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Analysis: Assignment 1Due on Friday, October 12, 2012

Lin 11:00am

A digital copy of this document can be found at http://pages.uoregon.edu/raies

Dan Raies

Last edited October 22, 2012

http://pages.uoregon.edu/raies

Contents

Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Part (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Exercise 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Part (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Exercise 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Exercise 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercise 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Exercise 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Exercise 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Exercise 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Exercise 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Exercise 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Exercise 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Dan Raies Analysis: Assignment 1 (Friday, October 12, 2012) Exercise 1

Exercise 1

Find the σ-algebra on R that is generated by the collection of all one-point subsets of R.

Solution to Exercise 1:

Let Σ be the set of all subsets S ⊆ R such that either S is countable or Sc is countable. Note that clearly

Σ contains all the singletons of R as a set of the form x is countable.

(i) Since ∅ is countable, ∅ ∈ Σ.

(ii) If a set S ∈ Σ then either S = (Sc)c

is countable or Sc is countable so Sc ∈ Σ.

(iii) Let Snn∈N ⊆ Σ.

Case 1: If Sn is countable for each n ∈ N then⋃n∈N Sn is the countable union of countable sets

so⋃n∈N Sn ∈ Σ.

Case 2: If there exists a k ∈ N such that Sk is uncountable then Sck must be countable which

forces (⋃n∈N

Sn

)c=⋂n∈N

Scn = Sk ∩

⋂n∈N\k

Sn

(1)

to be countable since the intersection of anything with a countable set is still countable. Hence⋃n∈N Sn ∈ Σ.

In both cases we have that⋃n∈N Sn ∈ Σ, as desired.

It follows that Σ is a σ-algebra.

Let Σ′ be any σ-algebra of R which contains the singletons in R. Let S be any countable subset of

R. Then S ∈ Σ′ as S =⋃x∈S x and Σ′ is a σ-algebra. Additionally, since S ∈ Σ′ we must have that

Sc ∈ Σ′. It follows that Σ ⊆ Σ′ so that Σ is the smallest σ-algebra of R generated by the singletons, as

desired.

of 25


Exercise 2

Show that B(R) is generated by the collection of all compact subsets of R.


Let B1 be the σ-algebra generated by the compact subsets of R. Since B(R) contains all of the open sets

and is closed under complementation, B(R) must contain all of the closed sets and hence must contain

all of the compact sets so that B1 ⊆ B(R). Choose an open set B ⊆ R. Then R \B is closed. If R \B is

also bounded then it is compact so that B ∈ B1. Even if R\B is not bounded then it is equal to the the

countable union of closed intervals which are compact. Since B1 is closed under countable union R\B is

in B1 and since B1 is closed under complementation, B ∈ B1 from whence it follows that B(R) ⊆ B1.

Finally, B1 = B(R), as desired.

of 25


Exercise 3

Let X be a set and E be a collection of subsets of X. Prove that there exists a unique monotone class M(E)

such that E ⊆M(E) and if M1 is another monotone class containing E then M1 ⊇M(E).

Lemma 1. If M1 and M2 are both monotone classes M1 ∩M2 is also a monotone class.

Proof. Let Ann∈N ⊆M1 ∩M2 be a sequence of sets such that An ⊆ An+1 for all n ∈ N. Since Ann∈N ⊆M1 and M1 is a monotone class then

⋃n∈NAn ∈M1. Similarly, since Ann∈N ⊆M2 and M2 is a monotone

class then⋃n∈NAn ∈M2. It follows that

⋃n∈NAn ∈M1 ∩M2.

Let Ann∈N ⊆M1 ∩M2 be a sequence of sets such that An ⊇ An+1 for all n ∈ N. Since Ann∈N ⊆M1

and M1 is a monotone class then⋂n∈NAn ∈ M1. Similarly, since Ann∈N ⊇ M2 and M2 is a monotone

class then⋂n∈NAn ∈M2. It follows that

⋂n∈NAn ∈M1 ∩M2.

Based on the above, M1 ∩M2 is a monotone class.


Let X be a set and E be a collection of subsets of X. Let M (E) be the collection of all monotone

classes which contain E. The collection M (E) is nonempty since P(X), the power set of X, is clearly

a monotone class which contains E. Now define

M(E) =⋂

M∈M(E)

M, (2)

that is M(E) is the intersection of all monotone classes which contain E. By Lemma 1, M(E) must be a

monotone class. It is clearly unique and must be contained in all other monotone classes which contain

E. The result then follows.

of 25


Exercise 4

Prove that a σ-ring is monotone and a monotone class which is a ring is always a σ-ring.

Part (a)

Prove that a σ-ring is a monotone class.

Solution to Exercise 4(a):

Let R be a σ-ring.

(i) Let Enn∈N ⊆ R such that En ⊆ En+1. Since R is closed under all countable unions it follows

that⋃n∈NEn ∈ R.

