions.)
In fact, both approaches lead to essentially the same theory. For any 1 ≤ p ≤ ∞,
if f ∶ E → R is measurable and νp (f ) < ∞, then the set B ∶= {x ∈ E ∶ ∣f (x)∣ = ∞} has
measure zero, and so we can always find g ∶ E → R such that g = f a.e. on E; for
example, we can choose g = f ⋅ χE∖B . Thus [f ] = [g].
So why not start with the vector space Lp (E, R) and use Proposition 6.3 to
define the quotient space Lp (E, R)? Well, perhaps the best answer we can give is
that Lp (E, R) does not define a vector space over R! Suppose
⎧ ⎧
⎪
⎪∞ if x ∈ Q ⎪
⎪−∞ if x = 0
f (x) = ⎨ and g(x) = ⎨ .
⎪
⎪0 if x ∈/ Q ⎪
⎪0 if x ≠ 0
⎩ ⎩
What should f + g(0) be? We haven’t defined ∞ + (−∞), and we won’t. Rats.
 6. Lp SPACES 105

So then where do these other authors get off defining “Lp (E)” in terms of equiv-
alence classes of extended real-valued functions? Well, for one thing, their construc-
tion of “Lp (E)” as often as not does not pass through Proposition 6.3. Instead,
they define two extended real-valued functions f and g to be equivalent if f = g a.e.
on E. While the set of extended real-valued functions does not form a vector space,
the equivalence classes defined do! In fact, up to isometric isomorphism, they form
exactly the same Lp -spaces we have defined. We like the seminorm-to-norm-via-
quotient-spaces approach we have used, in part because we do not have to wave our
hands at any point of our construction. There you go.
6.41. One obvious reason for studying Hölder’s Inequality is that it was required
to obtain Minkowski’s Inequality, which we needed to prove that each Lp (E, K) was
a linear space. The construction of the conjugate function f ∗ ∈ Lq (E, K) from
f ∈ Lp (E, K) might at first glance seem rather arcane, and the eager novice might
wonder why we would be interested in such a thing.
The answer lies in part in Functional Analysis - which is the study of normed
linear spaces and the continuous linear maps between them. Given a Banach space
(X, ∥ ⋅ ∥), one defines the dual space of X to be
X∗ ∶= B(X, K) = {x∗ ∶ X → K ∶ x∗ is linear and continuous}.
Elements of the dual space are called continuous linear functionals on X.
As we have seen in the Assignments, linear maps between normed linear spaces
are continuous if and only if they are bounded. The dual space carries a great deal of
information about the space X itself, and perhaps the most famous and important
theorem from Functional Analysis is the Hahn-Banach Theorem. This is not
actually a single result, but rather two classes of results, all referred to by that same
name. It is usually left up to the reader to recognize which version of the Hahn-
Banach Theorem is being invoked in any given application. We invite the reader to
consult the (free!) reference [3] for more details.
One of the important consequences of the Hahn-Banach Theorem is that the dual
space X∗ of X has sufficiently many functionals to separate points of X, meaning
that if x1 ≠ x2 ∈ X, then there exists x∗ ∈ X∗ such that x∗ (x1 ) ≠ x∗ (x2 ). In fact,
one can do better – given y ∈ X, one can find a linear functional y ∗ ∈ X∗ such that
∥y ∗ ∥X∗ = 1 and y ∗ (y) = ∥y∥. (Applying this to y = x1 − x2 above shows that the
corresponding y ∗ will separate x1 and x2 .)
What Hölder’s Inequality (Theorem 6.15) tells us is that with 1 < p < ∞ and
1 1
p + q = 1, given [g] ∈ Lq (E, K), we may define a linear functional

Φ[g] ∶ Lp (E, K) → K
[f ] ↦ ∫E f g
and that this linear map will be bounded with norm at most ∥[g]∥q . In the Assign-
ments, we verify that every linear functional φ ∈ Lp (E, K) is of the form φ = Φ[g]
for some [g] ∈ Lq (E, K), and that in fact ∥φ∥ = ∥[g]∥q . In other words, we are iden-
tifying Lq (E, K) with the dual space (Lp (E, K))∗ in an isometrically isomorphic
manner.
106 L.W. Marcoux Introduction to Lebesgue measure

The second half of Theorem 6.15 is just an example verifying the Corollary
to the Hahn-Banach Theorem that we mentioned above, namely: the functional
Φ[f ∗ ] ∈ (Lp (E, K))∗ is the norm-one functional which sends [f ] to ∥[f ]∥p .
For more general measure spaces (Z, µ) – which we have not discussed at all in
these notes – we still have that the dual of Lp (Z, µ) may be isometrically isomorphi-
cally identified with Lq (Z, µ) when 1 < p < ∞, but problems start to arise with the
dual spaces of L1 (Z, µ) and L∞ (Z, µ). Without getting into the details at all (the
interested reader may find them online), in the case where the space (Z, µ) is de-
composable, which includes the case where (Z, µ) is a σ-finite measure space, we
do have that (L1 (Z, µ))∗ = L∞ (Z, µ). In particular, this applies to the case of ℓ1 (I),
where I is any set equipped with counting measure. Thus we can always identify
(ℓ1 (I))∗ with ℓ∞ (I), which is a happy, happy situation. The dual of L∞ (Z, µ) tends
to be a real can of worms. For example, the dual of ℓ∞ (N) may be identified with
the so-called regular Borel measures on the Stone-Čech compactification βN
of the natural numbers. It’s big – very, very big.
6.42. The following diagram illustrates the relationship between the Lp -spaces
when the underlying measure space is a bounded interval. We use 1 to denote
constant function 1(x) = x for all x ∈ [a, b].

Inclusions of Lp -spaces

L1 ([a, b], K)

Lr ([a, b], K), 1 ≤ r < p < ∞

Lp ([a, b], K), 1 ≤ p < ∞

L∞ ([a, b], K)

[R∞ ([a, b], K)] [Simp([a, b], K)]

[C([a, b], K)] [Step([a, b], K)]

[K1]
 6. Lp SPACES 107

Exercises for Section 6.

Exercise 6.1. Let E ⊆ R and f ∶ E → R be a function. Define

⎧
⎪ 1 if f (x) ≥ 0
⎪
Ξf (x) = ⎨ , x ∈ E.
⎪
⎪−1 if f (x) < 0
⎩
Prove that if E is measurable and f ∈ L(E, R), then Ξf ∈ L(E, R) as well.

Exercise 6.2.
Let E ∈ M(R) satisfy mE < ∞ and φ ∈ Simp(E, K). Prove that for all 1 ≤ p ≤ ∞,
φ ∈ Lp (E, K).

Exercise 6.3.
Let E ∈ M(R) and 1 < p < ∞. Suppose that [f ] ∈ Lp (E, K) and [g] ∈ Lq (E, K).
Then
[f ] ⋅ [g] ∶= [f g]
is well-defined, and [f g] ∈ L1 (E, K).

Exercise 6.4. Assignment Question.

Let (X, ∥ ⋅ ∥) be a normed linear space. Prove that X is complete, and hence a
Banach space, if and only if every absolutely summable series in X is summable.
Here, a series ∑∞n=1 xn in X is said to be summable if
N
x ∶= lim ∑ xn
N →∞ n=1

exists in X, while the series is said to be absolutely summable if

N
lim ∑ ∥xn ∥ < ∞.
N →∞ n=1

Exercise 6.5.
Paragraph 6.29 and the subsequent paragraphs establish a set of relationships
between Lp -spaces for various values of p ∈ [1, ∞], as well as the spaces of (equiva-
lence classes) of step functions and simple functions, in the case where the underlying
measurable domain E ∈ M(R) has finite measure.
Determine what relations hold between these spaces in the case where mE = ∞.

Exercise 6.6.
Let E ∈ M(R), and 1 ≤ p ≤ ∞. Let D ⊆ Lp (E, K). Prove that the following are
equivalent:
(a) The set [D] ∶= {[f ] ∶ f ∈ D} is dense in Lp (E, K).
(b) For each g ∈ Lp (E, K) and ε > 0, there exists f ∈ D such that
νp (f − g) < ε.
108 L.W. Marcoux Introduction to Lebesgue measure

Exercise 6.7.
Recall that given E ∈ M(R) and 1 ≤ p ≤ ∞, we defined
Simpp (E, K) = Simp(E, K) ∩ Lp (E, K).
Prove that if mE < ∞ or if p = ∞, then Simpp (E, K) = Simp(E, K).

Exercise 6.8.
Let (X, d) be a metric space. Suppose that there exists an uncountable set
{xλ ∶ λ ∈ Λ} in X and a positive real number δ > 0 such that d(xα , xβ ) ≥ δ for all
α ≠ β ∈ Λ. Prove that (X, d) is not separable.

Exercise 6.9.
Let a < b ∈ R and let p(x) = p0 + p1 x + ⋯ + pm xm be a polynomial of degree
m in (C([a, b], R), ∥ ⋅ ∥sup ). Prove that given any ε > 0, there exists a polynomial
q(x) = q0 + q1 x + ⋯ + qm xm in (C([a, b], R), ∥ ⋅ ∥sup ) such that qk ∈ Q, 0 ≤ k ≤ m, and
∥p − q∥sup < ε.
Conclude that given a polynomial r(x) = r0 + r1 x + ⋯ + rm xm of degree m in
(C([a, b], C), ∥ ⋅ ∥sup ) and ε > 0, we can find a polynomial s(x) = s0 + s1 x + ⋯ + sm xm
with sk ∈ Q + iQ, 0 ≤ k ≤ m, such that
∥r − s∥sup < ε.

Exercise 6.10.
Recall that if A and B are sets, we define the symmetric difference of A and
B to be
A∆B ∶= (A ∪ B) ∖ (B ∩ A).
Let a < b be real numbers and suppose that a ≤ r1 < s1 ≤ b and a ≤ r2 < s2 ≤ b.
Suppose furthermore that either r1 ≠ r2 or that s1 ≠ s2 . Prove that the symmetric
difference
[r1 , s1 ] ∆ [r2 , s2 ]
contains a non-degenerate interval [u, v] ⊆ [a, b]. (By non-degenerate, we simply
mean that u < v.)

Exercise 6.11.
Let E ∈ M(R). Prove that the map:
Ω ∶ (C(E, K), ∥ ⋅ ∥sup ) → (L∞ (E, K), ∥ ⋅ ∥∞ )
f ↦ [f ]
is a linear isometry, and deduce that [C(E, K)] ∶= {[f ] ∶ f ∈ C(E, K)} is a closed
subspace Banach space of L∞ (E, K) which is (isometrically) isomorphic to C(E, K).
In other words, as Banach spaces, we can identify C(E, K) with its image
[C(E, K)] in L∞ (E, K).
 7. HILBERT SPACES 109

7. Hilbert spaces

Smoking kills. If you’re killed, you’ve lost a very important part of

your life.
Brooke Shields

7.1. We have seen in the previous section that if E ∈ M(R) and 1 ≤ p ≤ ∞, then
(Lp (E, R), ∥ ⋅ ∥p ) is a Banach space. The case where p = 2 is very special and merits
individual attention.
Let us recall the following definition.

7.2. Definition. An inner product on a K-vector space H is a function ⟨⋅, ⋅⟩ ∶

H × H → K which satisfies:
(a) ⟨x, x⟩ ≥ 0 for all x ∈ H;
(b) ⟨x, x⟩ = 0 if and only if x = 0;
(c) ⟨κx + y, z⟩ = κ⟨x, z⟩ + ⟨y, z⟩ for all x, y, z ∈ H and κ ∈ K; and
(d) ⟨x, y⟩ = ⟨y, x⟩ for all x, y ∈ K.
We say that x and y are orthogonal if ⟨x, y⟩ = 0, and we write x ⊥ y.
In the case where K = R, the “complex conjugate” appearing in (d) is obviously
superfluous.

7.3. Theorem. The Cauchy-Schwarz Inequality. Suppose that (H, ⟨⋅, ⋅⟩)
is an inner product space over K. Then
∣⟨x, y⟩∣ ≤ ⟨x, x⟩1/2 ⟨y, y⟩1/2
for all x, y ∈ H.
Proof. Let x, y ∈ H. If ⟨x, y⟩ = 0, then there is nothing to prove.
Suppose therefore that ⟨x, y⟩ ≠ 0. For any κ ∈ K,
0 ≤ ⟨x − κy, x − κy⟩
= ⟨x, x⟩ − κ⟨y, x⟩ − κ⟨x, y⟩ + ∣κ∣2 ⟨y, y⟩.
⟨x, y⟩
Setting κ = yields
⟨y, y⟩
∣⟨x, y⟩∣2 ∣⟨x, y⟩∣2 ∣⟨x, y⟩∣2
0 ≤ ⟨x, x⟩ − − + ,
⟨y, y⟩ ⟨y, y⟩ ⟨y, y⟩
which is equivalent to
∣⟨x, y⟩∣2 ≤ ⟨x, x⟩ ⟨y, y⟩,
as required.
◻
110 L.W. Marcoux Introduction to Lebesgue measure

7.4. Proposition. Let (H, ⟨⋅, ⋅⟩) be an inner product space. Then the map
∥x∥ ∶= ⟨x, x⟩1/2 , x∈H
defines a norm on H, called the norm induced by the inner product.
Proof. We shall prove that it defines a norm, and leave it to the reader to prove
that this is what it is called.

Let x, y ∈ H and κ ∈ K.
(a) Clearly ∥x∥ ≥ 0, as ⟨x, x⟩ ≥ 0.
(b) Note that ∥x∥ = 0 if and only if ∥x∥2 = ⟨x, x⟩ = 0, which happens if and only
if x = 0.
(c) ∥κx∥2 = ⟨κx, κx⟩ = ∣κ∣2 ⟨x, x⟩ = ∣κ∣2 ∥x∥2 , and thus
∥κx∥ = ∣κ∣ ∥x∥.
(d) From the Cauchy-Schwarz Inequality,
∥x + y∥2 = ⟨x + y, x + y⟩
= ⟨x, x⟩ + ⟨x, y⟩ + ⟨y, x⟩ + ⟨y, y⟩
≤ ∥x∥2 + ∣⟨x, y⟩∣ + ∣⟨y, x⟩∣ + ∥y∥2
= ∥x∥2 + 2∥x∥ ∥y∥ + ∥y∥2
= (∥x∥ + ∥y∥)2 .
Thus ∥x + y∥ ≤ ∥x∥ + ∥y∥.
This completes the proof.
◻
It follows that every inner product space (H, ⟨⋅, ⋅⟩) is also a normed linear space
(H, ∥⋅∥), using the norm induced by the inner product. Unless we explicitly mention
a different norm for H (which is highly unlikely), we shall always assume that this
is the norm to which we are referring. Also - let us not forget that every normed
linear space is also a metric space, using the metric induced by the norm.
As such, every inner product space is a metric space, under the metric induced
by the norm induced by the inner product.
Although a Hilbert space is technically an order pair consisting of a vector space
H and an inner product ⟨⋅, ⋅⟩, we shall typically speak informally of the Hilbert space
H.

7.5. Definition. A Hilbert space is a complete inner product space.

7.6. Examples.
(a) Let N ≥ 1 be an integer. Consider x = (xn )N N N
n=1 and y = (yn )n=1 ∈ H ∶= C .
The map
N
⟨x, y⟩ ∶= ∑ xn yn
n=1
 7. HILBERT SPACES 111

defines an inner product on H, called the standard inner product on H,

and H is complete with respect to the norm induced by this inner product.
Thus (H, ⟨⋅, ⋅⟩) is a Hilbert space.
(b) We can make this somewhat more general as follows. Fix an integer 1 ≤ N
and choose strictly positive real numbers ρ1 , ρ2 , . . . , ρN . We leave it to the
reader that CN becomes a Hilbert space when equipped with the inner
product
N
⟨x, y⟩ρ ∶= ∑ ρn xn yn .
n=1
2
(c) ℓ = {(xn )n ∶ xn ∈ K, n ≥ 1 and ∑∞n=1 ∣xn ∣
2
< ∞} is a Hilbert space, with the
inner product given by
⟨(xn )n , (yn )n ⟩ = ∑ xn yn .
n

Again, this is referred to as the standard inner product on ℓ2 .

7.7. Theorem. Let E ∈ M(R). The map

⟨⋅, ⋅⟩ L2 (E, K) × L2 (E, K) → K
([f ], [g]) ↦ ∫E f g
defines an inner product on L2 (E). Furthermore, the norm induced by this inner
product is the L2 -norm ∥ ⋅ ∥2 on L2 (E, K). Since (L2 (E, K), ∥ ⋅ ∥2 ) is complete, it is
a Hilbert space.
Proof. First observe that the map above is well-defined. That is, if [f1 ] = [f ] and
[g1 ] = [g], then f1 = f and g1 = g a.e. on E, so that f1 g1 = f g a.e. on E.
Furthermore, by Hölder’s Inequality,

∫ ∣f g∣ ≤ ∥[f ]∥2 ∥[g]∥2 < ∞,

E
and so [f g] = [f1 g1 ] ∈ L1 (E). The fact that f1 g1 = f g a.e. on E also implies that

∫ f1 g1 = ∫ f g ∈ K,
E E
and so the map is well-defined, as claimed.
Let [f ], [g], [h] ∈ L2 (E) and κ ∈ K. Then
(a)
⟨[f ], [f ]⟩ = ∫ ∣f ∣2 ≥ 0,
E
and equality occurs if and only if f = 0 a.e. on E, i.e. if and only if [f ] = 0.
1
(This also shows that ⟨[f ], [f ]⟩ 2 = ∥[f ]∥2 .)
(b)
⟨κ[f ] + [g], [h]⟩ = ∫ (κf + g)h
E

= κ ∫ f h + ∫ gh
E E
= κ⟨[f ], [h]⟩ + ⟨[g], [h]⟩.
112 L.W. Marcoux Introduction to Lebesgue measure

(c)
⟨[f ], [g]⟩ = ∫ f g = ∫ f g = ⟨[g], [f ]⟩.
E E
Thus ⟨⋅, ⋅⟩ is an inner product (this is the standard inner product on L2 (E, K)),
and the norm induced by this inner product is the ∥ ⋅ ∥2 -norm. The last statement
is clear.
◻

7.8. Recall that a subset E of an inner product space (H, ⟨⋅, ⋅⟩) is said to be or-
thogonal if x ≠ y in E implies that ⟨x, y⟩ = 0. Also, E is said to be an orthonormal
1
set if E is orthogonal and x ∈ E implies that ∥x∥ = 1 = ⟨x, x⟩ 2 .

7.9. Definition. Let H be a Hilbert space. An orthonormal basis (or Hilbert

space basis for H is a maximal orthonormal set in H. (Here, maximal is with
respect to inclusion.)
We shall abbreviate the phrase “orthonormal basis” to the acronym onb.

7.10. Remarks.
(a) By Zorn’s Lemma, every orthonormal set in H can be extended to an onb
for H.
(b) If H is infinite-dimensional, then an onb for H is not a Hamel basis for H.

7.11. Examples.
(a) Let N ≥ 1 be an integer and consider H = CN , equipped with the standard
inner product ⟨⋅, ⋅⟩. For 1 ≤ n ≤ N , define en = (δn,k )N
k=1 , where δa,b denotes
the Kronecker delta function. Then {en }N n=1 is an onb for H.
(b) Let N ≥ 1 be an integer and ρk = k, 1 ≤ k ≤ N . Set en = ( √1 δn,k )N k=1 , where
k
δa,b denotes the Kronecker delta function. Then {en }N N
n=1 is an onb for C ,
equipped with the inner product from Example 7.6 (b), namely
N
⟨x, y⟩ρ ∶= ∑ ρn xn yn .
n=1

(c) Generalizing example (a) above to infinite dimensions, let H = ℓ2 , equipped

with the standard inner product, and for n ≥ 1, let en = (δn,k )∞ k=1 , where
δa,b denotes the Kronecker delta function.Then {en }∞n=1 is an onb for H.
The proof is left as an exercise.
(d) Let H = L2 ([0, 2π], C), equipped with the inner product

⟨[f ], [g]⟩ ∶= ∫ f g.
[0,2π]

For n ∈ Z, define the continuous function

ξn ∶ [0, 2π] → C
θ ↦ √1 einθ .
2π
 7. HILBERT SPACES 113

Then [ξn ] ∈ L2 ([0, 2π], C) for all n ∈ Z. In the Assignments, we shall see
that {[ξn ]}n∈Z is an onb for L2 ([0, 2π], C).

We recall from Linear Algebra:

7.12. Theorem. The Gram-Schmidt Orthogonalisation Process

If H is a Hilbert space over K and {xn }∞ n=1 is a linearly independent set in H,
then we can find an orthonormal set {yn }∞ n=1 in H so that span{x1 , x2 , ..., xN } =
span {y1 , y2 , ..., yN } for all N ≥ 1.
Proof. We leave it to the reader to verify that setting y1 = x1 /∥x1 ∥, and recursively
defining
−1
xN − ∑N
n=1 ⟨xN , yn ⟩yn
yN ∶= −1
, N ≥1
∥xN − ∑N
n=1 ⟨xN , yn ⟩yn ∥
will do.
◻

7.13. Theorem. Let H be a Hilbert space and suppose that x1 , x2 , ..., xn ∈ H.

(a) [The Pythagorean Theorem] If the vectors are pairwise orthogonal, then
n n
∥ ∑ xj ∥2 = ∑ ∥xj ∥2 .
j=1 j=1

(b) [The Parallelogram Law ]

∥x1 + x2 ∥2 + ∥x1 − x2 ∥2 = 2 (∥x1 ∥2 + ∥x2 ∥2 ) .
Proof. Both of these results follow immediately from the definition of the norm in
terms of the inner product.
◻
It follows that if (X, ∥ ⋅ ∥) is a Banach space and the Parallelogram Law does
not hold in X, then there does not exist an inner product on X such that ∥ ⋅ ∥ is the
norm induced by the inner product. In particular, X is not an Hilbert space. Far
less obvious, but still true, is the fact that if the norm does satisfy the Parallelogram
Law, then there exists an inner product such that ∥ ⋅ ∥ is the norm induced by that
inner product. We shall not prove this here.

