Problem Set

Problem Set #2

Math 5322, Fall 2001

December 3, 2001

ANSWERS i Problem 1. [Problem 18, page 32] Let (X) be an algebra, σ the collection of countable unions of sets A ⊆ P A in , and σδ the collection of countable intersections of sets in σ. Let µ0 be A A A a premeasure on and µ∗ the induced outer measure. A a. For any E X and ε > 0 the exists A σ with E A and µ∗(A) ⊆ ∈ A ⊆ ≤ µ∗(E) + ε.

Answer: By definition,

∞ ∞ µ∗(E) = inf ( µ0(Ai) Ai ∞ ,E Ai ) . X | { }i=1 ⊆ A ⊆ [ i=1 i=1

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

∞ ∞ E Ai and µ0(Ai) < µ∗(E) + ε. ⊆ [ X i=1 i=1 Define a set A E by ⊇

∞ A = A . [ i i=1

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

∞ ∞ µ∗(A) µ∗(Ai) = µ0(Ai) < µ∗(E) + ε, ≤ X X i=1 i=1

since µ∗ = µ0 on . This completes the proof. A b. If µ∗(E) < then E is µ∗-measurable iff there exists B σδ with E B ∞ ∈ A ⊆ and µ∗(B E) = 0. \ Answer:

Since the σ-algebra of µ∗-measurable sets contains and is closed under M A countable unions and intersections, σδ . A ⊆ M For the first part of the proof, assume that E and µ∗(E) < . For each n N, we can apply the previous part of∈ the M problem to find∞ a set ∈ An σ such that E An and µ∗(An) < µ∗(E) + 1/n. Define B by ∈ A ⊆

∞ B = A . \ n n=1

1 Then E B and B σδ, since it is a countable intersection of sets in ⊆ ∈ A σ. For every n, we have A

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 are ⊆ \ in , so we can use the additivity of µ∗ on to get M M µ∗(B) = µ∗(E) + µ∗(B E). \

All three terms are finite and µ∗(E) = µ∗(B), so simple algebra shows µ∗(B E) = 0. \ For the second part of the proof, suppose that E X and there is a set ⊆ B σδ so that E B and µ∗(B E) = 0. As remarked above, σδ ∈ A ⊆ \ A ⊆ M so B is measurable. We showed in class that if F X and µ∗(F ) = 0, then ⊆ F is µ∗-measurable. Thus, in the present situation, B E is measurable. But E = B (B E) (check!). Since is closed under taking\ set differences, we conclude\ that\ E is measurable.M 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-finite ∞ we can find disjoint sets Fj , j N with µ0(Fj) < and X = ∞ Fj. ∈ A ∈ ∞ Sj=1 Set Ej = E Fj. Then the sets Ej are disjoint, µ∗(Ej) µ∗(Fj) = µ0(Fj) < ∩ ≤ and E = Ej. ∞ Sj As a first try, we might invoke the previous part of the problem to find sets Bj Ej with Bj σδ and µ∗(Bj Ej) = 0. The next step would be to ⊇ ∈ A \ try to define our desired set B as Bj. Unfortunately, this won’t work. Sj Since σδ is defined to be the collection of countable intersections of sets A in σ, there’s no obvious reason why σδ should be closed under taking A A countable unions. Thus, we don’t know that Bj will be in σδ. Sj A Note, however that σ is closed under countable unions. If Ai ∞ σ, A { }i=1 ⊆ A then for each i there is a sequence Ak ∞ such that  i k=1 ⊆ A