(ii) Let Enn∈N ⊆ R such that En ⊇ En+1. Note that E1 ⊆ En for all n ∈ N and R is closed under

set subtraction so E1 \En for all n ∈ N. Since R is also closed under countable union we have that

E1 \⋂n∈N

En =⋃n∈N

(E1 \ En) ∈ R. (3)

Since E1 ∈ R and E1 \⋂n∈NEn ∈ R it follows that

⋂n∈N

En = E1 \

(E1 \

⋂n∈N

En

)∈ R (4)

as desired.

Finally it follows that R is a monotone class.

Part (b)

Prove that a monotone class which is a ring is always a σ-ring.

Solution to Exercise 4(b):

Let M be a monotone class and a ring and choose Ann∈N ⊆M . Now let Bn =⋃nk=1An for each n ∈ N.

Since M is a ring and is closed under finite unions it follows that Bnn∈N ⊆M . Since Bn+1 = Bn∪An+1

for all n ∈ N we have Bn ⊆ Bn+1 for all n ∈ N. Then since M is a monotone class,⋃n∈NBn ∈M . It is

clear, though, that⋃n∈NAn =

⋃n∈NBn so that

⋃n∈NAn ∈M and M is a σ-ring.

of 25


Exercise 5

Suppose that µ is a finite measure on X,A .

Part (a)

Show that if A and B belong to A then

µ(A ∪B) = µ(A) + µ(B)− µ(A ∩B) . (5)


Choose A,B ∈ A and write A ∪B = (A \B) t (A ∩B) t (B \A), which is a disjoint union. Then

µ(A) + µ(B) = µ((A \B) t (A ∩B)) + µ((B \A) t (A ∩B))

= µ(A \B) + µ(A ∪B) + µ(B \A) + µ(A ∩B)

= µ((A \B) t (A ∩B) t (B \A)) + µ(A ∩B)

= µ(A ∪B) + µ(A ∩B) .

(6)

After rearranging Equation 6 we have that µ(A ∪B) = µ(A) + µ(B)− µ(A ∩B), as desired.

Part (b)

Show that if A, B, and C belong to A then

µ(A ∪B ∪ C) = µ(A) + µ(B) + µ(C)− µ(A ∩B)

− µ(A ∩ C)− µ(B ∩ C) + µ(A ∩B ∩ C)(7)


Choose A,B,C ∈ B(R). Using part (a) we have

µ(A ∪B ∪ C) = µ(A ∪ (B ∪ C))

= µ(A) + µ(B ∪ C)− µ(A ∩ (B ∪ C))

= µ(A) + µ(B) + µ(C)− µ(B ∩ C)− µ((A ∩B) ∪ (A ∩ C))

= µ(A) + µ(B) + µ(C)− µ(B ∩ C)

− (µ(A ∩B) + µ(A ∩ C)− µ((A ∩B) ∩ (A ∩ C)))

= µ(A) + µ(B) + µ(C)− µ(A ∩B)

− µ(A ∩ C)− µ(B ∩ C) + µ(A ∩B ∩ C)

(8)

as desired.

Part (c)

Find and prove a corresponding formula for the measure of the union of n sets.

Solution to Exercise 5(c):

For k, n ∈ N with k ≤ n define

Ωnk = S ⊆ 1, 2, . . . , n | |S| = k . (9)

Exercise 5(c) continued on next page. . . of 25

Dan Raies Analysis: Assignment 1 (Friday, October 12, 2012) Exercise 5(c)

As an example, if n = 10 and k = 3, one element of Ωnk is 3, 4, 9. I claim that for any collection of sets

Amnm=1 ⊆ A ,

µ

(n⋃

m=1

Am

)=

n∑m=1

(−1)m+1

∑S∈Ωn

m

µ

(⋂k∈S

Ak

) . (10)

I wish to prove the above claim by induction. Part (a) proved the base case so assume that Equation 10

holds for any collection of n sets in A and let Amn+1m=1 ⊆ A . Then

µ

(n+1⋃m=1

Am

)= µ

((n⋃

m=1

Am

)∪An+1

)

= µ

(n⋃

m=1

Am

)+ µ(An+1)− µ

((n⋃

m=1

Am

)∩An+1

)

=

n∑m=1

(−1)m+1

∑S∈Ωn

m

µ

(⋂k∈S

Ak

)+ µ(An+1)− µ

(n⋃

m=1

(Am ∩An+1)

)

=

n∑m=1

(−1)m+1

∑S∈Ωn

m

µ

(⋂k∈S

Ak

)+ µ(An+1)

−n∑

m=1

(−1)m+1

∑S∈Ωn

m

µ

(⋂k∈S

(Ak ∩An+1)

)=

n+1∑m=1

(−1)m+1

∑S∈Ωn+1

m

µ

(⋂k∈S

Ak

)

(11)

where the last equality in Equation 11 holds because

n+1⋃m=1

Ωn+1m =

(n⋃

m=1

Ωnm

)∪ n+ 1 ∪

(n⋃

m=1

C ∪ n+ 1 | C ∈ Ωnm

). (12)

By Equation 11 we have that

µ

(n+1⋃m=1

Am

)=

n+1∑m=1

(−1)m+1

∑S∈Ωn+1

m

µ

(⋂k∈S

Ak

) (13)

and hence by induction, Equation 10 holds for every n ∈ N.

of 25


Exercise 6

Let A be the σ-algebra of all subsets of N and let µ be the counting measure on (N,A ). Give a decreasing

sequence Akk∈N of sets in A such that µ(⋂

k∈NAk)6= limk→∞ µ(Ak).