7.14. Theorem. Let H be a Hilbert space, and K ⊆ H be a closed, non-empty

convex subset of H. Given x ∈ H, there exists a unique point y ∈ K which is closest
to x; that is,
∥x − y∥ = dist(x, K) ∶= min{∥x − z∥ ∶ z ∈ K}.
Proof. The proof is left as an Assignment question.
◻
114 L.W. Marcoux Introduction to Lebesgue measure

7.15. Theorem. Let H be a Hilbert space, and let M ⊆ H be a closed subspace.

Let x ∈ H, and m ∈ M. The following are equivalent:
(a) ∥x − m∥ = dist(x, M);
(b) The vector x − m is orthogonal to M, i.e., ⟨x − m, y⟩ = 0 for all y ∈ M.
Proof.
(a) implies (b): Suppose that ∥x−m∥ = dist (x, M), and suppose to the contrary
that there exists y ∈ M so that κ ∶= ⟨x − m, y⟩ =/ 0. There is no loss of
generality in assuming that ∥y∥ = 1. Consider z = m + κy ∈ M. Then
∥x − z∥2 = ∥x − m − κy∥2
= ⟨x − m − κy, x − m − κy⟩
= ∥x − m∥2 − κ⟨y, x − m⟩ − κ⟨x − m, y⟩ + ∣κ∣2 ∥y∥2
= ∥x − m∥2 − ∣κ∣2
< dist(x, M),
a contradiction. Hence x − m ∈ M⊥ .
(b) implies (a): Suppose that x − m ∈ M⊥ . If z ∈ M is arbitrary, then y ∶=
z − m ∈ M, so by the Pythagorean Theorem,
∥x − z∥2 = ∥(x − m) − y∥2 = ∥x − m∥2 + ∥y∥2 ≥ ∥x − m∥2 ,
and thus dist (x, M) ≥ ∥x − m∥. Since the other inequality is obvious, (a)
holds.
◻

7.16. Remarks.
(a) Given any non-empty subset S ⊆ H, let
S ⊥ ∶= {y ∈ H ∶ ⟨x, y⟩ = 0 for all x ∈ S}.
It is routine to show that S ⊥ is a norm-closed subspace of H. In particular,
⊥
(S ⊥ ) ⊇ span S,
the norm closure of the linear span of S.
(b) Recall from Linear Algebra that if V is a vector space and W is a (vector)
subspace of V, then there exists a (vector) subspace X ⊆ V such that
(i) W ∩ X = {0}, and
(ii) V = W + X ∶= {w + x ∶ w ∈ W, y ∈ X }.
We say that W is algebraically complemented by X . The existence of
such a X for each W says that every vector subspace of a vector space is
algebraically complemented. We shall write V = W +̇X to denote the fact
that X is an algebraic complement for W in V.
If X is a Banach space and Y is a closed subspace of X, we say that Y
is topologically complemented if there exists a closed subspace Z of X
such that Z is an algebraic complement to Y. The issue here is that both Y
and Z must be closed subspaces. It can be shown that the closed subspace
 7. HILBERT SPACES 115

c0 of ℓ∞ is not topologically complemented in ℓ∞ . This result is known as

Phillips’ Theorem (see the paper of R. Whitley [7] for a short but elegant
proof). We shall write X = Y ⊕ Z if Z is a topological complement to Y in
X.
Now let H be a Hilbert space and let M ⊆ H be a closed subspace of H.
We claim that H = M ⊕ M⊥ . Indeed, if z ∈ M ∩ M⊥ , then ∥z∥2 = ⟨z, z⟩ = 0,
so z = 0. Also, if x ∈ H, then we may let m1 ∈ M be the element satisfying
∥x − m1 ∥ = dist(x, M).
The existence of m1 is guaranteed by Theorem 7.14. By Theorem 7.15,
m2 ∶= x − m1 lies in M⊥ , and so x = m1 + m2 ∈ M + M⊥ . Since M and
M⊥ are closed subspaces of the Hilbert space H and they are algebraic
complements, we are done.
In this case, the situation is even stronger. Given a Banach space
X and a topologically complemented closed subspace Y of X, there is in
general no reason to expect a unique topological complement for Y. For
example, if X = R2 (equipped with your favourite norm – say ∥ ⋅ ∥∞ ), and
if Y denotes the x-axis in X, then any line passing through the origin and
not equal to the x-axis describes a closed subspace Z which is a topological
complement to Y. In our case, however, the space M⊥ above is unique
in that it is an orthogonal complement. That is, as well as being a
topological complement to M, every vector in M⊥ is orthogonal to every
vector in M.
(c) With M as in (b), we have H = M ⊕ M⊥ , so that if x ∈ H, then we may
write x = m1 + m2 with m1 ∈ M, m2 ∈ M⊥ in a unique way. Consider the
map:
P ∶ H → M ⊕ M⊥
x ↦ m1 ,
relative to the above decomposition of x. It is elementary to verify that P
is linear, and that P is idempotent, i.e., P = P 2 . We remark in passing
that m2 = (I − P )x, and that (I − P )2 = (I − P ) as well.
In fact, for x ∈ H, ∥x∥2 = ∥m1 ∥2 + ∥m2 ∥2 by the Pythagorean Theorem,
and so ∥P x∥ = ∥m1 ∥ ≤ ∥x∥, from which it follows that ∥P ∥ ≤ 1. If M =/
{0}, then choose m ∈ M with ∥m∥ =/ 0. Then P m = m, and so ∥P ∥ ≥ 1.
Combining these estimates, M =/ 0 implies ∥P ∥ = 1.
We refer to the map P as the orthogonal projection of H onto M.
The map Q ∶= (I − P ) is the orthogonal projection onto M⊥ , and we leave
it to the reader to verify that if M =/ H, then ∥Q∥ = 1.
(d) Let ∅ =/ S ⊆ H. We saw in (a) that S ⊥⊥ ⊇ span S. In fact, if we let
M = span S, then M is a closed subspace of H, and so by (b),
H = M ⊕ M⊥ .
It is routine to check that S ⊥ = M⊥ . Suppose that there exists 0 =/ x ∈ S ⊥⊥ ,
x ∈/ M. Then x ∈ H, and so we can write x = m1 + m2 with m1 ∈ M, and
116 L.W. Marcoux Introduction to Lebesgue measure

m2 ∈ M⊥ = S ⊥ (m2 =/ 0, otherwise x ∈ M). But then 0 =/ m2 ∈ S ⊥ and so

⟨m2 , x⟩ = ⟨m2 , m1 ⟩ + ⟨m2 , m2 ⟩
= 0 + ∥m2 ∥2
=/ 0.
This contradicts the fact that x ∈ S ⊥⊥ . It follows that S ⊥⊥ = span S.

7.17. Lemma. Let H be a Hilbert space over K and suppose that M ⊆ H is a

finite-dimensional linear manifold in H. Then M is norm-closed, and therefore a
subspace of H.
Proof. The proof of this Lemma is left to the Assignments.
◻

7.18. Proposition. Suppose that M is a finite-dimensional subspace of a

Hilbert space H over K. Suppose that 1 ≤ N is an integer and that {e1 , e2 , . . . , eN }
is an onb for M. If P is the orthogonal projection of H onto M, then
N
P x = ∑ ⟨x, en ⟩en , x ∈ H.
n=1
Proof. Suppose that M admits an orthonormal basis {ek }nk=1 . Let x ∈ H, and let
P denote the orthogonal projection onto M. By (b), P x is the unique element of
M so that x − P x lies in M⊥ . Consider the vector w = ∑nk=1 ⟨x, ek ⟩ek . Then
n
⟨x − w, ej ⟩ = ⟨x, ej ⟩ − ∑ ⟨⟨x, ek ⟩ek , ej ⟩
k=1
n
= ⟨x, ej ⟩ − ∑ ⟨x, ek ⟩⟨ek , ej ⟩
k=1
= ⟨x, ej ⟩ − ⟨x, ej ⟩ ∥ej ∥2
= 0.
It follows that x − w ∈ M⊥ , and thus w = P x. That is, P x = ∑nk=1 ⟨x, ek ⟩ek .
◻

7.19. Theorem. Bessel’s Inequality

If {en }∞
n=1 is an orthonormal set in a Hilbert space H, then for each x ∈ H,
∞
2 2
∑ ∣⟨x, en ⟩∣ ≤ ∥x∥ .
n=1
Proof. For each N ≥ 1, let PN denote the orthogonal projection of H onto
N
span {en }N
n=1 . Given x ∈ H, we have seen that ∥PN ∥ ≤ 1, and that PN x = ∑n=1 ⟨x, en ⟩en .
Hence
N
2 2 2
∑ ∣⟨x, en ⟩∣ = ∥PN x∥ ≤ ∥x∥
n=1
for all N ≥ 1, from which the result follows.
 7. HILBERT SPACES 117

◻
Before stating our next result, we recall the following from your previous Analysis
course.

7.20. Lemma. Let (X, d) be a separable metric space. Let δ > 0 be a real
number and Y ⊆ X be a set with the property that y, z ∈ Y with y ≠ z implies that
d(y, z) ≥ δ. Then Y is countable.
Proof. The proof is left to the exercises.
◻

7.21. Theorem. Let H be a separable Hilbert space, and suppose that E ⊆ H be

an orthonormal set. Then E is countable, say E = {en }∞
n=1 , and if x ∈ H, then
∞
∑ ⟨x, en ⟩en
n=1

converges in H.
Proof. First note that if x, y ∈ E with x ≠ y, then
1/2 √
∥x − y∥ = ⟨x − y, x − y⟩1/2 = (∥x∥2 + ∥y∥2 ) = 2.
By Lemma 7.20 above, E must be countable. Thus we may write E = {en }∞
n=1 .

Let x ∈ H and ε > 0. For each N ≥ 1, set

N
yN = ∑ ⟨x, en ⟩en .
n=1

(The astute reader will note that with MN ∶= span{e1 , e2 , . . . , eN } and PN defined
as the orthogonal projection of H onto MN , we have that yN ∶= PN x, N ≥ 1.
It results from Bessel’s Inequality that we can find N0 > 0 so that
∞
2 2
∑ ∣⟨x, ek ⟩∣ < ε .
k=N0 +1

If M ≥ N ≥ N0 , then (by the Pythagorean Theorem),

M
∥yM − yN ∥2 = ∥ ∑ ⟨x, en ⟩en ∥2
n=N +1
M
= ∑ ∣⟨x, en ⟩∣2
n=N +1
∞
2
≤ ∑ ∣⟨x, en ⟩∣
n=N0 +1
2
<ε .
118 L.W. Marcoux Introduction to Lebesgue measure

This shows that (yN )∞N =1 is a Cauchy sequence. Since H is complete, this Cauchy
sequence converges, i.e.
∞
∑ ⟨x, en ⟩en = lim yN ∈ H.
n=1 N →∞
◻

7.22. Theorem. Let E = {en }∞ n=1 be an orthonormal set in an infinite-

dimensional, separable Hilbert space H. The following are equivalent:
(a) The set E is an onb for H; i.e. E is a maximal orthonormal set in H.
(b) span E = H.
(c) For all x ∈ H, x = ∑∞n=1 ⟨x, en ⟩en .
(d) For all x ∈ H, ∥x∥2 = ∑∞ 2
n=1 ∣⟨x, en ⟩∣ . [Parceval’s Identity ]
Proof.
(a) implies (b): Let M = span E. If M =/ H, then M⊥ =/ {0}, so we can find
z ∈ M⊥ , ∥z∥ = 1. But then E ∪ {z} is an orthonormal set, contradicting the
maximality of E.
(b) implies (c): Let y = ∑∞n=1 ⟨x, en ⟩en , which exists by Theorem 7.21.
A routine calculation shows that ⟨y − x, en ⟩ = 0 for all n ≥ 1, so y − x
is orthogonal to M = span E = H. In particular, y − x is orthogonal to
y − x ∈ H, so that y − x = 0, i.e. y = x.
(c) implies (d): Let x ∈ H. By hypothesis, x = ∑∞ n=1 ⟨x, en ⟩en .
Thus
∞ ∞
∥x∥2 = ⟨ ∑ ⟨x, en ⟩en , ∑ ⟨x, ek ⟩ek ⟩
n=1 k=1
∞ ∞
= ∑ ∑ ⟨x, en ⟩ ⟨x, ek ⟩ ∣⟨en , ek ⟩∣ [Check!]
n=1 k=1
∞
= ∑ ∣⟨x, en ⟩∣2 .
n=1
(d) implies (a): If e ⊥ en for all n ≥ 1, then by hypothesis,
∞
∥e∥2 = ∑ ∣⟨e, en ⟩∣2 = 0,
n=1
so that E is a maximal orthonormal set in H, i.e. E is an onb in H.
◻
The appropriate notion of isomorphism in the category of Hilbert spaces involve
linear maps that preserve the inner product.
7.23. Definition. Two Hilbert spaces H1 and H2 are said to be isomorphic
if there exists a linear bijection U ∶ H1 → H2 so that
⟨U x, U y⟩ = ⟨x, y⟩
for all x, y ∈ H1 . We write H1 ≃ H2 to denote this isomorphism.
 7. HILBERT SPACES 119

We also refer to the linear maps implementing the above isomorphism as unitary
operators. Note that
∥U x∥2 = ⟨U x, U x⟩ = ⟨x, x⟩ = ∥x∥2
for all x ∈ H1 , so that unitary operators are isometries. Moreover, the inverse map
U −1 ∶ H2 → H1 defined by U −1 (U x) ∶= x is also linear, and
⟨U −1 (U x)U −1 (U y)⟩ = ⟨x, y⟩ = ⟨U x, U y⟩,
so that U −1 is also a unitary operator.
Note that if L ⊆ H1 is a closed subspace, then L is complete, whence U L is also
complete and hence closed in H2 .

7.24. Theorem. Any two separable, infinite-dimensional Hilbert spaces over K

are isomorphic.
Proof. The proof is left to the Assignments.
◻
120 L.W. Marcoux Introduction to Lebesgue measure

Appendix to Section 7.
7.25. As mentioned in the Appendix to Chapter 6, given E ∈ M(R) and 1 <
p < ∞, one may identify the dual space of Lp (E, R) with Lq (E, R) via a linear,
isometric isomorphism. Since the Lebesgue conjugate of p = 2 is q = 2, this gives
us an identification of the dual space of L2 (E, R) with itself, via a linear, isometric
isomorphism.
Once again, the case where p = 2 is special. In this case, we obtain a second
identification of the dual of a Hilbert H space with itself. Given y ∈ H, we define
the linear function Φy ∈ H∗ by setting
Φy (x) = ⟨x, y⟩, x ∈ H.
As we shall see in the Assignments, the map
ϱ ∶ H → H∗
y ↦ Φy
is a conjugate-linear isometric isomorphism. Here, conjugate-linear means that
ϱ(y + kz) = ϱ(y) + kϱ(z) for all y, z ∈ H and k ∈ K. Of course, when K = R, ϱ is linear.
There are reasons for preferring this identification of H∗ with H to the identi-
fication that we obtained in the Appendix to Chapter 6. Given Banach spaces X
and Y and an operator T ∈ B(X, Y), we may define an operator T ∗ ∈ B(Y∗ , X∗ ) via
T ∗ y ∗ (x) = y ∗ (T x) for all x ∈ X, y ∗ ∈ Y∗ . The identification of H with H∗ mentioned
in the paragraph above provides us with an involution on B(H), that is, a map
∗ ∶ B(H) → B(H) which satisfies
● (T ∗ )∗ = T for all T ∈ B(H);
● (T1 + kT2 )∗ = T1∗ + kT2∗ for all T1 , T2 ∈ B(H), k ∈ K;
● (T1 T2 )∗ = T2∗ T1∗ for all T1 , T2 ∈ B(H).
Moreover, one may check that ∥T ∗ ∥ = ∥T ∥ for all T ∈ B(H), and finally that
∥T ∗ T ∥ = ∥T ∥2 for all T ∈ B(H).
Thus B(H) becomes what is called an involutive Banach algebra using this
operation ∗, and the last equality is referred to as the C ∗ -equation. Involutive
Banach algebras whose involution satisfies the C ∗ -equation are called C ∗ -algebras.
The C ∗ -equation, as anodyne as it might at first appear, has untold consequences
for the structure and representation theory of the algebra. People have spent their
lives studying C ∗ -algebras. If you haven’t been moved to pity yet, your heart is
made of stone.
7.26. Theorem. The Riesz Representation Theorem. Let {0} =/ H be a
Hilbert space over K, and let φ ∈ H∗ . Then there exists a unique vector y ∈ H so
that
φ(x) = ⟨x, y⟩ for all x ∈ H.
Moreover, ∥φ∥ = ∥y∥.
Proof. Given a fixed y ∈ H, let us denote by βy the map βy (x) = ⟨x, y⟩. Our goal
is to show that H∗ = {βy ∶ y ∈ H}. First note that if y ∈ H, then βy (kx1 + x2 ) =
 7. HILBERT SPACES 121

⟨kx1 + x2 , y⟩ = k⟨x1 , y⟩ + ⟨x2 , y⟩ = kβy (x1 ) + βy (x2 ), and so βy is linear. Furthermore,

for each x ∈ H, ∣βy (x)∣ = ∣⟨x, y⟩∣ ≤ ∥x∥∥y∥ by the Cauchy-Schwarz Inequality. Thus
∥βy ∥ ≤ ∥y∥, and hence βy is continuous - i.e. βy ∈ H∗ .
It is not hard to verify that the map
Θ ∶ H → H∗
y ↦ βy
is conjugate-linear (if K = C), otherwise it is linear (if K = R). From the first
paragraph, it is also contractive. But [Θ(y)](y) = βy (y) = ⟨y, y⟩ = ∥y∥2 , so that
∥Θ(y)∥ ≥ ∥y∥ for all y ∈ H, and Θ is isometric as well. It immediately follows that
Θ is injective, and there remains only to prove that Θ is surjective.
Let φ ∈ H∗ . If φ = 0, then φ = Θ(0). Otherwise, let M = ker φ, so that
codim M = 1 = dim M⊥ , since H/M ≃ K ≃ M⊥ . Choose e ∈ M⊥ with ∥e∥ = 1.
Let P denote the orthogonal projection of H onto M, constructed as in Re-
mark 7.16. Then, as I − P is the orthogonal projection onto M⊥ , and as {e} is an
orthonormal basis for M⊥ , by Proposition 7.18, for all x ∈ H, we have
x = P x + (I − P )x = P x + ⟨x, e⟩e.
Thus for all x ∈ H,
φ(x) = φ(P x) + ⟨x, e⟩φ(e) = ⟨x, φ(e)e⟩ = βy (x),
where y ∶= φ(e)e. Hence φ = βy , and Θ is onto.
◻

7.27. Remark. The fact that the map Θ defined in the proof the Riesz Rep-
resentation Theorem above induces an isometric, conjugate-linear automorphism of
H is worth remembering.
122 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 7.

Exercise 7.1. Let H = ℓ2 , equipped with standard inner product. For n ≥ 1, let
en = (δn,k )∞ ∞
k=1 , where δa,b is the Kronecker delta function. Prove that {en }n=1 is an
onb for H.
 8. FOURIER ANALYSIS - AN INTRODUCTION 123

8. Fourier analysis - an introduction

The word “genius” isn’t applicable in football. A genius is a guy like

Norman Einstein.
Joe Theismann

8.1. One of the main achievements of linear algebra is that – in particular when
dealing with finite-dimensional vector spaces – one is able to reduce a great many
questions about abstract linear maps to very concrete and computation-friendly
questions about matrices. As we know, to each linear map T from an n-dimensional
vector space V over a field F to a m-dimensional vector space W over F we may
associate an m × n matrix [T ] ∈ Mm,n (F) as follows: we let BV ∶= {v1 , v2 , . . . , vn } and
BW = {w1 , w2 , . . . , wm } be bases for V and W respectively. Given x ∈ V, we may
express x as a linear combination of the elements of BV in a unique way, say
x = α1 v1 + α2 v2 + ⋯ + αn vn ,
where αi ∈ F, 1 ≤ i ≤ n. We write [x]BV to denote the corresponding n-tuple (αk )nk=1 .
Recall that the map
ΓV ∶ V → Fn
x ↦ [x]BV ,
n
is an isomorphism of V and F .
Similarly, every y ∈ W corresponds to a unique m-tuple [y]BW = (βj )m m
j=1 ∈ F ,
m m
so that y = ∑j=1 βj wj , and the map ΓW ∶ W → F defined by ΓW (y) = [y]BW is an
isomorphism of vector spaces.
The matrix that we then associate to T is [T ] ∶= [tij ], where, for 1 ≤ j ≤ n,
(ti,j )m
i=1 = [T vj ]BW . Obviously the matrix for T depends upon our choice of bases
BV for V and BW for W.
While T acts upon the elements of V, the matrix [T ] acts upon the coordinates
of vectors x ∈ V, and returns the coordinates of the vector T x; that is,
[T ][x]BV = [T x]BW ,
where [x]BV (resp. [T x]BW ) represents the coordinates of x (resp. T x) with respect
to the basis BV for V (resp. BW for W.)
In practice, finding the coordinates of a vector in a vector space V with re-
spect to a basis for a given vector space often reduces to solving a system of linear
equations. The situation is greatly simplified, however, if the vector space H is a
finite-dimensional Hilbert space over the field K, and the basis E ∶= {e1 , e2 , . . . , en }
in question is an orthonormal basis.
In this case, the coordinates of x ∈ H are deduced from the formula
x = ⟨x, e1 ⟩e1 + ⟨x, e2 ⟩e2 + ⋯ + ⟨x, en ⟩en .
Thus, for T ∶ H → H linear, the matrix of T corresponding to the basis E (for both
the domain and the codomain) is simply [T ] = [tij ], where tij = ⟨T ej , ei ⟩, 1 ≤ i, j ≤ n.
124 L.W. Marcoux Introduction to Lebesgue measure

8.2. When the vector spaces V and W above are infinite-dimensional, it is still
possible to associate to each linear map T ∶ V → W a (generalized) matrix as above;
the problem now lies in the fact that the bases involved are obviously infinite, and
there is no need for them to be countable. For example, it can be shown that if
(X, ∥ ⋅ ∥) is an infinite-dimensional Banach space, then any (Hamel) basis for X must
be uncountable. This greatly mitigates the usefulness of the (generalized) matrix
representation of linear maps from X to X.
Analysis, however, distinguishes itself from Algebra in its ubiquitous recourse to
approximation. We may ask whether or not we can find a suitable subset B of X,
preferably countable, such that every x ∈ X can be approximated in norm by finite
linear combinations of elements of B.
In the case of separable Hilbert spaces, we have already seen that this is the
case. When H is an infinite-dimensional separable Hilbert space over K, we have
seen that H admits a countable, maximal orthonormal set E = {en }∞ n=1 (which we
have dubbed an “orthonormal basis” for H – or a “Hilbert space basis” for H), such
that x ∈ H implies that
∞ N
x = ∑ ⟨x, en ⟩en ∶= lim ∑ ⟨x, en ⟩en .
n=1 N →∞ n=1

While E is certainly linearly independent, it is never a Hamel basis (see Exercise 1).

8.3. In particular, we saw in Theorem 7.7 that if E ∈ M(R), then L2 (E, C) is a

Hilbert space over C, when equipped with the inner product
⟨[f ], [g]⟩ = ∫ f g,
E
and that L2 ([−π, π], C) is separable (see Corollary 6.38).
Let us now investigate a different, but closely related example. Consider the
space
L2 (T, C) ∶= {f ∶ R → C ∶ f is measurable, 2π-periodic, and ∫ ∣f ∣2 < ∞}.
[−π,π)

We leave it as an exercise for the reader to prove that L2 (T, C) is a vector space
and that the function
ν2 ∶ L2 (T, C) → R
1/2
1 2
f ↦ ( 2π ∫[−π,π) ∣f ∣ )
defines a seminorm on L2 (T, C). Set N (T, C) ∶= {f ∈ L2 (T, C) ∶ ν2 (f ) = 0}. Arguing
as in Section 6.4, we see that [f ] = [g] in L2 (T, C) ∶= L2 (T, C)/N (T, C) if and only if
f = g a.e. on R, or equivalently f = g a.e. on [−π, π) (given that f, g are 2π-periodic
on R). By Proposition 6.3, we obtain a norm on L2 (T, C) by setting ∥[f ]∥2 ∶= ν2 (f ),
[f ] ∈ L2 (T, C).
Furthermore, the function
⟨⋅, ⋅⟩T ∶ L2 (T, C) × L2 (T, C) → C
1
([f ], [g]) ↦ 2π ∫[−π,π) f g
 8. FOURIER ANALYSIS - AN INTRODUCTION 125

defines an inner product on L2 (T, C), and ∥ ⋅ ∥2 is precisely the norm induced by
this inner product. A fortiori, L2 (T, C) is complete with respect to this norm, and
is therefore a Hilbert space. Finally, we leave it to the reader to verify that with
ξn (θ) = einθ , θ ∈ R, n ∈ Z, the set {[ξn ]}n∈Z forms an onb for L2 (T, C).
[f ]
Given n ∈ Z, we shall refer to the complex number αn ∶= ⟨[f ], [ξn ]⟩T as the
nth -Fourier coefficient of [f ] relative to the onb ([ξn ])n∈Z .
As we shall see in the Assignments, the map
U ∶ L2 (T, C) ↦ ℓ2 (Z, C)
[f ]
[f ] ↦ (αn )n∈Z
defines a unitary operator from the Hilbert space L2 (T, C) to ℓ2 (Z, C). In particular,
[f ] [g]
therefore, it is injective. Thus, if [f ], [g] ∈ L2 (T, C) and αn = αn for all n ∈ Z, then
f = g a.e. on R. In other words, an element [f ] ∈ L2 (T, C) is entirely determined
by its Fourier coefficients. Moreover, given any sequence (βn )n∈Z ∈ ℓ2 (Z, C), there
[f ]
exists [f ] ∈ L2 (T, C) such that αn = βn , n ∈ Z.
Let [f ] ∈ L2 (T, C). For each 1 ≤ N ∈ N, set
N
∆N ([f ]) = ∑ αn[f ] [ξn ].
n=−N
th
We shall say that ∆N ([f ]) is the N partial sum of the Fourier series of [f ]. It
follows from Theorem 7.22 that
[f ] = lim ∆N ([f ]),
N →∞
where the convergence is relative to the ∥ ⋅ ∥2 -norm defined above.
This is an entirely satisfactory state of affairs, and demonstrates clearly just how
well-behaved Hilbert spaces are. Our next goal is to try to extend this theory to
the L1 -setting. Here, we will quickly discover that things are far more complicated,
and for that reason perhaps also that much more interesting!
8.4. In trying to extend this theory beyond the Hilbert space setting, we shall
once again require some notations and definitions. We define:
● Trig(T, C) ∶= span{ξn ∶ n ∈ Z} = {∑N n=−N αn ξn ∶ αn ∈ C, 1 ≤ N ∈ N};
● C(T, C) ∶= {f ∶ R → C ∶ f is continuous and 2π-periodic};
● Simp(T, C) ∶= {f ∶ R → C ∶ f ∣[−π,π) is a simple function and f is 2π-periodic};
● Step(T, C) ∶= {f ∶ R → C ∶ f ∣[−π,π) is a step function and f is 2π-periodic};
● for 1 ≤ p < ∞,
Lp (T, C) ∶= {f ∶ R → C ∶ f is measurable, 2π-periodic, and ∫ ∣f ∣p < ∞};
[−π,π)
● and for p = ∞,
L∞ (T, C) = {f ∶ R → C ∶ f is measurable, 2π-periodic, and essentially bounded}.

