Lecture 2-6 Lebesgue Measure

This is taken from Chapter 2 of the book by Royden- Fitzpatrick,

H. L. Royden and P.M. Fitzpatrick, Real Analysis, Fourth Edition

2.7. The Cantor Set and The Cantor-Lebesgue Function

• We know that a countable set of real numbers has outer measure 0.

A Borel set is measurable.

These lead to the following questions.

(Question 1) If a set has the outer measure 0, is it countable?
(Question 2) If a set of real number is measurable, is it Borel measurable?

The Cantor set we will construct has outer measure 0.

We can see that the Cantor set is uncountable.

• (Cantor Set) We consider

I = [0, 1].

The first step is to divide I into three intervals of equal length 1/3.
We remove the interior of the middle interval to obtain the closed set C1 ,
1 2 1 2
C1 = I ∼ ( , ) = [0, ] ∪ [ , 1]..
3 3 3 3

We repeat the open middle one-third interval removal to each interval of

1 2
[0, ], [ , 1].
3 3
We obtain
1 2 1 2 7 8
C2 = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , 1].
9 9 3 3 9 9

We repeat the open middle one-third interval removal to each intervals in C2 , we

obtain C3 .

1
C3 has 23 closed intervals, each of length 33
.

We continue this process to obtain

Ck , k = 1, 2, · · · .

0
 We obtain the Cantor Set,
C = ∩∞
k=1 Ck .

The collection,
C1 , C2 , · · ·
has the following properties.
(C1) {Ck } is a descending sequence of closed sets. We have

Ck+1 ⊂ Ck
for each k = 1, 2, · · ·.
(C2) For each k = 1, 2, · · ·, Ck has 2k disjoint closed intervals of length 1/3k .

• (Proposition 19) C is a colsed set, uncountable and measure 0.

Proof:

Since each Ck is a closed set for each k, the intersection C is also a closed set. By
the Nested Set Theorem, C is not an empty set.

We have
2
m(Cn ) = ( )n
3
Hence
m(C) = lim m(Cn ) = 0.
n→∞

C is a closed set, since

C = ∩∞
n=1 Cn

and each Cn is a closed set.

We want to prove that C is uncountable.

We use the proof by contradiction to prove the result.

We assume C is countable and by enumeration is given as

C = {c1 , c2 , · · ·}.

We consider
1 2
C1 = [0, ] ∪ [ , 1].
3 3

1
F1 is one of
1 2
[0, ], [ , 1]
3 3
which does not contain c1 .

If
1
F1 = [0, ],
3
then
1
c1 ∈
/ [0, ].
3
We consider
1 1 2 1
C2 ∩ [0, ] = [0, ] ∩ [ , ].
3 9 9 3
F2 is one of
1 2 1
[0, ], [ , ].
9 9 3
which does not contain c2 .
It is possible that both
1 2 1
[0, ], [ , ],
9 9 3
do not contain c2 . In this case, we choose any one of them as F2 .

We have
F2 ⊆ F1
and
c2 ∈
/ F2 .

We can continue this process to obtain

F1 , F2 , · · · ,

such that for each n = 1, 2, · · ·, we have

(a) Fn+1 ⊆ Fn ,
(b) Fn ⊆ Cn ,
(c) cn ∈
/ Fn .
We consider
F = ∩∞
n=1 Fn .

By the Nested Theorem, F is not an empty set. We take

x0 ∈ F.

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 Since
Fn ⊆ Cn ⊆ ∪∞
i=1 Ci = C,

we have
F = ∩∞
i=1 Fi ⊆ C.

Therefore,
x0 ∈ C.
Then
x0 = cn0
for some positive integer n0 . This implies

x0 ∈
/ Fn0 .

Therefore,
/ ∩∞
x0 ∈ i=1 Fi = F,

this is a contradiction.

• (Cantor-Lebesgue Function) We define

On = [0, 1] ∼ Cn ,

for n = 1, 2, · · ·.

(a) On is the union of disjoint open intervals,

(1) 1 2
I1 = ( , ),
3 3
and
(2) 1 2 (2) 7 8
I1 = ( , ), I2 = ( , ),
9 9 9 9
for each
i = 2, 3, · · · , n,
1
there are 2i−1 open intervals of length 3i
, given by
(i) (i) (i)
I1 , I2 , · · · , I2i−1 .

The number of open subintervals in Cn is given by,

1 + 2 + 22 + · · · + 2n−1 = 2n − 1.

(b) We define
O = ∪∞
i=1 Ok

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We have
C = [0, 1] ∼ O.
O has measure 1,
m(O) = 1.

(c) We define the function ϕ on On with values

1 2 3 2n − 1
, , , · · · , ,
2n 2n 2n 2n
on 2n − 1 subintervals of On .
On On+1 , the number of subinterval is
2n+1 − 1,
which is equal to
1 + 2(2n − 1).
For evey subinterval of On , two intervals of length
1
3n+1
are added next to it with distance
1
3n+1
The
i − th
subinterval of On becomes
2i − th
subinterval of On+1 .
(d) ϕ is defined on
O = ∪∞
n=1 On .

