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Mathematics 105 — Spring 2004 — M. Christ Problem Set 8 Solutions (selecta)
VIII.A Let f : [a, b] × [c, d] → R and suppose that for each y ∈ [c, d], [a, b] 3 x 7 → f (x, y) is an integrable^1 function with respect to Lebesgue measure. Suppose also that the partial derivative ∂f /∂y exists for all (x, y) ∈ [a, b] × (c, d), and that its absolute value is bounded by a certain finite constant, which is independent of x, y. Prove that for each y ∈ (c, d), x 7 → ∂f (x, y)/∂y is a measurable function of x ∈ [a, b]. Prove that for all y ∈ (c, d),
d dy
[a,b]
f (x, y) dλ(x) =
[a,b]
∂f (x, y) ∂y
dλ(x).
Solution. Fix y ∈ (c, d). Then ∂f (x, y)/∂y = limn→∞ gn(x) where gn(x) = f^ (x,y+n
− (^1) )−f (x,y) 1 /n. Both x 7 → f (x, y) and x 7 → f (x, y + n−^1 ) are measurable functions of x ∈ [a, b], by hypothesis. Since linear combinations of measurable functions are measurable, so is each gn. We are given that gn(x) → ∂f (x, y)/∂y for every x ∈ [a, b]. Since a limit of measurable functions is measurable, it follows that ∂f (x, y)/∂y is a measurable function of x. For the second conlusion let M be a finite constant such that |∂f (x, y)/∂y| ≤ M for all (x, y) ∈ [a, b] × (c, d). Consider any sequence (hn) of nonzero real numbers tending to zero. Redefine
gn(x, y) = f^ (x,y+h hnn) −f^ (x,y). By the mean value theorem, for each (x, y) ∈ [a, b] × (c, d) there exists c = c(x, y, hn) such that gn(x, y) = (∂f /∂y)(x, c). Thus |gn(x, y)| ≤ M for all (x, y) ∈ [a, b] × (c, d). Moreover gn(x, y) → ∂f (x, y)/∂y as n → ∞, for all (x, y) ∈ [a, b] × (c, d), by definition of the partial derivative. Fix y ∈ (c, d). We may apply the dominated convergence theorem to the functions x 7 → gn(x, y), using the integrable function M χ[a,b] as the Lebesgue dominator, to conclude that
∫^ [a,b]^ gn(x, y)^ dλ(x)^ → [a,b]
∂f (x,y) ∂y dλ(x). Define now F (y) =
[a,b]
∂f (x,y) ∂y dλ(x). The associated difference quotients are
F (y + h) − F (y) h
= h−^1
[a,b]
f (x, y + h) dλ(x) −
[a,b]
f (x, y) dλ(x)
[a,b]
( (^) f (x, y + h) − f (x, y) h
dλ(x).
Thus we’ve proved that
lim n→∞
F (y + hn) − F (y) hn
[a,b]
∂f (x, y) ∂y dλ(x).
Since this holds for any sequence (hn) of nonzero real numbers tending to zero, with a limit indepen- dent of the sequence, we’ve proved that limh→ 0 F^ (y+h h)− F^ (y)exists and equals
[a,b]
∂f (x,y) ∂y dλ(x).
VIII.B∫ Let (E, A, μ) be any measure space and let f ∈ L^1 (E, μ). Consider the mapping Lf (A) =
A f dμ, where the domain is the collection of all sets^ A^ ∈ A. Give an example to show that, for general measure spaces, the parameter δ in problem 3.3.16 cannot be chosen to depend on ε and ‖f ‖L^1 alone. Solution. To prove that δ cannot be chosen to depend on ε and ‖f ‖L^1 alone, for a particular measure space (E, A, μ), it suffices to show that for any c > 0 there exists ε > 0 such that for any sufficiently small δ > 0, there exist a nonnegative function f ∈ L^1 satisfying ‖f ‖L 1 = c and a set A ∈ A satisfying μ(A) < δ such that
A f dμ > ε. I’ll prove this under one assumption on (E,^ A, μ): For any δ > 0 there exists A ∈ A satisfying 0 < μ(A) < δ. (For measure spaces which don’t satisfy
(^1) I carelessly wrote “measurable” in the original problem statement. We need to assume that the integrals that appear in the problem statement are defined!