Let A = P(N) and let µ be the counting measure on (N,A ). Define the sequence Nkk∈N such that

Nk = n ∈ N | n ≥ k for all k ∈ N. (14)

Note that Nkk∈N is a decreasing sequence and µ(Nk) = ∞ for all k ∈ N so that limk→∞ µ(Nk) = ∞.

Additionally,⋂k∈NNk = ∅ so that µ

(⋂k∈NNk

)= 0. It follows that Nkk∈N is a decreasing sequence

such that

µ

(⋂k∈N

Nk

)6= limk→∞

µ(Nk) (15)

as desired.

of 25


Exercise 7

Let µ be a measure on measurable space (X,A ) and let Enn∈N ⊆ A .

Lemma 2. Let (X,A , µ) be a measure space and let Eii∈N ⊆ A .

(i) if Ei ⊆ Ei+1 for all i ∈ N then

limi→∞

µ(Ei) = µ

(⋃i∈N

Ei

)(16)

and

(ii) if Ei ⊇ Ei+1 for all i ∈ N and µ(Ei) <∞ for all i ∈ N then

limi→∞

µ(Ei) = µ

(⋂i∈N

Ei

). (17)

I need Lemma 2 for the solution below. I’m omitting the proof because it’s rather trivial but I wanted

to state the result explicitly since it was not mentioned in class.

Part (a)

Prove that if there is a k ∈ N such that µ(⋃∞n=k En) <∞ then

µ

⋂j∈N

∞⋃n=j

En

≥ lim supn→∞

µ(En) . (18)


It is clear that, without loss of generality, we can assume k = 1 as both sides of Equation 18 are limits

which ignore the first finite number of terms. For convenience I will define

lim supEn =⋂j∈N

∞⋃n=j

En

. (19)

Define Fj =⋃∞n=j En for all j ∈ N. It then follows that Fjj∈N is a monotone sequence which

decreases to lim supEn. Since µ(⋃

n∈NEn)<∞ it follows that µ(En) <∞ for each n ∈ N so that

limj→∞

µ(Fj) = µ(lim supEn) . (20)

Now choose j ∈ N. Since En ⊆ Fj for each n ≥ j we have that µ(En) ≤ µ(Fj) from which we have

that supn≥j µ(En) ≤ µ(Fj). It then follows that

lim supn→∞

µ(En) = limj→∞

(infn≥j

µ(En)

)≤ limj→∞

µ(Fj) . (21)

Finally

lim supn→∞

µ(En) ≤ µ(lim supEn) (22)

as desired.

Exercise 7 continued on next page. . . of 25

Dan Raies Analysis: Assignment 1 (Friday, October 12, 2012) Exercise 7(b)

Part (b)

Prove that

µ

⋃j∈N

∞⋂n=j

Ej

≤ lim infn→∞

µ(En) . (23)


For convenience I will define

lim inf En =⋃j∈N

∞⋂n=j

En

. (24)

Now define Fj =⋂∞n=j En for all j ∈ N. We then have that Fjj∈N is a monotone sequence which

increases to lim inf En. It then follows that

limj→∞

µ(Fj) = µ(lim inf En) . (25)

Now choose j ∈ N. Since Fj ⊆ En for all n ≥ j we have that µ(Fj) ≤ µ(En) from which we have

that infn≥j µ(En) ≥ µ(Fj). It then follows that

lim infn→∞

µ(En) = limj→∞

(supn≥j

µ(En)

)≥ limj→∞

µ(Fj) . (26)

Finally

lim infn→∞

µ(En) ≥ µ(lim inf En) (27)

as desired.

of 25


Exercise 8

Let (X,A , µ) be a measure space and define µ• : A → [0,∞) by

µ•(A) = sup µ(B) | B ⊆ A,B ∈ A , µ(B) <∞ . (28)

Lemma 3. Let A ∈ A such that µ(A) <∞. Then µ(A) = µ•(A).

Proof. Suppose A ∈ A is any set such that µ(A) <∞. Since µ is a measure µ(A′) ≤ µ(A) whenever A′ ⊆ A.

It is clear that A ⊆ A, A ∈ A , and µ(A) <∞ so that

µ•(A) = sup µ(B) | B ⊆ A,B ∈ A , µ(B) <∞ = µ(A) (29)

as desired.

Part (a)

Show that µ• is a measure on (X,A ).


First observe that µ•(∅) = 0 as the only set in A contained in ∅ is the empty set itself and µ(∅) = 0.

Now suppose that Aii∈N ⊆ A such that Ai ∩Aj whenever i 6= j and write A =⋃i∈NAi.

