Integration

1 Partitions
Let a ≤ b be real numbers. A partition of the interval [a, b] is a finite subset P of [a, b] containing
both a and b. We’ll write P0 for the smallest element of P , P1 for the second smallest, and so forth.
In other words, P = {P0 , P1 , . . . , Pn } and

a = P0  P1  · · ·  Pn = b.

We define the mesh of P by

mesh(P ) := max{Pk − Pk−1 : k = 1, . . . , n}.

If P and Q are both partitions, we say that Q is a refinement of P if Q ⊇ P . Clearly we then have
mesh(Q) ≤ mesh(P ).

Example 1.1. The following are all partitions of the unit interval [0, 1].

P := {0, 1}
1
 
Q := 0, , 1
2
1 2
 
R := 0, , , 1
2 3
1 1 5
 
S := 0, , , , 1
4 3 6
Notice that Q, R and S are all refinements of P , R is also a refinement of Q, S is not a refinement
of either Q or R. We have mesh(P ) = 1, and the mesh of all of remaining partitions is 1/2.

If P and Q are both partitions of [a, b], a common refinement of P and Q is a partition which
is a refinement of both P and Q simultaneously. Notice that the union P ∪ Q is always a common
refinement of P and Q. In other words, common refinements always exist.

Remark 1.2. Observe the following properties of the relation “is a refinement of” on the set of
all partitions of [a, b].

(1) Any partition P is a refinement of itself.

(2) If P is a refinement of a Q, and Q is also a refinement of P , then P = Q.

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(3) If P is a refinement of Q, and Q is a refinement of R, then P is a refinement of R.

(4) If P and Q are partitions, there exists a partition R which is a refinement of both P and Q.

The first three properties say exactly that the relation “is a refinement of” is reflexive, antisymmet-
ric, and transitive. In other words, the relation “is a refinement of” is an example of a partial order,
and the set of all partitions, equipped with this relation, is a partially ordered set. Notice that this
partial order isn’t totally ordered, since there are partitions neither of which is a refinement of the
other. But, even though its not a totally ordered set, it is a directed set because it also satisfies
property 4 listed above.
Anyway, this is all just terminology and we won’t use this terminology again. We will, however,
see some of the above properties recur as we start proving things about integrals.

2 Upper and Lower Darboux Sums

Let f : [a, b] → R be a bounded function and let P be a partition of [a, b]. We define the upper
Darboux sum of f with respect to P to be the quantity
n
X
U (f, P ) := (Pk − Pk−1 ) sup f ([Pk−1 , Pk ])
k=1

and analogously we define the lower Darboux sum of f with respect to P to be the quantity
n
X
L(f, P ) := (Pk − Pk−1 ) inf f ([Pk−1 , Pk ]).
k=1

These quantities have a very natural interpretation in terms of areas of rectangles. I’ll draw a
picture in class.
The following lemma says that taking refinements “squeezes” the Darboux sums closer together.

Lemma 2.1. Suppose P is a partition of [a, b] and Q is a refinement of P . Then

L(f, P ) ≤ L(f, Q) ≤ U (f, Q) ≤ U (f, P ).

Proof. The middle inequality is obvious. We’ll prove the inequality L(f, Q) ≤ L(f, P ), and the
proof that U (f, P ) ≤ U (f, Q) is analogous. First consider the case when Q has exactly one more
point than P . In other words, the elements of Q are

P0  P1  Pk−1  Qk  Pk  · · ·  Pn .

Then the fact that [Pk−1 , Pk ] contains both [Pk−1 , Qk ] and [Qk , Pk ] means that

inf f ([Pk−1 , Qk ]), inf f ([Qk , Pk ]) ≥ inf f ([Pk−1 , Pk ]).

We want to show that L(f, Q) − L(f, P ) ≥ 0. Notice that the only terms that don’t cancel out in

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this difference are the following.

inf f ([Pk−1 , Qk ])(Qk − Pk−1 ) + inf f ([Qk , Pk ])(Pk − Qk ) − inf f ([Pk−1 , Pk ])(Pk − Pk−1 )

To see that this is nonnegative, observe the following.

inf f ([Pk−1 , Pk ])(Pk − Pk−1 ) = inf f ([Pk−1 , Pk ])(Pk − Qk + Qk − Pk−1 )

= inf f ([Pk−1 , Pk ])(Qk − Pk−1 ) + inf f ([Pk−1 , Pk ])(Pk − Qk )
≤ inf f [Pk−1 , Qk ](Qk − Pk−1 ) + inf f ([Qk , Pk ])(Pk − Qk )

This completes the proof. In general, if Q has m more elements than P , then we just apply the
above argument m times, adding one extra element of Q each time.