this assumption, there is no such counterexample; if δ > 0 is sufficiently small then μ(A) < δ implies μ(A) = 0 and hence
A |f^ |^ d μ^ = 0^ < ε^ for any^ f^ ∈^ L
(^1) and any ε > 0.) Let c > 0 be given. Define ε = c/2. Let δ > 0 be given. Choose any set A ∈ A satisfying 0 < μ(A) < δ, and define f = cμ(A)−^1 χA. Then f is nonnegative, ‖f ‖L 1 = c, yet
A f^ = cμ(A)−^1 μ(A) = c > ε.
3.3.16 Let (E, A, ν) be a measure space f ∈ L^1 be a nonnegative integrable function. Define μ(A) =
A f dν^ for all^ A^ ∈ A.^ (i) Show that^ μ^ is a measure.^ (ii) Prove that^ μ^ is^ absolutely continuous, that is, that for any ε > 0 there exists δ > 0 such that
A f dν < ε^ for all measurable sets satisfying ν(A) < δ. Solution. (i) μ(∅) =
∅ f dν^ = 0 since^ ν(∅) = 0,^ μ(A) is defined for all elements^ A^ of the^ σ-algebra A, and μ(A) ≥ 0 for all A ∈ A. So all that remains to be proved is that if sets Aj ∈ A are pairwise disjoint, then μ(∪j Aj ) =
j μ(Aj^ ).^ (Here^ j^ ranges either over^ N^ =^ {^1 ,^2 ,^3 ,^ · · · }^ or over { 1 , 2 , · · · , N } for some N ∈ N. For the sake of simplicity of notation, I’ll discuss only the former case.) Define gj = f χAj. For any positive integer n, the characteristic function of ∪Nj=1Aj equals ∑N j=1 χAj. Therefore
μ(∪nj=1Aj ) =
A 1 ∪···∪An
f dν =
f · χA 1 ∪···∪An dν =
f ·
∑^ n
j=
χAj dν =
∑^ n
j=
Aj
f dν =
∑^ n
j=
μ(Aj )
because the integral of any finite sum of integrable functions equals the sum of the corresponding integrals. To prove the corresponding identity for infinite sums, let A = ∪∞ j=1Aj and note that
∑^ n
j=
χAj (x) = χA 1 ∪···∪An (x) → χA(x) as n → ∞ for all x ∈ E.
Define gj = f · χA 1 ∪···∪An. These are measurable functions, 0 ≤ gn(x) ≤ gn+1(x) ≤ f χA(x) for all x, and we have just shown that limn→∞ gn(x) = f χA(x) for all x. Therefore by the monotone convergence theorem,
gn dν →
f χA dν as n → ∞. The right-hand side here is μ(A), while the left-hand side is μ(∪nj=1Aj ) =
∑n j=1 μ(Aj^ ). Thus we have shown that limn→∞
∑n j=1 μ(Aj^ ) =^ μ(A), that is, that^
j=1 μ(Aj^ ) =^ μ(∪
∞ j=1Aj^ ). (ii): To prove that μ(A) → 0 as ν(A) → 0 (in the precise specified in the problem statement) consider first the case where f is bounded. Fix some finite constant such that 0 ≤ f (x) ≤ M for all x ∈ E. Then
A f dν^ ≤^
A M dν^ =^ M ν(A), so the conclusion obviously holds. Next let f be an arbitrary nonnegative integrable function, and let ε > 0 be given. I’ll show in the next paragraph that there exists a bounded nonnegative measurable function g such that ‖f − g‖L^1 < ε/2. Granting this, choose M ∈ (0, ∞) so that 0 ≤ g(x) ≤ M for all x. (M depends on g, but we have already chosen g in terms of ε.) Define δ = ε/ 2 M. If A ∈ A satisfies ν(A) < δ then
μ(A) =
A
f dν =
A
g dν +
A
(f − g) dν.
The two terms on the right-hand side are both favorable, but for different reasons. Firstly
A
g dν ≤ M ν(A) ≤ M δ = ε/ 2.
