Transcript
Page 1: Answers for Problem Set #2

Problem Set

Problem Set #2

Math 5322, Fall 2001

December 3, 2001

ANSWERS

Page 2: Answers for Problem Set #2

i

Page 3: Answers for Problem Set #2

Problem 1. [Problem 18, page 32]Let A ⊆ P(X) be an algebra, Aσ the collection of countable unions of sets

in A, and Aσδ the collection of countable intersections of sets in Aσ. Let µ0 bea premeasure on A and µ∗ the induced outer measure.

a. For any E ⊆ X and ε > 0 the exists A ∈ Aσ with E ⊆ A and µ∗(A) ≤µ∗(E) + ε.

Answer :

By definition,

µ∗(E) = inf

{ ∞∑i=1

µ0(Ai) | {Ai }∞i=1 ⊆ A, E ⊆∞⋃i=1

Ai

}.

Thus, given ε > 0, we can find some sequence {Ai }∞i=1 ⊆ A so that

E ⊆∞⋃i=1

Ai and∞∑i=1

µ0(Ai) < µ∗(E) + ε.

Define a set A ⊇ E by

A =∞⋃i=1

Ai.

Since A is a countable union of sets in A, A ∈ Aσ. Since µ∗ is subadditive,

µ∗(A) ≤∞∑i=1

µ∗(Ai) =∞∑i=1

µ0(Ai) < µ∗(E) + ε,

since µ∗ = µ0 on A. This completes the proof.

b. If µ∗(E) <∞ then E is µ∗-measurable iff there exists B ∈ Aσδ with E ⊆ Band µ∗(B \ E) = 0.

Answer :

Since the σ-algebraM of µ∗-measurable sets contains A and is closed undercountable unions and intersections, Aσδ ⊆M.

For the first part of the proof, assume that E ∈ M and µ∗(E) < ∞. Foreach n ∈ N, we can apply the previous part of the problem to find a setAn ∈ Aσ such that E ⊆ An and µ∗(An) < µ∗(E) + 1/n. Define B by

B =∞⋂n=1

An.

1

Page 4: Answers for Problem Set #2

Then E ⊆ B and B ∈ Aσδ, since it is a countable intersection of sets inAσ. For every n, we have

µ∗(E) ≤ µ∗(B) ≤ µ∗(An) < µ∗(E) + 1/n.

Since n is arbitrary, we must have µ∗(B) = µ∗(E).

Since E ⊆ B, B is the disjoint union of E and B \ E. These three sets arein M, so we can use the additivity of µ∗ on M to get

µ∗(B) = µ∗(E) + µ∗(B \ E).

All three terms are finite and µ∗(E) = µ∗(B), so simple algebra showsµ∗(B \ E) = 0.

For the second part of the proof, suppose that E ⊆ X and there is a setB ∈ Aσδ so that E ⊆ B and µ∗(B \E) = 0. As remarked above, Aσδ ⊆Mso B is measurable. We showed in class that if F ⊆ X and µ∗(F ) = 0, thenF is µ∗-measurable. Thus, in the present situation, B \ E is measurable.But E = B\(B\E) (check!). SinceM is closed under taking set differences,we conclude that E is measurable. This completes the proof.

c. If µ0 is σ-finite, the restriction µ∗(E) <∞ in (b) is superfluous.

Answer :

First suppose that E is µ∗-measurable and µ∗(E) =∞. Since µ0 is σ-finitewe can find disjoint sets Fj ∈ A, j ∈ N with µ0(Fj) <∞ and X =

⋃∞j=1 Fj .

Set Ej = E∩Fj . Then the sets Ej are disjoint, µ∗(Ej) ≤ µ∗(Fj) = µ0(Fj) <∞ and E =

⋃j Ej .

As a first try, we might invoke the previous part of the problem to find setsBj ⊇ Ej with Bj ∈ Aσδ and µ∗(Bj \ Ej) = 0. The next step would be totry to define our desired set B as

⋃j Bj . Unfortunately, this won’t work.

Since Aσδ is defined to be the collection of countable intersections of setsin Aσ, there’s no obvious reason why Aσδ should be closed under takingcountable unions. Thus, we don’t know that

⋃j Bj will be in Aσδ.

Note, however that Aσ is closed under countable unions. If {Ai }∞i=1 ⊆ Aσ,then for each i there is a sequence

{Aki}∞k=1⊆ A such that

Ai =∞⋃k=1

Aki .

But then if A =⋃iAi, we have

A =∞⋃i=1

[ ∞⋃k=1

Aki

]=

⋃(i,k)∈N×N

Aki ,

a countable union of elements of A (check!). Thus, A ∈ Aσ.

2

Page 5: Answers for Problem Set #2

So, we proceed as follows. For each n ∈ N, we invoke part (a) to get a setAnj ∈ Aσ such that Ej ⊆ Anj and

µ∗(Anj ) < µ∗(Ej) +1n2j

.