Case 1: Suppose that µ•(Ak) = ∞ for at least one k ∈ N. Since Ak ⊆ A we have that ∞ = µ•(Ak) ≤µ•(A) by definition and it is clear that ∞ = µ•(Ak) ≤

∑i∈N µ

•(Ai) so that µ•(⋃

i∈NAi)

=∑i∈N µ

•(Ai).

Case 2: Suppose that µ•(Ai) <∞ for all i ∈ N and that µ•(A) <∞. Then, by Lemma 3,

µ•

(⋃i∈N

Ai

)= µ

(⋃i∈N

Ai

)=∑i∈N

µ(Ai) =∑i∈N

µ•(Ai) (30)

so that µ•(⋃

i∈NAi)

=∑i∈N µ

•(Ai).

Case 3: Suppose that µ•(Ai) < ∞ for all i ∈ N and that µ•(A) = ∞. Let Bi =⋃ik=1Ak. Since Ai is

µ•-finite for every i ∈ N it follows that Bi must also be µ•-finite. Since Bi is a monotonic sequence

which increases to⋃i∈NAi we have that

µ•(A) = sup µ(B) | B ⊆ A,B ∈ A , µ(B) <∞= limi→∞

µ(Bi)

= limi→∞

i∑k=1

µ(Ak)

= limi→∞

i∑k=1

µ•(Ak)

=∑k∈N

µ•(Ak)

(31)

so that µ•(⋃

i∈NAi)

=∑i∈N µ

•(Ai).

In any case, µ•(⋃

i∈NAi)

=∑i∈N µ

•(Ai) so that µ• is a measure, as desired.


Dan Raies Analysis: Assignment 1 (Friday, October 12, 2012) Exercise 8(b)

Part (b)

Show that if µ is σ-finite then µ• = µ.


If µ is σ-finite then there exists a countable set Xii∈N such that Xi is µ-finite for each i ∈ N and

X =⋃i∈NXi. Then for any A ∈ A we have that A =

⋃i∈N (Xi ∩A) and µ

(⋃ik=1 (Xi ∩A)

)is finite for

all i ∈ N so that

µ•(A) = sup µ(B) | B ⊆ A,B ∈ A , µ(B) <∞

= limi→∞

µ

(i⋃

k=1

(Xi ∩A)

)

= µ

(⋃k∈N

(Xi ∩A)

)= µ(A) .

(32)

Hence µ•(A) = µ(A) as desired.

Part (c)

Find µ• if X is nonempty and µ is the measure defined such that µ(∅) = 0 and µ(A) =∞ if A ∈ A \ ∅.

Solution to Exercise 8(c):

If A ∈ A note that clearly ∅ ⊆ A. Also, since the only set in A with finite µ-measure is ∅ so the only

set B ⊆ A such that B ∈ A and µ(B) <∞ is B = ∅. Thus µ•(A) = µ(∅) = 0 for all sets A ∈ A .

of 25


Exercise 9

Let C be a countable subset of R. Using only the definition of λ∗, show that λ∗(C) = 0.


Let C be a countable subset of R and write C = cnn∈N. Fix ε > 0 and for n ∈ N let

In =(cn −

ε

2n+1, cn +

ε

2n+1

). (33)

Now, observe that

λ(C) = inf

∑n∈N

(bn − an)

∣∣∣∣∣ C ⊆ ⋃n∈N

(an, bn)

≤∑n∈N

λ(In) =∑n∈N

ε

2n= ε. (34)

By letting ε→ 0 we have that λ(C) = 0, as desired.

of 25


Exercise 10

Show that for each subset A of R there is a Borel subset B ∈ B(R) that includes A and satisfies λ(B) = λ∗(A).


Let A be a subset of R. If λ(A) =∞ then R ∈ B(R) and λ(R) = λ∗(A) so we can assume that λ(A) <∞.

By definition,

λ∗(A) = inf

∑k∈N

(bk − ak)

∣∣∣∣∣ A ⊆ ⋃k∈N

(ak, bk)

. (35)

Notice that a set of the form⋃k∈N (ak, bk) must be in B(R) and hence for each n ∈ N we can find a

set Bn ∈ B(R) such that A ⊆ Bn and λ∗(A) + 1n ≥ λ(Bn). Now let B =

⋂n∈NBn and notice that

B ∈ B(R).

Since A ⊆ B and B ∈ B(R) we have

λ∗(A) ≤ λ∗(B) = λ(B) . (36)

Additionally, by construction,

λ∗(A) +1

n≥ λ(Bn) ≥ λ(B) for all n ∈ N. (37)

Since Equation 37 holds for all n ∈ N it follows, by letting n→∞, that λ∗(A) ≥ λ(B).

Finally, λ∗(A) = λ(B) as desired.

of 25


Exercise 11

Lemma 4. If E ⊆ Rd is a Lebesgue measurable set and T : Rd → Rd is an invertible linear transformation

then T (E) is Lebesgue measurable and λ(T (E)) = |detT |λ(E).