Secondly
|
A
(f − g) dν| ≤
A
|f − g| dν ≤
E
|f − g| dν = ‖f − g‖L 1 < ε/ 2.
Apology. In order both to simplify notation and focus on the most essential ideas I’ve decided to write the solution out only in dimension one, and hence have formulated the result that way. Solution. Let N ∈ N be arbitrary. For 0 ≤ j ≤ N define tj = a + j(b − a)/N. Define the intervals Ij = [tj− 1 , tj ) for 1 ≤ j < N and IN = [tN − 1 , tN ]. (Both tj and Ij depend also on N , and this dependence could be indicated by writing tNj and IjN , but I hope there’s little danger of confusion now that the issue has been pointed out.) For each N, j choose xj ∈ Ij , and set cj = f (xj ). (Again,
these depend also on N .) Since f is Riemann integrable,
∫ (^) b a f^ (x)^ dx^ = limN^ →∞
∑N
j=1 cj^ (b^ −^ a)/N^. Consider the function fN (x) =
∑N
j=1 cj^ χIj (x).^ From the continuity of^ f^ it follows easily that fN (x) → f (x) as N → ∞, for all x ∈ [a, b]. Therefore by the bounded convergence theorem, the Lebesgue integral of fN converges to the Lebesgue integral of f over [a, b] as N → ∞. Now the Lebesgue integral of fN equals
∑N
j=1 cj^ λ(Ij^ ), which clearly equals^
∑N
j=1 cj^ (b^ −^ a)/N^ ; and we’ve already noted that these quantities converge to the Riemann integral of f.
3.3.20(i) Let fn, gn be measurable functions on a measure space (E, A, μ). Suppose that gn ∈ L^1 , g ∈ L^1 , and gn → g in L^1 norm. Suppose that fn ≤ gn μ-a.e. for all n. Prove that lim supn→∞
fn dμ ≤
lim supn→∞ fn dμ. Solution. Two things are apparent: Fatou’s lemma ought to come into play, and the hypothesis fn ≤ gn (almost everywhere with respect to μ) can be written as gn − fn ≥ 0 almost everywhere. Thus we get (^) ∫
lim inf(gn − fn) dμ ≤ lim inf
(gn − fn) dμ,
where lim inf means of course lim infn→∞. Since
gn dμ →
∫ g dμ, the right-hand side equals g dμ − lim sup
fn dμ, which is certainly encouraging. But the left-hand side is a problem; conver- gence of {gn} to g in L^1 norm does not imply any sort of pointwise or almost-everywhere convergence, as we discussed in class. The next thought that occurs to us is that some subsequence of {gn} does converge to g, almost everywhere with respect to μ. However, after passing to such a subsequence, we would be considering only some fn, not all fn, so we couldn’t possibly reach any conclusion about lim sup
fn dμ. The fix is surprisingly easy: At the outset choose a subequence fnk such that limk→∞
fnk dμ equals lim supn→∞
fn dμ. (Here we allow the possibility that the lim sup equals +∞, in which case we mean by this identity that
fnk dμ → +∞ as k → ∞. The lim sup of any sequence of extended real numbers equals the limit of some subsequence.) Next, choose a further subsequence (nkj ) of (nk) so that gnkj → g μ-a.e. To simplify notation set Fj = fnkj and Gj = gnkj. Then ∫ Fj dμ → lim supn→∞
fn dμ, and Gj → g almost everywhere. By invoking Fatou as above we find that ∫ lim inf j→∞ (Gj − Fj ) dμ ≤ lim inf j→∞
(Gj − Fj ) dμ. (1)
For any x for which Gj (x) → g(x) we have lim inf(Gj − Fj )(x) = g(x) − lim sup Fj (x). On the other hand
Gj dμ →
g dμ since Gj → g in L^1 norm, so
lim inf
(Gj − Fj ) dμ =
g dμ − lim sup
Fj dμ.