Since Anj and Ej are measurable and have finite measure, this gives us

µ∗(Anj \ Ej) <1n2j

.

We then define

An =∞⋃j=1

Anj .

As remarked above, An ∈ Aσ. Next, we claim that

An \ E ⊆∞⋃j=1

(Anj \ Ej).(∗)

To see this, suppose that x ∈ An \ E. Since x ∈ An, there is some j suchthat x ∈ Anj . Since x /∈ E and Ej ⊆ E, x /∈ Ej . Thus x ∈ Anj \ Ej . Thiscompletes the proof of the claim.

From (∗), we have

µ∗(An \ E) ≤∞∑j=1

µ∗(Anj \ Ej) <∞∑j=1

1n2j

=1n.

Now define

B =∞⋂n=1

An.

Then, B ∈ Aσδ and E ⊆ B. For every n we have

µ∗(B \ E) ≤ µ∗(An \ E) <1n,

and so µ∗(B \ E) = 0, which completes the first part of the proof.

The proof of the converse implication is the same as in part (b).

Problem 2. [Problem 25, page 39]Complete the proof of Theorem 1.19.Thus, we want to prove that the following conditions on a set E ⊆ R are

equivalent

3

Page 6: Answers for Problem Set #2

a. E ∈Mµ

b. E = V \N1, where V is a Gδ set and µ(N1) = 0.

c. E = H ∪N2, where H is an Fσ set and µ(N2) = 0.

Here µ is a Lebesgue-Stieltjes measure on R and Mµ is its domain (the µ∗-measurable sets, where µ∗ is the outer measure used in the construction of µ).

Answer :The proof of the implication (a) =⇒ (b) is pretty similar to what we did inthe last problem. In general, a countable union of Gδ sets is not a Gδ set. But,we can start with Proposition 1.18.

So, suppose that E ∈ Mµ. Let {Fj } be a sequence of disjoint measurablesets such that R =

⋃j Fj and µ(Fj) <∞. We can take the Fj ’s to be bounded

intervals, for example. Let Ej = E∩Fj , so the Ej ’s are disjoint measurable setswhose union is E and each Ej has finite measure.

By Proposition 1.18, for each n ∈ N and j we can find an open set Unj suchthat Ej ⊆ Unj and

µ(Unj ) < µ(Ej) +1n2j

.

Since Unj = Ej ∪ (Unj \ Ej) (disjoint union), we have

µ(Unj ) = µ(Ej) + µ(Unj \ Ej).

All the terms in this equation are finite, so we can rearrange the equation to get

µ(Unj \ Ej) = µ(Unj )− µ(Ej) <1n2j

.

Define a set Un by Un =⋃j U

nj . Then E ⊆ Un and Un is open, since any

union of open sets is open. We have

Un \ E ⊆∞⋃j=1

(Unj \ Ej)

(check), and so

µ(Un \ E) ≤∞∑j=1

µ(Unj \ Ej) <∞∑j=1

1n2j

=1n.

Now we define

V =∞⋂n=1

Un,

4

Page 7: Answers for Problem Set #2

so V is a Gδ set and V ⊇ E. For any n, we have

µ(V \ E) ≤ µ(Un \ E) <1n,

so we must have µ(V \ E) = 0. We have

E = V \ (V \ E) = V \N1

where V is a Gδ set and N1 = V \E is a nullset. This completes the proof that(a) =⇒ (b).

We next prove that (a) =⇒ (c). We’ll give a direct proof, which is easierthan the proof of (a) =⇒ (b). So suppose that E ∈ Mµ, let {Fj } be apartition of R into sets of finite measure and let Ej = E ∩ Fj .

Fix j for the moment. For each n ∈ N we can apply Proposition 1.18 to geta compact set Kn ⊆ Ej such that

µ(Ej)−1n< µ(Kn)

and so

µ(Ej \Kn) <1n.

Set Hj =⋃nKn. Since each Kn is closed, Hj is an Fσ set and Hj ⊆ Ej . For

each n,

Ej \Hj ⊆ Ej \Kn

so

µ(Ej \Hj) ≤ µ(Ej \Kn) <1n.

Thus, we have µ(Ej \ Hj) = 0, where Hj ⊆ Ej is an Fσ set. Thus, we haveEj = Hj ∪Nj where Nj = Ej \Hj is a nullset.

Taking the union over j, we have

E =( ∞⋃j=1

Hj

)∪( ∞⋃j=1

Nj

).

Since Fσ is closed under countable unions (by an argument similar to the lastproblem), the first set on the right of this equation is an Fσ set. Of course acountable union of nullsets is a nullset, so the second set on the right of the lastequation is a nullset. This completes the proof of (a) =⇒ (c).

Next, we prove that (b) =⇒ (a). This is pretty easy. Suppose thatE = V \ N1 where V is a Gδ set and N1 is a nullset. Of course a nullset ismeasurable. We know that Mµ contains the Borel sets, and hence the Gδ sets.Since the σ-algebra Mµ is closed under complements and finite intersections,we conclude that E ∈Mµ.