Corollary 2.2. If P and Q are any partitions of [a, b], then L(f, P ) ≤ U (f, Q).

Proof. Let R be a common refinement of P and Q. Then lemma 2.1 tells us that

L(f, P ) ≤ L(f, R) ≤ U (f, R) ≤ U (f, Q).

3 Integrability
For a bounded function f : [a, b] → R, we define the following.

U (f ) := inf{U (f, P ) : P is a partition of [a, b]}

L(f ) := sup{L(f, P ) : P is a partition of [a, b]}

Lemma 3.1. For any partition P of [a, b], we have

L(f, P ) ≤ L(f ) ≤ U (f ) ≤ U (f, P ).

In particular, L(f ) and U (f ) are real numbers.

Proof. Both the first and third inequalities are trivial. We only need to show that L(f ) ≤ U (f ).
Fix a partition Q. Then corollary 2.2 guarantees that L(f, Q) ≤ U (f, R) for all partitions R,
which means that L(f, Q) is a lower bound for the same set for which U (f ) is an infimum. Thus
L(f, Q) ≤ U (f ). But now, since Q is an arbitrary partition, we see that U (f ) is an upper bound
for the same set for which L(f ) is a supremum, so L(f ) ≤ U (f ).

If L(f ) = U (f ), then the function f is said to be integrable and the common value L(f ) = U (f )
is called the integral of f on [a, b] and is denoted
Z b Z b
f or f (x) dx.
a a

Proposition 3.2. Let f : [a, b] → R be a bounded function. Then f is integrable if and only if, for
each ε 0, there exists a partition P of [a, b] such that U (f, P ) − L(f, P ) ≤ ε.

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Proof. Suppose first that f is integrable and fix ε 0. Then, by definition of supremums and
infimums, there exist partitions P and Q such that
ε ε
L(f, P ) ≥ L(f ) + and U (f, Q) ≤ U (f ) + .
2 2
Let R be a common refinement of P and Q. Then, using lemma 2.1, we see that

U (f, R) − L(f, R) ≤ U (f, Q) − L(f, P ) ≤ U (f ) − L(f ) + ε = ε.

Conversely, fix some ε 0 and let P be a partition such that U (f, P ) − L(f, P ) ≤ ε. Observe that

U (f ) ≤ U (f, P ) = U (f, P ) − L(f, P ) + L(f, P ) ≤ ε + L(f, P ) ≤ ε + L(f ).

Since ε 0 is arbitrary, we conclude that U (f ) ≤ L(f ), which proves that U (f ) = L(f ) using the
middle inequality of lemma 3.1.

Our next task is to prove that a large class of functions are automatically integrable.

Lemma 3.3. Let f : [a, b] → R be a function.

(a) If f is monotonic, then it is integrable.

(b) If f is continuous, then it is integrable.

Proof. We’ll prove that f is integrable when it is monotonically increasing, and the monotonically
decreasing case is analogous. If f (a) = f (b), then f is a constant function and we are done
immediately, so suppose that f (a)  f (b). Notice that

f (a) ≤ f (x) ≤ f (b)

for all x ∈ [a, b], so f is bounded. We’ll show that f is integrable using proposition 3.2. Fix ε 0
and let P be any partition such that
ε
mesh(P ) ≤ .
f (b) − f (a)

Then
n
X
U (f, P ) − L(f, P ) = (sup f ([Pk−1 , Pk ]) − inf f ([Pk−1 , Pk ]))(Pk − Pk−1 )
k=1
Xn
= (f (Pk ) − f (Pk−1 ))(Pk − Pk−1 )
k=1
n
X ε
≤ (f (Pk ) − f (Pk−1 )) ·
k=1 f (b) − f (a)
n
ε X
= (f (Pk ) − f (Pk−1 ))
f (b) − f (a) k=1
= ε.

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This completes the proof of (a). For (b), if f is continuous, the fact that [a, b] is compact means
that f is automatically bounded and uniformly continuous. We’ll again use proposition 3.2 to prove
integrability. Fix ε 0 and let δ 0 be such that |x − y| ≤ δ implies
ε
|f (x) − f (y)| ≤ .
b−a

Fix any partition P such that mesh(P ) ≤ δ. Since f assumes its maximum and minimum on
[Pk−1 , Pk ] for all k, we see that
ε
sup f ([Pk−1 , Pk ]) − inf f ([Pk−1 , Pk ]) ≤ .
b−a
This means that n
X ε
U (f, P ) − L(f, P ) ≤ · (Pk − Pk−1 ) = ε.
k=1 b − a

Lemma 3.4. Let f and g be integrable functions [a, b] → R and let c be a real number.
(a) cf is integrable and
Z b Z b
cf = c f.
a a

(b) f + g is integrable and

Z b Z b Z b
(f + g) = f+ g.
a a a

(c) |f | is integrable, where |f | is given by x 7→ |f (x)|.