I am not always sure what I am allowed to assume. However, I think it is fair to assume that an invertible

linear map T takes intervals of volume A in Rd to parallelpipeds of area |detT |A in Rd. This is a fairly easy

result to prove in an undergraduate calculus class. I will use this fact in the proof below of Lemma 4.

Proof. Let E ⊆ Rd be a Lebesgue measurable set and let T : Rd → Rd be an invertible linear transformation.

Since T is a homeomorphism (with the usual topology on Rd) it must map Borel sets to Borel sets and hence

must map Lebesgue measurable sets to Lebesgue measurable sets so that T (E) is measurable.

For ε > 0 choose intervals Ikk∈N in Rd such that E ⊆⋃k∈N Ik and

∑k∈N λ(Ik) < λ(E) + ε. Since T is

a bijection it follows that T (E) ⊆⋃k∈N T (Ik). Then since λ(T (Ik)) = |detT |λ(Ik) for every k ∈ N, we have

that

λ(T (E)) ≤∑k∈N

λ(T (Ik)) = |detT |∑k∈N

λ(Ik) < |detT | (λ(E) + ε) (38)

and letting ε→ 0 in Equation 38 yields

λ(T (E)) ≤ |detT |λ(E) . (39)

If δ > 0 then since E is measurable, choose an open set G such that E ⊆ G and λ(G \ E) < δ. Since

G is an open set G must also be an Fσ set (by Proposition 1.1.5) which means that we can write G as the

countable union of non-overlapping closed intervals1 in Rd so write G =⋃k∈N Jk where Jk is an interval

for each k ∈ N and λ(Jp ∩ Jq) = 0. Since T is a bijection we have that T (G) =⋃k∈N T (Jk) and since the

intervals Jk are non-overlapping it follows that

λ(T (G)) =∑k∈N

λ(T (Jk)) = |detT |∑k∈N

λ(Jk) = |detT |λ(G) . (40)

Since E ⊆ G it is clear that |detT |λ(E) ≤ |detT |λ(G) = λ(L(G)). Additionally,

λ(T (G)) ≤ λ(T (E)) + λ(T (G \ E)) ≤ λ(T (E)) + |detT | δ (41)

so that after substitution, |detT |λ(E) ≤ λ(T (E)) + |detT | δ. Letting δ → 0 yields

|detT |λ(E) ≤ λ(T (E)) . (42)

Combining Equation 39 and Equation 42 yields |detT |λ(E) = λ(T (E)), as desired.

1Note that non-overlapping is different than disjoint. Disjoint intervals intersect trivially whereas non-overlapping intervals

intersect to a set of measure zero.


Dan Raies Analysis: Assignment 1 (Friday, October 12, 2012) Exercise 11(a)

Part (a)

Prove that every straight line in R2 has Lebesgue measure zero.


For n ∈ Z consider the set Ln =

(n+ x, 0) ∈ R2∣∣ 0 ≤ x ≤ 1

. Let ε > 0 and write Iε =

(n− ε, n+ 1 + ε)× (−ε, ε). We have that Iε is an interval and Ln ⊆ Iε so that

λ(Ln) ≤ inf λ(Iε) | ε > 0 = limε→0

(2ε+ 4ε2

)= 0. (43)

Now consider the line L =

(x, 0) ∈ R2∣∣ x ∈ R

and observe that L =

⋃n∈Z Ln. Then

λ(L) = λ

(⋃n∈Z

Ln

)≤∑n∈Z

λ(Ln) = 0 (44)

so that L has measure zero.

Now, by Lemma 4, the image of L under any invertible linear transformation must also have measure

zero from which it follows that every line through the origin must have measure zero. We showed in class

that translation preserves measure so the translation of any line through the origin must have measure

zero. Finally, every line in R2 must have measure zero.

Part (b)

Prove that every circle in R2 has Lebesgue measure zero.

Lemma 5. Let E ⊆ R and let E =

(x1 − x2, y1 − y2) ∈ R2∣∣ (x1, y1) , (x2, y2) ∈ E

. Then λ(E) = 0 if and

only if E contains an interval of R2 around (0, 0).

In class we proved the analog of Lemma 5 in R and the proof of Lemma 5 mimics that proof exactly. The

proof of part (b) can be done similarly to that of part (a) but I think the proof which employs Lemma 5 is

more powerful and interesting.


Let C =

(x, y) ∈ R2∣∣ x2 + y2 = 1

be the unit circle in R2. Now let CI = (x, y) ∈ C | x, y ≥ 0

and write CI =

(x1 − x2, y1 − y2) ∈ R2∣∣ (x1, y1) , (x2, y2) ∈ CI

. It is clear that CI cannot contain an

interval around the origin because otherwise there would be a point in CI which is a distance different

from 1 away from the origin. It follows from Lemma 5 that λ(CI) = 0. Now let CII be the points in

C which are in the second quadrant, and define CIII and CIV similarly. It is clear that CII , CIII , and

CIV are the images of CI under linear transformations so that each set has measure zero by Lemma 4.