5

Page 8: Answers for Problem Set #2

The proof that (c) =⇒ (a) is similar. Suppose E = H ∪N2 where H is anFσ set and N2 is a nullset. Then N2 ∈ Mµ and H ∈ Mµ since Mµ containsthe Borel sets and hence the Fσ sets. Since Mµ is closed under finite unions,we conclude that E ∈Mµ.

This completes the proof.Several people observed that once you know (a) =⇒ (c), you can use

this and de Morgan’s laws to prove (a) =⇒ (b) (or vice-versa). Perhaps thesimplest proof of our Proposition would be to prove (a) =⇒ (c) as above, thenuse de Morgan’s laws to get (a) =⇒ (b), and then do the reverse implicationsas above.

Problem 3. [Problem 26, page 39]Prove Proposition 1.20 (Use Theorem 1.18.)Thus, we want to prove that if E ∈ Mµ and µ(E) < ∞, then for every

ε > 0 there is a set A that is a finite [disjoint] union of open intervals such thatµ(E 4A) < ε

Here µ is a Lebesgue-Stieltjes measure on R. The symbol 4 denotes thesymmetric difference of the sets, i.e., E 4A = (E \A) ∪ (A \ E).

Answer :Suppose E ∈Mµ with µ(E) <∞ and let ε > 0 be given.

By Theorem 1.18 we can find an open set U ⊆ R such that E ⊆ U andµ(U) < µ(E)+ε. Since U and E have finite measure µ(U\E) = µ(U)−µ(E) < ε.

Since U is an open subset of R, U can be written as a countable disjointunion of open intervals. Suppose first that the number of intervals is infinite,say

U =∞⋃k=1

Ik,

where each Ik is an open interval. We have

∞∑k=1

µ(Ik) = µ(U) <∞.

Since this series converges, there is some n such that

∞∑k=n+1

µ(Ik) < ε.(∗)

We define A by

A =n⋃k=1

Ik.

6

Page 9: Answers for Problem Set #2

If there are only finitely many intervals in U , we can label them as I1, I2, . . . , Inand let Ik = ∅ for k ≥ n+ 1. Then we define A = U and (∗) still holds.

We now have E \A ⊆ U \A so

µ(E \A) ≤ µ(U \A)= µ(U)− µ(A)

=∞∑k=1

µ(Ik)−n∑k=1

µ(Ik)

=∞∑

k=n+1

µ(Ik) < ε.

On the other hand, A \ E ⊆ U \ E, so

µ(A \ E) ≤ µ(U \ E) < ε.

Thus,

µ(E 4A) ≤ µ(E \A) + µ(A \ E) < 2ε.

Since ε > 0 was arbitrary, the proof is complete.

Problem 4. [Problem 28, page 39]Let F be increasing and right continuous, and let µF be the associated

measure. Then µF ({ a }) = F (a) − F (a−), µF ([a, b)) = F (b−) − F (a−),µF ([a, b]) = F (b)− F (a−) and µF ((a, b)) = F (b−)− F (a).

Answer :Recall that µF is constructed so that on the algebra of h-intervals it takes thevalues

µF ((a, b]) = F (b)− F (a),

and (hence) that µF is finite on bounded subsets of R.To calculate µF ({ a }), note that

{ a } =∞⋂n=1

(a− 1/n, a].(A)

Certainly a is in the right-hand side and any number bigger that a is not. Ifx < a then x < a − 1/n for sufficiently large n, so x is not in the intersectionon the right-hand side of (A).

The intervals (a−1/n, a] form a decreasing sequence of sets, so by continuityfrom above (Theorem 1.8d) we have

µF ({ a }) = limn→∞

µF ((a− 1/n, a]) = limn→∞

[F (a)− F (a− 1/n)].

7

Page 10: Answers for Problem Set #2

Since F is increasing,

limn→∞

F (a− 1/n) = limx→a−

F (x) = sup {F (x) | x < a } = F (a−).

Thus, we have

µF ({ a }) = F (a)− F (a−).(B)

Next, consider µF ([a, b]). We have [a, b] = { a } ∪ (a, b] (disjoint union), so

µF ([a, b]) = µF ({ a }) + µF ((a, b])= F (a)− F (a−) + F (b)− F (a)= F (b)− F (a−)

so

µF ([a, b]) = F (b)− F (a−)(C)

Next, consider µF ((a, b)), where we have to allow a and/or b to be infinity.If b is infinity, then (a,∞) is an h-interval and the definition of µF gives

µF ((a,∞)) = F (∞)− F (a).