Insert all this information into (1). Since g ∈ L^1 we may subtract its integral from both sides and rearrange to deduce
lim sup
Fj dμ ≤
lim sup Fj dμ. (2)
We have arranged that the left-hand side equals lim supn→∞
fn dμ. For the right-hand side we’ve
made no such arrangement^2 but luck or providence is on our side:
lim sup j→∞
Fj (x) = lim sup j→∞
fnkj (x) ≤ lim sup n→∞
fn(x)
simply because the lim sup of a subsequence never exceeds the lim sup of the whole sequence. In- serting these two pieces of information into (2) completes the proof.
3.3.20(ii) Let the functions gn, g satisfy the hypotheses of part (i) of this problem. Suppose that fn is measurable, |fn| ≤ gn μ-almost everywhere and that fn → f μ-almost everywhere. Show that ‖fn − f ‖L 1 → 0. Solution. f is measurable. If fn → f almost everywhere, we have proved this. If fn → f in measure then some subsequence fnk converges to f almost everywhere, whence the same conclusion. |f | ≤ g almost everywhere, and consequenty f ∈ L^1. Indeed, since gn → g in L^1 norm, there exists a subsequence gnk which converges almost everywhere to g. By passing to a further sub- sub-subsequence, still denoted (nk), we may also arrange that fnk → f almost everywhere, whence |fnk (x)| → |f (x)| for almost every x. Whenever x ∈ E is a point for which both these convergence statements hold, and for which |fn(x)| ≤ gn(x) for all n, it follows that |f (x)| ≤ g(x). Since a countable union of null sets is again a null set, it follows that |f (x)| ≤ g(x) for almost every x. Now we set things up for an invocation of part (i). Define Fn = |fn − f | and Gn = gn + g. Then Gn → G = 2g in L^1 norm, and Fn ≤ Gn almost everywhere. By part (i),
lim sup
|fn − f | dμ ≤
lim sup |fn − f | dμ.
If fn → f almost everywhere then the right-hand side is 0, whence lim sup
|fn − f | dμ = 0; this means that ‖fn − f ‖L 1 → 0. If we merely know that fn → f in measure then we revert to the very beginning of the argument, and choose a subsequence (fnk ) such that ‖fnk − f ‖L 1 → lim supn→∞ ‖fn − f ‖L 1 as k → ∞. Next choose a sub-subsequence, again denoted (fnkj ), which converges to f almost everywhere. Apply
the result proved in the preceding paragraphs to conclude that lim supj→∞ ‖fnkj − f ‖L^1 = 0. But we’ve arranged that limk→∞ ‖fnk − f ‖L 1 exists. Therefore it equals the corresponding lim sup over any subsequence. Thus lim supn→∞ ‖fn − f ‖L 1 = limk→∞ ‖fnk − f ‖L 1 = 0.
3.3.23 Let (E, ρ) be a metric space, let μ, ν be two finite measures defined on the Borel σ-algebra BE. Suppose that
ϕ dμ =
ϕ dν for all bounded, uniformly continuous functions ϕ : E → R. Show that μ = ν (that is, μ(A) = ν(A) for all A ∈ BE ). Solution. The natural way to attack the problem is this: Consider any A ∈ BE , and let ε > 0; we aim to prove that |μ(A) − ν(A)| < ε. Note that χA ∈ L^1 (μ) ∩ L^1 (μ) since both μ, ν are finite measures. We know that there exist a bounded uniformly continuous function ϕ such that
|χA − ϕ| dν < ε/2, and likewise a bounded uniformly continuous ψ such that
|χA − ψ| dμ < ε/2. If we had ϕ ≡ ψ then we’d be in business:
∣ ∣μ(A) − ν(A)
χA dμ −
χA dν
(χA − ϕ) dμ −
(χA − ϕ) dν
would equal
(χA − ψ) dμ −
(χA − ϕ) dν
∣ (^) if ϕ = ψ
since
ϕ dν =
ϕ dμ by hypothesis. We could then deduce that
∣ ∣μ(A) − ν(A)
(χA − ψ) dμ
(χA − ϕ) dν
|χA − ψ| dμ +
|χA − ϕ| dν < ε.
(^2) And we can’t; if you try to do this you could well end up choosing a completely different sequence of indices for each point x ∈ E, and if E is uncountable this is hopeless...