It is extended to the Cantor set ,

C = [0, 1] ∼ O,
by the
ϕ(0) = 0,
and for
x ∈ C, x 6= 0,
by
ϕ(x) = sup{ϕ(t); t ∈ O, t ≤ x}.

We say ϕ is the Cantor-Lebesgue function.

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• (Proposition 20: The Cantor-Lebesgue function has the following properties.
(a) It is an increasing continuous function from [0, 1] onto [0, 1].
(b) Its derivative exists on O and ϕ0 = 0 on O, with

m(O) = 1.

Proof:

For x0 ∈ O, x0 is in an unique open subinterval Ix of O. Then ϕ is constant on Ix

In this case, ϕ is continuous at x0 .

We consider
x0 ∈ [0, 1] ∼ O = C,
the Cantor set.

Since
O = ∪∞
k=1 Ok ,

and
C= [0, 1] ∼ O
= [0, 1] ∩ Oc
= [0, 1] ∩ (∪∞
k=1 Ok )
c
∞ c
= [0, 1] ∩ (∩k=1 Ok ),
we have
x0 ∈ [0, 1] ∩ Okc
for any k. The set
[0, 1] ∩ Okc
1
is a disjoint union of closed interval of length 3k
. Therefore, there is a unique
ak < bk , such that
ak ≤ x 0 ≤ b k ,
and
1
b k − ak = ,
3k
and
1
ϕ(bk ) − ϕ(ak ) = .
2k
We also have
ϕ(ak ) ≤ ϕ(x0 ) ≤ ϕ(b).
This implies ϕ is continuous at x0 .

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 Since ϕ is increasing, continuous and

ϕ(0) = 0, ϕ(1) = 1,

by IntermediateValue Thoerem, ϕ is an onto funtion from [0, 1] to [0, 1].

• (Proposition 21) Let ϕ be the Cantor-Lebesgue function. We define

ψ(x) = x + ϕ(x), x ∈ [0, 1].

(i) ψ is strickly increasing continuous function from [0, 1] onto [0, 2].
(ii) ψ(C) is a measurable set of positive measure.
(iii) ψ maps a measurable set, a subset of the Cantor set, onto a nonmeasurable set.

Proof:

We say ψ is stricly increasing if ψ is increasing and

ψ(x) < ψ(y)

if x < y.

Since ψ is strickly increasing, continuous and

ψ : [0, 1] → [0, 2]

is onto. From
[0, 1] = C ∪ O,
a disjoint union, we have
[0, 2] = ψ(C) ∪ ψ(O).

Since
ψ −1 : [0, 2] → [0, 1]
is also continuous, we have
ψ(C)
is also closed set,
ψ(O)
is open.

In the following, we want to show

m(ψ(O)) = 1.

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 This implies
m(ψ(C)) = 1.

We write
O = ∪∞
k=1 Ik .
Each Ik is an open interval.
On each Ik , ϕ is constant. Hence ψ, one each Ik , is the same as the identity mapping
by adding a constant. This implies
ψ(Ik )
is an intervals with length m(Ik ). Since ψ is one-to-one, we can conclude
∞
X
m(ψ(O)) = m(Ik ) = m(O) = 1.
k=1

Since
m(ψ(C)) = 1,
by Vitali ’s Thorem, ψ(C) contains a nonmeasurable set W . The set
ψ −1 (W ),
is a subset of C. Hence it has measure 0, therefore is a measurable set.
We can conclude that ψ maps a measurable set to a nonmeasurable set.
• (Proposition 22) There is a subset of C which is not Borel measurable.

Proof:
In Proposition 21, there is a subset of C given by A such that ψ(A) is nonmeasurable.
We want to show that A is not Borel measurable. We use the following result that
f −1 (B)
is Borel measurable if
f :R→R
is a continuous function and B is a Borel subset of R. We apply this result to
f = ψ −1
and B = A. Assume A is Borel measurable. Then
f −1 (A) = ψ(A)
is Borel measurable, and hence is measurable. This is a contradiction to the fact
that
ψ(A)
is not measurable. This proves that A is not Borel measurable, but is measurable.

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• (Lemma 23) Assume
f :R→R
is continuous. Then for each Borel subset A of R, we have

f −1 (A)

is a Borel subset of R.

Proof:

We recall B(R) the Borel σ-algebra of R is the the smallest σ-algebra containing

O,

the collection of all open subsets of R.

To prove f −1 (A) is a Borel subset of R, if A is a Borel seubset of R, we consider

G = {B ∈ B(R); f −1 (B) ∈ B(R)}.

We want to prove
G = B(R).
This result implies for any A ∈ B(R), we have

A ∈ G.

From the definition of G, we can conclude

f −1 (A) ∈ B(R).

Hence the proof is complete.