Then, since C = CI ∪ CII ∪ CIII ∪ CIV , we have

λ(C) = λ(CI ∪ CII ∪ CIII ∪ CIV ) ≤ λ(CI) + λ(CII) + λ(CIII) + λ(CIV ) = 0. (45)

Every circle in R2 centered at the origin is the image of C under a linear transformation and so must

have measure zero by Lemma 4. Since every circle in R2 is the translation of a circle centered at the

origin it follows that every circle in R2 has measure zero, as desired.

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Exercise 12

Let A be a subset of Rd. Show that the following are equivalent:

(i) A is Lebesgue measurable.

(ii) A is the union of an Fσ set and a set of Lebesgue measure zero.

(iii) There is an Fσ set B that satisfies λ∗(A∆B) = 0.


I proceed to show that (i)⇔ (ii) and (ii)⇔ (iii) which proves the desired result.

(i) ⇒ (ii): Assume that A is Lebesgue measurable. For every n ∈ N there exists an open set Bn such

that A ⊆ Bn and λ∗(A) + 1/n ≥ λ∗(Bn). It then follows that λ∗(A \Bn) ≤ 1/n for all n ∈ N.

Define B =⋂n∈NBn which is clearly an Fσ set. Now, for all n ∈ N,

0 ≤ λ∗(A \B) = λ∗

(A \

⋃n∈N

Bn

)≤ λ∗(A \Bn) ≤ 1/n (46)

so that, by letting n → ∞, we have λ∗(A \B) = 0. Finally we have that A = B ∪ (A \B) where

B is an Fσ set and A \B has Lebesgue measure zero.

(ii) ⇒ (i): Assume that A = F ∪ Z where F is an Fσ set and λ∗(Z) = 0. Then F and Z must both be

Lebesgue measurable so that A is Lebesgue measurable.

(ii) ⇒ (iii): Assume that A = F ∪ Z where F is an Fσ set and λ∗(Z) = 0. Let B = A \ Z and note

that B is an Fσ set. Then

λ∗(A∆B) = λ∗((A \B) ∪ (B \A)) = λ∗(Z ∪ ∅) = 0. (47)

We now have that B is an Fσ set such that λ∗(A∆B) = 0, as desired.

(iii) ⇒ (ii): Assume that there is an Fσ set B such that λ∗(A∆B) = 0. Then

0 = λ∗(A∆B) = λ∗((A \B) t (B \A)) = λ∗(A \B) + λ∗(B \A) . (48)

It follows that λ∗(A \B) = 0 = λ∗(B \A). Finally A = B ∪ (B \A) where B is an Fσ set and

B \A has Lebesgue measure zero.

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Exercise 13

Show that for each number α that satisfies 0 < α < 1 there is a closed subset D of [0, 1] such that λ(D) = α

and includes no non-empty open set.


Choose a fixed but arbitrary real number α ∈ (0, 1) and let δ = 1− α. Define a sequence Dk∞k=0 such

that Dk ⊆P(R). Let D0 = [0, 1] and for k > 0 define

Dk =

[a,a+ b

2− δ

2 · 3k

]∪[a+ b

2+

δ

2 · 3k, b

] ∣∣∣∣ [a, b] ∈ Dk−1

. (49)

Note that Dk is a collection of disjoint closed subintervals of [0, 1] for each k. Now define a sequence

Dk∞k=0 such that Dk = ∪Dk is the union of all the intervals in Dk. Then let D =⋂∞k=0Dk.

We start by showing that D has measure α. Since Dk is the union of disjoint closed intervals each

Dk is closed and since D is the countable intersection of closed sets, D is closed and hence must be

measurable. Since Dk ⊂ Dk−1 for each k ≥ 0 we have that λ(D) = limk→∞ λ(Dk), hence it suffices to

find λ(Dk) for each k ≥ 0. Observe that D1 is the union of two subintervals of [0, 1] and the number of

subintervals doubles each time k increases by 1 so that Dk is the union of 2k such subintervals. Choose

k ≥ 0 and consider E = Dk−1 \ Dk. Now Dk is attained by removing one interval from each of the

intervals in Dk−1 so E is the collection of 2k−1 closed intervals. By the construction we have that each

of the closed intervals which make up E have length δ3−k. It follows that λ(E) = δ2k−13−k and Dk is a

disjoint union of Dk−1 and E. Combining all of things we have

λ(D0) = 1 and

λ(Dk) = λ(Dk)− δ2k−12−k whenever k > 0.(50)

After expanding λ(Dk) to a sum and simplifying we have

λ(Dk) = 1− δ

3

k∑n=0

(2

3

)n

= 1− δ

3

1−

(2

3

)k+1

1−(

2

3

)

= 1− δ + δ

(2

3

)k+1

.

(51)

Finally,

λ(D) = limk→∞

λ(Dk) = limk→∞

(1− δ + δ

(2

3

)k+1)

= 1− δ = α (52)

and the measure of D is α.