∞ A = Ak. i [ i k=1

But then if A = Ai, we have Si ∞ ∞ A = Ak = Ak, [ [ i  [ i i=1 k=1 (i,k) N N ∈ ×

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

2 So, we proceed as follows. For each n N, we invoke part (a) to get a set n n ∈ A σ such that Ej A and j ∈ A ⊆ j 1 µ (An) < µ (E ) + . ∗ j ∗ j n2j n Since Aj and Ej are measurable and have finite measure, this gives us

n 1 µ∗(A Ej) < . j \ n2j We then define

∞ An = An. [ j j=1

n As remarked above, A σ. Next, we claim that ∈ A

n ∞ n ( ) A E (A Ej). ∗ \ ⊆ [ j \ j=1

To see this, suppose that x An E. Since x An, there is some j such n ∈ \ ∈ n that x Aj . Since x / E and Ej E, x / Ej. Thus x Aj Ej. This completes∈ the proof of∈ the claim. ⊆ ∈ ∈ \ From ( ), we have ∗

n ∞ n ∞ 1 1 µ∗(A E) µ∗(A Ej) < = . \ ≤ X j \ X n2j n j=1 j=1

Now define

∞ B = An. \ n=1

Then, B σδ and E B. For every n we have ∈ A ⊆ n 1 µ∗(B E) µ∗(A E) < , \ ≤ \ n

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 a. E µ ∈ M 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 µ is its domain (the µ∗- M 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 in ⇒ the 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 µ. Let Fj be a sequence of disjoint measurable ∈ M { } sets such that R = Fj and µ(Fj) < . We can take the Fj’s to be bounded Sj ∞ intervals, for example. Let Ej = E Fj, so the Ej’s are disjoint measurable sets ∩ whose union is E and each Ej has finite measure. n By Proposition 1.18, for each n N and j we can find an open set Uj such n ∈ that Ej U and ⊆ j 1 µ(U n) < µ(E ) + . j j n2j

n n Since U = Ej (U Ej) (disjoint union), we have j ∪ j \ n n µ(U ) = µ(Ej) + µ(U Ej). j j \ All the terms in this equation are finite, so we can rearrange the equation to get

n n 1 µ(U Ej) = µ(U ) µ(Ej) < . j \ j − n2j n n n n n Define a set U by U = j Uj . Then E U and U is open, since any union of open sets is open. WeS have ⊆

n ∞ n U E (U Ej) \ ⊆ [ j \ j=1

(check), and so

n ∞ n ∞ 1 1 µ(U E) µ(U Ej) < = . \ ≤ X j \ X n2j n j=1 j=1

Now we define

∞ V = U n, \ n=1

4 so V is a Gδ set and V E. For any n, we have ⊇ 1 µ(V E) µ(U n E) < , \ ≤ \ n 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 easier ⇒ than the proof of (a) = (b). So suppose that E µ, let Fj be a ⇒ ∈ M { } partition 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 get ∈ a compact set Kn Ej such that ⊆ 1 µ(Ej) < µ(Kn) − n and so 1 µ(Ej Kn) < . \ n

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

Ej Hj Ej Kn \ ⊆ \ so 1 µ(Ej Hj) µ(Ej Kn) < . \ ≤ \ n

Thus, we have µ(Ej Hj) = 0, where Hj Ej is an Fσ set. Thus, we have \ ⊆ Ej = Hj Nj where Nj = Ej Hj is a nullset. Taking∪ the union over j, we\ have

∞ ∞ E =  Hj  Nj. [ ∪ [ j=1 j=1

Since Fσ is closed under countable unions (by an argument similar to the last problem), the first set on the right of this equation is an Fσ set. Of course a countable union of nullsets is a nullset, so the second set on the right of the last equation is a nullset. This completes the proof of (a) = (c). Next, we prove that (b) = (a). This is pretty⇒ easy. Suppose that ⇒ E = V N1 where V is a Gδ set and N1 is a nullset. Of course a nullset is \ measurable. We know that µ contains the Borel sets, and hence the Gδ sets. M Since the σ-algebra µ is closed under complements and finite intersections, M we conclude that E µ. ∈ M