Proof. For (a), there is nothing to show when c = 0. First consider the case when c 0. Let P be
a partition of [a, b]. Clearly

sup cf ([Pk−1 , Pk ]) = c · sup f ([Pk−1 , Pk ])

which means that U (cf, P ) = cU (f, P ). Thus it follows that U (cf ) = cU (f ). Similarly, we also
have L(cf ) = cL(f ). Then
U (cf ) = cU (f ) = cL(f ) = L(cf )
so cf is integrable and Z b Z b
cf = c f.
a a

Now to prove the case when c  0, it suffices to consider the case when c = −1, since then for
general c  0, we have
Z b Z b Z b Z b Z b
cf = −(−c)f = − (−c)f = −(−c) f =c f
a a a a a

using the case c = −1 for the second equality and the case c 0 for the third. Thus we assume
that c = −1. Then its easy to see that U (−f, P ) = −L(f, P ) for all partitions P . Thus

U (−f ) = inf{U (−f, P ) : P is a partition} = − sup{L(f, P ) : P is a partition} = −L(f ).

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Similarly, we also get L(−f ) = −U (f ). Thus, since f is integrable, we have

U (−f ) = −L(f ) = −U (f ) = L(−f )

so −f is integrable and Z b Z b
(−f ) = − f.
a a

This completes the proof of (a).

For (b), we’ll use proposition 3.2 again. Fix ε 0. Since f is sintegrable, there exists a partition
with respect to which the upper Darboux sum and lower Darboux sum of f differ by at most ε/2.
There exists another partition which does the same thing for g, and then by taking a common
refinement of these two partitions, we find a partition P such that
ε ε
U (f, P ) − L(f, P ) ≤ and U (g, P ) − L(g, P ) ≤ .
2 2
Observe that for any S ⊆ [a, b], we have

inf(f + g)(S) ≥ inf f (S) + inf g(S)

which implies that

L(f + g, P ) ≥ L(f, P ) + L(g, P )
and similarly we have
U (f + g, P ) ≤ U (f, P ) + U (g, P ).
Thus
U (f + g, P ) − L(f + g, P ) ≤ (U (f, P ) + U (g, P )) − (L(f, P ) + L(g, P )) ≤ ε.
Thus f + g is integrable by proposition 3.2. Moreover, observe that
Z b Z b Z b
(f + g) ≤ U (f + g, P ) ≤ U (f, P ) + U (g, P ) ≤ L(f, P ) + L(g, P ) + ε ≤ f+ g+ε
a a a

and similarly
Z b Z b Z b
(f + g) ≥ L(f + g, P ) ≥ L(f, P ) + L(g, P ) ≥ U (f, P ) + U (g, P ) − ε = f+ g − ε.
a a a

Combining these inequalities, we find that

Z b Z b Z b Z b Z b
f+ g−ε≤ (f + g) ≤ f+ g+ε
a a a a a

so, since ε is arbitrary, we are done.

For (c), first observe that

sup |f | (S) − inf |f | (S) ≤ sup f (S) − inf f (S)

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for any subset S ⊆ [a, b]. Indeed, observe that for any x, y ∈ S, we have

|f (x)| − |f (y)| ≤ |f (x) − f (y)| ≤ |sup f (S) − inf f (S)| = sup f (S) − inf f (S).

Now taking the supremum over all x ∈ S and then the infimum over all y ∈ S to complete the
proof. Now notice that this inequality tells us that

U (|f | , P ) − L(|f | , P ) ≤ U (f, P ) − L(f, P )

for any partition P . Thus, for any ε 0, there exists a partition P such that U (f, P ) − L(f, P ) ≤ ε
since f is integrable, and then we have

U (|f | , P ) − L(|f | , P ) ≤ U (f, P ) − L(f, P ) ≤ ε

so proposition 3.2 guarantees integrability.