Observe that
Trig(T, C) ⊆ C(T, C) ⊆ Lp (T, C), 1 ≤ p ≤ ∞.
126 L.W. Marcoux Introduction to Lebesgue measure

As was the case with p = 2 above, for each 1 ≤ p < ∞, the set Lp (T, C) forms a
vector space over C, and the map
νp ∶ Lp (T, C) → R
1/p
1 p
f ↦ ( 2π ∫[−π,π) ∣f ∣ )
defines a seminorm on Lp (T, C).
Moreover, if p = ∞, we repeat the arguments of Chapter 6. That is, for f ∈
L∞ (T, C), we set
ν∞ (f ) ∶= ess sup(f ) ∶= inf{δ > 0 ∶ m{θ ∈ [−π, π) ∶ ∣f (θ)∣ > δ} = 0},
and verify that this defines a seminorm on L∞ (T, C).
Appealing to Proposition 6.3, for each 1 ≤ p ≤ ∞, we obtain a norm ∥ ⋅ ∥p
on Lp (T, C) ∶= Lp (T, C)/Np (T, C), where Np (T, C) ∶= {f ∈ Lp (T, C) ∶ νp (f ) = 0}.
Perhaps unsurprisingly, we find that [f ] = [g] ∈ Lp (T, C) if and only if f = g a.e. on
R (equivalently f = g a.e. on [−π, π), because of 2π-periodicity). The details are
left to the reader (see the Exercises below).
It is also a simple but important exercise (which we again leave to the reader)
to verify that for f ∈ C(T, C),
∥[f ]∥∞ = ∥f ∥sup ∶= sup{∣f (θ)∣ ∶ −π ≤ θ < π}.
Note that the supremum on the right hand side of this equation exists as a finite
number since f ∈ C(T, C) implies that f is continuous on R, and hence f is bounded
on [−π, π] ⊇ [−π, π).

Given any function f ∶ [−π, π) → C, denote by fˇ ∶ R → C its 2π-periodic exten-

sion, that is: fˇ(θ) = f (θ), θ ∈ [−π, π), and fˇ(θ + 2π) = fˇ(θ), θ ∈ R. Clearly fˇ always
exists and is uniquely defined by f .
8.5. Theorem. Let 1 ≤ p ≤ ∞. The map
Ξp ∶ Lp ([−π, π), C) → Lp (T, C)
[f ] ↦ [fˇ ]
is an isometric isomorphism.
Proof. See the Exercises.
◻
It follows that all of our favourite results from the previous Chapters hold in
Lp (T, C); for example, Hölder’s Inequality holds in Lp (T, C) and says that if 1 ≤ p <
∞ and q is the Lebesgue conjugate of p, then for [f ] ∈ Lp (T, C) and [g] ∈ Lq (T, C)
we have that [f g] ∈ L1 (T, C) and
∥[f g]∥1 ≤ ∥[f ]∥p ∥[g]∥q .
A second result which carries over is that [Simp(T, C)] is dense in Lp (T, C) (in
the ∥ ⋅ ∥p -norm) for all 1 ≤ p ≤ ∞. Again, the explicit proof of this is left to the
exercises. Note that the fact that [Step(T, C)] and [C(T, C)] are dense in Lp (T, C)
for all 1 ≤ p < ∞ is not an immediate consequence of Theorem 8.5, since, for example,
 8. FOURIER ANALYSIS - AN INTRODUCTION 127

a function f which lies in C(T, C) must satisfy f (−π) = limθ→π f (θ). Despite this,
[Step(T, C)] and [C(T, C)] are dense in Lp (T, C) for all such p, and this is also left
to the exercises. These facts will prove useful below.

Of course, this raises the question of why we even bother with Lp (T, C), given
that it is isomorphic to Lp ([−π, π), C). This would be a good time for the reader to
consult the Appendix, even if this is not something the reader typically does.

8.6. Definition. For f ∈ L1 (T, C) and n ∈ Z, we refer to

1
f̂(n) ∶= ∫ f ξn
2π [−π,π)
as the nth -Fourier coefficient of f . We also refer to

∑ f̂(n) ξn
n∈Z

as the Fourier series of f in L1 (T, C).

8.7. Remark. Observe that if f, g ∈ L1 (T, C) and f = g a.e. on [−π, π), then

f̂(n) = ̂
g (n) for all n ∈ Z.

In other words, if we set nth -Fourier coefficient of [f ] ∈ L1 (T, C) to be

αn[f ] ∶= f̂(n), n ∈ Z,
then this is well-defined. We then define
[f ]
∑ αn [ξn ]
n∈Z

to be the Fourier series of [f ].

The perspicacious reader (hopefully you) will have observed that at no point
have we said anything about convergence of the above series. Indeed, at this stage,
the series notation is strictly formal, and is meant only as a shorthand to represent
[f ] ∞
the sequence of partial sums (∑N n=−N αn [ξn ])N =0 . The remainder of the course is
devoted to examining in what sense the series (i.e. the sequence of partial sums)
above converges.
We also point out that we may extend the notion of a Fourier coefficient to
non-integer powers of eiθ . That is, for f ∈ L1 (T, C) and r ∈ R, we define
1
f̂(r) = ∫ f ξr ,
2π [−π,π)
where ξr (θ) = eirθ for all θ ∈ R.
128 L.W. Marcoux Introduction to Lebesgue measure

[f ]
8.8. In the case p = 2, we have more than once seen that (αn )n∈Z ∈ ℓ2 (Z, C).
While nothing so nice holds for [f ] ∈ L1 (T, C), the situation is not altogether hope-
less.
First note that ∣ξr (θ)∣ = 1 for all θ ∈ R and all r ∈ R. As such, for f ∈ L1 (T, C),
we have
1
∣f̂(r)∣ = ∣ ∫ f ξr ∣
2π [−π,π)
1
≤ ∫ ∣f ξr ∣
2π [−π,π)
1
= ∫ ∣f ∣
2π [−π,π)
= ν1 (f )
= ∥[f ]∥1 .

As before, if f, g ∈ L1 (T, C) and f = g a.e., then f̂(r) = ̂

g (r) for all r ∈ R, and so
[f ] [f ]
we may define αr ∶= f̂(r), r ∈ R. It trivially follows that supr∈R ∣αr ∣ ≤ ∥[f ]∥1 for
[f ]
all [f ] ∈ L1 (T, C), and more specifically (αn )n∈Z ∈ ℓ∞ (Z, C). In fact, we can do
better.

8.9. Theorem. The Riemann-Lebesgue Lemma.

Let f ∈ L1 (T, C). Then

lim f̂(r) = 0 = lim f̂(r).

r→∞ r→−∞

In particular,

(αn[f ] )n∈Z ∈ c0 (Z, C).

Proof. The key to the proof of this result is to notice that it is really quite simple
to prove when f ∣[−π,π) is the characteristic function of an interval. But Lebesgue
integration is linear, and the span of these (2π-periodic extensions of) characteristic
functions of intervals is Step(T, C). So the result will hold for this class as well.
Given this, one simply appeals to the density of [Step(T, C)] in L1 (T, C) to obtain
the full result. As a rule, life is not great, but some parts of it really are.
● Suppose first that f0 is the characteristic function of an interval, say

f0 = χ[s,t] ,

where −π ≤ s < t < π. Let f ∶= fˇ0 denote the 2π-periodic extension of f0

to R, so that f ∈ Step(T, C). Then, keeping in mind that every bounded,
Riemann-integrable function over a bounded interval is Lebesgue integral,
 8. FOURIER ANALYSIS - AN INTRODUCTION 129

and that the Lebesgue and Riemann integrals coincide (see Theorem 5.24),
1
f̂(r) = ∫ χ[s,t] ξr
2π [−π,π)
1
= ∫ e−irθ
2π [s,t]
1 t
−irθ
= ∫ e dθ
2π s
θ=t
1 e−irθ
= ]
2π −ir θ=s
1 e−irt − e−irs
= [ ].
2π −ir
From this it easily follows that
2 1
∣f̂(r)∣ ≤ = ,
2π∣r∣ π∣r∣
whence
lim f̂(r) = 0 = lim f̂(r).
r→∞ r→−∞
● Next, suppose that f ∈ Step(T, C). Let f0 ∶= f ∣[−π,π) , and write f0 =
∑Mk=1 βk χHk as a disjoint representation, where each Hk = [sk , tk ] is a subin-
terval of [−π, π).
The result is now a simple consequence of the argument above, com-
bined with the linearity of the Lebesgue integral, and is left to the reader.
● Finally, let [f ] ∈ L1 (T, C), ε > 0 be arbitrary, and choose g ∈ Step(T, C)
such that ∥[f ] − [g]∥1 < ε/2.
Then
1
f̂(r) = ∫ f ξr
2π [−π,π)
1 1
= ∫ (f − g)ξr + ∫ gξr
2π [−π,π) 2π [−π,π)
= f̂ − g(r) + ̂
g (r).
But as we have seen, ∣f̂ − g(r)∣ ≤ ν1 (f − g) = ∥[f − g]∥1 = ∥[f ] − [g]∥1 < ε/2
for all r ∈ R. Since g ∈ Step(T, C), from the previous case, we see that
we may choose N > 0 such that ∣r∣ ≥ N implies that ∣̂ g (r)∣ < ε/2, and thus
∣r∣ ≥ N implies that
ε ε
∣f̂(r)∣ ≤ ∣f̂
− g(r)∣ + ∣̂
g (r)∣ < + = ε.
2 2
This proves that limr→∞ f (r) = 0 = limr→−∞ f̂(r), completing the proof
̂
of the first statement of the Theorem. The second statement is an easy
consequence of the first, and is left to the reader.
◻
130 L.W. Marcoux Introduction to Lebesgue measure

[f ]
8.10. We began by recalling that [f ] ∈ L2 (T, C) if and only if (αn )n∈Z ∈
ℓ2 (Z, C). We then defined the Fourier coefficients of [f ] ∈ L1 (T, C) in exactly the
same way as for elements of L2 (T, C), and we have succeeded in showing that
(αn[f ] )n∈Z ∈ c0 (Z, C).
So far, however, we have not shown that this is an “if and only if” statement, and
one reason for this is that it is not. We shall see by Chapter 12 that the map
Λ ∶ (L1 (T, C), ∥ ⋅ ∥1 ) → (c0 (Z, C), ∥ ⋅ ∥∞ )
[f ]
[f ] ↦ (αn )n∈Z
is a continuous, injective linear map, but that it is not surjective. Indeed, linearity
is a simple result which is left to the exercises, while continuity of Λ is an easy
consequence of the estimate of Section 8.8. In analogy to the situation for L2 (T, C),
it is tempting to ask whether or not the range of Λ is ℓ1 (Z, C). In Chapter 12, we
shall discover that this is overly optimistic.
We are left with a number of questions. Let [f ] ∈ L1 (T, C).
[f ]
(a) Does the Fourier series ∑n∈Z αn [ξn ] converge, and if so, in which sense?
Pointwise (a.e.)? Uniformly? In the L1 -norm?
(b) If the Fourier series does converge in some sense, does it converge back to
f?
(c) Is [f ] completely determined by its Fourier series? That is, if [f ], [g] ∈
[f ] [g]
L1 (T, C) and αn = αn for all n ∈ Z, is [f ] = [g]? (We know that this was
true for [f ], [g] ∈ L2 (T, C).)
These are some of the questions we will consider in the following sections.
 8. FOURIER ANALYSIS - AN INTRODUCTION 131

Appendix to Section 8.
8.11. So where does the notation L1 (T, C) come from, given that we are dealing
with 2π-periodic functions on R? The issue lies in the fact that we are really inter-
ested in studying functions on T ∶= {z ∈ C ∶ ∣z∣ = 1}, but that we have not yet defined
what we mean by a measure on that set. We are therefore identifying [−π, π) with
T via the bijective function ψ(θ) = eiθ . Thus, an alternative approach to this would
be to say that a subset E ⊆ T is measurable if and only if ψ −1 (E) ⊆ [−π, π) is
Lebesgue measurable. In order to “normalise” the measure of T (i.e. to make its
measure equal to 1), we simply divide Lebesgue measure on [−π, π) by 2π.
This still doesn’t quite explain why we are interested in 2π-periodic functions
on R, rather than just functions on [−π, π), though. Here is the “kicker ”. The
unit circle T ⊆ C has a very special property, namely, that it is a group. Given
θ0 ∈ T, we can “rotate” a function f ∶ T → C in the sense that we set g(θ) = f (θ ⋅ θ0 ).
Observe that rotation along T corresponds to translation (modulo 2π) of the interval
[−pi, π). The key is the irritating “modulo 2π” problem. If we don’t use modular
arithmetic, and if a function g is only defined on [−π, π), we can not “translate” it,
since the new function need no longer have as its domain: [−π, π). We get around
this by extending the domain of g to R and making g 2π-periodic. Then we may
translate g by any real number τs○ (g)(θ) ∶= g(θ − s), which has the effect that if we
set f (eiθ ) = g(θ), then g(θ −s) = f (eiθ ⋅e−is ). That is, translation of g under addition
corresponds to rotation of f under multiplication.
The last thing that we need to know is that such translations of functions will
play a crucial role in our study of Fourier series of elements of L1 (T, C). Aside from
being a Banach space, L1 (T, C) can be made into an algebra under convolution.
While our analysis will not take us as far as that particular result, we will still need
to delve into the theory of convolutions of continuous functions with functions in
L1 (T, C). This will provide us with a way of understanding how and why various
series associated to the Fourier series of an element [f ] ∈ L1 (T, C) converge or
diverge. Since convolutions are defined as averages under translation by the group
action, and since T is a group under multiplication and R is a group under addition,
our identification of (T, ⋅) with ([−π, π), +) (using modular arithmetic) is not an
unreasonable way of doing things.
We have been speaking in vague generalities. The next four sections will hope-
fully add meaning to the above statements.
132 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 8.

Exercise 8.1.
Let H be an infinite-dimensional Hilbert space over K, and let E be an onb for
H.
(a) Prove that E is linearly independent.
(b) Prove that E is not a Hamel basis for H.

Exercise 8.2.
Let 1 ≤ p ≤ ∞. Prove that [f ] = [g] in Lp (T, C) if and only if f = g a.e. on R.

Exercise 8.3.
(a) Prove that the map
Φ ∶ L2 ([−π, π], C) → L2 ([−π, π), C)
[f ] ↦ [f ∣[−π,π) ]
is an isomorphism of Hilbert spaces.
(b) Recall from Example[5.11] that if we set
1
ξn (θ) = √ einθ , θ ∈ [−π, π], n ∈ Z,
2π
then ([ξn ])n∈Z is an onb for L2 ([−π, π], C). Let ψn ∶= ξn ∣[−π,π) , n ∈ Z, and
prove that ([ψn ])n∈Z is an onb for L2 ([−π, π), C).

Exercise 8.4.
Let 1 ≤ p ≤ ∞. Prove that the map
Ξp ∶ Lp ([−π, π), C) → Lp (T, C)
[f ] ↦ [fˇ]
is a well-defined, isometric isomorphism of Banach spaces.

Exercise 8.5.
Let 1 ≤ p < ∞.
(a) Prove that [Simp(T, C)] is dense in (Lp (T, C), ∥ ⋅ ∥p ).
(b) Prove that [Step(T, C)] is dense in (Lp (T, C), ∥ ⋅ ∥p ).
(c) Prove that [C(T, C)] is dense in (Lp (T, C), ∥ ⋅ ∥p ).
(d) Prove that [Simp(T, C)] is dense in (L∞ (T, C), ∥ ⋅ ∥∞ ).

Exercise 8.6.
Let f ∈ C(T, C). Prove that
∥[f ]∥∞ = ∥f ∥sup ∶= sup{∣f (θ)∣ ∶ θ ∈ [−π, π)}.
 8. FOURIER ANALYSIS - AN INTRODUCTION 133

Exercise 8.7. Prove that the map

Λ ∶ (L1 (T, C), ∥ ⋅ ∥1 ) → (c0 (Z, C), ∥ ⋅ ∥∞ )
[f ]
[f ] ↦ (αn )n∈Z
is linear and continuous.

Exercise 8.8.
Let f ∈ L1 (T, C). Prove (or disprove) that the function
f̂ ∶ R → C
1
r → 2π ∫[−π,π) f ξr
is continuous on R.
134 L.W. Marcoux Introduction to Lebesgue measure

9. Convolution

Reporter: “Where do you go from here, Mike?”

“I don’t know, man, I mean, I might just fade into Bolivian.”

Mike Tyson

9.1. Recall that an algebra B is a vector space over a field F which also happens
to be a ring. A Banach algebra A is a Banach space over K which is simultane-
ously an algebra, and for which multiplication is jointly continuous by virtue of its
satisfying the inequality
∥ab∥ ≤ ∥a∥ ∥b∥
for all a, b ∈ A.
For example, (C(X, K), ∥ ⋅ ∥sup ) is a Banach algebra for each locally compact,
Hausdorff topological space X, as is Mn (K) ≃ B(Kn ) (for each n ≥ 1), when equipped
with the operator norm.
So far, we have seen that L1 (T, C) is a Banach space, but we have not investi-
gated any multiplicative structure on it. One can equip L1 (T, C) with an operation
∗ under which it becomes a Banach algebra. Indeed, given f, g ∈ L1 (T, C), we set
1
g ◇ f (θ) ∶= ∫ g(s)f (θ − s)dm(s),
2π [−π,π)
and we refer to this as the convolution of g and f . One shows that [f1 ] = [f2 ] and
[g1 ] = [g2 ] in L1 (T, C) yields that g1 ◇ f1 = g2 ◇ f2 almost everywhere, allowing one
to define
[g] ∗ [f ] ∶= [g ◇ f ]
for all [f ], [g] ∈ L1 (T, C).
It is not entirely clear, a priori, that g ◇ f (θ) ∈ C for any θ ∈ R, and it is
even less clear that g ◇ f ∈ L1 (T, C). Nevertheless, it is true. The proof of this,
however, requires a bit more measure theory than we have developed so far. The
key ingredient we are missing is Fubini’s Theorem, which is stated in the Appendix
to this Chapter.
What is easier to prove, however, and what we shall prove is that we can turn
L1 (T, C) (and consequently L1 (T, C)) into a left module over C(T, C) using con-
volution. That is, given g ∈ C(T, C) and f ∈ L1 (T, C), we shall set
1
g ◇ f (θ) ∶= ∫ g(s)f (θ − s)dm(s),
2π [−π,π)
and we shall prove that g ◇ f ∈ C(T, C) ⊆ L1 (T, C). Assuming this for the moment,
if f1 ∈ L1 (T, C) and f1 = f a.e. on R, then clearly g ◇ f (θ) = g ◇ f1 (θ) for all θ ∈ R,
whence g ◇ f = g ◇ f1 , and this allows us to define
g ∗ [f ] = [g ◇ f ], [f ] ∈ L1 (T, C).
 9. CONVOLUTION 135

The advantage to convolving with continuous functions only is that we shall be

able to reformulate the convolution as a L1 (T, C) Riemann integral (in the sense
of Chapter 1), and this will allow us gather more information about the continuity
properties of this operation, and ultimately about convergence properties of Fourier
series.
We invite the reader to consult the Appendix for more information on modules,
if necessary.
We begin with a Lemma.
9.2. Lemma. Let f ∈ L1 (T, C) and s ∈ R be fixed.
(a)
∫ f =∫ τs○ (f ),
[−π,π) [−π,π)
where τs○ (f )(θ) = f (θ − s).
(b) If h(θ) ∶= f (−θ), θ ∈ R, then

∫ h=∫ f.
[−π,π) [−π,π)

(c) Define φf,θ ∶ R → C by φf,θ (s) = f (θ − s). Then φf,θ ∈ L1 (T, C) and
ν1 (φf,θ ) = ν1 (f ).
That is,
1 1
∫ ∣f (θ − s)∣dm(s) = ∫ ∣f (t)∣dm(t).
2π [−π,π) 2π [−π,π)
Proof. This is an Assignment question.
◻

9.3. Definition. Let f ∈ L1 (T, C) and g ∈ C(T, C). We define the convolution
of f by g to be the function
g◇f ∶ R → C
1
θ ↦ 2π ∫[−π,π) g(s)f (θ − s)dm(s).

The alert reader (hopefully you) will have observed that there is a problem with
this definition. For one thing - how do we know that g ◇ f (θ) exists as a complex
number for each θ ∈ R? Let us resolve this issue immediately.
Fix θ ∈ R. Then
1
∣g ◇ f (θ)∣ = ∣∫ g(s)f (θ − s)dm(s)∣
2π [−π,π)
1
≤ ∫ ∣g(s)∣∣φf,θ (s)∣dm(s)
2π [−π,π)
≤ ∥g∥sup ν1 (φf,θ )
= ∥g∥sup ν1 (f ) < ∞.
Thus g ◇ f is indeed a complex-valued function.
136 L.W. Marcoux Introduction to Lebesgue measure

The following extremely useful computation will be used repeated below.

9.4. Lemma. Let f ∈ L1 (T, C), g ∈ L∞ (T, C). If θ ∈ R, then

∫ g(s)f (θ − s)dm(s) = ∫ g(θ − t)f (t)dm(t).

[−π,π) [−π,π)

In particular, this holds if f ∈ L1 (T, C) and g ∈ C(T, C).

Proof. This is an Assignment question.
◻

9.5. Remark. In light of Lemma 9.4, for f ∈ L1 (T, C) and g ∈ C(T, C), we shall
define the convolution of g by f to be
1
f ◇ g(θ) = ∫ g(θ − t)f (t)dm(t).
2π [−π,π)
In so doing, we have guaranteed that f ◇ g(θ) = g ◇ f (θ) for all θ ∈ R, and so
henceforth we shall simply refer to this function as the convolution of f and g.

9.6. Proposition. Let g ∈ C(T, C) and f ∈ L1 (T, C). Then g ◇ f ∈ C(T, C).
Proof. Note that since g is continuous on R and 2π-periodic, it is in fact uniformly
continuous on R. (See the exercises below.)
Let ε > 0 and choose δ > 0 such that ∣x − y∣ < δ implies that ∣g(x) − g(y)∣ < ε.
Let θ0 , θ ∈ R and suppose that ∣θ − θ0 ∣ < δ. Then, by noting that

∣g(θ − s) − g(θ0 − s)∣ < ε for all s ∈ R,

and by applying Lemma 9.4 above, we get:

1
∣g ◇ f (θ) − g ◇ f (θ0 )∣ = ∣∫ g(s) (f (θ − s) − f (θ0 − s)) dm(s)∣
2π [−π,π)
1
= ∣∫ (g(θ − s) − g(θ0 − s)) f (s) dm(s)∣
2π [−π,π)
1
≤ ∫ ∣g(θ − s) − g(θ0 − s)∣∣f (s)∣ dm(s)
2π [−π,π)
1
≤ ∫ ε ∣f (s)∣ dm(s)
2π [−π,π)
= ν1 (f ) ε.

From this it clearly follows that g ◇ f is (uniformly) continuous.

The fact that g ◇ f is 2π-periodic is clear from the 2π-periodicity of both g and
f.
◻
 9. CONVOLUTION 137

9.7. Remark. Suppose that g ∈ C(T, C), f1 , f2 ∈ L1 (T, C) and that [f1 ] = [f2 ] ∈
L1 (T, C); i.e. that f1 = f2 a.e. on R. Then, since φg,θ f1 = φg,θ f2 a.e. on R (where
φg,θ is defined as in Lemma 9.2(c) above), we find that for each θ ∈ R,
g ◇ f1 (θ) = f1 ◇ g(θ)
1
= ∫ g(θ − t)f1 (t)dm(t)
2π [−π,π)
1
= ∫ g(θ − t)f2 (t)dm(t)
2π [−π,π)
= f2 ◇ g(θ)
= g ◇ f2 (θ).

This allows us to extend our notion of convolution as follows:

9.8. Definition. Given g ∈ C(T) and [f ] ∈ L1 (T), we define the convolution

of g and [f ] to be
g ∗ [f ] ∶= [g ◇ f ],
where g ◇ f ∈ L1 (T, C) is the convolution of Definition 9.3.