The proposed formula

µF ((a, b)) = F (b−)− F (a)(D)

is correct because F (∞) = limx→∞ F (x). If a = −∞ and b = ∞ then (a, b) =(−∞,∞) is an h-interval and the definition of µF gives

µF ((−∞,∞)) = F (∞)− F (−∞),

which fits into formula (D) again.If both a and b are finite, we have (a, b) = (a, b] \ { b }, so

µF ((a, b)) = µF ((a, b])− µF ({ b })= F (b)− F (a)− [F (b)− F (b−)]= F (b−)− F (a),

so (D) holds.Finally, consider intervals of the form [a, b), where b might be infinity. We

can write [a, b) = { a } ∪ (a, b) (disjoint union) so

µF ([a, b)) = µF ({ a }) + µF (a, b)= F (a)− F (a−) + F (b−)− F (a)= F (b−)− F (a−),

Note that this is valid if F (b−) = F (∞) =∞.

Problem 5. [Problem 29, page 39]Let E be a Lebesgue measurable set.

8

Page 11: Answers for Problem Set #2

a. If E ⊆ N , where N is the nonmeasurable set described in Section 1.1, thenm(E) = 0.

Answer :

Recall the construction in Section 1.1. We define an equivalence relation ∼on [0, 1) by x ∼ y if x − y ∈ Q. We let N be a set that contains exactlyone element from each equivalence class (using the Axiom of Choice). SetR = Q ∩ [0, 1). For any set S ⊆ [0, 1) and r in R we define

Sr = {x+ r | x ∈ S ∩ [0, 1− r) } ∪ {x+ r − 1 | x ∈ S ∩ [1− r, 1) } .

We can then argue that if S is measurable, m(Sr) = m(S), using thetranslation invariance of Lebesgue measure. We also argued that

[0, 1) =⋃r∈R

Nr, (disjoint union).

If N was measurable, we would have

1 = m([0, 1)) =∞∑n=1

m(N),

which is impossible since the right-hand side can only be 0 or ∞.

For the first part of the present problem, we want to show that if E ⊆ Nis Lebesgue measurable, then m(E) = 0. Well, for each r ∈ R we haveEr ⊆ Nr, as defined above. Since the sets Nr are pairwise disjoint,

F =⋃r∈R

Er

is a countable disjoint union of measurable sets contained in [0, 1). Thus,we have

1 ≥ m(F ) =∑r∈R

m(Er) =∞∑i=1

m(E).

If m(E) > 0, the sum on the right would be∞, so we must have m(E) = 0.

b. If m(E) > 0, then E contains a non-measurable set. (It suffices to assumethat E ⊆ [0, 1). In the notation of Section 1.1, E =

⋃r∈RE ∩Nr.)

Answer :

Briefly, if E ⊆ R and m(E) > 0, we can write

E =∞⋃

n=−∞E ∩ [n, n+ 1), (disjoint union).

9

Page 12: Answers for Problem Set #2

Since m(E) > 0, at least one of the sets E ∩ [n, n + 1) must have nonzeromeasure. Select one such set F = E ∩ [n, n + 1). If F contains a nonmea-surable set, so does E. The set F −n is contained in [0, 1) and has the samemeasure as F . If we prove that every subset of [0, 1) of nonzero measurecontains a nonmeasurable set, then F − n will contain a non-measurableset A. But then A+ n ⊆ F is nonmeasurable (if A+ n was measurable, sowould be (A+ n)− n = A).

Thus, it will suffice to consider the case where E ⊆ [0, 1) and m(E) > 0.

First, observe that the process we’ve defined above that sends S to Sr isinvertible. Indeed, S0 = S and for t ∈ (0, 1) we define ϕt : [0, 1)→ [0, 1) by

ϕt(x) =

{x+ t, x ∈ [0, 1− t)x+ t− 1, x ∈ [1− t, 1).

Then St = ϕt(S). It is easy to check that ϕt([0, 1− t)) = [t, 1) and ϕt([1−t, 1)) = [0, t). It is then easy to check that the inverse of ϕt is ϕ1−t. Thus

(St)1−t = S.

From the first part of the problem, we can conclude that if F is measurableand F ⊆ Nr for some r then m(F ) = 0. To see this, note that if F ⊆ Nrthen F1−r ⊆ N . As we’ve argued before, F1−r is measurable andm(F1−r) =m(F ). From the first part of the problem, m(F1−r) = 0, so we concludethat m(F ) = 0.

Now suppose that E ⊆ [0, 1) and m(E) > 0. The interval [0, 1) is thedisjoint union of the sets Nr for r ∈ R, so E is the disjoint union of the setsE ∩Nr. If all of the sets E ∩Nr were measurable, we would have

m(E) =∑r∈R

m(E ∩Nr) =∑r∈R

0 = 0,

which contradicts the assumption that m(E) > 0. Thus at least one of thesets E ∩ Nr must be nonmeasurable and we conclude that E contains anonmeasurable set.

Before going into the problems on Page 48ff. of the book, lets recall some ofthe definitions involved.

Let (X,M) be a measurable space, soM⊆ P(X) is a σ-algebra. If A ⊆ X,we can define a σ-algebra M|A ⊆ P(A) by

M|A = {E ∩A | E ∈M} .(0.1)

If A itself is in M, we have

M|A = {E | E ⊆ A, E ∈M} .