5 The proof that (c) = (a) is similar. Suppose E = H N2 where H is an ⇒ ∪ Fσ set and N2 is a nullset. Then N2 µ and H µ since µ contains ∈ M ∈ M M the Borel sets and hence the Fσ sets. Since µ is closed under finite unions, M we conclude that E µ. This completes the∈ M 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 the simplest proof of our Proposition would be⇒ to prove (a) = (c) as above, then use de Morgan’s laws to get (a) = (b), and then do the⇒ reverse implications as above. ⇒

Problem 3. [Problem 26, page 39] Prove Proposition 1.20 (Use Theorem 1.18.) Thus, we want to prove that if E µ and µ(E) < , then for every ε > 0 there is a set A that is a finite [disjoint]∈ M union of open∞ intervals such that µ(E A) < ε Here4 µ is a Lebesgue-Stieltjes measure on R. The symbol denotes the symmetric difference of the sets, i.e., E A = (E A) (A E).4 4 \ ∪ \ Answer: Suppose E µ with µ(E) < and let ε > 0 be given. By Theorem∈ M 1.18 we can find∞ an open set U R such that E U and µ(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− disjoint union of open intervals. Suppose first that the number of intervals is infinite, say

∞ U = I , [ k k=1 where each Ik is an open interval. We have

∞ µ(Ik) = µ(U) < . X ∞ k=1 Since this series converges, there is some n such that

∞ ( ) µ(Ik) < ε. ∗ X k=n+1

We define A by

n A = I . [ k k=1

6 If there are only finitely many intervals in U, we can label them as I1,I2,...,In and 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) − n ∞ = µ(Ik) µ(Ik) X − X k=1 k=1

∞ = µ(I ) < ε. X k k=n+1 On the other hand, A E U E, so \ ⊆ \ µ(A E) µ(U E) < ε. \ ≤ \ Thus,

µ(E A) µ(E A) + µ(A E) < 2ε. 4 ≤ \ \ 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 the values

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) a = (a 1/n, a]. \ { } n=1 − Certainly a is in the right-hand side and any number bigger that a is not. If x < a then x < a 1/n for sufficiently large n, so x is not in the intersection on the right-hand side− of (A). The intervals (a 1/n, a] form a decreasing sequence of sets, so by continuity from above (Theorem− 1.8d) we have

µF ( a ) = lim µF ((a 1/n, a]) = lim [F (a) F (a 1/n)]. n n { } →∞ − →∞ − −

7 Since F is increasing, lim F (a 1/n) = lim F (x) = sup F (x) x < a = F (a ). n − x a { | } − →∞ → − Thus, we have

(B) µF ( a ) = F (a) F (a ). { } − − 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 a. If E N, where N is the nonmeasurable set described in Section 1.1, then m(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 exactly∼ one element from∼ each− equivalence∈ class (using the Axiom of Choice). Set R = 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 the translation invariance of Lebesgue measure. We also argued that

[0, 1) = N , (disjoint union). [ r r R ∈ If N was measurable, we would have

∞ 1 = m([0, 1)) = m(N), X n=1 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 N is Lebesgue measurable, then m(E) = 0. Well, for each r R we⊆ have ∈ Er Nr, as defined above. Since the sets Nr are pairwise disjoint, ⊆ F = E [ r r R ∈ is a countable disjoint union of measurable sets contained in [0, 1). Thus, we have

∞ 1 m(F ) = m(Er) = m(E). ≥ X X r R i=1 ∈ 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 assume

that E [0, 1). In the notation of Section 1.1, E = r R E Nr.) ⊆ S ∈ ∩ Answer: Briefly, if E R and m(E) > 0, we can write ⊆