We now wish to show that D contains no open intervals. Choose x ∈ D. Since D =⋂∞k=0Dk it

follows that x ∈ Dk for each k ≥ 0. Since Dk = ∪Dk there must exist an interval in Dk which contains x;

denote this interval Xk. That is, Xk ∈ Dk and x ∈ Xk. Note also that since Dk is the union of disjoint

intervals, Xk is the largest connected subset of Dk which contains x. It follows that X =⋂∞k=0Xk is the

largest connected subset of D which contains x. We have already shown that

λ(Xk) =1− δ + δ

(23

)k+1

2k(53)



for every k ≥ 0. Since Xk is a decreasing sequence of closed intervals and limk→∞ λ(Xk) = 0 it follows

by the Cantor Intersection Theorem that X contains a single point, namely x. Hence, for any given

x ∈ D, the largest connected component of D which contains x is the singleton x. Now suppose that

there exists an open subset B ⊆ D and choose any x ∈ B. Since B is open there must exist an open

interval I such that x ∈ I ⊆ B ⊆ D and hence the largest connected component of D which contains x

must contain I, which is a contradiction. Hence D cannot contain any open subsets.

Finally, D ⊆ [0, 1] with λ(D) = α and D includes no open sets. Since α ∈ (0, 1) was chosen arbitrarily

such a set exists for any α ∈ (0, 1).

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Exercise 14

Show that if B is a subset of R that satisfies λ∗(B) > 0 then B includes a set that is not Lebesgue measurable.


Choose any set B ⊆ R such that λ∗(B) > 0. By proposition 1.4.9 of Cohn’s text there exists a set A ⊆ Rsuch that every Lebesgue measurable set contained in either A or Ac has measure zero. It is clear that

B = (B ∩A) ∪ (B ∩Ac) and that B ∩ A ⊆ A and B ∩ Ac ⊆ Ac. Suppose that both B ∩ A and B ∩ Acare Lebesgue measurable. Then, by the assumption on A, both B ∩ A and B ∩ Ac must have measure

zero which then implies that B has measure zero which is a contradiction. Hence at least one of either

B ∩A ⊆ B or B ∩Ac ⊆ B must not be Lebesgue measurable and the assertion is proved.

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Exercise 15

Show that there is a Lebesgue measurable subset of R2 whose projection on R under the map (x, y) 7→ x is

not Lebesgue measurable.


Let F ⊆ R be any set that is not Lebesgue measurable; we showed in class that such a set must exist.

Now let F ′ = (x, x) | x ∈ F so that clearly the projection of F ′ to R in the first coordinate is F . In

exercise 11(a) we showed that the set L = (x, x) | x ∈ R is a Lebesgue measurable set with measure

zero. Since F ′ is a subset of L and L has measure zero by Exercise 11 we have that F ′ also has measure

zero. It follows that F ′ is a Lebesgue measurable subset of R2 whose projection onto R is not Lebesgue

measurable.

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Exercise 16

Let X be a set and R be a ring on X. Denote by S(R) the σ-ring generated by R and denote by M(R) the

monotone class generated by R. Prove that S(R) = M(R).

Lemma 6. Suppose that X is a set and R is a collection of subsets of R. If both

(i) X \A ∈ R whenever A ∈ R and

(ii) A ∩B ∈ R whenever A,B ∈ R

then R is a ring.

Proof. Choose E1, E2 ∈ R. Then E1∪E2 = X \ [(X \ E1) ∩ (X \ E2)] ∈ R and E1 \E2 = E1∩ (X \ E2) ∈ R.

It follows that R is a ring.


We have already shown in Referencesex4 that S(R) is a monotone class and since M(R) is the smallest

monotone class containing R it follows that M(R) ⊆ S(R). In order to show that S(R) ⊆ R(R) it suffices

to show that M(R) is a σ-ring since S(R) is the smallest σ-ring containing R.

I start by showing that M(R) is closed under complements. Write

N0 = A ∈M(R) | X \A ∈M(R) . (54)

Since R is a ring and R ⊆ M(R) it follows that R ⊆ N0. Let Ann∈N ⊆ N0 be an increasing sequence

and write A =⋃n∈NAn. Since N0 ⊆ M(R) and M(R) is a monotone class it follows that A ∈ M(R).

Since each An ∈ N0 we have X \An ∈M(R) by the definition of N0. Then

X \A =⋂n∈N

(X \An) (55)

and X \Ann∈N is a decreasing sequence in the monotone class M(R) so X \ A ∈ M(R). Then, since

A ∈M(R) and X \A ∈M(R) it follows that A ∈ N0 by the definition of N0 so that N0 is closed under

countable increasing unions. Let Ann∈N ⊆ N0 be a decreasing sequence and write A =⋂n∈NAn. Since

N0 ⊆ M(R) and M(R) is a monotone class it follows that A ∈ M(R). Since each An ∈ N0 we have

X \An ∈M(R) by the definition of N0. Then

X \A =⋃n∈N

(X \An) (56)

and X \Ann∈N is an increasing sequence in the monotone class M(R) so X \A ∈M(R). Then, since

A ∈ M(R) and X \ A ∈ M(R) it follows that A ∈ N0 by the definition of N0 so that N0 is closed

under countable decreasing intersections. Thus N0 is a monotone class containing R. Since M(R) is the

smallest monotone class containing R we must have that M(R) ⊆ N0 but N0 was defined to be a subset

of M(R) so that M(R) = N0. Hence M(R) is closed under complementation.