10

Page 13: Answers for Problem Set #2

If (Y,N ) is a measurable space and f : X → Y , we say that f is measurableon A ⊆ X if f |A : A → Y (the restriction of f to A) is measurable for theσ-algebra M|A, which means that (f |A)−1(N) = f−1(N) ∩ A ∈ M|A for allN ∈ N . If A ∈ M, then f is measurable on A iff f−1(N) ∩ A ∈ M for allN ∈ N . Clearly, if f is a measurable function X → Y , then f is measurable onevery subset of A of X since, in this case, f−1(N)∩A ∈M|A for all N , becausef−1(N) ∈M.

The following proposition is pretty easy, but useful.

Proposition 0.1. Let (X,M) and (Y,N ) be measurable spaces and let f : X →Y be a function. Let {Eα }α∈A be a countable (finite or infinite) collection ofmeasurable subsets of X such that

X =⋃α∈A

Eα.

Then, f is measurable if and only if f is measurable on each Eα, i.e., f ismeasurable if and only if for each N ∈ N , f−1(N) ∩ Eα ∈M for all α ∈ A.

Proof. If f is measurable, then f−1(N) ∈M for allN ∈ N . Since each Eα ∈M,we have f−1(N) ∩ Eα ∈M.

Conversely, suppose that for each N ∈ N , f−1(N) ∩ Eα is measurable forall α. Then

f−1(N) =⋃α∈A

f−1(N) ∩ Eα

is a countable union of sets inM, so f−1(N) ∈M. Thus, f is measurable.

Problem 6. [Problem 1, page 48]Let (X,M) be a measurable space. Let f : X → R and Y = f−1(R). Then

f is measurable iff f−1({−∞}) ∈M, f−1({∞}) ∈M and f is measurable onY .

Answer :Suppose first that f is measurable. Then f is measurable on Y , as discussedabove. The sets {−∞} and {∞} are closed sets, hence Borel sets, in R, sof−1({−∞}) and f−1({∞}) are measurable.

For the second part of proof suppose f−1({−∞}) and f−1({∞}) are mea-surable and f is measurable on Y . The sets f−1({−∞}), f−1({∞}) andY form a partition of X. Since f−1({−∞}) and f−1({∞}) are measurable,Y = X \ (f−1({−∞}) ∪ f−1({∞})) is measurable. To show f is measurableit will suffice to show it is measurable on the three sets in our partition. Ofcourse, f is measurable on Y by assumption. On the set f−1({−∞}), f isconstant and a constant function is always measurable. Similarly, f is constant,and hence measurable, on f−1({∞}). This completes the proof.

11

Page 14: Answers for Problem Set #2

Problem 7. [Problem 2, page 48]Let (X,M) be a measurable space. Suppose that f, g : X → R are measur-

able.

a. fg is measurable (where 0 · (±∞) = 0).

Answer :

We’ll consider two solutions, in slightly different spirits.

First Solution. Since f is measurable, we can partition X into the mea-surable sets F1 = f−1(−∞), F2 = f−1((−∞, 0)), F3 = f−1(0), F4 =f−1((0,∞)) and F5 = f−1(∞). (f−1(a) is the same thing as f−1({ a }).) Wehave a similar partition G1 = g−1(−∞), G2 = g−1((−∞, 0)), G3 = g−1(0),G4 = g−1((0,∞)), G5 = g−1(∞) for g. Taking pairwise intersections, we geta partition of X into the 25 measurable sets (don’t panic!) Fi∩Gj . To showfg is measurable it will suffice to prove that fg is measurable on each of thesets Fi ∩Gj .Observe that fg is constant on the set F1 ∩ G1 = f−1(−∞) ∩ g−1(−∞).Indeed fg is constant on all of the sets F1 ∩ Gj , j = 1, . . . , 5. Similarly, fgis constant on F5 ∩ Gj , j = 1, . . . , 5 and Fi ∩ G1, i = 1, . . . 5 and Fi ∩ G5,i = 1, . . . , 5.

This leaves the nine sets Fi ∩Gj , i, j = 2, 3, 4 to consider. But the union ofthese sets is S = f−1(R) ∩ g−1(R). Of course, f and g are real valued on Sand so fg is measurable on S by Proposition 2.6 (p. 45). This completes thesolution.

Second Solution We try to follow the proof of Proposition 2.6 on page 45of the book. Thus, we define F : X → R × R by F (x) = (f(x), g(x)). Asdiscussed in the book, this is measurable.

Next we define ψ : R × R → R by ψ(x, y) = xy (with the 0 · (±∞) = 0convention). Note that ψ is not continuous at the four points (0,±∞),(±∞, 0).

Nonetheless, we claim that ψ is Borel measurable. To see this, define A ⊆R× R by

A = { (0,−∞), (0,∞), (−∞, 0), (∞, 0) } .