In the rest, we want to prove

G = B(R).
To prove this result, we consider the following two statements:
(a) G contains O, the collection of all open subsets of R.
(b) G is a σ-algebra of R.
We want to show if (a) and (b) hold, then

G = B(R).

This is because that G is a σ-algebra such that

O ⊆ G.

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But
B(R)
is the smallest σ-algebra of R such that

O ⊆ B(R).

Therefore, we can conclude

B(R) ⊆ G.
But, by the definition of G, we have

G ⊆ B(R).

Therefore, we have
G = B(R).

To prove (a), we recall that, by the property that f is a continuous function,

f −1 (O)

is an open subset of R, if O is an open subset of R. This proved that (a) holds..

To prove (b), we need to prove the following conditions

(b1) ∅, R ∈ G.
(b2) Assume A ∈ G. Then Ac ∈ G.
(b3) Assume A1 , A2 , · · · are elements of G. Then

∪∞
i=1 Ai ,

is an element of G.

(b1) is easy to see. We omit it.

(b2) follows from the relation,

f −1 (Ac ) = (f −1 (A))c ,

where Ac in f −1 (Ac ) is given by

Ac = R ∼ A,

and
(f −1 (A))c = R ∼ f −1 (A).
To prove
f −1 (Ac ) = (f −1 (A))c ,

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we consider two steps.
(i) f −1 (Ac ) ⊆ (f −1 (A))c .
(ii) (f −1 (A))c ⊆ f −1 (Ac ).

To prove (i), we take

x0 ∈ f −1 (Ac ),
we want to prove
x0 ∈ (f −1 (A))c .

The condition
x0 ∈ f −1 (Ac ),
implies
f (x0 ) ∈ Ac .
That is,
f (x0 )
is not in A. Therefore,
/ f −1 (A).
x0 ∈
which is the same
x0 ∈ (f −1 (A))c .
This proves (i).

To prove (ii), we take

x1 ∈ (f −1 (A))c .
We want to prove
x1 ∈ f −1 (Ac ).

The condition
x1 ∈ (f −1 (A))c ,
implies
f (x1 ) ∈
/ A.
Then
f (x1 ) ∈ Ac .
This implies
x1 ∈ f −1 (Ac ).
This also proves (ii).

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 (b3) follows by using the following relation:

f −1 (∪∞ ∞
i=1 Ai ) = ∪i=1 f
−1
(Ai ).

This can be proved in two steps.

(iii) f −1 (∪∞ ∞
i=1 Ai ) ⊆ ∪i=1 f
−1
(Ai ).
(iv) ∪∞
i=1 f
−1
(Ai ) ⊆ f −1 (∪∞
i=1 Ai ).

We omit the details of the proof of (iii) and (iv) , which are similar to the proof of
(i) and (ii).

Using
f −1 (∪∞ ∞
i=1 Ai ) = ∪i=1 f
−1
(Ai ),
assume
A1 , A2 , · · ·
are elements of G. Then
f −1 (Ai ) ∈ B(R)
for each i = 1, 2, 3, · · ·. We have

f −1 (∪∞ ∞
i=1 Ai ) = ∪i=1 f
−1
(Ai ) ∈ B(R).

Therefore, by the definition of G, we have

∪∞
i=1 Ai ∈ G.

This proves (b3).

Homework 2-6

The book Royden-Fitzpatrick is referred to the following book.

H. L. Royden and P.M. Fitzpatrick (2010), Real Analysis, Fourth Edition

Deadline: In two weeks after we have finished the discussion of this lecture note.

1. Reading assignment: Read P49-P53 in the book Royden-Fitzpatrick and also

this lecture note. Do the following.
(a) Take notes the important concepts, results.
(b) Mention the materials in this part you have questions. Find the answers of your
question(s) from other sources (such as book or internet).

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2. Let
f : R → R.
Take
A1 , A2 , · · ·
subsets of R. Prove
f −1 (∪∞ ∞
i=1 Ai ) = ∪i=1 f
−1
(Ai ),
and
f −1 (∩∞ ∞
i=1 Ai ) = ∩i=1 f
−1
(Ai ).
3. Assume the function
f :R→R
satisfies the condition that there is a c and
|f (u) − f (v)| ≤ c|u − v|
for evey u, v ∈ R.
(a) Prove f is a continuous function.
(b) Assume A has outer measure 0. Prove f (A) has outer measure 0.
(Hint: For any  > 0, take a sequence of open intervals
I1 , I2 , · · ·
such that
A ⊆ ∪∞
i=1 Ii
and ∞
X
`(Ii ) < .
i=1
For each i, take xi ∈ Ii ∩ A if
Ii ∩ A 6= ∅.
and define
Ji = (yi − c`(Ii ), yi + c`(Ii )).
If
Ii ∩ A = ∅,
take
Ji = ∅.
Prove
f (A) ⊆ ∪i = 1∞ Ji .
Using this to prove
m∗ (f (A)) ≤ 2c.
)
4. Problem 39 in P 53.

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