We also define the convolution operator with kernel g to be the map:

Cg ∶ L1 (T, C) → L1 (T, C)
[f ] ↦ g ∗ [f ].
Observe that if [f1 ] and [f2 ] ∈ L1 (T, C), and if κ ∈ C, then
Cg (κ[f1 ] + [f2 ]) = g ∗ [κf1 + f2 ]
1
= ∫ g(s)(κf1 (θ − s) + f2 (θ − s))dm(s)
2π [−π,π)
1 1
=κ ∫ g(s)f1 (θ − s)dm(s) + ∫ g(s)f2 (θ − s)dm(s)
2π [−π,π) 2π [−π,π)
= κg ∗ [f1 ] + g ∗ [f2 ]
= κCg ([f1 ]) + Cg ([f2 ]),
proving that Cg is a linear map on L1 (T, C). Given that Cg is a linear map and
(L1 (T, C), ∥ ⋅ ∥1 ) is a Banach space, it is of interest to determine whether or not Cg
is bounded, and if so, then what is its norm? As we shall see - the answer to this
question will be intimately related to the question of convergence of Fourier series
of elements of L1 (T, C).
A direct computation of the norm of Cg as defined above is rather difficult, using
the tools currently at our disposition. Our strategy, therefore, is to reformulate Cg
as a vector-valued Riemann integral on L1 (T, C), as developed in Chapter 1. In
fact, we shall be able to extend this notion of convolution beyond the Banach space
L1 (T, C). For this, we shall need the concept of a homogeneous Banach space, which
we now define.
138 L.W. Marcoux Introduction to Lebesgue measure

9.9. Homogeneous Banach spaces. Let f ∈ L1 (T, C), and let s ∈ R. Con-
sider the function
τs○ (f ) ∶ R → C
θ ↦ f (θ − s).
One should think of τs○ as translating f by s. The superscript ○ above the τs is to
indicate that we are acting on functions. When acting on elements of L1 (T, C), we
shall drop this superscript.
As we shall see in the Assignments, the fact that M(R) is invariant under transla-
tion, that Lebesgue measure is translation-invariant, and that the set of 2π-periodic
functions is again invariant under translation implies that
τs○ (f ) ∈ L1 (T, C)
as well. Furthermore, if [f ] = [g] ∈ L1 (T, C), then [τs○ (f )] = [τs○ (g)], as is easily
verified. Thus we define the operation of translation by s on L1 (T, C) via
τs ([f ]) ∶= [τs○ (f )].

9.10. Definition. A homogeneous Banach space over T is a linear man-

ifold B in L1 (T) equipped with a norm ∥ ⋅ ∥B with respect to which (B, ∥ ⋅ ∥B ) is a
Banach space, and satisfying
(a) ∥[f ]∥1 ≤ ∥[f ]∥B for all [f ] ∈ B.
(b) [Trig(T)] ⊆ B;
(c) B is invariant under translation; that is, for all [f ] ∈ B and s ∈ R,
τs [f ] = [τs○ (f )] ∈ B;
(d) for all [f ] ∈ B and s ∈ R, ∥τs [f ]∥B = ∥[f ]∥B ; and
(e) for each [f ] ∈ B, the map
Ψ[f ] ∶ R → B
s ↦ τs ([f ])
is continuous.
The idea that a linear manifold M of a Banach space X might not be closed in
the ambient norm, but that (M, ∥⋅∥M ) might be complete in its own norm and hence
a Banach space might seem a bit strange at first. Come to think of it, however, we
have already seen this multiple times! Note that each of the spaces Lp (T, C) is
dense in L1 (T, C), 1 ≤ p < ∞, and that each is complete using the corresponding
∥ ⋅ ∥p norm.

9.11. Example. Recall that [C(T, C)] ⊆ L∞ (T, C) is a subset of L1 (T, C) and
that it is clearly a linear manifold. Furthermore, for f ∈ C(T, C), it follows from
Example 6.19 (b) (see also Exercise 6.11) that
∥[f ]∥∞ = ∥f ∥sup ∶= sup{∣f (θ)∣ ∶ θ ∈ [−π, π)},
and that ([C(T, C)], ∥ ⋅ ∥∞ ) is a Banach space. We claim that it is in fact a homo-
geneous Banach space over T.
 9. CONVOLUTION 139

(a) Let f ∈ C(T, C). Then

1 1
∥[f ]∥1 = ∫ ∣f ∣ ≤ ∫ ∥f ∥sup = ∥f ∥sup = ∥[f ]∥∞ .
2π [−π,π) 2π [−π,π)
(b) Since, for each n ∈ Z, ξn ∈ C(T, C), and since the latter is a linear manifold,
we have that [Trig(T, C)] ⊆ [C(T, C)].
(c) If f ∈ C(T, C) and s ∈ R, then clearly τs○ (f ) ∈ C(T, C), so that τs [f ] ∈
[C(T, C)]. It follows that [C(T, C)] is translation-invariant.
(d) Given f ∈ C(T, C) and s ∈ R,
∥τs [f ]∥∞ = ∥[τs○ (f )]∥∞
= ∥τs○ (f )∥sup
= sup{∣f (θ − s)∣ ∶ θ ∈ R}
= sup{∣f (θ)∣ ∶ θ ∈ R}
= ∥f ∥sup
= ∥[f ]∥∞ .
(e) Let [f ] ∈ [C(T, C)], and assume without loss of generality that f ∈ C(T, C).
Then f ∶ R → C is continuous and 2π-periodic. From this we easily conclude
that f is uniformly continuous on R. Let ε > 0, and choose δ > 0 such that
x, y ∈ R and ∣x − y∣ < δ implies that ∣f (x) − f (y)∣ < 2ε .
Let s0 ∈ R be fixed. If ∣s − s0 ∣ < δ, then
∥Ψ[f ] (s) − Ψ[f ] (s0 )∥∞ = ∥τs [f ] − τs0 [f ]∥∞
= ∥τs○ (f ) − τs○0 (f )∥sup
= sup ∣f (θ − s) − f (θ − s0 )∣
θ∈[−π,π)
ε
≤
2
< ε,
since ∣(θ − s) − (θ − s0 )∣ = ∣s − s0 ∣ < δ for all θ ∈ [−π, π).
Thus Ψ[f ] is continuous at s0 . But s0 ∈ R was arbitrarily chosen, so Ψ[f ]
is continuous on R. Since [f ] ∈ [C(T, C)] was also arbitrary, this completes
the proof of the fact that ([C(T, C), ∥ ⋅ ∥∞ ) is a homogeneous Banach space
over T.

9.12. Example. Let 1 ≤ p < ∞. We claim that (Lp (T, C), ∥ ⋅ ∥p ) is a homoge-
neous Banach space over T.
(a) Let f ∈ Lp (T, C) and let q denote the Lebesgue conjugate of p; i.e. p1 + 1q = 1.
Recall from Proposition 4.10 that there exists a measurable function
u ∶ R → T such that f = u ⋅ ∣f ∣. Note that u ∈ Lq (T, C); indeed, the fact
that f is 2π-periodic implies that u is, and u has already been seen to be
140 L.W. Marcoux Introduction to Lebesgue measure

measurable. Moreover,
1/q 1/q
1 1
∥[u]∥q = ( ∫ ∣u∣q ) = ( ∫ 1) = 1.
2π [−π,π) 2π [−π,π)

Of course, ∥[u]∥q = ∥[u]∥q = 1 as well. By Hölder’s Inequality,

1
∥[f ]∥1 = ∫ ∣f ⋅ u∣ ≤ ∥[f ]∥p ∥[u]∥q ≤ ∥[f ]∥p .
2π [−π,π)

(b) Observe that [Trig(T, C)] ⊆ [C(T, C)] ⊆ Lp (T, C) ⊆ L1 (T, C).
(c) and (d) Let [f ] ∈ Lp (T, C) and s ∈ R. As noted in Section 9.9, τs [f ] ∈
L1 (T, C), and in particular τs○ (f ) is measurable.
Moreover, using Lemma 9.2:
1/p
1
∥τs [f ]∥p = ( ∫ ∣f (θ − s)∣p dm(s))
2π [−π,π)
1/p
1
=( ∫ ∣f (θ)∣p dm(θ))
2π [−π,π)
= ∥[f ]∥p < ∞.

In particular, τs [f ] ∈ Lp (T, C).

(e) Let [f ] ∈ Lp (T, C) and s0 ∈ R. Fix ε > 0. Since [C(T, C)] is dense in
(Lp (T, C), ∥ ⋅ ∥p ), we can find h ∈ C(T, C) such that
ε
∥[f ] − [h]∥p < .
3
By Example 9.11, the map s ↦ τs [h] is continuous with respect to the
∥ ⋅ ∥∞ -norm on [C(T, C)]. Thus we may choose δ > 0 such that ∣s − s0 ∣ < δ
implies that
ε
∥τs [h] − τs0 [h]∥∞ < .
3
By the triangle inequality,

∥τs [f ] − τs0 [f ]∥p ≤ ∥τs [f ] − τs [h]∥p + ∥τs [h] − τs0 [h]∥p + ∥τs0 [h] − τs0 [f ]∥p .

Let us estimate each of the terms on the right-hand side of this inequality.
Now (since translation is isometric on Lp (T, C) from above)
ε
∥τs [f ] − τs [h]∥p = ∥τs [f − h]∥p = ∥[f − h]∥p = ∥[f ] − [h]∥p < ,
3
and similarly,
ε
∥τs0 [f ] − τs0 [h]∥p < .
3
 9. CONVOLUTION 141

Moreover,
1/p
1
∥τs [h] − τs0 [h]∥p = ( ∫ ∣τs○ (h) − τs○0 (h)∣p )
2π [−π,π)
1/p
1
≤( ∫ ∥τs [h] − τs0 [h]∥p∞ )
2π [−π,π)
1/p
1
≤( ∫ (ε/3)p )
2π [−π,π)
ε
= .
3
Substituting these estimates into the inequality above shows that for
∣s − s0 ∣ < δ,
ε ε ε
∥τs [f ] − τs0 [f ]∥p < + + = ε,
3 3 3
and thus translation is continuous on (Lp (T, C), ∥ ⋅ ∥p ).
Hence (Lp (T, C), ∥ ⋅ ∥p ) is a homogeneous Banach space over T when 1 ≤ p < ∞.

9.13. Example. When p = ∞, the situation is rather different. We shall now

prove that (L∞ (T, C), ∥ ⋅ ∥∞ ) is not a homogeneous Banach space over T.
We leave it to the exercises for the reader to show that
(a) ∥[f ]∥1 ≤ ∥[f ]∥∞ for all [f ] ∈ L∞ (T, C).
(b) [Trig(T, C)] ⊆ L∞ (T, C);
(c) for all [f ] ∈ L∞ (T, C) and s ∈ R,

τs [f ] ∈ L∞ (T, C); and

(d) for all [f ] ∈ L∞ (T, C) and s ∈ R,

∥τs [f ]∥∞ = ∥[f ]∥∞ .

The failure of L∞ (T, C) to be a homogeneous Banach space over T comes down to

the fact that translation is not continuous in L∞ (T, C).
To see this, consider the function f0 ∶= χ[0,π) ∈ L∞ ([−π, π)), and let f = fˇ0 ∈
L∞ (T, C) be its 2π-periodic extension, as defined in Section 8.4.
If −π < s < 0, then τs○ (f )(θ) − τ0○ (f )(θ) = 1 − 0 = 1 for all θ ∈ (s, 0), and thus

∥τs [f ] − τ0 [f ]∥∞ ≥ 1.

(It is in fact equal to 1.) In particular,

lim ∥τs [f ] − τ0 [f ]∥∞ = 1 ≠ 0 = ∥τ0 [f ] − τ0 [f ]∥∞ ,

s→0

and so the map s ↦ τs [f ] is not continuous at 0.

142 L.W. Marcoux Introduction to Lebesgue measure

9.14. Let g ∈ C(T, C) and [f ] ∈ L1 (T, C). We have defined (see Definition 9.8)
the convolution of g and [f ] to be g ∗ [f ] ∶= [g ◇ f ], where
1
g ◇ f (θ) = ∫ g(s)f (θ − s) dm(s).
2π [−π,π)
That is, we define g ∗[f ] by first defining g ◇f pointwise, using Lebesgue integration.
Now consider that Example 9.12 shows that (L1 (T, C), ∥ ⋅ ∥1 ) is a homogeneous
Banach space over T. As such, the function
β ∶ R → L1 (T, C)
s ↦ g(s)τs [f ]
is continuous. By Theorem 1.14,
1 π 1 π
∫ β(s) ds = ∫ g(s)τs [f ] ds
2π −π 2π −π
exists in L1 (T, C), and is obtained as an ∥ ⋅ ∥1 -limit of Riemann sums (β, PN , PN∗ ) ∈
L1 (T, C) using partitions PN of [−2π, 2π] with corresponding choices PN∗ of test
values for PN .
If we fix g ∈ C(T, C), then we obtain a map
Γg ∶ L1 (T, C) → L1 (T, C)
1 π
[f ] ↦ 2π ∫−πg(s)τs [f ] ds.
We leave it to the reader to verify that Γg is linear.
The reader may have noticed that there is a striking resemblance between the
operators Cg and Γg . After all, τs [f ] = [τs○ (f )], where τs○ (f )(θ) = f (θ − s), θ ∈ R.
Our next goal is to show that in fact, Γg = Cg , so that for each [f ] ∈ L1 (T, C),
1 π
Cg [f ] = g ∗ [f ] = [g ◇ f ] = ∫ g(s)τs [f ] ds = Γg [f ].
2π −π
This is not an obvious nor a trivial result: the two constructions are entirely different.
Quite frankly, some of us – and you know who you are – don’t deserve this.
9.15. In Example 9.11, we showed that ([C(T, C)], ∥ ⋅ ∥∞ ) is a homogeneous Ba-
nach space over T. It follows that given f ∈ C(T, C), the map s ↦ τs [f ] is continuous,
or equivalently, the map s ↦ τs○ (f ) is continuous from (R, ∣ ⋅ ∣) to (C(T, C), ∥ ⋅ ∥sup ).

Before presenting our general result, we shall need a crucial Lemma.

9.16. Lemma. Let f, g ∈ (C(T), ∥ ⋅ ∥sup ). Let
1 π
Γ○g (f ) ∶= ○
∫ g(s)τs (f ) ds,
2π −π
taken as a Banach space Riemann integral in (C(T), ∥⋅∥sup ) in the sense of Chapter 1.
Then
1
Γ○g (f )(θ) = g ◇ f (θ) = ∫ g(s)f (θ − s) dm(s) for all θ ∈ R.
2π [−π,π)
Proof. Perhaps the most difficult part of this proof is to first ensure that we
understand the difference between Γ○g (f ) and g◇f . On the one hand, (C(T), ∥⋅∥sup ) is
 9. CONVOLUTION 143

a Banach space and the map β ∶ R → C(T, C) defined by β(s) ∶= g(s)τs○ (f ) ∈ C(T, C)
is continuous. By Theorem 1.14, Γ○g (f ) ∈ C(T, C) exists as a ∥ ⋅ ∥sup -limit of Riemann
sums S(β, PN , PN∗ ) ∈ (C(T, C), ∥ ⋅ ∥sup ). In fact, as we saw there, we may suppose
without loss of generality that for each N ≥ 1, PN ∈ P([−π, π]) is a regular partition
2π
of [−π, π] into 2N subintervals of equal length N , and PN∗ = PN ∖ {−π}, so that PN∗
2
is a set of test values for PN .
Meanwhile, g◇f is the convolution of g and f defined pointwise through Lebesgue
integration, as in Definition 9.3.
Let us temporarily fix θ0 ∈ R and define a function γ(= γθ0 ) ∶ R → K via:

γ(s) = g(s)f (θ0 − s), s ∈ R.

In the present case where g and f are both continuous, the map γ is easily seen to
also be continuous, and thus both bounded and Riemann integrable on [−π, π). By
Theorem 5.24 therefore,
1
g ◇ f (θ0 ) ∶= ∫ g(s)f (θ0 − s) dm(s)
2π [−π,π)
1
= ∫ γ(s) dm(s)
2π [−π,π)
1 π
= ∫ γ(s) ds
2π −π
1 π
= ∫ g(s)f (θ0 − s) ds,
2π −π

where the last two integrals are Riemann integrals.

Since (C, ∣ ⋅ ∣) is a Banach space, the argument of Theorem 1.14 also shows that
with PN and PN∗ defined as above (alternatively, from first-year Calculus),
1 π
g ◇ f (θ0 ) = ∫ γ(s) ds
2π −π
= lim S(γ, PN , PN∗ ).
N →∞

Finally,

∥Γ○g (f ) − S(β, PN , PN∗ )∥sup ≥ ∣Γ○g (f )(θ0 ) − S(β, PN , PN∗ )(θ0 )∣

2N
= ∣Γ○g (f )(θ0 ) − ∑ (β(pn ))(θ0 )(pn − pn−1 )∣
n=1
2N
= ∣Γ○g (f )(θ0 ) − ∑ (g(pn ))f (θ0 − pn )(pn − pn−1 )∣
n=1
144 L.W. Marcoux Introduction to Lebesgue measure

2N
= ∣Γ○g (f )(θ0 ) − ∑ γ(pn )(pn − pn−1 )∣
n=1
= ∣Γg (f )(θ0 ) − S(γ, PN , PN∗ )∣.
○

But limN →∞ ∥Γ○g (f ) − S(β, PN , PN∗ )∥sup = 0, so

0 = lim ∣Γ○g (f )(θ0 ) − S(γ, PN , PN∗ )∣.
N →∞
That is,
Γ○g (f )(θ0 ) = lim S(γ, PN , PN∗ ) = g ◇ f (θ0 ).
N →∞
Since θ0 ∈ R was arbitrary, Γ○g (f ) = g ◇ f .
◻

9.17. Theorem. Let g ∈ C(T, C) and [f ] ∈ L1 (T, C). Let

1 π
Γg [f ] ∶= ∫ g(s)τs [f ] ds,
2π −π
where the integral is a Banach space Riemann integral in (L1 (T, C), ∥ ⋅ ∥1 ) in the
sense of Chapter 1. Then
Γg [f ] = g ∗ [f ] = [g ◇ f ] = Cg [f ].
Proof. Since [C(T, C)] is ∥ ⋅ ∥1 -dense in L1 (T, C), we can find a sequence (fm )∞
m=1
in C(T, C) such that
lim ∥[fm ] − [f ]∥1 = 0.
n→∞
For each m ≥ 1, define
1 π
Γg [fm ] ∶= ∫ g(s)τs [fm ] ds
2π −π
as a Riemann integral in (L1 (T, C), ∥ ⋅ ∥1 ) in the sense of Chapter 1.
Because fm ∈ C(T, C) for each m ≥ 1, we have that the map s ↦ g(s)τs○ (f ) is
continuous, and thus the integral
1 π
Γ○g (fm ) ∶= ○
∫ g(s)τs (fm ) ds,
2π −π
converges in (C(T, C), ∥ ⋅ ∥sup ) by Lemma 9.16. Equivalently, the Banach space
Riemann integral
1 π
[Γ○g (fm )] ∶= ∫ g(s)τs [fm ] ds
2π −π
converges in ([C(T, C)], ∥ ⋅ ∥∞ ). But ∥[h]∥1 ≤ ∥[h]∥∞ for all h ∈ C(T, C), and thus
[Γ○g (fm )] = Γg [fm ],
as the integral also converges (to the same element) in (L1 (T, C), ∥ ⋅ ∥1 ). (More
details may be found in the Appendix – Remark 9.39.)
 9. CONVOLUTION 145

Step One.
First we shall show that Γg [fm ] = g ∗ [fm ] for all m ≥ 1.
By Lemma 9.16, Γ○g (fm ) = g ◇ fm for all m ≥ 1. Thus

Γg [fm ] = [Γ○g (fm )] = [g ◇ fm ] = g ∗ [fm ], m ≥ 1.

Step Two.
Next we show that g ∗ [f ] = limm→∞ g ∗ [fm ] in (L1 (T, C), ∥ ⋅ ∥1 ).
Now, for all m ≥ 1 and all θ ∈ R,

∣g ◇ (f − fm )(θ)∣ ≤ ν∞ (g) ν1 (f − fm ) = ∥g∥sup ∥[f ] − [fm ]∥1 ,

as we saw in the paragraph following Definition 9.3. Thus for all m ≥ 1,

∥g ∗ [f ] − g ∗ [fm ]∥1 = ∥g ∗ [f − fm ]∥1

1
= ∫ ∣g ◇ (f − fm )(θ)∣dm(θ)
2π [−π,π)
1
≤ ∫ ∥g∥sup ∥[f ] − [fm ]∥1 dm(θ)
2π [−π,π)
= ∥g∥sup ∥[f ] − [fm ]∥1 ,

from which it easily follows that

g ∗ [f ] = lim g ∗ [fm ] in (L1 (T, C), ∥ ⋅ ∥1 ).

m→∞

Step Three.
We now show that Γg [f ] = limm→∞ Γg [fm ] in (L1 (T, C), ∥ ⋅ ∥1 ).
Indeed,
1 π
∥Γg [f ] − Γg [fm ]∥1 = ∥ ∫ g(s)τs ([f − fm ]) ds∥
2π −π 1
1 π
≤ ∫ ∣g(s)∣ ∥τs ([f − fm ])∥1 ds
2π −π
1 π
≤ ∥g∥sup ∫ ∥[f ] − [fm ]∥1 ds
2π −π
= ∥g∥sup ∥[f ] − [fm ]∥1 .

As before, it easily follows that

Γg [f ] = lim Γg [fm ] in (L1 (T, C), ∥ ⋅ ∥1 ).

m→∞

Step Four.
Finally (!) we see that in (L1 (T, C), ∥ ⋅ ∥1 ) we have that

Γg [f ] = lim Γg [fm ] = lim g ∗ [fm ] = g ∗ [f ] = Cg [f ].

m→∞ m→∞

◻
146 L.W. Marcoux Introduction to Lebesgue measure

Viewing Γg as a map from L1 (T, C) into itself, we have shown that Γg = Cg ; in

other words, the two notions of “convolution” agree. But in fact, when g ∈ C(T, C),
we may define
ΓB
g ∶ B → B
1 π
[f ] ↦ 2π ∫−π g(s)τs [f ] ds
as a map on any homogeneous Banach space B over T. Also, Cg [f ] = [g ◇ f ] ∈ B,
Let us now show that it always agrees with convolution.

9.18. Theorem. Let (B, ∥⋅∥B ) be a homogeneous Banach space over T, [f ] ∈ B,

and let g ∈ C(T, C). Then
1 π
ΓB
g [f ] ∶= ∫ g(s)τs [f ]ds
2π −π
converges in B.
(a) Furthermore
ΓB
g [f ] = g ∗ [f ] = [g ◇ f ] = Cg [f ],

where – as always – for all θ ∈ R,

1
g ◇ f (θ) = ∫ g(s)f (θ − s) dm(s).
2π [−π,π)
(b) Moreover,
∥g ∗ [f ]∥B ≤ ν1 (g) ∥[f ]∥B .
Proof.
(a) Since (B, ∥ ⋅ ∥B ) is a Banach space, and since – for fixed [f ] ∈ B – the
function β ∶ R → B defined by β(s) ∶= g(s)τs [f ] is continuous, we see from
Theorem 1.14 that
1 π 1 π
ΓBg [f ] ∶= ∫ β(s) ds = ∫ g(s)τs [f ] ds
2π −π 2π −π
exists in B (i.e. the Riemann sums corresponding to the integral converge
in B).
As always, using the arguments of Theorem 1.14, we see that if we set
PN to be a regular partition of [−π, π] into 2N subintervals of equal length,
and if PN∗ = PN ∖ {π} is a corresponding set of test values for PN , then
∗
lim ∥ΓB
g [f ] − S(β, PN , PN )∥B = 0.
N →∞

But ∥[h]∥1 ≤ ∥[h]∥B for all [h] ∈ B, as B is a homogeneous Banach space

over T. Thus
∗
lim ∥ΓB
g [f ] − S(β, PN , PN )∥1 = 0.
N →∞
In other words,
1 π
ΓB
g [f ] = ∫ g(s)τs [f ]ds in (L1 (T, C), ∥ ⋅ ∥1 ).
2π −π
 9. CONVOLUTION 147

L (T,C)
Phrased another way, ΓB
g [f ] = Γg
1
[f ]. By Theorem 9.17,
L1 (T,C)
ΓB
g [f ] = Γg [f ] = g ∗ [f ] = Cg [f ].
(b) Recall that – in any homogeneous Banach space over T – we defined the
continuous map Φ[f ] ∶ R → B via Ψ[f ] (s) = τs [f ], and we have that
∥Ψ[f ] (s)∥B = ∥[f ]∥B for all s ∈ R.
Next, consider that
1 π
∥g ∗ [f ]∥B = ∥∫ g(s)τs [f ] ds∥
2π −π B
1 π
= ∥∫ g(s)Ψ[f ] (s) ds∥
2π −π B
1 π
≤ ∫ ∣g(s)∣ ∥Ψ[f ] (s)∥B ds
2π −π
1 π
= ∫ ∣g(s)∣ ∥[f ]∥B ds
2π −π
= ∥[f ]∥B ν1 (g).
◻

9.19. Remark. The conclusion of Theorem 9.18 (a) is actually somewhat

stronger than it might at first appear. We showed in Proposition 9.6 that if g ∈
C(T, C) and f ∈ L1 (T, C), then g ◇ f ∈ C(T, C), and thus g ∗ [f ] ∶= [g ◇ f ] ∈ [C(T, C)].
So why should it lie in B? After all, there is no reason why B should contain all con-
tinuous functions (despite the fact that it contains all trigonometric polynomials).
What we have just shown is that g ∗ [f ] ∈ B even if B doesn’t contain [C(T, C)]. In
other words, convolution (at least by a continuous function) places us in the smaller
space B (when we start in the smaller space).
It can in fact be shown that if [g] ∈ L1 (T, C), [f ] ∈ B ⊆ L1 (T, C), and if [g] ∗ [f ]
is defined as in Section 9.1, then [g] ∗ [f ] ∈ B, but that is beyond the scope of the
course.