This is a closed subset of R×R, hence a Borel subset. Thus, B = R×R \Ais Borel.

Recall from our discussion in class that if Z is a metric space, and W is aBorel subset of Z,

(BZ)|W = {E ⊆W | E ∈ BZ } = BW .

Thus, the principal of the Proposition we proved above applies: If Z is theunion of two Borel sets V and W then f : Z → R is Borel measurable if andonly if f is Borel measurable on V and W .

12

Page 15: Answers for Problem Set #2

Thus, in order to show that ψ is Borel measurable, it will suffice to show itis Borel measurable on A and B. On A, ψ is constant (with value 0) andso Borel measurable. On B, ψ is continuous, and hence Borel measurable.This proves the claim that ψ is Borel measurable.

To complete the solution, let h = fg. Then h = ψ ◦ F . Let B ⊆ R be aBorel set. Since ψ is Borel measurable, ψ−1(B) is Borel in R × R. Since Fis measurable, F−1(ψ−1(B)) ∈M. Thus,

h−1(B) = (ψ ◦ F )−1(B) = F−1(ψ−1(B)) ∈M,

which shows that h is measurable.

b. Fix a ∈ R and define h(x) = a if f(x) = −g(x) = ±∞ and h(x) = f(x)+g(x)otherwise. Then h is measurable.

Answer :

First Solution. We can partition X into the measurable sets

f−1(−∞) ∩ g−1(−∞)(A.1)

f−1(−∞) ∩ g−1(R)(A.2)

f−1(−∞) ∩ g−1(∞)(A.3)

f−1(R) ∩ g−1(−∞)(A.4)

f−1(R) ∩ g−1(R)(A.5)

f−1(R) ∩ g−1(∞)(A.6)

f−1(∞) ∩ g−1(−∞)(A.7)

f−1(∞) ∩ g−1(R)(A.8)

f−1(∞) ∩ g−1(∞)(A.9)

and it will suffice to prove that h is measurable on each of these sets. Onthe sets (A.3) and (A.7), h is constant, with value a, by our definition. Onthe sets (A.1), (A.2), (A.4), (A.6), (A.8), and (A.9), h is constant (withvalue either ±∞). Finally, on the set f−1(R) ∩ g−1(R) in (A.5), both fand g are real valued and h coincides with f + g, which is measurable byProposition 2.6. This completes the solution.

Second Solution. As in the first part of the problem, define F : X → R×Rby F (x) = (f(x), g(x)). As discussed in the book, this is measurable. LetA ⊆ R be defined by

A = { (−∞,∞), (∞,−∞) } ,

13

Page 16: Answers for Problem Set #2

which is closed, and hence Borel, in R×R. Let B = R×R \A, which is alsoBorel. Define ψ : R× R→ R by

ψ(x, y) =

{a, (x, y) ∈ Ax+ y, (x, y) ∈ B.

This function is not continuous at the points in A, but it is nonetheless Borelmeasurable. Of course, ψ is continuous, hence Borel measurable, on B, andψ is constant, hence Borel measurable, on A. Thus, ψ is Borel measurableon R× R.

Since ψ is Borel measurable, ψ◦F is measurable (as discussed in the solutionof the first part of the problem), but h = ψ ◦ F , so the proof is complete.

Problem 8. [Problem 3, page 48]Let (X,M) be a measurable space.If { fn } is a sequence of measurable functions on X, then {x | lim fn exists }

is a measurable set.

Answer :The problem is not stated very well, since it leaves some ambiguity about wherethe values of the fn’s are supposed to be. Going by the previous problems, I’dsay we should consider functions with values in R.

Define E = {x | lim fn exists }. By Proposition 2.7, the functions g, h : X →R defined by

g(x) = lim infn→∞

fn(x)

h(x) = lim supn→∞

fn(x)

are measurable and, of course,

E = {x ∈ X | g(x) = h(x) } ,

so it will suffice to prove that the set where these two measurable functions areequal is measurable.

For a first attempt, one could try to define ϕ(x) = h(x) − g(x), claim thatϕ is measurable and observe that E = ϕ−1(0). It’s not quite that easy, sincethere may be points where h(x)− g(x) is undefined, e.g., if h(x) = g(x) =∞.

We can take care of this in the spirit of some previous solutions. We can

14

Page 17: Answers for Problem Set #2

partition X into the following measurable sets

S1 = g−1(−∞) ∩ h−1(−∞)(A.1)

S2 = g−1(−∞) ∩ h−1(R)(A.2)

S3 = g−1(−∞) ∩ h−1(∞)(A.3)

S4 = g−1(R) ∩ h−1(−∞)(A.4)

S5 = g−1(R) ∩ h−1(R)(A.5)

S6 = g−1(R) ∩ h−1(∞)(A.6)

S7 = g−1(∞) ∩H−1(−∞)(A.7)

S8 = g−1(∞) ∩ h−1(R)(A.8)

S9 = g−1(∞) ∩ h−1(∞)(A.9)

It will suffice to show that E ∩ Sj is measurable for j = 1, . . . , 9 (since thenE is a finite union of measurable sets).