∞ E = E [n, n + 1), (disjoint union). [ n= ∩ −∞

9 Since m(E) > 0, at least one of the sets E [n, n + 1) must have nonzero measure. 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 same measure as F . If we prove that− every subset of [0, 1) of nonzero measure contains a nonmeasurable set, then F n will contain a non-measurable set A. But then A + n F is nonmeasurable− (if A + n was measurable, so would 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 is invertible. Indeed, S0 = S and for t (0, 1) we define ϕt : [0, 1) [0, 1) by ∈ → x + t, x [0, 1 t) ϕt(x) = ( ∈ − 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 measurable and F Nr for some r then m(F ) = 0. To see this, note that if F Nr ⊆ ⊆ then F1 r N. As we’ve argued before, F1 r is measurable and m(F1 r) = − ⊆ − − m(F ). From the first part of the problem, m(F1 r) = 0, so we conclude that m(F ) = 0. − Now suppose that E [0, 1) and m(E) > 0. The interval [0, 1) is the ⊆ disjoint union of the sets Nr for r R, so E is the disjoint union of the sets ∈ E Nr. If all of the sets E Nr were measurable, we would have ∩ ∩

m(E) = m(E Nr) = 0 = 0, X ∩ X r R r R ∈ ∈ which contradicts the assumption that m(E) > 0. Thus at least one of the sets E Nr must be nonmeasurable and we conclude that E contains a nonmeasurable∩ set.

Before going into the problems on Page 48ff. of the book, lets recall some of the definitions involved. Let (X, ) be a measurable space, so (X) is a σ-algebra. If A X, M M ⊆ P ⊆ we can define a σ-algebra A (A) by M| ⊆ P (0.1) A = E A E . M| { ∩ | ∈ M } If A itself is in , we have M A = E E A, E . M| { | ⊆ ∈ M }

10 If (Y, ) is a measurable space and f : X Y , we say that f is measurable N → on A X if f A : A Y (the restriction of f to A) is measurable for the ⊆ | → 1 1 σ-algebra A, which means that (f A)− (N) = f − (N) A A for all M| | 1 ∩ ∈ M| N . If A , then f is measurable on A iff f − (N) A for all N ∈ N. Clearly,∈ if Mf is a measurable function X Y , then f ∩is measurable∈ M on ∈ N 1 → every subset of A of X since, in this case, f − (N) A A for all , because 1 ∩ ∈ M| N f − (N) . The following∈ M proposition is pretty easy, but useful. Proposition 0.1. Let (X, ) and (Y, ) be measurable spaces and let f : X M N → Y be a function. Let Eα α A be a countable (finite or infinite) collection of measurable subsets of {X such} ∈ that

X = E . [ α α A ∈

Then, f is measurable if and only if f is measurable on each Eα, i.e., f is 1 measurable if and only if for each N , f − (N) Eα for all α A. ∈ N ∩ ∈ M ∈ 1 Proof. If f is measurable, then f − (N) for all N . Since each Eα , 1 ∈ M ∈ N ∈ M we have f − (N) Eα . ∩ ∈ M 1 Conversely, suppose that for each N , f − (N) Eα is measurable for all α. Then ∈ N ∩

1 1 f − (N) = f − (N) Eα [ ∩ α A ∈ 1 is a countable union of sets in , so f − (N) . Thus, f is measurable. M ∈ M

Problem 6. [Problem 1, page 48] 1 Let (X, ) be a measurable space. Let f : X R and Y = f − (R). Then M 1 1 → f is measurable iff f − ( ) , f − ( ) and f is measurable on Y . { −∞ } ∈ M { ∞ } ∈ M Answer: Suppose first that f is measurable. Then f is measurable on Y , as discussed above. The sets and are closed sets, hence Borel sets, in R, so 1 { −∞1 } { ∞ } f − ( ) and f − ( ) are measurable. { −∞ } { ∞ } 1 1 For the second part of proof suppose f − ( ) and f − ( ) are mea- { −∞1 } { ∞1 } surable and f is measurable on Y . The sets f − ( ), f − ( ) and 1 {1 −∞ } { ∞ } Y form a partition of X. Since f − ( ) and f − ( ) are measurable, 1 1 { −∞ } { ∞ } Y = X (f − ( ) f − ( )) is measurable. To show f is measurable it will suffice\ to{ −∞ show } it∪ is measurable{ ∞ } on the three sets in our partition. Of 1 course, f is measurable on Y by assumption. On the set f − ( ), f is constant and a constant function is always measurable. Similarly, {f −∞is constant, } 1 and hence measurable, on f − ( ). This completes the proof. { ∞ }