9.20. Theorem. Let g ∈ C(T, C), and let

Cg ∶ ([C(T, C)], ∥ ⋅ ∥∞ ) → ([C(T, C)], ∥ ⋅ ∥∞ )
be the convolution operator corresponding to g, so that Cg [h] = g ∗ [h]. Then ∥Cg ∥ =
ν1 (g) = ∥[g]∥1 .
Proof. By Theorem 9.18 (b), we see that for all [f ] ∈ ([C(T, C)], ∥ ⋅ ∥∞ ),
∥Cg [f ]∥∞ = ∥g ∗ [f ]∥∞ ≤ ν1 (g) ∥[f ]∥∞ ,
whence ∥Cg ∥ ≤ ν1 (g). There remains to show that ∥Cg ∥ ≥ ν1 (g).
Let f ∈ C(T, C), with ∥[f ]∥∞ ≤ 1. Then g ∗ [f ] ∈ [C(T, C)], so
∥Cg [f ]∥∞ = ∥g ∗ [f ]∥∞ = ∥g ◇ f ∥sup ≥ ∣g ◇ f (0)∣.
148 L.W. Marcoux Introduction to Lebesgue measure

Next, using the fact that the Lebesgue and Riemann integrals of bounded, continu-
ous functions are equal, we find that
1
g ◇ f (0) = ∫ g(s)f (0 − s)dm(s)
2π [−π,π)
1 π
= ∫ g(s)f (−s)ds.
2π −π
Case One. Suppose that g is invertible; i.e. g(s) ≠ 0 for all s ∈ R. Choose

g(−s)
f (s) = , s ∈ R,
∣g(−s)∣
so that f is continuous and ∥[f ]∥∞ = ∥f ∥sup = 1.
Furthermore,
∥Cg ∥ ≥ ∥Cg [f ]∥∞
≥ ∣Cg (f )(0)∣
1 π
= ∣∫ g(s)f (−s)ds∣
2π −π
1 π
= ∫ ∣g(s)∣ds
2π −π
= ∥[g]∥1 .

Case Two. We shall make use of the following result, which was once a bonus
question for the Assignments, but has now been relegated to the Appendix (see
Theorem 9.40): let g ∈ C(T, C) and ε > 0. Then there exists h ∈ C(T, C) such that
h(s) ≠ 0 for all s ∈ R and ∥h − g∥sup < ε.

From this it immediately follows that

∣∥[g]∥1 − ∥[h]∥1 ∣ ≤ ∥[g] − [h]∥1 ≤ ∥[g] − [h]∥∞ < ε.
But then
∥Cg − Ch ∥ = ∥Cg−h ∥ ≤ ∥[g − h]∥1 < ε.
By Case One, ∥Ch ∥ = ∥[h]∥1 , whence
∥Cg ∥ ≥ ∥Ch ∥ − ∥Cg−h ∥ ≥ ∥[h]∥1 − ε > ∥[g]∥1 − 2ε.
Since ε > 0 was arbitrary,
∥Cg ∥ ≥ ∥[g]∥1 ,
completing the proof.
◻
We now establish a similar result for convolution by a continuous function g
acting on L1 (T, C).
 9. CONVOLUTION 149

9.21. Theorem. Let g ∈ C(T, C), and let Cg ∶ L1 (T, C) → L1 (T, C) be the
convolution operator corresponding to g, so that Cg [f ] = g ∗ [f ]. Then ∥Cg ∥ =
ν1 (g) = ∥[g]∥1 .
Proof. By Theorem 9.18 (b), we see that for all [f ] ∈ L1 (T, C),
∥Cg [f ]∥1 = ∥g ∗ [f ]∥1 ≤ ν1 (g) ∥[f ]∥1 ,
whence ∥Cg ∥ ≤ ν1 (g).
There remains to show that ∥Cg ∥ ≥ ν1 (g), or equivalently, that ∥Cg ∥ ≥ ∥[g]∥1 .
To that end, we consider the functions fn = nπχ[−1/n,1/n] , n ≥ 1.
Clearly [fn ] ∈ L1 (T, C) and ∥[fn ]∥1 = 1 for all n.
Moreover, for all θ ∈ [−π, π), and hence for all θ ∈ R, we have (using Lemma 9.2)
1
(g ◇ fn )(θ) = ∫ g(s)fn (θ − s)dm(s)
2π [−π,π)
1
= ∫ g(θ − t)fn (t)dm(t)
2π [−π,π)
1
= (nπ) ∫ g(θ − t)dm(t)
2π [−1/n, 1/n]
n
= ∫ g(θ − t)dm(t).
2 [−1/n, 1/n]
Note also that
n
g(θ) = ∫ g(θ) dm(t),
2 [−1/n,1/n]
since g(θ) acts as a constant in this integral.
Thus
n
∣g ◇ fn (θ) − g(θ)∣ = ∣ ∫ g(θ − t) − g(θ) dm(t)∣
2 [−1/n, 1/n]
n
≤ ∫ ∣g(θ − t) − g(θ)∣dm(t)
2 [−1/n, 1/n]
Let ε > 0 and choose 1 ≤ N ∈ N such that ∣x−y∣ < N1 implies that ∣g(x)−g(y)∣ < ε.
This is possible because g is uniformly continuous on R (by virtue of being continuous
on [−π, π) and 2π-periodic).
For n > N , t ∈ [− n1 , n1 ] implies that ∣(θ − t) − θ∣ ≤ n1 < δ, and so ∣g(θ − t) − g(θ)∣ < ε.
But then for all n ≥ N and for all θ ∈ R,
n
∣g ◇ fn (θ) − g(θ)∣ ≤ ∫ ε dm(t) = ε.
2 [−1/n, 1/n]
Hence
1
ν1 (g ◇ fn − g) = ∫ ∣g ◇ fn (θ) − g(θ)∣ dm(θ)
2π [−π,π)
1
≤ ∫ ε dm(θ)
2π [−π,π)
= ε.
150 L.W. Marcoux Introduction to Lebesgue measure

From this it follows that in L1 (T, C),

lim Cg ([fn ]) = [g],
n→∞
and so
lim ∥Cg ([fn ])∥1 = ∥[g]∥1 .
n→∞
But ∥[fn ]∥1 = 1 for all n ≥ 1, and thus
∥Cg ∥ ≥ sup ∥Cg ([fn ])∥1 = ∥[g]∥1 ,
n≥1
completing the proof.
◻

In the next couple of chapters, we shall explore the connection between convo-
lution operators and convergence of Fourier series.
 9. CONVOLUTION 151

Appendix to Section 9.

9.22. As mentioned at the start of the Chapter, an algebra A over a field F is

a vector space over F which is also a ring. In other words, as well as being a vector
space, the algebra A must also admit a multiplication, denoted by ⋅, which satisfies
the following properties:
(i) (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c) for all a, b, c ∈ A;
(ii) a ⋅ (b + c) = a ⋅ b + a ⋅ c for all a, b, c ∈ A; and
(iii) (a + b) ⋅ c = a ⋅ c + b ⋅ c for all a, b, c ∈ A.
It often transpires that we simply write ab instead of a ⋅ b for the product in an
algebra. We shall do so in the examples below.
If there is an element e ∈ A such that e ⋅ a = a = a ⋅ e for all a ∈ A, then we say that
A is a unital algebra, and that e is the multiplicative identity for A. There is
a simple construction which allows us to embed any algebra A into a unital algebra
B, however in practice this construction is not always natural. In fact, it is precisely
in the case that interests us, namely in the case where A = L1 (T, C) equipped with
convolution as the “multiplication” that the construction leads to an identity which
operates as a point mass at zero - and there is no natural way of thinking of this as
a nice “function” on R.
We are not deterred by the lack of a multiplicative identity in our algebras.
Below we shall discuss what is often a suitable and perfectly acceptable alternative.

9.23. Example. Since the notion of an algebra is an algebraic one (a most

reasonable state of affairs), it seems only fair to begin with an example which will
warm the cockles of every algebraist’s heart:
Let F be a field. We define

F[x] ∶= {p0 + p1 x + p2 x2 + ⋯ + pn xn ∶ pj ∈ F, 1 ≤ j ≤ n, n ≥ 1}.

Then F[x] is the set of all polynomials with coefficients in F in an indeterminate x;

using the usual operations governing addition, scalar multiplication, and multipli-
cation of polynomials, F[x] becomes an algebra over F.

9.24. Example. Let n ≥ 1 be an integer. Then A ∶= Mn (C) is a unital

algebra over C, using the usual notions of matrix multiplication, scalar multi-
plication, and addition. The identity under multiplication is the identity matrix
In ∶= diag(1, 1, 1, . . . , 1).

9.25. Example. Let X be a Banach space, and set A = B(X). Then A is a

unital algebra, where the multiplication in B(X) refers to the composition of linear
maps. That is, (AB)x = A(Bx) for all x ∈ X, A, B ∈ A.
The multiplicative identity in this case is again the identity map Ix = x, x ∈ X.
152 L.W. Marcoux Introduction to Lebesgue measure

9.26. Example. Consider A ∶= ℓ∞ (N, C). Clearly A is a vector space. We may

define a product on A by setting
(xn )n ⋅ (yn )n ∶= (xn yn )n .
Using this product, A is a unital algebra. The space B ∶= c0 (N, C) is a subalgebra
of A. That is, it is a subset of A and is an algebra using the operations inherited
from A.

9.27. Closely related to the notion of an algebra is the notion of a module.

In its most general incarnation, a (left) module is an abelian group which admits a
left-action by a ring. We are only interested in a special case of this phenomenon,
and so we shall only define the concept of a module in the setting of interest to us.
For more information about modules in general, we refer the reader to [1] (just one
amongst a cornucopia of books that deal with the subject).
9.28. Definition. Let M be a vector space over K, and let A be an algebra over
K. We say that M is a left-A module if there exists an operation ○ ∶ A × M → M
which satisfies:
(i) (a + b) ○ m = a ○ m + b ○ m for all a, b ∈ A, m ∈ M;
(ii) a ○ (m + n) = a ○ m + a ○ n for all a ∈ A, m, n ∈ M; and
(iii) (ab) ○ m = a ○ (b ○ m) for all a, b ∈ A and m ∈ M.
If A is unital, we also ask that 1 ○ m = m for all m ∈ M.
As the reader might imagine, if B is an algebra, there exists also the concept of M
being a right−B module. We leave the definition to the reader’s vivid imagination.
Finally, M is said to be a A − B bimodule if it is a left-A module and a right-B
module. If A = B, we refer to M simply as an A bimodule.

9.29. Example. Let A = M3 (C) and B = M7 (C). Then M ∶= M3×7 (C) becomes
an A − B bimodule, using usual matrix multiplication on the left by elements of A,
and usual matrix multiplication on the right by elements of B.

9.30. Example.
(a) Let A = ℓ∞ (N, C) and M = c0 (N, C). Then M is a A bimodule, where we
define (an )n ⋅ (mn )n ∶= (an mn )n = (mn )n ⋅ (an )n for all a = (an )n ∈ A and
(mn )n ∈ M.
(b) We can also set M = ℓ∞ (N, C) and A = c0 (N, C). Using the same operations
as above, ℓ∞ (N, C) becomes a c0 (N, C) bimodule.

9.31. Our approach to convolution has been to show that L1 (T, C) is a bimodule
over C(T, C), using convolution ∗ as our “multiplication” operation. It is relatively
straightforward to prove that conditions (i) and (ii) of Definition 9.34 hold. What is
left obvious (and what is left as an assignment exercise) is that condition (iii) holds
as well.
For this, it is worth observing that given g, h ∈ C(T, C) and f ∈ L1 (T, C), h ◇ f ∈
C(T, C), and thus g ◇ (h ◇ f ) may be realised as a Riemann integral, instead of
 9. CONVOLUTION 153

a Lebesgue integral. This is crucial in letting us circumvent the use of Fubini’s

Theorem, which is the raison-d’être of our approach.

9.32. Definition. We say that a Banach space (A, ∥ ⋅ ∥) is a Banach algebra

if it is also an algebra, and if
∥ab∥ ≤ ∥a∥ ∥b∥ for all a, b ∈ A.

The norm condition for the product of a and b above ensures that multiplication
is jointly continuous; i.e. the map
µ∶ A×A → A
(a, b) ↦ ab
is continuous.

Some of the above examples of algebras are actually examples of Banach alge-
bras.

9.33. Example. If X is a Banach space, then (B(X), ∥ ⋅ ∥) is a Banach algebra,

as
∥AB∥ = sup ∥ABx∥
x∈X,∥x∥≤1
≤ sup ∥A∥ ∥Bx∥
x∈X,∥x∥≤1
≤ sup ∥A∥ ∥B∥
x∈X,∥x∥≤1
= ∥A∥ ∥B∥.

9.34. We leave it as an exercise for the reader to prove that (C(K, C), ∥⋅∥sup ) is a
unital Banach algebra, whenever K is a compact, Hausdorff space. Here, functions
are added and multiplied pointwise, and (κf )(x) = κ(f (x)) for all κ ∈ C, f ∈
C(K, C).

9.35. As mentioned above, one can extend the notion of convolution to obtain
a product operation on L1 (T, C), under which the latter becomes a Banach algebra.
Alas, this algebra is non-unital: that is, it does not admit an identity element under
this operation.
In the study of Banach algebras, and more specifically of C ∗ -algebras of oper-
ators on a Hilbert space, one often comes across the situation where the algebra
is non-unital, but does admit the “next best thing” to a unit, namely a bounded
approximate unit.
154 L.W. Marcoux Introduction to Lebesgue measure

9.36. Definition. Let (A, ∥ ⋅ ∥) be a Banach algebra. An approximate unit

for A is a net (eλ )λ∈Λ in A such that
lim ∥eλ a − a∥ + ∥aeλ − a∥ = 0
λ

for all a ∈ A.
We say that (eλ )λ∈Λ is a bounded approximate unit if there exists M > 0
such that ∥eλ ∥ ≤ M for all λ ∈ Λ.
Typically, when the algebra in question is separable (i.e. A admits a countable
dense set), the approximate unit may be chosen to be a sequence, rather than a net.
We shall come across examples of these in Chapter 11, hidden under the guise
of summability kernels.

9.37. Example. Let

A = c0 (N, C) = {(wn )∞
n=1 ∶ wn ∈ C for all n ≥ 1 and lim wn = 0}.
n→∞

Then A is a Banach space using pointwise operations, and we can also give
it a multiplication operation in a similar way: that is, we set (wn )∞ ∞
n=1 ⋅ (zn )n=1 =
∞
(wn zn )n=1 . It is routine to verify that (A, ∥ ⋅ ∥∞ ) is a Banach algebra.
It is clearly non-unital. However, if we set en = (1, 1, . . . , 1, 0, 0, 0, . . .) for each
n ≥ 1 (where there are n terms which are equal to 1 and all remaining terms are
equal to 0), then (en )∞ n=1 is easily seen to be a bounded approximate identity for A.

9.38. There are also discrete versions of convolution; let g ∈ ℓ∞ (Z, C) and f ∈
ℓ1 (Z, C). We may define the discrete convolution of g and f as follows:
∞ ∞
g ∗ f (n) = ∑ g(m)f (n − m) = ∑ g(n − m)f (m).
m=−∞ m=−∞
For those who are interested, this operation turns (ℓ1 (Z, C), ∥⋅∥1 ) into a commu-
tative Banach algebra, using discrete convolution as the multiplication operation.
Is it unital?

9.39. Remark. There is a minor subtlety in the proof of Theorem 9.17 that
goes along the following lines:
Fix 1 ≤ m an integer. We defined Γ○g (fm ) as a Riemann integral in the Banach
space (C(T, C), ∥ ⋅ ∥sup ). Thus, for an appropriate sequence (PN )N of partitions of
[−π, π], we have that
Γ○g (fm ) = lim S(βm , PN , PN∗ ),
N →∞
where βm (s) = g(s)τs○ (fm ), s ∈ [−π, π).
But given any Riemann sum S(βm , Q, Q∗ ) of the form
M M
∗ ∗ ○
∑ βm (qk )(qk − qk−1 ) = ∑ g(qk )τqk∗ (fm )(qk − qk−1 )
k=1 k=1
 9. CONVOLUTION 155

in C(T, C), its image in [C(T, C)] is

M
[S(βm , Q, Q∗ )] = ∑ g(qk∗ )τqk∗ [fm ](qk − qk−1 ).
k=1

Since the map h → [h] from (C(T, C), ∥ ⋅ ∥sup ) to ([C(T, C)], ∥ ⋅ ∥∞ ) is a bijective
linear isometry, it follows that the image [Γ○g (fm )] of Γ○g (fm ) is
[Γ○g (fm )] = lim [S(βm , PN , PN∗ )],
N →∞

and that this convergence is with respect to the ∥ ⋅ ∥∞ norm. On the other hand,
looking how each [S(βm , PN , PN∗ )] is defined, we see that the latter limit is precisely
how we defined Γg ([fm ]) ∈ ([C(T, C), ∥ ⋅ ∥∞ ), and thus
[Γ○g (fm )] = Γg [fm ], m ≥ 1.

Next, each [S(βm , PN , PN∗ )] ∈ [C(T, C)] ⊆ L1 (T, C), and Γg [fm ] ∈ [C(T, C)] ⊆
L1 (T, C) as well. As remarked in the proof of Theorem 9.17, ∥[h]∥1 ≤ ∥[h]∥∞ for all
[h] ∈ [C(T, C)], and thus
0 ≤ lim ∥[Γ○g (fm )] − [S(βm , PN , PN∗ )]∥1 ≤ lim ∥[Γ○g (fm )] − [S(βm , PN , PN∗ )]∥∞ = 0,
N →∞ N →∞
proving that
Γg ([fm ]) = [Γ○g (fm )] = lim [S(βm , PN , PN∗ )],
N →∞
with the convergence taking place in (L1 (T, C), ∥ ⋅ ∥1 ).

The following result was required for Case Two of Theorem 9.20.

9.40. Theorem. If f ∈ C(T, C) and ε > 0, then there exists g ∈ C(T, C) so that
g(x) =/ 0 for all x ∈ [−π, π] and ∥f − g∥sup < ε.
Remark: Just in case you’ve forgotten - here’s a note to remind you that this is
very much a complex phenomenon. If we replace complex functions by real-valued
functions, the corresponding assertion is false.
A Banach algebra with the property that the invertible elements are dense is
said to have topological stable rank one. As you might imagine, there is a
notion of higher topological stable ranks. When X is a compact topological space,
the topological stable rank of C(X, K) is supposed to be a measure of the dimension
of X.
Proof. (I) Let ε > 0. By the Stone-Weierstrass Theorem, we can find a polynomial
p0 so that
sup ∣p0 (x) − f (x)∣ < ε/100.
x∈[a,b]
51
By adding a constant function κ1 to p0 with ∣κ∣ ≤ 100 ε if necessary, we can assume
that for p = p0 + κ1,
min(∣p(a)∣, ∣p(b)∣) ≥ ε/2.
156 L.W. Marcoux Introduction to Lebesgue measure

Observe that
1
∣p(a) − p(b)∣ = ∣p0 (a) − p0 (b)∣ ≤ ∣p0 (a) − f (a)∣ + ∣f (a) − f (b)∣ + ∣f (b) − p0 (b)∣ ≤ ε,
50
and that
52
sup ∣f (x) − p(x)∣ ≤ ε.
x∈[a,b] 100

(II)
Let p(x) = α0 (x−α1 )(x−α2 )⋯(x−αn ) for the appropriate choices of αi , 0 ≤ i ≤ n.
If we impose the norm ∥b∥∞ ∶= supx∈[a,b] ∣b(x)∣ for b ∈ C[x] ⊆ C([a, b], K), then it is
clear that the map
Φ∶ (Cn , ∥ ⋅ ∥∞ ) → (C[x], ∥ ⋅ ∥∞ )
(β1 , β2 , ..., βn ) ↦ (x − β1 )(x − β2 )⋯(x − βn )
is continuous. Thus, given η > 0 we can find β1 , β2 , ..., βn ∈ C ∖ R so that βk − αk is
sufficiently small, 1 ≤ k ≤ n, to guarantee that with q(x) = α0 (x−β1 )(x−β2 )⋯(x−βn ),
we have ∥p − q∥∞ < η. Since βk ∈/ R for all 1 ≤ k ≤ n, it follows that q has no real
roots - i.e. q(x) =/ 0 for all x ∈ [a, b].
Observe that ∥p − q∥∞ < η implies that
1
∣q(a) − q(b)∣ ≤ ∣q(a) − p(a)∣ + ∣p(a) − p(b)∣ + ∣p(b) − q(b)∣ ≤ η + ε + η.
50
1
In particular, if we choose η = 100 ε, then
1
∣q(a) − q(b)∣ ≤ ε.
25

(III)
Now we are in the situation where
● q ∶ [a, b] → C is a polynomial and q(x) =/ 0 for all x ∈ [a, b],
1
● ∣q(a)∣ ≥ ∣p(a)∣ − η ≥ 12 ε − 100 49
ε = 100 49
ε, and similarly ∣q(b)∣ ≥ 100 ε, and
●
1
∣q(a) − q(b)∣ ≤ ε.
25
Since q is continuous at b, we can find δ > 0 so that b − δ ≤ x ≤ b implies that
1
∣q(x) − q(b)∣ < ε.
25
Let r ∶ [a, b] → C be the continuous function on [a, b] defined by
● r(x) = 0 if x ∈ [a, b − δ],
x − (b − δ)
● r(x) = (q(a) − q(b)) , x ∈ [b − δ, b].
δ
 9. CONVOLUTION 157

1
Observe that ∣r(x)∣ ≤ ∣q(a) − q(b)∣ ≤ 25 ε for all x ∈ [b − δ, b], and hence for all
x ∈ [a, b]. Let g(x) = q(x) + r(x), x ∈ [a, b]. Clearly g is continuous since each of q
and r are continuous, and for x ∈ [b − δ, b] we have
1 1 41
∣g(x)∣ = ∣q(x) + r(x)∣ ≥ ∣q(x)∣ − ∣r(x)∣ ≥ (∣q(b)∣ − ε) − ε ≥ ε.
25 25 100
In particular, g(x) =/ 0 for all x ∈ [b − δ, b]. Of course, g(x) = q(x) =/ 0 for all
x ∈ [a, b − δ], and thus g(x) =/ 0 for all x ∈ [a, b].
Also,
g(a) = q(a) = q(b) + r(b) = g(b).
1
Finally, as we have seen above, ∣r(x)∣ ≤ ∣q(a) − q(b)∣ ≤ 25 ε for all x ∈ [a, b], so
sup ∣f (x) − g(x)∣ ≤ sup ∣f (x) − p(x)∣ + sup ∣p(x) − q(x)∣ + sup ∣q(x) − g(x)∣
x∈[a,b] x∈[a,b] x∈[a,b] x∈[a,b]
52 1 1
≤ ε+ ε+ ε
100 100 25
< ε.
This completes the proof.
◻

9.41. Finally, we end this appendix with a statement of Fubini’s Theorem, as

promised.
9.42. Theorem. Fubini’s Theorem
Let (X, µX ) and (Y, µY ) be σ−finite measure spaces, and suppose that X × Y is
given the product measure µX×Y ∶= µX × µY . If f ∶ X × Y → K lies in L1 (X × Y, K),
then

∫ (∫ f (x, y)dµY (y)) dµX (x) = ∫ (∫ f (x, y)dµX (x)) dµY (y),
X Y Y X
and these coincide with
∫ f (x, y)dµX×Y (x, y).
X×Y
158 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 9.

Exercise 9.1. Let h ∶ R → C be a 2π-periodic and continuous function. Prove that

h is uniformly continuous on R.

Exercise 9.2. Prove that L∞ (T, C) ⊆ L1 (T, C) and that

(a) ∥[f ]∥1 ≤ ∥[f ]∥∞ for all [f ] ∈ L∞ (T, C);
(b) [Trig(T, C)] ⊆ L∞ (T, C);
(c) for all [f ] ∈ L∞ (T, C) and s ∈ R,
τs [f ] ∈ L∞ (T, C); and
(d) for all [f ] ∈ L∞ (T, C) and s ∈ R,
∥τs [f ]∥∞ = ∥[f ]∥∞ .