In case (A.1), we have E ∩ S1 = S1, which is measurable.In case (A.2), E ∩ S2 = ∅, since g(x) 6= h(x) for all x ∈ S2.In case (A.3), E ∩ S3 = ∅.In case (A.4), E ∩ S4 = ∅.In case (A.5), both g and h are real-valued on S5, so E ∩S5 = (h− g)−1(0),

which is measurable, since h− g is measurable on S5 by Proposition 2.6.In case (A.6), E ∩ S6 = ∅.In case (A.7), E ∩ S7 = ∅.In case (A.8), E ∩ S8 = ∅.Finally, in case (A.9), E ∩ S9 = S9, which is measurable.This completes the solution.Alternate Solution. Just for fun, here’s another way to do it. Since g ≤ h,

X is the disjoint union of E = {x | g(x) = h(x) } and F = {x | g(x) < h(x) }.Since the complement of a measurable set is measurable, it will suffice to showthat F is measurable.

We claim

F =⋃r∈Q

{x | g(x) < r } ∩ {x | r < h(x) } .(∗)

To see this, suppose first that p ∈ F . Then g(p) < h(p) so there is some rationals so that g(p) < s < h(p). But then

p ∈ {x | g(x) < s } ∩ {x | s < h(x) } ,

which is one of the sets in the union in (∗).Conversely, if p is in the union in (∗), then there is some r ∈ Q so that

p ∈ {x | g(x) < r } ∩ {x | r < h(x) } .

15

Page 18: Answers for Problem Set #2

But then g(p) < r and r < h(p), so g(p) < h(p), i.e., p ∈ F . This completes theproof of the claim.

Of course, each of the sets

{x | g(x) < r } ∩ {x | r < h(x) } = g−1([−∞, r)) ∩ h−1((r,∞])

is measurable, so (∗) shows that F is a countable union of mensurable sets,hence measurable.

Problem 9. [Problem 4, page 48]Let (X,M) be a measurable space.If f : X → R and f−1((r,∞]) ∈M for each r ∈ Q, then f is measurable.

Answer :As remarked in the book on page 45, the Borel algebra B

Rof R is generated

by the rays (a,∞] for a ∈ R. Hence, to show f is measurable, it will suffice toshow that f−1((a,∞]) is measurable for each a ∈ R.

To do this, let a ∈ R be fixed but arbitrary. Since the rationals are dense inR, we can find a sequence of rationals { rn } that decrease to a, i.e., rn → a andthe rn’s form a decreasing sequence.

We claim that

(a,∞] =∞⋃n=1

(rn,∞].(∗)

To see this, first suppose x ∈ (a,∞]. Then x > a and we can find an openinterval U around a that does not contain x. Since rn ↘ a, there is some Nsuch that rn ∈ U for n ≥ N . But then rn < x, so x ∈ (rn,∞], for n ≥ N . Thus,x is in the union on the right hand side of (∗).

Conversely, if x is in the union in (∗), then x ∈ (rn,∞] for some n. But thena < rn < x, so x ∈ (a,∞]. This completes the proof of the claim.

From (∗), we conclude that

f−1((a,∞]) =∞⋃n=1

f−1((rn,∞]).

Since rn is rational, f−1((rn,∞]) is measurable by our hypothesis. Thus,f−1((a,∞]) is a countable union of measurable sets, and so is measurable. Thiscompletes the proof.

Problem 10. [Problem 9, page 48]Let f : [0, 1]→ [0, 1] be the Cantor function and let g(x) = f(x) + x.

16

Page 19: Answers for Problem Set #2

a. g is a bijection from [0, 1] to [0, 2] and h = g−1 is continuous from [0, 2] to[0, 1].

Answer :

Since the Cantor function is continuous, g is continuous.

Recall that the Cantor function is nondecreasing, i.e., if x < y then f(x) ≤f(y). But then if x < y, g(x) = f(x) + x < f(y) + y = g(y). Thus, g isstrictly increasing and hence one-to-one. Since f(0) = 0 and f(1) = 1, wehave g(0) = 0 and g(1) = 2. If x ∈ [0, 1], 0 = g(0) < g(x) < g(1) = 2, sog maps [0, 1] into [0, 2]. By the intermediate value theorem, every point in[0, 2] is in the image of g. Thus, g is a bijection from [0, 1] to [0, 2].

The fact that h = g−1 is continuous is a general fact about strictly monotonefunctions that follows from the intermediate value theorem. For complete-ness, we’ll give a proof here, but you should probably skip it on a firstreading.

Lemma 0.2. Let f : I → R be a continuous strictly increasing function,where I ⊆ R is an interval. Then the image of an interval J ⊆ I is aninterval of the same type (i.e., open, closed, etc.) In particular, f(I) is aninterval of the same type as I.