11 Problem 7. [Problem 2, page 48] Let (X, ) be a measurable space. Suppose that f, g : X R are measur- able. M → 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- 1 1 1 surable sets F1 = f − ( ), F2 = f − (( , 0)), F3 = f − (0), F4 = 1 −∞1 1 −∞ 1 f − ((0, )) and F5 = f − ( ). (f − (a) is the same thing as f − ( a ).) We ∞ ∞ 1 1 { } 1 have a similar partition G1 = g− ( ), G2 = g− (( , 0)), G3 = g− (0), 1 1 −∞ −∞ G4 = g− ((0, )), G5 = g− ( ) for g. Taking pairwise intersections, we get ∞ ∞ a partition of X into the 25 measurable sets (don’t panic!) Fi Gj. To show fg is measurable it will suffice to prove that fg is measurable∩ on each of the sets Fi Gj. ∩ 1 1 Observe that fg is constant on the set F1 G1 = f − ( ) g− ( ). ∩ −∞ ∩ −∞ Indeed fg is constant on all of the sets F1 Gj, j = 1,..., 5. Similarly, fg ∩ is 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 of 1 ∩ 1 these sets is S = f − (R) g− (R). Of course, f and g are real valued on S and so fg is measurable on∩ S by Proposition 2.6 (p. 45). This completes the solution. Second Solution We try to follow the proof of Proposition 2.6 on page 45 of the book. Thus, we define F : X R R by F (x) = (f(x), g(x)). As discussed in the book, this is measurable.→ × Next we define ψ : R R R by ψ(x, y) = xy (with the 0 ( ) = 0 convention). 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 A is Borel. × × \ Recall from our discussion in class that if Z is a metric space, and W is a Borel subset of Z,

( Z ) W = E W E Z = W . B | { ⊆ | ∈ B } B Thus, the principal of the Proposition we proved above applies: If Z is the union of two Borel sets V and W then f : Z R is Borel measurable if and only if f is Borel measurable on V and W . →

12 Thus, in order to show that ψ is Borel measurable, it will suffice to show it is Borel measurable on A and B. On A, ψ is constant (with value 0) and so 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 a 1 ◦ ⊆ Borel set. Since ψ is Borel measurable, ψ− (B) is Borel in R R. Since F 1 1 × is measurable, F − (ψ− (B)) . Thus, ∈ M 1 1 1 1 h− (B) = (ψ F )− (B) = F − (ψ− (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

1 1 f − ( ) g− ( )(A.1) −∞ ∩ −∞ 1 1 f − ( ) g− (R)(A.2) −∞ ∩ 1 1 f − ( ) g− ( )(A.3) −∞ ∩ ∞ 1 1 f − (R) g− ( )(A.4) ∩ −∞ 1 1 f − (R) g− (R)(A.5) ∩ 1 1 f − (R) g− ( )(A.6) ∩ ∞ 1 1 f − ( ) g− ( )(A.7) ∞ ∩ −∞ 1 1 f − ( ) g− (R)(A.8) ∞ ∩ 1 1 f − ( ) g− ( )(A.9) ∞ ∩ ∞

and it will suffice to prove that h is measurable on each of these sets. On the sets (A.3) and (A.7), h is constant, with value a, by our definition. On the sets (A.1), (A.2), (A.4), (A.6), (A.8), and (A.9), h is constant (with 1 1 value either ). Finally, on the set f − (R) g− (R) in (A.5), both f and g are real±∞ valued and h coincides with f +∩g, which is measurable by Proposition 2.6. This completes the solution. Second Solution. As in the first part of the problem, define F : X R R by F (x) = (f(x), g(x)). As discussed in the book, this is measurable.→ × Let A R be defined by ⊆ A = ( , ), ( , ) , { −∞ ∞ ∞ −∞ }