Lemma 3.5. If (fn )n∈N is a uniformly convergent sequence of integrable functions [a, b] → R and
f = lim fn , then f is integrable and
Z b Z b
f = lim fn .
a n→∞ a

Proof. Let εn := kfn − f ksup so that

fn (x) − εn ≤ f (x) ≤ fn (x) + εn

for all n ∈ N and x ∈ [a, b]. This means that, for any partition P , we have

L(fn − εn , P ) ≤ L(f, P ) and U (f, P ) ≤ U (fn + εn , P )

so taking the supremum and the infimum over all P , we find that
Z b Z b
(fn − εn ) = L(fn − εn ) ≤ L(f ) ≤ U (f ) ≤ U (fn + εn ) = (fn + εn ). (1)
a a

This means that

Z b Z b Z b
0 ≤ U (f ) − L(f ) ≤ (fn + εn ) − (fn − εn ) = (2εn ) = 2εn (b − a).
a a a

Now observe that this is true for all n and that lim εn = 0, so we conclude that U (f ) = L(f ). Thus
f is integrable. Now observe that (1) tells us that
Z b ! Z b Z b !
fn − εn (b − a) ≤ f≤ fn + εn (b − a).
a a a

In other words, Z 
 b Z b 

 f − fn  ≤ εn (b − a)
 a a 

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so letting n tend to infinity completes the proof.

4 Properties of Integrals
Lemma 4.1. Let f and g be integrable functions [a, b] → R. If f (x) ≤ g(x) for all x ∈ [a, b], then
Z b Z b
f≤ g.
a a

Proof. Observe that h = g − f is integrable due to lemma 3.4 and h(x) ≥ 0 for all x ∈ X. This
clearly means that L(h, P ), U (h, P ) ≥ 0 for all partitions P , so
Z b Z b Z b
g− f= h ≥ 0.
a a a

Corollary 4.2. If f : [a, b] → R is integrable, then

Z 
 b  Z b
f ≤ |f | ,
 

 a  a

where |f | denotes the function x 7→ |f (x)|.

Proof. We know from lemma 3.4 that |f | is integrable. Observe that − |f | (x) ≤ f (x) ≤ |f | (x) for
all x. This means that Z b Z b Z b
− |f | ≤ f≤ |f |
a a a

using lemma 4.1, and this proves the result.

Lemma 4.3. If f : [a, b] → R is continuous and nonnegative and

Z b
f = 0,
a

then f (x) = 0 for all x ∈ [a, b].

Proof. For any partition P , the fact that f (x) ≥ 0 means that L(f, P ) ≥ 0. This means that
Z b
0 ≤ L(f, P ) ≤ L(f ) = f =0
a

so actually L(f, P ) = 0. But

n
X
L(f, P ) = inf f ([Pk−1 , Pk ])(Pk − Pk−1 )
k=1

and Pk − Pk−1 0, so actually we must have

inf f ([Pk−1 , Pk ]) = 0.

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This is true for any partition P . Then for any compact interval [s, t] contained in [a, b], we can
consider the partition P = {a, s, t, b} and the above shows that inf f ([s, t]) = 0.
Now suppose for a contradiction that f (x) 6= 0 for some x. Since f is nonnegative, we have
f (x) 0. Since f is continuous, there exists some δ 0 such that f (y) 0 whenever |x − y| ≤ δ.
But then we clearly cannot have inf f ([x − δ, x + δ]) = 0, since f must achieve its minimum on
the compact set [x − δ, x + δ] but its also strictly positive on this entire compact set. This is a
contradiction.

5 Sample Problems
Problem 1. Show that the following function f : [0, 1] → R is not integrable.

1 if x ∈ Q, and
f (x) =
0 if x ∈
/ Q.

Problem 2. Give an example of a sequence of integrable functions (fn )n∈N which converges point-
wise to a non-integrable function.

Hint. Find a sequence of integrable functions converging to the function in problem 1.

Problem 3. Let f : [a, b] → R be a function and c ∈ [a, b] some value such that f is integrable on
[a, c] and [c, b]. Then f is integrable on [a, b] and
Z b Z c Z b
f= f+ f.
a a c

Proof. Fix ε 0. Then there exists a partition P of [a, c] such that U (f, P ) − L(f, P ) ≤ ε/2.
Similarly there exists a partition Q of [c, b] such that U (f, Q) − L(f, Q) ≤ ε/2. But then observe
that R = P ∪ Q is a partition of [a, b], and clearly

U (f, R) = U (f, P ) + U (f, Q) and L(f, R) = L(f, P ) + L(f, Q)

so then
U (f, R) − L(f, R) = U (f, P ) − L(f, P ) + U (f, Q) − L(f, Q) ≤ ε.
Thus f is integrable. Moreover, observe that
Z b Z c Z b
f = U (f ) ≤ U (f, R) = U (f, P ) + U (f, Q) ≤ L(f, P ) + L(f, Q) + ε ≤ f+ f +ε
a a c

and analogously we also have Z c Z b Z b

f+ f −ε≤ f
a c a

so, since ε 0 is arbitrary, we are done.

Problem 4. Let f : [a, b] → R be a continuous function. Prove that there exists some c ∈ [a, b]

9
such that
1 Zb
f (c) = f.
b−a a
Hint. Use the intermediate value theorem.

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