Proof of Lemma. Let (a, b) ⊆ I be an open interval. If x ∈ (a, b) thena < x < b and f(a) < f(x) < f(b), since f is strictly increasing. Thus,f((a, b)) ⊆ (f(a), f(b)). On the other hand, if y ∈ (f(a), f(b)) then there isan x ∈ (a, b) such that f(x) = y, by the intermediate value theorem. Thus,f maps the open interval (a, b) onto the open interval (f(a), f(b)).

Suppose that [a, b) ⊆ I is a half-open interval. By the previous part of theproof (a, b) gets mapped onto (f(a), f(b)), and a certainly gets mapped ontof(a). Thus, f maps [a, b) onto [f(a), f(b)). The other types of intervals aredealt with in a similar fashion.

Proposition 0.3. Let f : I → R be a continuous strictly increasing function,where I ⊆ R is an interval, and let J = f(I). Then the inverse functionf−1 : J → I is continuous.

Proof. We need to show that if U ⊆ I is open relative to I, then (f−1)−1(U) =f(U) is open relative to J . Every open set in J can be written as a union ofbounded relatively open intervals, and the bijection f preserves unions, so itwill suffice to show that the image of a bounded relatively open interval isopen in J .

If (a, b) ⊆ I, then (a, b) is open relative to I, and f((a, b)) = (f(a), f(b)) (bythe lemma) which is open relative to J . If I has a lower endpoint, call it a,then intervals of the form [a, b) ⊆ I are open relative to I. By the lemma,

17

Page 20: Answers for Problem Set #2

f(a) is the lower endpoint of J , and f([a, b)) = [f(a), f(b)), which is openrelative to J . The case where I has an upper endpoint is dealt with similarly.

There are similar results (with the obvious modifications) for strictly de-creasing functions.

b. If C is the Cantor set, m(g(C)) = 1.

Answer :Recall that A = [0, 1] \ C is the union of a countably infinite collection ofdisjoint open intervals { Ik }∞k=1 such that

m([0, 1] \ C) =∞∑k=1

m(Ik) = 1

(which is why C has measure zero). Also recall that the Cantor function isdefined to be constant on each of the closed intervals Ik.If Ik = (ak, bk), then Cantor function f takes some constant value α on[ak, bk]. Then we have

g((ak, bk)) = (g(ak), g(bk))= (ak + f(ak), bk + f(bk))= (ak + α, bk + α),

which is an interval of the same length as Ik. Thus,

m(g(A)) =∞∑k=1

m(g(Ik)) =∞∑k=1

m(Ik) = 1.

Since [0, 1] = A ∪ C (disjoint union), we have

[0, 2] = g([0, 1]) = g(A) ∪ g(C), (disjoint union)

and so

2 = m([0, 2]) = m(g(A)) +m(g(C)) = 1 +m(g(C)),

from which we conclude that m(g(C)) = 1.

c. By Exercise 29 of Chapter 1, g(C) contains a Lebesgue nonmeasurable setA. Let B = g−1(A). Then B is Lebesgue measurable but not Borel.

Answer :Since A ⊆ g(C), we have B = g−1(A) ⊆ C. Thus, B is a subset of theLebesgue nullset C, and so is Lebesgue measurable (since Lebesgue measureis complete). On the other hand, if B was Borel, so would be g(B) =(g−1)−1(B), since g−1 is continuous. But g(B) = A and A is not Lebesguemeasurable, let alone Borel. Thus, B is not Borel.

18

Page 21: Answers for Problem Set #2

d. There exists a Lebesgue measurable function F and a continuous function Gon R so that F ◦G is not Lebesgue measurable.

Answer :

Let the sets A ⊆ [0, 2] and B ⊆ C ⊆ [0, 1] be as in the last part of theproblem.

Let F = χB, which is Lebesgue measurable since B is Lebesgue measurableand let G : [0, 2] → [0, 1] be g−1, which is continuous. The function F ◦G : [0, 2]→ R takes only the values 0 and 1 and

1 = (F ◦G)(x) ⇐⇒ 1 = (χB ◦ g−1)(x)

⇐⇒ 1 = χB(g−1(x))

⇐⇒ g−1(x) ∈ B⇐⇒ x ∈ g(B) = A

⇐⇒ 1 = χA(x)

Thus, F ◦ G = χA, which is not Lebesgue measurable, since the set A isnonmeasurable.

Actually, this example doesn’t quite fulfill the requirements of the problem,since G is defined only on [0, 2], not all of R. To fix this, extend the Cantorfunction f from [0, 1] to R by defining f(x) = 0 for x < 0 and f(x) = 1for x > 1. Then define g on R by g(x) = f(x) + x. Now g is a strictlyincreasing function R → R, so g−1 : R → R is continuous. On [0, 1] this gagrees with our old g, so the above example will work with G = g−1 : R→ R

and F = χB (defined on all of R).

19


Recommended