Exercise 9.3. Consider the function χ[0,π) ∈ L∞ ([−π, π)), and let f = χ
̂ [0,π) ∈
L∞ (T, C) be its 2π-periodic extension, as defined in Paragraph 8.4.
If −π < s < 0, prove that
∥τs [f ] − τ0 [f ]∥∞ = 1.

Exercise 9.4. Let (X, ∥ ⋅ ∥) be a Banach space and g ∶ R → X be a 2π-periodic,

continuous function.
Prove that for all s ∈ R,
π π
∫ g(θ)dθ = ∫ g(θ − s)ds.
−π −π

Exercise 9.5. Let g ∈ C(T, C). Prove that the convolution operator Cg ∶ L1 (T, C) →
L1 (T, C) defined by Cg [f ] = g ∗ [f ] is linear.

Exercise 9.6. What does Fubini’s Theorem say when X = Y = N and µX = µY

denote counting measure on N? That is, µX (E) denotes the cardinality of E ⊆ N?
(It should seem like a very familiar result!)

Exercise 9.7. (Assignment Exercise)

Let g, h ∈ C(T, C) and f ∈ L1 (T, C). Prove that for all θ ∈ R,
(g ◇ (h ◇ f ))(θ) = ((g ◇ h) ◇ f )(θ).
Conclude that L1 (T, C) is a bimodule over C(T, C).

Exercise 9.8. Let f ∈ L1 (T, C) and g ∈ L∞ (T, C). Define

g◇f ∶ R → C
1
θ ↦ 2π ∫[−π,π) g(s)f (θ − s)dm(s).
Prove that ∣g ◇ f (θ)∣ ≤ ν∞ (g)ν1 (f ) for all θ ∈ R, and that g ◇ f is continuous on
R.
 10. THE DIRICHLET KERNEL 159

10. The Dirichlet kernel

I just want to thank everyone who made this day necessary.

Yogi Berra

10.1. We began our discussion of Fourier series by noting that if [f ] ∈ L2 (T, C),
then the sequence (∆N ([f ]))∞ N =1 of partial sums of the Fourier series of [f ] converges
in the ∥ ⋅ ∥2 -norm to [f ] (see Paragraph 8.3). Our goal was to see to what extent we
could extend these results to elements [f ] ∈ L1 (T, C). Somewhere along the way,
we seem to have been distracted by the concept of convolution. Let us show that
all roads lead to the Dirichlet kernel, which we now define.

10.2. Definition. For each n ∈ Z, recall that ξn ∈ C(T, C) is the function

ξn (θ) = einθ . For N ≥ 1, we define the Dirichlet kernel of order N to be

N
DN = ∑ ξn .
n=−N

We mention in passing that this use of the word “kernel ” has nothing to do with
the null space of any linear map. It is just another example of the overuse of certain
terminology in mathematics.

10.3. Let f ∈ L1 (T, C). For each N ≥ 1, we shall define

N N
∆○N (f ) = ∑ αn[f ] ξn = ∑ f̂(n)ξn .
n=−N n=−N

It is clear that ∆○N (f ) ∈ C(T, C), being a finite linear combination of {ξn }N
n=−N .
[f ] [g]
If f = g a.e. on R, then αn = αn for all n ∈ Z, and thus ∆N (f ) = ∆○N (g) for
○

all N ≥ 1. We may therefore define

∆N ([f ]) = [∆○N (f )], N ≥ 1.

Thus ∆N ([f ]) is the N th -partial sum of the Fourier series of [f ]. Note that when
[f ] ∈ L2 (T, C), this definition coincides with our previous definition.
160 L.W. Marcoux Introduction to Lebesgue measure

Now, for N ≥ 1, f ∈ L1 (T, C) and θ ∈ R,

N
∆○N (f )(θ) = ∑ αn[f ] ξn
n=−N
N
1
= ∑ ( ∫ f (s)ξn (s) dm(s)) ξn (θ)
n=−N 2π [−π,π)
N
1
= ∑ ∫ f (s)ein(θ−s) dm(s)
n=−N 2π [−π,π)
N
1
= ∑ ∫ f (θ − s)eins dm(s) by our Assignments
n=−N 2π [−π,π)
N
1 ins
= ∫ ∑ f (θ − s)e dm(s)
2π [−π,π) n=−N
1
= ∫ DN (s)f (θ − s) dm(s)
2π [−π,π)
= (DN ◇ f )(θ).

In other words, ∆○N (f ) = DN ◇ f , or

∆N ([f ]) = DN ∗ [f ] = CDN ([f ]), N ≥ 1.
We have expressed the N th -partial sum of the Fourier series of [f ] ∈ L1 (T, C)
as the convolution of the Dirichlet kernel DN of order N with [f ]. Suddenly our
peripatetic meanderings through the land of convolution do not seem as unwarranted!
The question of whether or not these partial sums converge to [f ] in L1 (T, C)
now becomes a question of whether or not limN →∞ CDN ([f ]) = [f ] in L1 (T, C). In
order to answer this question, we shall examine the nature of the Dirichlet kernel a
bit closer, and borrow a couple of results from our previous real analysis courses.
10.4. Theorem. Let N ≥ 1 be an integer, and let DN denote the Dirichlet
kernel of order N . Then
(a) DN (−θ) = DN (θ) ∈ R for all θ ∈ R.
1 π
(b) 2π ∫−π DN (θ) dθ = 1.
(c) For 0 ≠ θ ∈ [−π, π),
sin((N + 12 )θ)
DN (θ) = .
sin( 12 θ)
Also, DN (0) = 2N + 1.
4 N 1
(d) ∥[DN ]∥1 = ν1 (DN ) ≥ 2 ∑ .
π n=1 n
 10. THE DIRICHLET KERNEL 161

Proof.
(a) For all θ ∈ R and n ≥ 1, ξ−n (θ) + ξn (θ) = e−inθ + einθ = 2 cos(nθ). Thus
N N
DN (θ) = ∑ ξn (θ) = 1 + 2 ∑ cos(nθ) ∈ R.
n=−N n=1

From this it is also clear that DN (−θ) = DN (θ) for all θ ∈ R.

(b) Now
π π N π
1 1 1
∫ DN (θ) dθ = ∫ 1 dθ + 2 ∑ ∫ cos(nθ) dθ.
2π −π 2π −π n=1 2π −π
π
We leave it as an exercise for the reader to show that ∫−π cos(nθ) dθ = 0,
n ≥ 1, and so
1 π 1 π
∫ DN (θ) dθ = ∫ 1 dθ = 1.
2π −π 2π −π
(c) Let θ ∈ R. Set ρ(θ) ∶= e−iθ/2 − eiθ/2 = (−2i) sin( 12 θ). A routine calculation
shows that ρ ⋅ DN involves a telescoping sum, so that
1
ρ(θ)DN (θ) = e−i(N + 2 )θ − ei(N + 2 )θ = (−2i) sin ((N + )θ) .
1 1

2
If 0 ≠ θ ∈ [−π, π), then
(−2i) sin ((N + 12 )θ) sin((N + 12 )θ)
DN (θ) = = .
ρ(θ) sin( 12 θ)
Meanwhile,
N
DN (0) = 1 + 2 ∑ cos(0) = 2N + 1.
n=1
(d) Since DN is an even, continuous function, and since ∣ sin(x)∣ ≤ ∣x∣, 0 ≤ x ≤ π,
1
ν1 (DN ) = ∫ ∣DN ∣
2π [−π,π)
1 π
= ∫ ∣DN (θ)∣ dθ
2π −π
1 π
= ∫ ∣DN (θ)∣ dθ
π 0
1 π sin((N + 1 )θ)
2
= ∫ ∣ 1
∣ dθ
π 0 sin( 2 θ)
1 π ∣ sin((N + 1 )θ)∣
2
≥ ∫
π 0 ∣ 12 θ∣
2 π ∣ sin((N + 1 )θ)∣
2
= ∫ .
π 0 ∣θ∣
162 L.W. Marcoux Introduction to Lebesgue measure

We are dealing with Riemann integration here, and so substitution of vari-

ables is permissible. Let λ = (N + 12 )θ, so that dλ = (N + 12 )dθ. Then
2 π ∣ sin((N + 1 )θ)∣
2
ν1 (DN ) ≥ ∫
π 0 ∣θ∣
2 (N + 2 )π
1
∣ sin λ∣ 1
≥ ∫ dλ
π 0 ∣λ/(N + 2 )∣ (N + 12 )
1

2 N π ∣ sin λ∣
≥ ∫ dλ
π 0 ∣λ∣
2 N nπ ∣ sin λ∣
= ∑∫ dλ
π n=1 (n−1)π ∣λ∣
2 N nπ ∣ sin λ∣
≥ ∑∫ dλ
π n=1 (n−1)π nπ
2 N 1 nπ
≥ 2 ∑ ∫ ∣ sin λ∣ dλ
π n=1 n (n−1)π

But on any interval of the form [(n − 1)π, nπ], the sine function does not
change sign (i.e. it is either always non-positive on such an interval, or
always non-negative), and so an easy calculation shows that
nπ nπ
∫ ∣ sin λ∣ dλ = ∣∫ sin λ dλ∣ = 2.
(n−1)π (n−1)π
From this we see that
4 N 1
ν1 (DN ) ≥ ∑ ,
π 2 n=1 n
as required.
◻
We invite the reader to skip to the Appendix to see the graphs of D2 , D5 and
D10 .
The next result follows immediately from Theorems 9.18 and 9.20, together with
Theorem 10.4 (d) and the fact that the harmonic series ∑∞ 1
n=1 n diverges.

10.5. Corollary. For each N ≥ 1, let DN denote the Dirichlet kernel of order
N.
(a) If CDN ∈ B(([C(T, C)], ∥ ⋅ ∥∞ )) is the convolution operator corresponding to
DN , N ≥ 1, then
lim ∥CDN ∥ = ∞.
N →∞
(b) If CDN ∈ B((L1 (T, C), ∥ ⋅ ∥1 )) is the convolution operator corresponding to
DN , N ≥ 1, then
lim ∥CDN ∥ = ∞.
N →∞
 10. THE DIRICHLET KERNEL 163

As mentioned above, in order to exploit the connection between the Dirichlet

kernel and convolution, we shall require a couple of results from a previous real
analysis course. We start by recalling a definition.

10.6. Definition. Let (X, d) be a metric space and H ⊆ X. We say that H

is nowhere dense (or meager, or thin) if G ∶= X ∖ H is dense in X. In other
words, the interior of H is empty.
Note: Here, H refers to the closure of H in X.

10.7. Examples. We think of nowhere dense subsets of metric spaces as being

“small”, as the alternate terminology “meager” and ”thin” suggest.
(a) The set H = Z is nowhere dense in R, as is easily verified.
(b) The Cantor set C is nowhere dense in X = [0, 1], equipped with the stan-
dard metric inherited from R. This is left as an exercise for the reader.
(c) The set H = Q of rational numbers is not nowhere dense in R, as X ∖ H =
R ∖ R = ∅, which is as far from being dense in R as any set can get.

10.8. Definition. We say that a subset H of a metric space (X, d) is of the

first category in (X, d) if there exists a sequence (Fn )∞
n=1 of closed, nowhere dense
sets in X such that
H ⊆ ∪∞ n=1 Fn .
Otherwise, H is said to be of the second category.
We assume that the reader is familiar with the following result, whose proof we
consign to the Appendix.

10.9. Theorem. The Baire Category Theorem

A complete metric space (X, d) is of the second category. That is, X is not a
countable union of closed, nowhere dense sets in X.

10.10. Examples.
(a) It follows from the Baire Category Theorem that (R, d) is of the second
category, where d represents the standard metric d(x, y) = ∣x − y∣, x, y ∈ R.
(b) Let δ denote the discrete metric on Q, so that
⎧
⎪1 if p ≠ q
⎪
δ(p, q) = ⎨
⎪
⎪0 if p = q.
⎩
Writing Q = {qn }∞ n=1 , which we may do since Q is denumerable, we find
that for each n ≥ 1, Fn ∶= {qn } is a closed set, and clearly Q = ∪∞
n=1 Fn .
However, Fn is not nowhere dense, since as well as being closed, Fn is
open, and qn ∈ Fn = int(Fn ) ≠ ∅.
This is just as well, since (Q, δ) is a complete metric space, and hence
of the second category by the Baire Category Theorem. In other words,
we knew that Q was not a countable union of closed, nowhere dense sets
in (Q, δ).
164 L.W. Marcoux Introduction to Lebesgue measure

(c) The sets Fn = {qn } from (b) are closed and nowhere dense in (R, d), where
d is the standard metric from (a). Since Q = ∪∞
n=1 Fn , we see that Q is is of
the first category in (R, d).

10.11. Remark. An alternate form of the Baire Category Theorem says that if
(X, d) is a complete metric space, and if (Gn )∞
n=1 is a countable collection of dense,
open sets in X, then
∩∞
n=1 Gn ≠ ∅.
We shall not require this below.
The second result from real analysis which we shall recall is the following.

10.12. Theorem. The Banach-Steinhaus Theorem, aka The Uniform

Boundedness Principle
Let (X, d) be a complete metric space and ∅ ≠ F ⊆ C(X, R). Suppose that for all
x ∈ X, there exists a constant κx > 0 such that
∣f (x)∣ ≤ κx , f ∈ F.
Then there exists a non-empty open set G ⊆ X and κ > 0 such that
∣f (x)∣ ≤ κ, x ∈ G, f ∈ F.
Proof. See the Appendix to this Chapter.
◻
There is a stronger version of the Banach-Steinhaus Theorem which applies to
the setting of linear operators on Banach spaces. Before stating and proving it, we
remind the reader that a Banach space is a complete metric space under the metric
induced by the norm.

10.13. Theorem. The Banach-Steinhaus Theorem, aka The Uniform

Boundedness Principle for Operators
Let (X, ∥⋅∥X ) and (Y, ∥⋅∥Y ) be Banach spaces and suppose that ∅ ≠ F ⊆ B(X, Y).
Let H ⊆ X be a subset of the second category in X, and suppose that for each x ∈ H,
there exists a constant κx > 0 such that
∥T x∥Y ≤ κx , T ∈ F.
Then F is bounded; that is,
sup ∥T ∥ < ∞.
T ∈F
Proof. For each n ≥ 1, set
Fn ∶= {x ∈ X ∶ ∥T x∥Y ≤ n for all T ∈ F}.
If (xk )∞
k=1 is a sequence in Fn and if x = limk→∞ xk exists in X, then by the continuity
of each T ∈ F,
T x = lim T xk ,
k→∞
 10. THE DIRICHLET KERNEL 165

from which it easily follows that

∥T x∥ = lim ∥T xk ∥ ≤ n.
k→∞
Hence x ∈ Fn , proving that Fn is closed.
By hypothesis,
H ⊆ ∪∞n=1 Fn .
Our hypothesis says that H is of the second category, and hence there must exist
1 ≤ N such that int(FN ) = int(FN ) ≠ ∅.
Let x0 ∈ int(FN ), and choose δ > 0 such that
B(x0 , δ) ∶= {x ∈ X ∶ ∥x − x0 ∥X < δ} ⊆ FN .
Then ∥T x∥Y ≤ N for all x ∈ B(x0 , δ) and T ∈ F.
Suppose that w ∈ X and ∥w∥ ≤ 1. Then x0 − 2δ w ∈ B(x0 , δ), and so for all T ∈ F,
δ
∥T (x0 − w)∥Y ≤ N.
2
δ
By the triangle inequality, 2 ∥T w∥Y − ∥T x0 ∥Y ≤ N . Thus
2
∥T w∥Y ≤ (N + ∥T x0 ∥Y ).
δ
2
Setting K ∶= δ (N + ∥T x0 ∥Y ), we see that
sup ∥T ∥ ≤ K < ∞.
T ∈F
◻

10.14. Corollary. Let (X, ∥⋅∥X ) and (Y, ∥⋅∥Y ) be Banach spaces and let (Tn )∞
n=1
be an unbounded sequence in B(X, Y), i.e. supn≥1 ∥Tn ∥ = ∞.
Let H = {x ∈ X ∶ supn≥1 ∥Tn x∥ < ∞}. Then H is of the first category in X, and
J ∶= X ∖ H is of the second category.
Proof. Note that 0 ∈ H, so that H ≠ ∅.
If H were of the second category, then – by the Banach-Steinhaus Theorem for
Operators, Theorem 10.13 above – {Tn }∞ n=1 would be bounded, a contradiction.
Thus H is of the first category. Let J = X ∖ H. By definition, for all x ∈ J,
sup ∥Tn x∥Y = ∞.
n≥1
We claim that J is of the second category.
Indeed, suppose otherwise. Then we could choose sequences (Kn )∞ ∞
n=1 and (Ln )n=1
of closed, nowhere dense sets in X such that
H ⊆ ∪∞
n=1 Kn and J ⊆ ∪∞
n=1 Ln .
But then
X = H ∪ J = ∪∞ ∞
n=1 Kn ⋃ ∪n=1 Ln
must be of the first category, contradicting the Baire Category Theorem, as X is a
complete metric space.
◻
166 L.W. Marcoux Introduction to Lebesgue measure

10.15. It is perhaps worth noting that this result is much, much better than it
might appear on the surface. The statement that supn≥1 ∥Tn ∥ = ∞ is the statement
that for each n ≥ 1, there exists xn ∈ X with ∥xn ∥X = 1 such that limn→∞ ∥Tn xn ∥Y =
∞. A priori, it is not clear that there should exist any x ∈ X such that
lim ∥Tn x∥Y = ∞.
n→∞

The above Corollary not only says that such a vector x ∈ X exists; it asserts that
this is true for a very large set of x’s, in the sense that the set H of x’s for which it
fails is a set of the first category in X.
We are finally prepared to answer the question of whether or not the partial
sums of the Fourier series of an element [f ] of L1 (T, C) necessarily converge to [f ]
in the ∥ ⋅ ∥1 -norm. As we shall now see, an easy application of Corollary 10.14 shows
that this almost never happens. (Here we use “almost never” informally, to refer
to the notion of sets of the first category, and not in the sense of sets of Lebesgue
measure zero!). What is more, essentially the same argument shows that the partial
sums of the Fourier series of an element [f ] of [C(T, C)] rarely converge in the ∥ ⋅ ∥∞
norm. These are dark times indeed for the sequence of partial sums of a Fourier
series.

10.16. Theorem. The unbearable lousiness of being a Dirichlet kernel

(a) Let
K∞ ∶= {[f ] ∈ [C(T, C)] ∶ [f ] = lim ∆N ([f ]) in ([C(T, C)], ∥ ⋅ ∥∞ )}.
N →∞

Then K∞ is a set of the first category in ([C(T, C)], ∥ ⋅ ∥∞ ), whose comple-

ment [C(T, C)] ∖ K∞ is a set of the second category.
(b) Let
K1 ∶= {[f ] ∈ L1 (T, C) ∶ [f ] = lim ∆N ([f ]) in (L1 (T, C), ∥ ⋅ ∥1 )}.
N →∞

Then K1 is a set of the first category in (L1 (T, C), ∥⋅∥1 ), whose complement
L1 (T, C) ∖ K1 is a set of the second category in L1 (T, C).
Proof. The proofs of both of these parts are almost identical. We shall prove (b),
and leave the proof of (a) as an exercise.
Recall that ∆N ([f ]) = DN ∗ [f ] = CDN ([f ]), N ≥ 1, where CDN ∈ B(L1 (T, C))
is the convolution operator corresponding to DN , as described in Theorem 9.20.
Furthermore, by Corollary 10.5,
lim ∥CDN ∥ = ∞.
N →∞

Of course, if [h] ∈ L1 (T, C) and [h] = limN →∞ ∆N ([h]) = limN →∞ CDN ([h]),
then (CDN ([h]))∞N =1 is bounded in L1 (T, C), and thus it is clear that K1 ⊆ H1 ,
where
H1 ∶= {[f ] ∈ L1 (T, C) ∶ sup ∥CDN ([f ])∥1 < ∞}.
N ≥1
 10. THE DIRICHLET KERNEL 167

By Corollary 10.14, H1 is a set of the first category in L1 (T, C), and J1 ∶= L1 (T, C) ∖
H1 is a set of the second category in L1 (T, C).
In particular, for any [f ] ∈ J1 , we have that the partial sums (∆N ([f ]))∞N =1 fail
to converge to [f ], as the sequence is not even bounded.
◻
168 L.W. Marcoux Introduction to Lebesgue measure

Appendix to Section 10.

10.17. It is instructive to look at the graph of the Dirichlet kernels of various

orders. Two things worth noticing are that
● first, the amplitude of the function is increasing near 0; this is clear since
each DN is continuous, and DN (0) = 2N + 1, N ≥ 1.
● Each of the functions DN spends a lot of time being negative, and a lot of
time being positive. This accounts for the fact that the integrals of DN are
bounded, but that the integrals of ∣DN ∣ are not.

Figure 3. The graph of D2 .

Figure 4. The graph of D5 .

10. THE DIRICHLET KERNEL 169

Figure 5. The graph of D10 .

170 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 10.

Exercise 10.1.
Prove that the Cantor set is nowhere dense in [0, 1], where [0, 1] is equipped
with the standard metric d(x, y) = ∣x − y∣ inherited from R.

Exercise 10.2.
Fill in the details of the proof of Theorem 10.4(c) by proving that with ρ(θ) ∶=
(−2i) sin( 12 θ),
1
ρ(θ)DN (θ) = e−i(N + 2 )θ − ei(N + 2 )θ = (−2i) sin ((N + )θ) .
1 1

Exercise 10.3.
Fill in the details of the proof of Theorem 10.16 (a), namely: let
K∞ ∶= {[f ] ∈ [C(T, C)] ∶ [f ] = lim ∆N ([f ]) in ([C(T, C)], ∥ ⋅ ∥∞ )}.
N →∞
Prove that K∞ is a set of the first category in ([C(T, C)], ∥ ⋅ ∥∞ ), whose complement
[C(T, C)] ∖ K∞ is a set of the second category.
 11. THE FÉJER KERNEL 171

11. The Féjer kernel

I’ve never really wanted to go to Japan. Simply because I don’t like

eating fish. And I know that’s very popular out there in Africa.
Britney Spears

11.1. We have seen in the last Chapter that if [f ] ∈ L1 (T, C), and if ∆N ([f ]) =
[f ]
∑N
n=−N αn [ξn ], then (∆N ([f ]))∞N =1 almost never converges to [f ].
Not all is lost. In this Chapter we shall replace partial sums (∆N ([f ]))∞ N =1 of
the Fourier series of [f ] by weighted partial sums which will converge to [f ].

11.2. Definition. Let (X, ∥ ⋅ ∥X ) be a Banach space, and (xn )∞

n=0 be a sequence
in X. The N th -Cesàro mean of the sequence is
1
σN ∶= (x0 + x1 + ⋯ + xN −1 ), N ≥ 1.
N

The next Proposition is routine, and its proof is left to the exercises.

11.3. Proposition. Suppose that X is a Banach space and (xn )∞ n=0 is a se-
quence in X. Let (σN )∞
N =1 denote the sequence of Cesàro means of (x n ) ∞
n=0 .
If x = limn→∞ xn exists, then x = limN →∞ σN .

11.4. Definition. Let f ∈ L1 (T, C). The N th -Cesàro sum of the Fourier
series of f is the N th -Cesàro mean of the sequence (∆○n (f ))∞
n=0 . Thus

○ 1
σN (f ) = (D0 ◇ f + D1 ◇ f + ⋯ + DN −1 ◇ f )
N
= FN ◇ f,

where FN ∶= N1 (D0 + D1 + ⋯ + DN −1 ) is called the Féjer kernel of order N .

We also define the N th -Cesàro sum of the Fourier series of [f ] is the N th -
Cesàro mean of the sequence (∆n [f ]))∞n=0 , namely

1
σN [f ] ∶= (D0 ∗ [f ] + D1 ∗ [f ] + ⋯ + DN −1 ∗ [f ])
N
= FN ∗ [f ]
= [FN ◇ f ]
○
= [σN (f )].
172 L.W. Marcoux Introduction to Lebesgue measure

11.5. Remark. We remark that the fact that Dn ∈ C(T, C) for all n ≥ 0 implies
○
that FN ∈ C(T, C) for all N ≥ 1. By Proposition 9.6, it follows that σN (f ) ∈ C(T, C) ⊆
L1 (T, C) for all f ∈ L1 (T, C).
Furthermore, for all θ ∈ R,
○ 1
σN (f )(θ) = ∫ FN (s)f (θ − s) dm(s)
2π [−π,π)
1
= ∫ FN (θ − s)f (s) dm(s)
2π [−π,π)
As seen in Theorem 9.18, the fact that FN ∈ C(T, C) also implies that for every
homogeneous Banach space B and [f ] ∈ B, we have
σN [f ] = FN ∗ [f ] ∈ B.
As a special case of this phenomenon, σN [f ] = FN ∗ [f ] ∈ Lp (T, C) for all [f ] ∈
Lp (T, C).