13 which is closed, and hence Borel, in R R. Let B = R R A, which is also Borel. Define ψ : R R R by × × \ × → a, (x, y) A ψ(x, y) = ( ∈ x + y, (x, y) B. ∈ This function is not continuous at the points in A, but it is nonetheless Borel measurable. Of course, ψ is continuous, hence Borel measurable, on B, and ψ is constant, hence Borel measurable, on A. Thus, ψ is Borel measurable on R R. × Since ψ is Borel measurable, ψ F is measurable (as discussed in the solution of the first part of the problem),◦ but h = ψ F , so the proof is complete. ◦

Problem 8. [Problem 3, page 48] Let (X, ) be a measurable space. M 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 where the values of the fn’s are supposed to be. Going by the previous problems, I’d say 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 inf fn(x) n →∞ h(x) = lim sup fn(x) n →∞ 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 are equal is measurable. For a first attempt, one could try to define ϕ(x) = h(x) g(x), claim that 1 − ϕ is measurable and observe that E = ϕ− (0). It’s not quite that easy, since there 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 partition X into the following measurable sets

1 1 S1 = g− ( ) h− ( )(A.1) −∞ ∩ −∞ 1 1 S2 = g− ( ) h− (R)(A.2) −∞ ∩ 1 1 S3 = g− ( ) h− ( )(A.3) −∞ ∩ ∞ 1 1 S4 = g− (R) h− ( )(A.4) ∩ −∞ 1 1 S5 = g− (R) h− (R)(A.5) ∩ 1 1 S6 = g− (R) h− ( )(A.6) ∩ ∞ 1 1 S7 = g− ( ) H− ( )(A.7) ∞ ∩ −∞ 1 1 S8 = g− ( ) h− (R)(A.8) ∞ ∩ 1 1 S9 = g− ( ) h− ( )(A.9) ∞ ∩ ∞

It will suffice to show that E Sj is measurable for j = 1,..., 9 (since then E 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) = h(x) for all x S2. ∩ ∅ 6 ∈ In case (A.3), E S3 = . ∩ ∅ In case (A.4), E S4 = . ∩ ∅ 1 In case (A.5), both g and h are real-valued on S5, so E S5 = (h g)− (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 show} that F is measurable. We claim

( ) F = x g(x) < r x r < h(x) . ∗ [ { | } ∩ { | } r Q ∈ To see this, suppose first that p F . Then g(p) < h(p) so there is some rational s 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 But then g(p) < r and r < h(p), so g(p) < h(p), i.e., p F . This completes the proof of the claim. ∈ Of course, each of the sets

1 1 x g(x) < r x r < h(x) = g− ([ , r)) h− ((r, ]) { | } ∩ { | } −∞ ∩ ∞ is measurable, so ( ) shows that F is a countable union of mensurable sets, hence measurable. ∗

Problem 9. [Problem 4, page 48] Let (X, ) be a measurable space. M 1 If f : X R and f − ((r, ]) for each r Q, then f is measurable. → ∞ ∈ M ∈ Answer: As remarked in the book on page 45, the Borel algebra R of R is generated by the rays (a, ] for a R. Hence, to show f is measurable,B it will suffice to 1 ∞ ∈ show that f − ((a, ]) is measurable for each a R. To do this, let a∞ R be fixed but arbitrary.∈ Since the rationals are dense in ∈ R, we can find a sequence of rationals rn that decrease to a, i.e., rn a and { } → the rn’s form a decreasing sequence. We claim that

∞ ( ) (a, ] = (r , ]. [ n ∗ ∞ n=1 ∞

To see this, first suppose x (a, ]. Then x > a and we can find an open ∈ ∞ interval U around a that does not contain x. Since rn a, there is some N & such 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 then ∗ ∈ ∞ a < rn < x, so x (a, ]. This completes the proof of the claim. From ( ), we∈ conclude∞ that ∗