Next I wish to show that M(R) is closed under finite intersections. I first show that A ∩B ∈M(R)

whenever A ∈ R and B ∈M(R). Fix A ∈ R and define

N1 = B ∈M(R) | A ∩B ∈M(R) . (57)

Since R is a ring, R ⊆ N1. Let Bnn∈N ⊆ N1 be an increasing sequence and write B =⋃n∈NBn. Since

N1 ⊆M(R) and M(R) is a monotone ring, B ∈M(R). Since Bn ∈ N1 we have that A∩Bn ∈M(R) by

definition for all n ∈ N and

A ∩B =⋃n∈N

(A ∩Bn) (58)



where A ∩Bnn∈N ⊆ N1 is an increasing sequence in the monotone class M(R) so that A∩B ∈M(R).

Because B ∈M(R) and A∩B ∈M(R) we have that B ∈ N1 by definition and hence N1 is closed under

countable increasing unions. Let Bnn∈N ⊆ N1 be a decreasing sequence and write B =⋂n∈NBn. Since

N1 ⊆M(R) and M(R) is a monotone ring, B ∈M(R). Since Bn ∈ N1 we have that A∩Bn ∈M(R) by

definition for all n ∈ N and

A ∩B =⋂n∈N

(A ∪Bn) (59)

where A ∩Bnn∈N ⊆ N1 is an increasing sequence in the monotone class M(R) so that A∩B ∈M(R).

Because B ∈M(R) and A∩B ∈M(R) we have that B ∈ N1 by definition and hence N1 is closed under

countable decreasing intersections. Thus N1 is a monotone class which contains M(R) from which it

follows that N1 = M(R). Since the choice of A was arbitrary we have that A ∩ B ∈ M(R) whenever

A ∈ R and B ∈M(R).

Now choose C ∈ M(R) and let N2 = D ∈M(R) | C ∩D ∈M(R). We know that D ∩ C ∈ M(R)

whenever D ∈ R so that R ⊆ N2. By an argument identical to the one used in the previous paragraph we

have that N2 is a monotone class so that N2 = M(R) and hence M(R) is closed under finite intersection.

Since M(R) is closed under finite intersection and complementation, M(R) is a ring by Lemma 6.

We have already shown that a monotone class which is also a ring is a σ-ring so that M(R) is a σ-ring.

As previously discussed, the result follows.

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Exercise 17

Let (X,A , µ) be a measurable space. Suppose that there is a sequence Enn∈N such that X =⋃n∈NEn.

Define

µ∗(E) = inf µ(F ) | E ⊆ F ∈ A (60)

for all E ⊆ X. Show that µ∗ is an outer measure.


Let (X,A , µ) be a measurable space such that there exists a sequence Enn∈N such that X =⋃n∈NEn

and define µ∗ : P(R)→ [0,∞) as in Equation 60. There are three conditions which must be satisfied in

order to verify that µ∗ is an outer measure.

(i) I first wish to ensure that µ∗(∅) = 0. We have that

µ∗(∅) = inf µ(F ) | ∅ ⊆ F ∈ A (61)

but since ∅ ∈ A , µ(∅) = 0, and µ(F ) ≥ 0 for all F ∈ A it follows that µ∗(∅) = 0 as desired.

(ii) Now I wish to show that µ∗(A) ≤ µ∗(B) whenever A ⊆ B ⊆ X. Let SA be the set of all sets in A

which contain A and let SB be the set of all sets in A which contain B. It then follows that

µ∗(A) = infF∈SA

µ(F ) and

µ∗(B) = infF∈SB

µ(F ) .(62)

Since SA ⊆ SB as A ⊆ B we then have that

µ∗(A) = infF∈SA

µ(F ) ≤ infF∈SB

µ(F ) = µ∗(B) (63)

as desired.

(iii) Finally I wish to verify that µ∗(⋃

n∈NAn)≤∑n∈N µ

∗(An) whenever Ann∈N ⊆ P(X). To this

end, fix Ann∈N ⊆ P(X) and let ε > 0. For each n ∈ N there exists a set Fn ∈ A such that

An ⊆ Fn and µ(Fn) ≤ µ∗(An) + ε2n . Observe that

⋃kn=1An ⊆

⋃kn=1 Fn for all k ∈ N from whence

it follows

µ∗

(k⋃

n=1

An

)≤ µ

(k⋃

n=1

Fn

)≤

k∑n=1

µ(Fn) ≤k∑

n=1

(µ∗(An) +

ε

2n

)= ε+

k∑n=1

µ∗(An) . (64)

Letting n→∞ in Equation 64 yields

µ∗

(⋃n∈N

An

)≤ ε+

∑n∈N

µ∗(An) (65)

and since Equation 65 holds for every ε > 0 we have that µ∗(⋃

n∈NAn)≤∑n∈N µ

∗(An), as desired.

It finally follows that µ∗ is an outer measure, as desired.

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