11.6. Theorem. For each 1 ≤ N ∈ N,

(a) FN is a 2π-periodic, even, continuous function;
(b) If 0 ≠ θ ∈ [−π, π), then
N 2
1 1 − cos(N θ) 1 ⎛ sin( 2 θ) ⎞
FN (θ) = ( )= ,
N 1 − cos θ N ⎝ sin( 12 θ) ⎠
while FN (0) = N . In particular, FN (θ) ≥ 0 for all θ ∈ R;
1 π 1 π
(c) ν1 (FN ) = 2π ∫−π ∣FN (θ)∣ dθ = 2π ∫−π FN (θ) dθ = 1.
(d) For all 0 < δ ≤ π,
−δ π
lim (∫ ∣FN (θ)∣ dθ + ∫ ∣FN (θ)∣ dθ) = 0;
N →∞ −π δ
(e) For 0 < ∣θ∣ < π,
π2
0 ≤ FN (θ) ≤ .
N θ2
Proof.
(a) That FN is 2π-periodic, even and continuous follows immediately from the
fact that each Dn is, 0 ≤ n ≤ N − 1.
(b) First,
1 N −1
FN (0) = ( ∑ Dn (0))
N n=0
1 N −1
= ( ∑ (2n + 1))
N n=0
1 (N − 1)N
= (2 + N)
N 2
= N.
 11. THE FÉJER KERNEL 173

For 0 ≠ θ ∈ [−π, π),

1 N −1
FN = ∑ Dn
N n=0
1 N −1 n
= ∑ ( ∑ ξk )
N n=0 k=−n
1
= (ξ−N +1 + 2ξ−N +2 + ⋯ + (N − 1)ξ−1 + N ξ0
N
+(N − 1)ξ1 + ⋯ + 2ξN −2 + ξN −1 ) .

Let ρ = (2 − (ξ−1 + ξ1 ))N . As was the case with the Dirichlet kernel, we
observe that the product ρ ⋅ FN involves a telescoping sum, and that

ρ(θ) FN (θ) = 2 − (ξ−N (θ) + ξN (θ)) = 2 − 2 cos (N θ).

That is,
1 1 − cos (N θ)
FN (θ) =
N 1 − cos θ
2
i θ N
−i θN
1 (e 2 − e 2 )
=
N (ei θ2 − e−i θ2 )2

N 2
1 ⎛ sin( 2 θ) ⎞
= .
N ⎝ sin( 21 θ) ⎠

In particular, FN ≥ 0.
π
(c) Keeping in mind that 0 ≠ n implies that ∫−π ξn (θ) dθ = 0, we get:
1 π
ν1 (FN ) = ∫ ∣FN (θ)∣ dθ
2π −π
1 π
= ∫ FN (θ) dθ
2π −π
1 1 π N −1 n
= ∫ ( ∑ ( ∑ ξk (θ))) dθ
2π N −π n=0 k=−n
1 N −1 n 1 π
= ∑ ∑ ∫ ξk (θ) dθ
N n=0 k=−n 2π −π
1 N −1 1 π
= ∑ ∫ ξ0 (θ) dθ
N n=0 2π −π
1 N −1
= ∑1
N n=0
= 1.
174 L.W. Marcoux Introduction to Lebesgue measure

(d) Let 0 < δ ≤ π. For δ ≤ ∣θ∣ ≤ π, we have that

θ δ
∣sin ∣ ≥ sin ,
2 2
and so
2
−δ 1 −δ ⎛ sin( N θ) ⎞ 1 1
2
∫ FN (θ) dθ = ∫ 1
dθ ≤ (π − δ) .
−π N −π ⎝ sin( 2 θ) ⎠ N (sin 2δ )2
Thus
−δ
lim ∫ FN (θ) dθ = 0.
N →∞ −π
Similarly,
π
lim ∫ FN (θ) dθ = 0,
N →∞ δ
from which (d) easily follows.
(e) Finally, we leave it as a routine calculus exercise for the reader to verify
that on the interval [− π2 , π2 ], we have
2
∣θ∣ ≤ ∣ sin θ∣,
π
and so
N 2
1 ⎛ sin( 2 θ) ⎞
FN (θ) =
N ⎝ sin( 12 θ) ⎠
R R2
1 RRRR 1 RRRR
≤ R R
N RRRR π2 2θ RRRR
R R
π2
= .
N θ2
◻
Our goal is to show that the Cesàro sums of the Fourier series of an element [f ]
of L1 (T, C) converge in the ∥ ⋅ ∥1 -norm back to [f ]. There is nothing unique about
the Féjer kernel, however. Let us examine a more general phenomenon, of which
the Féjer kernel is but an example.

11.7. Definition. A summability kernel is a sequence (kn )∞

n=1 of 2π-periodic,
continuous, complex-valued functions on R satisfying:
1 π
(a) 2π ∫−π kn = 1 for all n ≥ 1;
1 π
(b) supn≥1 ν1 (kn ) = supn≥1 2π ∫−π ∣kn ∣ < ∞; and
(c) for all 0 < δ ≤ π,
−δ π
lim (∫ ∣kn ∣ + ∫ ∣kn ∣) = 0.
n→∞ −π δ
If, in addition, kn ≥ 0 for all n ≥ 1, we say that (kn )∞
n=1 is a positive
summability kernel.
 11. THE FÉJER KERNEL 175

11.8. Theorem. The Féjer kernel (FN )∞ N =1 is a positive summability kernel.

Proof. This is the content of Theorem 11.6.
◻

11.9. Examples.
(a) For each 1 ≤ n ∈ N, consider the piecewise linear function
kn● ∶ [−π, π) → R
⎧
⎪
⎪ 0 if θ ∈ [−π, −1 1
n ] ∪ [ n , π)
⎪
⎪
θ ↦ ⎨n + n2 θ if θ ∈ ( −1 n , 0]
⎪
⎪
⎪ 2 1
⎩n − n θ if θ ∈ (0, n ).
⎪
For 1 ≤ n ∈ N, let kn be the 2π-periodic function on R whose restriction to
the interval [−π, π) coincides with kn● .
Then (kn )∞n=1 is a positive summability kernel. The details are left to
the reader.
(b) For each 1 ≤ n ∈ N, consider the piecewise linear function
rn● ∶ [−π, π) → R
⎧
⎪
⎪ 0 if θ ∈ [−π, 0] ∪ [ n2 , π)
⎪
⎪
θ ↦ ⎨n2 θ if θ ∈ (0, n1 ]
⎪
⎪
⎪ 2 1 1 2
⎩n − n (θ − n ) if θ ∈ ( n , n ).
⎪
For 1 ≤ n ∈ N, let rn be the 2π-periodic function on R whose restriction to
the interval [−π, π) coincides with rn● .
Then (rn )∞ n=1 is a positive summability kernel. The details are left to
the reader.

11.10. Theorem. Let (B, ∥ ⋅ ∥B ) be a homogeneous Banach space over T and

let (kn )∞
n=1 be a summability kernel. If [f ] ∈ B, then
lim ∥kn ∗ [f ] − [f ]∥B = 0,
n→∞

and so [f ] = limn→∞ kn ∗ [f ] in B.
Proof. The result is trivial if [f ] = 0. Let 0 ≠ [f ] ∈ B. Recall that
(a) the function
Ψ[f ] ∶ R → B
s ↦ τs [f ]
is continuous, and that
(b) τs is isometric for all s ∈ R. In particular, ∥τs [f ]∥B = ∥[f ]∥B for all s ∈ R.
That τ0 [f ] = [f ] is clear from the definition of τ0 . Let M ∶= supn≥1 ν1 (kn ) < ∞, as
(kn )∞
n=1 is a summability kernel.
Let ε > 0 and choose δ > 0 such that ∣s − 0∣ < δ implies that
ε
∥τs [f ] − τ0 [f ]∥B < .
2M
176 L.W. Marcoux Introduction to Lebesgue measure

This is possible by (a) above. Next, choose 1 ≤ N ∈ N such that n ≥ N implies that
1 −δ 1 π ε
∫ ∣kn (s)∣ds + ∫ ∣kn (s)∣ds < .
2π −π 2π δ 4∥[f ]∥B
Then n ≥ N implies that
1 π
∥kn ∗ [f ] − [f ]∥B = ∥ ∫ kn (s)(τs [f ] − τ0 [f ])ds∥B
2π −π
1 −δ
≤ ∫ ∣kn (s)∣ ∥τs [f ] − τ0 [f ]∥B ds
2π −π
1 δ
+ ∫ ∣kn (s)∣ ∥τs [f ] − τ0 [f ]∥B ds
2π −δ
1 π
+ ∫ ∣kn (s)∣ ∥τs [f ] − τ0 [f ]∥B ds.
2π δ
Now ∥τs [f ] − τ0 [f ]∥B ≤ ∥τs [f ]∥B + ∥τ0 [f ]∥B = 2∥[f ]∥B , and thus for n ≥ N , we have
1 −δ
∥kn ∗ [f ] − [f ]∥B ≤ ∫ ∣kn (s)∣(2∥[f ]∥B ) ds
2π −π
1 δ ε
+ ∫ ∣kn (s)∣ ds
2π −δ 2M
1 −δ
+ ∫ ∣kn (s)∣(2∥[f ]∥B ) ds
2π −π
ε ε
≤ 2∥[f ]∥B +M
4∥[f ]∥B 2M
= ε.
In other words,
lim kn ∗ [f ] = [f ].
n→∞
◻

11.11. Corollary.
(a) For each f ∈ (C(T, C), ∥ ⋅ ∥sup ),
○
lim σN (f ) = f.
N →∞

(b) Let 1 ≤ p < ∞. For each [g] ∈ (Lp (T, C), ∥ ⋅ ∥p ),

lim σN [g] = [g].
N →∞
Proof.
(a) By Example 9.11, we know that ([C(T, C)], ∥⋅∥∞ ) is a homogeneous Banach
space and that the map
Γ ∶ (C(T, C), ∥ ⋅ ∥sup ) → ([C(T, C)], ∥ ⋅ ∥∞ )
f ↦ [f ]
is an isometric isomorphism of Banach spaces.
 11. THE FÉJER KERNEL 177

Let f ∈ C(T, C). By Theorem 11.8, (FN )∞ N =1 is a (positive) summability

kernel, and by definition, σN ([f ]) = FN ∗ [f ]. By Theorem 11.10 above,

lim ∥FN ◇ f − f ∥sup = lim ∥FN ∗ [f ] − [f ]∥∞ = 0.

N →∞ N →∞

(b) Again, by Example 9.12, for each 1 ≤ p < ∞, (Lp (T, C), ∥ ⋅ ∥p ) is a homo-
geneous Banach space over T. Let 1 ≤ p < ∞ and [f ] ∈ Lp (T, C). Since
(FN )∞N =1 is a (positive) summability kernel, and since σN ([f ]) = FN ∗ [f ]
for all N ≥ 1,

lim σN ([f ]) = lim FN ∗ [f ] = [f ]

N →∞ N →∞

by Theorem 11.10.
◻

We are now in a position to show that the Fourier coefficients of an Lp (T, C)-
function completely determine that function (almost everywhere).

[f ] [g]
11.12. Corollary. Let 1 ≤ p < ∞. If [f ], [g] ∈ Lp (T, C) and αn = αn for all
n ∈ Z, then [f ] = [g].
[f ] [g]
Proof. It is clear that if αn = αn for all n ∈ Z, then σN [f ] = σN ([g]) for all N ≥ 1.
By Corollary 11.11,

[f ] = lim σN ([f ]) = lim σN ([g]) = [g].

N →∞ N →∞

11.13. Local structure and Féjer kernels. Corollaries 11.11 and 11.12 tell
us that Féjer kernels are nice enough to recover an element of Lp (T, C) from its
Fourier coefficients. As we have emphasized in these notes, however, these elements
are equivalence classes of functions in Lp (T, C), and not functions themselves. In
other words, the aforementioned Corollaries say that we can recover functions in
Lp (T, C) almost everywhere on R. We now turn our attention to the functions
themselves, and study in what sense (if any) the convolution of Lp (T, C) functions
with Féjer kernels converge pointwise.

11.14. Definition. Given f ∈ L1 (T, C) and θ ∈ R, we set

f (θ − t) + f (θ + t)
ωf (θ) ∶= lim+ ,
t→0 2
provided that the limit exists. When it does exist, we shall refer to this value as the
average value of f at θ.
178 L.W. Marcoux Introduction to Lebesgue measure

11.15. Theorem. [Féjer’s Theorem.] Let f ∈ L1 (T, C).

(a) If θ ∈ R and ωf (θ) exists, then

lim FN ◇ f (θ) = ωf (θ).

N →∞

(b) Suppose that there exists a closed interval [a, b] ⊆ [−π, π) such that f is
continuous on [a, b] (in particular, f is continuous from the right at a and
from the left at b), then

(FN ◇ f )∞
N =1

converges uniformly to f on [a, b].

Proof.
(a) Let ε > 0 and choose δ > 0 such that 0 < ∣s∣ < δ implies that

f (θ − s) + f (θ + s)
∣ωf (θ) − ∣ < ε.
2

Now

1
∣σN (f )(θ) − ωf (θ)∣ = ∣ ∫ FN (s)f (θ − s) dm(s) − ωf (θ)∣
2π [−π,π)
1
=∣ ∫ FN (s)(f (θ − s) − ωf (θ)) dm(s)∣
2π [−π,π)
1
≤∣ ∫ FN (s)(f (θ − s) − ωf (θ)) dm(s)∣
2π [−δ,δ]
1
+∣ ∫ FN (s)(f (θ − s) − ωf (θ)) dm(s)∣ .
2π [−π,−δ)∪(δ,π]

By our work in the Assignments,

∫ FN (s)(f (θ − s) − ωf (θ)) dm(s) = ∫ FN (−s)(f (θ + s) − ωf (θ)) dm(s).

[−δ,δ] [−δ,δ]

But FN is even, so FN (−s) = FN (s), s ∈ [−δ, δ]. Thus

1
∫ FN (s)(f (θ − s) − ωf (θ)) dm(s) =
2π [−δ,δ]
1 f (θ − s) + f (θ + s)
∫ FN (s) ( − ωf (θ)) dm(s).
2π [−δ,δ] 2
 11. THE FÉJER KERNEL 179

Thus
1
∣ ∫ FN (s)(f (θ − s) − ωf (θ)) dm(s)∣ ≤
2π [−δ,δ]
1 f (θ − s) + f (θ + s)
∫ FN (s) ∣ − ωf (θ)∣ dm(s)
2π [−δ,δ] 2
1
≤ ∫ FN (s)ε dm(s)
2π [−δ,δ]
1
≤ ε( ∫ FN (s) dm(s))
2π [−π,π)
= ε.
Meanwhile, for δ ≤ ∣s∣ ≤ π,
π2 π2
0 ≤ FN (s) ≤ ≤ ,
N s2 N δ 2
and so
1
∣ ∫ FN (s)(f (θ − s) − ωf (θ)) dm(s)∣
2π [−π,−δ)∪(δ,π]
π2 1
≤ 2
( ∫ ∣f (θ − s)∣ + ∣ωf (θ)∣ dm(s))
N δ 2π [−π,−δ)∪(δ,π]
π2 1
≤ 2
( ∫ ∣f (θ − s)∣ dm(s) + ∣ωf (θ)∣)
N δ 2π [−π,π)
π2
≤ (∥[f ]∥1 + ∣ωf (θ)∣),
N δ2
which converges to 0 as N tends to ∞.
Thus limN →∞ σN (f )(θ) = ωf (θ).
(b) The proof is essentially identical to that above.
First note that the continuity of f at θ for θ ∈ [a, b] implies that ωf (θ) =
f (θ) for all θ ∈ [a, b]. But f continuous on [a, b] implies that f is uniformly
continuous on [a, b], and so we may find a single δ > 0 such that 0 ≤ ∣s∣ > δ
implies that
f (θ − s) + f (θ + s)
∣f (θ) − ∣ < ε, θ ∈ [a, b].
2
For this δ > 0, and for all θ ∈ [a, b], the estimates from part (a) above show
that
1 f (θ − s) + f (θ + s)
∣σN (f )(θ) − f (θ)∣ ≤ ∫ FN (s) ∣ − f (θ)∣ dm(s)
2π [−δ,δ] 2
1
+ ∫ FN (s)(∣f (θ − s)∣ + ∣f (θ)∣) dm(s)
2π [−π,−δ)∪(δ,π]
π2
<ε+ (∥[f ]∥1 + ∣f (θ)∣) ,
N δ2
180 L.W. Marcoux Introduction to Lebesgue measure

and thus (σN (f ))∞

N =1 converges uniformly to f on [a, b].
◻

11.16. Corollary.
(a) If f ∈ L1 (T, C) and f is continuous at θ0 ∈ R, then
lim FN ◇ f (θ0 ) = f (θ0 ).
N →∞

(b) If f ∈ C(T, C), then

(FN ◇ f )∞
N =1
converges uniformly to f on R.
(c) Let f ∈ L1 (T, C) and θ0 ∈ R. If f is continuous at θ0 and (DN ◇ f (θ0 ))∞
N =1
converges, then
lim DN ◇ f (θ0 ) = f (θ0 ).
N →∞
Proof.
(a) If f is continuous at θ0 , then f (θ0 ) = ωf (θ0 ) and so by Theorem 11.15,
lim FN ◇ f (θ0 ) = ωf (θ0 ) = f (θ0 ).
N →∞

(b) This is exactly Corollary 11.11 (a).

(c) Again, if f is continuous at θ0 , then f (θ0 ) = ωf (θ0 ).
Moreover, if (DN ◇ f (θ0 ))∞ N =1 converges to some value β ∈ C, then
(FN ◇ f (θ0 ))∞N =1 converges to β by Proposition 11.3. But (a) shows that
β = f (θ0 ), completing the proof.
◻

11.17. Culture. Let f ∈ L1 (T, C). A point θ ∈ R is called a Lebesgue-point

of f if
1 f (θ − s) + f (θ + s)
lim+ ∫ ∣ − f (θ)∣ dm(s) = 0.
h→0 h [0,h] 2
It can be shown that almost every real number θ is a Lebesgue point of f .
A modification of the proof of Féjer’s Theorem yields

The Lebesgue-Féjer Theorem.

If θ ∈ R is a Lebesgue point for f ∈ L1 (T, C), then
f (θ) = lim FN ◇ f (θ).
N →∞

In particular,
f (θ) = lim FN ◇ f (θ)
N →∞
almost everywhere on R.
 11. THE FÉJER KERNEL 181

11.18. So far we have seen that while it is extremely rare for the partial sums
of the Fourier series of a continuous function f , or of an element [g] of L1 (T, C) to
converge to f uniformly (or to [g] in the ∥ ⋅ ∥1 -norm), nevertheless, this is always
the case regarding the Cesàro sums of the Fourier series.
Furthermore, Féjer’s Theorem shows that if f ∈ L1 (T, C) and if the average
value ωf (θ) exists at θ ∈ R, then

ωf (θ) = lim σN (f )(θ) = lim FN ◇ f (θ).

N →∞ N →∞

Our next goal is to show that if f is sufficiently “smooth” at a point θ0 (in a

sense which we shall now make explicit), then in fact the partial sums of f at θ0
converge to f (θ0 ).

11.19. Definition. Suppose that f ∶ R → C is measurable and that θ0 ∈ R. We

say that f is locally Lipschitz at θ0 if there exist M > 0 and δ > 0 such that

∣f (θ0 + s) − f (θ0 )∣ < M ∣s∣ for all 0 ≤ ∣s∣ < δ.

11.20. Example. Suppose that f ∶ R → C is measurable and that f admits

left- and right-sided derivatives at θ0 ∈ R, in the sense that there exist y1 , y2 ∈ C
such that
f (θ0 + s) − f (θ0 ) f (θ0 + s) − f (θ0 )
lim− ∣ − y1 ∣ = 0 = lim+ ∣ − y1 ∣ .
s→0 s s→0 s

Let 0 < ε < 1, and choose δ > 0 such that

f (θ0 + s) − f (θ0 )
● −δ < s < 0 implies that ∣ − y1 ∣ < ε, and
s
f (θ0 + s) − f (θ0 )
● 0 < s < δ implies that ∣ − y2 ∣ < ε.
s
(That is, find δ1 > 0, δ2 > 0 that work for y1 and y2 respectively, and let δ =
min(δ1 , δ2 ).)
Then −δ < s < δ implies that

∣f (θ0 + s) − f (θ0 )∣ < (ε + max(∣y1 ∣, ∣y2 ∣)) ∣s∣ < M ∣s∣,

where M = 1 + max(∣y1 ∣, ∣y2 ∣) is a fixed constant.

Thus f is locally Lipschitz at θ0 .

In particular, if f is actually differentiable at θ0 ∈ R, then f is locally Lipschitz

at θ0 .
182 L.W. Marcoux Introduction to Lebesgue measure

11.21. Theorem. Let f ∈ L1 (T, C) and suppose that f is locally Lipschitz at

θ0 ∈ R. Then
lim sN (f )(θ0 ) = lim DN ◇ f (θ0 ) = f (θ0 ).
N →∞ N →∞
π
Proof. Fix M > 0 and 0 < δ < as in the definition of locally Lipschitz such that
2
0 ≤ ∣s∣ < δ implies that
∣f (θ0 + s) − f (θ0 )∣ ≤ M ∣s∣.
Now,
1
sN (f )(θ0 ) = ∫ DN (s)f (θ0 − s)dm(s),
2π [−π,π)
and
1
1= ∫ DN ,
2π [−π,π)
and so
1
∣sN (f )(θ0 ) − f (θ0 )∣ = ∣ ∫ (f (θ0 + s) − f (θ0 ))DN (s) dm(s)∣
2π [−π,π)
1
≤ ∫ ∣f (θ0 + s) − f (θ0 )∣∣DN (s)∣ dm(s)
2π [−δ,δ]
1
∣∫ (f (θ0 + s) − f (θ0 ))DN (s) dm(s)∣
2π [−π,−δ]∪[δ,π)

π 2
Now, since 0 ≤ ∣s∣ < δ ≤ , we have that ∣s∣ ≤ ∣ sin s∣, and thus
2 π
1 s
∣s∣ ≤ ∣ sin ∣, 0 ≤ ∣s∣ < δ.
π 2
Thus
1
∫ ∣f (θ0 + s) − f (θ0 )∣∣DN (s)∣ dm(s)
2π [−δ,δ]
1 ∣f (θ0 + s) − f (θ0 )∣ 1
≤ ∫ ∣ sin(N + )s∣ dm(s)
2π [−δ,δ] ∣s∣/π 2
1 M ∣s∣
≤ ∫ dm(s)
2 [−δ,δ] ∣s∣
= M δ,
independent of N !!!
Next we consider δ ≤ ∣s∣ ≤ π. Observe that
f (θ0 + s) − f (θ0 ) 1
(f (θ0 + s) − f (θ0 ))DN (s) = sin((N + )s)
sin(s/2) 2
f (θ0 + s) − f (θ0 ) eis/2 iN s e−is/2 −iN s
= ( e − e ).
sin(s/2) 2i 2i
 11. THE FÉJER KERNEL 183

Define
f (θ0 + s) − f (θ0 ) eis/2
g1 (s) = χ[−π,−δ)∪(δ,π)
sin(s/2) 2i
and
f (θ0 + s) − f (θ0 ) e−is/2
g2 (s) = χ[−π,−δ)∪(δ,π) .
sin(s/2) 2i
Then g1 , g2 are measurable (why?).
Also,
1 1 ∣f (θ0 + s) − f (θ0 )∣ 1
∫ ∣g1 ∣ = ∫
2π [−π,π) 2π [−π,−δ)∪(δ,π) ∣ sin s/2∣ 2
1 ∣f (θ0 + s)∣ + ∣f (θ0 )∣
≤
4π ∣ sin δ/2∣
1
= (∥[f ]∥1 + 2π∣f (θ0 )∣)
2π∣ sin δ/2∣
< ∞.
Thus g1 ∈ L1 (T, C), and similarly, g2 ∈ L1 (T, C).
By the Riemann-Lebesgue Lemma 8.9, we have that
1
lim ∫ g1 (s)eiN s dm(s) = 0
N →∞ 2π [−π,π)

and
1
lim ∫ g2 (s)e−iN s dm(s) = 0.
N →∞ 2π [−π,π)
From this it easily follows that
1
lim ∫ (f (θ0 + s) − f (θ0 ))DN (s) dm(s) = 0.
N →∞ 2π [−π,−δ)∪(δ,π)

Thus, given ε > 0, we may choose δ > 0 and N ≥ 1 such that

ε
(i) M δ < , and
2
(ii) for all n ≥ N we have
1 ε
∣ ∫ (f (θ0 + s) − f (θ0 ))DN (s) dm(s)∣ < .
2π [−π,−δ)∪(δ,π) 2
Then n ≥ N implies that
∣sN (f )(θ0 ) − f (θ0 )∣ < ε,
or in other words,
lim sN (f )(θ0 ) = f (θ0 ).
N →∞
◻
184 L.W. Marcoux Introduction to Lebesgue measure

11.22. Example. Consider the function f (θ) = ∣θ∣, θ ∈ [−π, π), extended 2π-
periodically to all of R. Clearly f is continuous on R.