∞ f 1((a, ]) = f 1((r , ]). − [ − n ∞ n=1 ∞

1 Since rn is rational, f − ((rn, ]) is measurable by our hypothesis. Thus, 1 ∞ f − ((a, ]) is a countable union of measurable sets, and so is measurable. This completes∞ 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 1 a. g is a bijection from [0, 1] to [0, 2] and h = g− 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 ≤is strictly increasing and hence one-to-one. Since f(0) = 0 and f(1) = 1, we have g(0) = 0 and g(1) = 2. If x [0, 1], 0 = g(0) < g(x) < g(1) = 2, so g 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]. 1 The fact that h = g− is continuous is a general fact about strictly monotone functions that follows from the intermediate value theorem. For complete- ness, we’ll give a proof here, but you should probably skip it on a first reading.

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 an interval of⊆ the same type (i.e., open, closed, etc.) In particular, f⊆(I) is an interval of the same type as I.

Proof of Lemma. Let (a, b) I be an open interval. If x (a, b) then a < 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 is an 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 the proof (a, b) gets mapped⊆ onto (f(a), f(b)), and a certainly gets mapped onto f(a). Thus, f maps [a, b) onto [f(a), f(b)). The other types of intervals are dealt 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 function 1 ⊆ f − : J I is continuous. → 1 1 Proof. We need to show that if U I is open relative to I, then (f − )− (U) = f(U) is open relative to J. Every⊆ open set in J can be written as a union of bounded relatively open intervals, and the bijection f preserves unions, so it will suffice to show that the image of a bounded relatively open interval is open in J. If (a, b) I, then (a, b) is open relative to I, and f((a, b)) = (f(a), f(b)) (by the 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 f(a) is the lower endpoint of J, and f([a, b)) = [f(a), f(b)), which is open relative 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 of \ disjoint open intervals Ik ∞ such that { }k=1 ∞ m([0, 1] C) = m(Ik) = 1 \ X k=1 (which is why C has measure zero). Also recall that the Cantor function is defined 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)) = m(g(I )) = m(I ) = 1. X k X k k=1 k=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 set 1 A. Let B = g− (A). Then B is Lebesgue measurable but not Borel. Answer: 1 Since A g(C), we have B = g− (A) C. Thus, B is a subset of the Lebesgue⊆ nullset C, and so is Lebesgue measurable⊆ (since Lebesgue measure is complete). On the other hand, if B was Borel, so would be g(B) = 1 1 1 (g− )− (B), since g− is continuous. But g(B) = A and A is not Lebesgue measurable, let alone Borel. Thus, B is not Borel.

18 d. There exists a Lebesgue measurable function F and a continuous function G on 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 the problem. ⊆ ⊆ ⊆

Let F = χB, which is Lebesgue measurable since B is Lebesgue measurable 1 and let G: [0, 2] [0, 1] be g− , which is continuous. The function F G: [0, 2] R takes→ only the values 0 and 1 and ◦ → 1 1 = (F G)(x) 1 = (χB g− )(x) ◦ ⇐⇒ ◦ 1 1 = χB(g− (x)) ⇐⇒ 1 g− (x) B ⇐⇒ ∈ x g(B) = A ⇐⇒ ∈ 1 = χA(x) ⇐⇒

Thus, F G = χA, which is not Lebesgue measurable, since the set A is nonmeasurable.◦ 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 Cantor function f from [0, 1] to R by defining f(x) = 0 for x < 0 and f(x) = 1 for x > 1. Then define g on R by g(x) = f(x) + x. Now g is a strictly 1 increasing function R R, so g− : R R is continuous. On [0, 1] this g → → 1 agrees with our old g, so the above example will work with G = g− : R R → and F = χB (defined on all of R).

19