Figure 6. The graph of f .

Since f is continuous and linear on (−π, 0) ∪ (0, π), it is locally Lipschitz there.
It is also easy to see that f is locally Lipschitz at θ = nπ, n ∈ Z with Lipschitz
constant M = 1.
We are preparing to do the unthinkable: we will calculate sN (f ), N ≥ 1. To
[f ]
do this, we must first calculate αn , n ∈ Z. To quote Fourier himself, “buckle up,
cupcake”.
We first point out that f is Riemann integrable and bounded over [−π, π),
and that multiplying f by ξn = einθ (for any value of n ∈ Z) doesn’t change this
fact. Because of this, and by virtue of Theorem 5.24, in calculating the Fourier
coefficients of [f ], we may always replace the Lebesgue integrals which appear by
Riemann integrals.
Case 1: n = 0. Observe that
1 1 π 1 π 1 2 π
[f ]
α0 = ∫ f= ∫ f (θ) dθ = ∫ ∣θ∣ dθ = π = .
2π [−π,π) 2π −π 2π −π 2π 2

Case 1: n ≠ 0.
In this case,
1
αn[f ] = ∫ f ξn
2π [−π,π)
1 π
−inθ
= ∫ ∣θ∣e dθ
2π −π
1 0 1 π
−inθ −inθ
= ∫ (−θ)e dθ + ∫ θe dθ.
2π −π 2π 0
These integrals are easily found using integration by parts, and we leave them
as exercises for the reader. The answers are:
 11. THE FÉJER KERNEL 185

⎧
⎪0 0 ≠ n an even integer
⎪
⎪
⎪
αn[f ] = ⎨
⎪
⎪
⎪
⎪ −2
⎪
⎪ 0 ≠ n an odd integer.
⎩ n2 π
Note that e−inθ + einθ = 2 cos(nθ) for all n ≥ 1. Thus – taking into account that
we only wish to sum over odd n’s below –
π −2
sN ([f ]) = + ∑ ( ) 2 cos((2n − 1)θ)
2 1≤2n−1≤N (2n − 1)2 π
π 4 1
= − ∑ cos((2n − 1)θ).
2 π 1≤2n−1≤N (2n − 1)2
But
1 1
∣ ∑ 2
cos((2n − 1)θ)∣ ≤ ∑ 2
1≤2n−1≤N (2n − 1) 1≤2n−1≤N (2n − 1)
N
1
≤∑ 2
.
n=1 n
But this last series converges, which implies that (sN ([f ]))∞ n=1 converges in the
∥ ⋅ ∥∞ -norm, as ([C(T, C)], ∥ ⋅ ∥∞ ) is complete. In other words,
N
sN (f ) = ∑ αn[f ] ξn
n=−N
converges uniformly in (C(T, C), ∥ ⋅ ∥sup ) to a continuous function.

Since f is globally Lipschitz, by Theorem 11.21,

f (θ) = ∣θ∣ = lim sN (f )(θ)
N →∞
for all θ ∈ [−π, π). That is,
π 4 ∞ 1
f (θ) = − ∑ cos((2n − 1)θ).
2 π n=1 (2n − 1)2

It is time to have fun:

(a) Let θ = 0. Then
π 4 ∞ 1
0 = ∣0∣ = − (∑ cos((2n − 1)0))
2 π n=1 (2n − 1)2
π 4 ∞ 1
= − (∑ ).
2 π n=1 (2n − 1)2
Thus
π2 ∞ 1 1 1 1 1
=∑ 2
= 2 + 2 + 2 + 2 + ⋯.
8 n=1 (2n − 1) 1 3 5 7
186 L.W. Marcoux Introduction to Lebesgue measure

(b) Let θ = π2 . Then

π π π 4 ∞ 1 π
= ∣ ∣ = − (∑ cos((2n − 1) ))
2 2 2 π n=1 (2n − 1)2 2
π 4 ∞ 1
= − (∑ ⋅ 0) .
2 π n=1 (2n − 1)2
Thus
π π
= ,
2 2
a relatively well-known result.
⎧ 1
π π ⎪ √2 if n = 1 mod 4 or n = 0 mod 4
⎪
(c) Let θ = 4 . Now cos((2n − 1) 4 ) = ⎨ −1 .
⎪
⎪ √ if n = 2 mod 4 or n = 3 mod 4
⎩ 2
Thus
π π π 4 ∞ 1 π
= ∣ ∣ = − (∑ 2
cos((2n − 1) ))
4 4 2 π n=1 (2n − 1) 4
π 4 1 1 1 1 1 1 1
= − √ ( 2 − 2 + 2 − 2 + 2 − 2 + ⋯) .
2 π 2 1 3 5 7 9 11
In other words,
√ 2
2π 1 1 1 1 1 1
= ( 2 − 2 + 2 − 2 + 2 − 2 + ⋯) .
16 1 3 5 7 9 11
Clearly life does not get better than this. Alas.
 11. THE FÉJER KERNEL 187

Appendix to Section 11.

11.23. It is also instructive to look at the graph of the Féjer kernels of various
orders. Two things worth noticing are that
● first, the amplitude of the function is increasing near 0; this is clear since
each FN is continuous, and FN (0) = N , N ≥ 1.
● For each δ > 0, the functions are becoming uniformly close to zero when
δ < ∣θ∣ < π.

Figure 7. The graph of K2 .

Figure 8. The graph of K5 .

188 L.W. Marcoux Introduction to Lebesgue measure

Figure 9. The graph of K10 .

Figure 10. The graph of K20 .

11.24. Remark. In fact, it can be shown that if k ∶ R → C is piecewise-

continuous on [−π, π) and k is 2π-periodic, then for all f ∈ L1 (T, C), k ◇ f is con-
tinuous on R.
Using this, one may extend the definition of a summability kernel to include
piecewise-continuous functions, as opposed to continuous functions.

11.25. Examples.
(a) For 1 ≤ n ∈ N, let kn be the 2π-periodic function on R whose restriction to
the interval [−π, π) coincides with
nπχ[− 1 , 1 ] .
n n

Then (kn )∞
n=1 is a positive summability kernel. The details are left to
the reader.
 11. THE FÉJER KERNEL 189

(b) For 1 ≤ n ∈ N, let kn be the 2π-periodic function on R whose restriction to

the interval [−π, π) coincides with
2nπχ[0, 1 ] .
n

Then (kn )∞
n=1 is a positive summability kernel. Again, the details are
left to the reader.
190 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 11.

Exercise 11.1.
Prove the claim of Proposition 11.3, namely: suppose that X is a Banach space
and (xn )∞ ∞
n=0 is a sequence in X. Let (σN )N =1 denote the sequence of Cesàro means
∞
of (xn )n=0 .
Prove that if x = limn→∞ xn exists, then x = limN →∞ σN .

Exercise 11.2.
Prove that the sequences (kn )n and (rn )n listed in Examples 11.9 are indeed
positive summability kernels.
 12. WHICH SEQUENCES ARE SEQUENCES OF FOURIER COEFFICIENTS? 191

12. Which sequences are sequences of Fourier coefficients?

I’ve learned about his illness. Let’s hope it’s nothing trivial.
Irving S. Cobb

12.1. Given [f ] ∈ L1 (T, C), we have defined the Fourier series of [f ] to be

[f ]
∑ αn [ξn ].
n∈Z

The Riemman-Lebesgue Lemma is the statement that

(αn[f ] )n∈Z ∈ c0 (Z, C).
It is a natural question to ask, therefore, whether every sequence (βn )n∈Z ∈
c0 (Z, C) is the sequence of coefficients of some [f ] ∈ L1 (T, C). What is clear from
our work on Hilbert spaces is that every (γn )n∈Z ∈ ℓ2 (Z, C) is the set of Fourier
coefficients of some [f ] ∈ L2 (T, C), namely [f ] = ∑n∈Z γn [ξn ].

Our approach to this problem will be via Operator Theory. Recall from Chap-
ter 8 that we defined the map
Λ ∶ (L1 (T, C), ∥ ⋅ ∥1 ) → (c0 (Z, C), ∥ ⋅ ∥∞ )
[f ]
[f ] ↦ (αn )n∈Z .
Since Lebesgue integration is linear, so is Λ. Also, as was shown in paragraph 8.8
∣αn[f ] ∣ ≤ ∥[f ]∥1 for all n ∈ Z,
so
∥Λ([f ])∥∞ = sup{∣αn[f ] ∣ ∶ n ∈ Z} ≤ ∥[f ]∥1 .
This is precisely the statement that the operator Λ is bounded, with ∥Λ∥ ≤ 1.
By Corollary 11.12, if [f ], [g] ∈ L1 (T, C) and Λ([f ]) = Λ([g]), then [f ] = [g],
and thus Λ is injective.
The question of whether or not every sequence in c0 (Z, C) is the sequence of
Fourier coefficients of some element of L1 (T, C) is therefore the question of whether
or not Λ is surjective.
The result we shall need from Functional Analysis is the Inverse Mapping The-
orem. To get this result, we will first require a lemma, and some notation.
Given a Banach space (Z, ∥ ⋅ ∥Z ) and a real number r > 0, we denote the closed
ball of radius r centred at the origin by
Zr = {z ∈ Z ∶ ∥z∥Z ≤ r}.
For z0 ∈ Z and ε > 0, we shall denote by B Z (z0 , ε) = {z ∈ Z ∶ ∥z − z0 ∥ < ε} the open
ball of radius ε in Z, centred at z0 .
192 L.W. Marcoux Introduction to Lebesgue measure

12.2. Lemma. Let X and Y be Banach spaces and suppose that T ∈ B(X, Y).
If Y1 ⊆ T Xm for some m ≥ 1, then Y1 ⊆ T X2m .
Proof. First observe that Y1 ⊆ T Xm implies that Yr ⊆ T Xrm for all r > 0.
Choose y ∈ Y1 . Then there exists x1 ∈ Xm so that ∥y − T x1 ∥ < 1/2. Since
y − T x1 ∈ Y1/2 ⊆ T Xm/2 , there exists x2 ∈ Xm/2 so that ∥(y − T x1 ) − T x2 ∥ < 1/4. More
generally, for each n ≥ 1, we can find xn ∈ Xm/2n−1 so that
n
1
∥y − ∑ T xj ∥ < .
j=1 2n
Since X is complete and ∑∞
n=1 ∥xn ∥ ≤ ∑∞
n=1
m
2n−1
= 2m, we have x = ∑∞
n=1 xn ∈ X2m .
By the continuity of T ,
∞ N
T x = T ( ∑ xn ) = lim T ( ∑ xn ) = y.
n=1 N →∞ n=1
Thus y ∈ T X2m ; i.e. Y1 ⊆ T X2m .
◻

12.3. Theorem. The Open Mapping Theorem.

Let X and Y be Banach spaces and suppose that T ∈ B(X, Y) is a surjection.
Then T is an open map - i.e. if G ⊆ X is open, then T G ⊆ Y is open.
Proof. Since T is surjective, Y = T X = ∪∞ ∞
n=1 T Xn = ∪n=1 T Xn . Now Y is a complete
metric space, and so by the Baire Category Theorem 10.9 , there exists m ≥ 1
so that the interior int(T Xm ) =/ ∅. As T Xm is dense in T Xm , we can choose y ∈
int(T Xm ) ∩ T Xm .
Let δ > 0 be such that B Y (y, δ) = y + B Y (0, δ) ⊆ int(T Xm ) ⊆ (T Xm ). Then
Y
B (0, δ) ⊆ −y + T Xm ⊆ T Xm + T Xm ⊆ T X2m . (This last step uses the linearity of T .)
Thus Yδ/2 ⊆ B Y (0, δ) ⊆ T X2m . By Lemma 12.2 above,
Yδ/2 ⊆ T X4m ,
or equivalently,
T Xr ⊃ Yrδ/8m
for all r > 0.
Suppose that G ⊆ X is open and that y ∈ T G, say y = T x for some x ∈ G. Since
G is open, we can find ε > 0 so that x + B X (0, ε) ⊆ G. Thus
T G ⊇ T x + T (B X (0, ε))
⊇ y + T Xε/2
⊇ y + Yεδ/16m
⊇ y + B Y (0, εδ/16m)
= B Y (y, εδ/16m).
Thus y ∈ T G implies that y ∈ int T G, and so T G is open.
◻
 12. WHICH SEQUENCES ARE SEQUENCES OF FOURIER COEFFICIENTS? 193

12.4. Corollary. The Inverse Mapping Theorem

Let X and Y be Banach spaces and suppose that T ∈ B(X, Y) is a bijection. Then
T −1 is continuous, and so T is a homeomorphism.
Proof. That T −1 is linear is basic linear algebra.
If G ⊆ X is open, then (T −1 )−1 (G) = T G is open in Y by the Open Mapping
Theorem above. Hence T −1 is continuous.
◻

12.5. Theorem. The map

Λ ∶ (L1 (T, C), ∥ ⋅ ∥1 ) → (c0 (Z, C), ∥ ⋅ ∥∞ )
[f ]
[f ] ↦ (αn )n∈Z .
is not surjective.
Proof. As we saw in paragraph 12.1, Λ is continuous, linear and injective. If it
were surjective, then by the Inverse Mapping Theorem,
Λ−1 ∶ c0 (Z, C) → L1 (T, C)
[f ]
(αn )n∈Z ↦ [f ]
would be continuous.
Let DN = ∑Nn=−N ξn be the Dirichlet kernel of order N , and let dN ∶= Λ([DN ]),
N ≥ 1.
Then dN = (..., 0, 0, . . . , 0, 1, 1, . . . , 1, 1, 0, 0, . . .), with the 1’s appearing for the
indices −N ≤ k ≤ N . Clearly ∥dN ∥∞ = 1, each dN is finitely supported, but by
Theorem 10.4,
lim ∥Λ−1 (dN )∥1 = lim ∥[DN ]∥1 = ∞.
N →∞ N →∞
−1
Thus Λ is not continuous, and so Λ is not surjective.
That is, there exist sequences (βn )n∈Z ∈ c0 (Z, C) which are not the Fourier
coefficients of any element of L1 (T, C).
◻
[f ]
12.6. Of course, the fact that [f ] ∈ L2 (T, C) if and only if (αn )n∈Z ∈ ℓ2 (Z, C)
makes it tempting to conjecture that perhaps the range of the map Λ from Theo-
rem 12.5 should be ℓ1 (Z, C). Tempting, but alas, false.
⎧
⎪ 1 if n ≥ 1
⎪
The sequence βn = ⎨ n is clearly in ℓ2 (Z, C), and thus [f ] ∶= ∑n∈Z βn [ξn ]
⎪
⎪0 if n ≤ 0
⎩
converges in L2 (T, C) ⊆ L1 (T, C). On the other hand,
Λ([f ]) = (βn )n∈Z
is definitely not in ℓ1 (Z, C).
194 L.W. Marcoux Introduction to Lebesgue measure

As stated in the book by Katznelson [2], p. 23,

The only spaces, defined by conditions of size or smoothness of
the functions, for which we obtain ... [a] complete characteri-
sation, that is, a necessary and sufficient condition expressed in
terms of order of magnitude, for a sequence {an } to be the Fourier
coefficients of a function in the space, are L2 (T, C) and its “deriva-
tives”. (Such as the space of absolutely continuous functions with
derivatives in L2 (T, C).)

So - much like the enigma surrounding the Cadbury Caramilk bar, the mystery
persists.
12. WHICH SEQUENCES ARE SEQUENCES OF FOURIER COEFFICIENTS? 195

Appendix to Section 12.

196 L.W. Marcoux Introduction to Lebesgue measure

Exercises for Section 12.

Exercise 12.1.
∞
ξn (θ) − ξ−n (θ) sin nθ
Keeping in mind that sin nθ = , θ ∈ R, the series ∑ can be
2i n=1 n
expressed as the Fourier series of a function in L2 (T, C). Which function?
 Bibliography

[1] I.N. Herstein. Topics in Algebra, 2nd ed. John Wiley and Sons, New York, Chichester, Brisbane,
Toronto, Singapore, 1975.
[2] Y. Katznelson. An introduction to harmonic analysis, Third edition. Cambridge Mathematical
Library. Cambridge University Press, Cambridge, 2004.
[3] L.W. Marcoux. An introduction to Functional Analysis. Online notes – available at
http://www.math.uwaterloo.ca/∼lwmarcou/, 2012.
[4] J.C. Oxtoby. Measure and Category: A survey of the analogies between topological and measure
spaces, volume 2 of Graduate Texts in Mathematics. Springer-Verlag, New York-Berlin, 1971.
[5] R.M. Solovay. A model of set-theory in which every set of reals is lebesgue measurable. Ann. of
Math., 92:1–56, 1970.
[6] G. Vitali. Sul problema della misura dei gruppi di punti di una retta. Bologna, Tip. Gamberini
e Parmeggiani, 1905.
[7] R. Whitley. Projecting m onto c0 . The Amer. Math. Monthly, 73:285–286, 1966.
[8] Stephen Willard. General Topology. Addison-Wesley Publishing Co., Reading, Mass.-London-
Don Mills, Ont., 1970.

197
 Index

C ∗ -algebras, 120 orthonormal, 112

C ∗ -equation, 120 Bernstein sets, 24
L1 (E, K), 84 Berra, Yogi, 159
Lp -norm, 89 Bessel’s Inequality, 116
Lp -space, 89 Bierce, Ambrose, ii
L1 (E, R), 68 bimodule, 152
L1 (E, R), 68 Borel sets, 32
∥ ⋅ ∥sup , 2 bounded approximate unit, 154
σ-additive, 21
σ-algebra, 28 Cantor (middle thirds) set, 35
generated by a set, 40 Cantor set, 34, 62
σ-algebra of Borel sets, 32 Capote, Truman, ii
σ-algebra of sets, 28 Carathéodory, 27, 33
σ-finite measure space, 106 Cauchy Criterion, 5, 72
σ-subadditivity, 17 Cauchy’s Integral Formula, 11
onb, 112 Cauchy-Schwarz Inequality, 3, 109, 121
characteristic function, 8, 43
absolutely summable, 107
compactification
Aleksandrov compacitification, 53
of a topological space, 52
algebra, 134, 151
complement
Banach, 153
orthogonal, 115
unital, 151
complemented
algebra of sets, 27
algebraically, 114
algebraically complemented, 114
topologically, 114
almost everywhere, 62
cone
approximate unit, 154
real, 50
bounded, 154
Assignment 7, 135 conjugate
Assignment Question, 33, 35–37 Lebesgue, 85, 89
Axiom of Choice, 22, 24 conjugate function, 89, 96
conjugate-linear, 120
Baire Category Theorem, 163 continuous linear functionals, 105
Banach algebra, 134, 153 convergence
involutive, 120 pointwise, 9
Banach algebras, 11 convolution, 131, 134, 135, 137
Banach space, 2 discrete, 154
homogeneous (over T), 138 convolution operator, 137
basis countable subadditivity, 17
Hamel, 112 cover
Hilbert space, 112 by open intervals, 16

199
200 INDEX

decomposable, 106 inner product, 109

Dirichlet kernel, 159, 160, 162 integral
discrete convolution, 154 Riemann, 5
disjoint representation (of a simple involution, 120
function), 58 involutive Banach algebra, 120
Dominated Convergence Theorem, 78 isomorphism of Hilbert spaces, 118
dual space, 105, 120
kernel, 137
essential supremum, 92 Kronecker delta function, 112
Euclidean norm, 11
extended real numbers, 47, 104
Lebesgue conjugate, 85, 89, 120, 126, 139
extended real-valued function, 48
Lebesgue conjugate function, 86
Fatou’s Lemma, 76, 78 Lebesgue integrable (complex functions), 69
Fatou’s Theorem, 76 Lebesgue integral, 68
finite intersection property, 35 Lebesgue integral (non-negative functions),
first category 60
set of the, 163 Lebesgue measurable function, 42
fixed point, 38 Lebesgue measurable set, 27
Fourier coefficient, 125, 127 Lebesgue measurable sets, 28
Fourier series, 95, 125, 127 Lebesgue measure, 32
Fubini’s Theorem, 134 Lec, Stanislaw J., ii
function left module, 134, 152
characteristic, 43 Lehrer, Tom, 57
indicator, 43 Lemma
Lebesgue measurable, 42 Fatou’s, 76, 78
Fundamental Theorem of Calculus, 61, 71 length
of an interval, 16
Goldwyn, Samuel, ii Levant, Oscar, 1
Gram-Schmidt Orthogonalisation Process, linear functionals
113 continuous, 105
linear manifold, 99
Hölder’s Inequality, 86, 88, 89, 95, 105, 111,
126, 140
Marx, Groucho, ii
Hadas, Moses, ii
meager, 163
Hahn-Banach Theorem, 105
measurable
Hamel basis, 112, 124
extended real-valued function, 48
harmonic series, 162
measurable sets, 28
Hercules, 45
measurable, Lebesgue, 27
Hilbert space, 110
measure, 10
Hilbert space basis, 112
homogeneous Banach space, 138, 139, 141 Lebesgue, 32
outer, 16
idempotent, 115 regular Borel, 106
increasing sequence, 9 metric
indicator function, 8, 43 induced by the norm, 2
Inequality Milligan, Spike, 27
Bessel’s, 116 Minkowski’s Inequality, 87, 89, 105
Cauchy-Schwarz, 3, 109 module, 152
Hölder’s, 86, 88, 89, 95, 105, 111, 126, 140 left, 134, 152
Minkowski, 87, 89 Monotone Convergence Theorem, 63, 66, 75,
Minkowski’s, 89, 105 76, 90
Triangle, 87, 89 monotone increasing, 16
Young’s, 85 multiplicative identity, 151
 INDEX 201

Nelson, Craig T., 83 standard inner product on L2 (E), 112

norm, 1 Stone-Čech compactification, 52, 106
Euclidean, 11 subalgebra, 152
operator, 3 subbase, 52
normed linear space, 1 summability kernel, 154
Novak, Ralph, ii summable
nowhere dense, 163 absolutely, 107
summable series, 107
one-point compactification, 53 supremum
operator norm, 3 essential, 92
operators supremum norm, 2
unitary, 119 symmetric difference, 108
orthogonal, 109, 112
orthogonal complement, 115 ternary expansion, 35
orthogonal projection, 115 test values (for a partition), 4
orthonormal basis, 112, 123 Theismann, Joe, 123
orthonormal set, 112 Theorem
outer measure, 16 Dominated Convergence, 78
Oxtoby, John, 24 Fubini, 134
Fundamental Theorem of Calculus, 61
Parallelogram Law, 113 Hahn-Banach, 105
Parceval’s Identity, 118 Monotone Convergence, 75, 76, 90
Parker, Dorothy, ii, 16 Montone Convergence, 66
partition, 4 Phillips’, 115
Phillips’ Theorem, 115 Pythagorean, 113, 117
pointwise convergence, 9 Riesz Representation (for Hilbert spaces),
power set, 16 120
Pythagorean Theorem, 113, 117 Schröder-Bernstein, 36, 38
The Monotone Convergence Theorem, 63
quotient space, 84
Whitley’s proof of Phillips’, 115
real cone, 50 thin, 163
regular Borel measures, 106 topologically complemented, 114
Riemann integral, 5, 71, 137 translation invariant, 20
Riemann-Lebesgue Lemma, 128 Triangle Inequality, 87, 89
Riesz Representation Theorem, 120 two-point compactification, 52
Tyson, Mike, 134
Schröder-Bernstein Theorem, 36, 38
second category uniform convergence, 3
set of the, 163 unitary operators, 119
seminorm, 1, 83
Vitali set, 24
separable (topological space), 102
Vitali’s set, 22, 26, 37
sequence
Vitali, Giuseppe, 24
increasing, 9
Shields, Brooke, 109 Weierstraß Approximation Theorem, 102
simple function, 49
standard form, 49 Yarborough, Cale, 42
simple functions, 10 Young’s Inequality, 85
Sisyphus, 45
Zermelo-Fraenkel axioms, 24
Solovay, Robert, 24
standard form, 57
standard form (simple funtion), 49
standard inner product on Cn , 111
standard inner product on ℓ2 , 111