Download Solutions to Math 245A Homework 04: Properties of Measurable Functions and Sets and more Assignments Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity!
Jeffrey Hellrung Wednesday, November 09, 2005 Math 245A, Homework 04 Chapter 2, # 9, 10, 12, 13, 14, 15, 16, 19, 20
- Let f : [0, 1] → [0, 1] be the Cantor function (§1.5), and let g(x) = f (x) + x.
a. g is a bijection from [0, 1] to [0, 2], and h = g−^1 is continuous from [0, 2] to [0, 1]. b. If C is the Cantor set, m(g(C)) = 1. c. By Exercise 29 of Chapter 1, g(C) contains a Lebesgue nonmeasurable set A. Let B = g−^1 (A). Then B is Lebesgue measurable but not Borel. d. There exist a Lebesgue measurable function F and a continuous function G on R such that F ◦ G is not Lebesgue measurable.
Solution
a. Since f is increasing and continuous and x 7 → x is strictly increasing and continuous, it follows that g is strictly increasing and continuous. Since g(0) = f (0) + 0 = 0 and g(1) = f (1) + 1 = 2, g maps [0, 1] bijectively onto [0, 2]. It follows that h = g−^1 : [0, 2] → [0, 1] exists and is continuous. b. We know that [0, 1]\C =
⋃^ ∞
n=
An,
where each of the An’s are disjoint, and each An is a disjoint union of 2n^ intervals of length 3 −(n+1). Since An ∩ C = ∅, we have f constant on each subinterval of each An. It follows that g(An) is a disjoint union of 2n^ intervals of length 3−(n+1), since, as just mentioned f is constant on each subinterval, hence the image of each subinterval under g is just translation. Therefore, m(g(An)) = m(An), hence m(g([0, 1]\C)) = m([0, 1]\C) = 1. We have thus shown that
m(g(C)) = m(g([0, 1])) − m(g([0, 1]\C)) = 2 − 1 = 1.
c. If A ⊂ g(C) is a Lebesgue nonmeasurable set, certainly B = g−^1 (A) ⊂ g−^1 (g(C)) = C, which is a Lebesgue null set, hence the completion of the Lebesgue measure implies that B is a Lebesgue null set as well. However, B cannot be a Borel set, since A = g(B) = h−^1 (B) is not a Borel set, and h, being continuous, is Borel measurable. d. Let B be the Lebesgue set from c., F = χB , and G = h. Then
(F ◦ G)−^1 ({ 1 }) = G−^1
F −^1 ({ 1 })
= G−^1 (B) = g(B) = A,
where A is the Lebesgue nonmeasurable set from c. inducing B. Therefore F ◦ G is not Lebesgue measurable.
- Prove Proposition 2.11.
2.11 Proposition. The following implications are valid iff the measure μ is complete:
a. If f is measurable and f = g μ-a.e., then g is measurable. b. If fn is measurable for n ∈ N and fn → f μ-a.e., then f is measurable.
Solution Suppose first that μ is complete, and that f is measurable and f = g μ-a.e. Set N = {x ∈ X | f (x) 6 = g(x)}; then μ(N ) = 0. Now for any measurable set E on the range space,
f −^1 (E) = (f |X\N )−^1 (E) ∪ (f |N )−^1 (E),
g−^1 (E) = (g|X\N )−^1 (E) ∪ (g|N )−^1 (E), so g−^1 (E) and f −^1 (E) differ only by null sets, since g|X\N = f |X\N and both (g|N )−^1 (E) and (f |N )−^1 (E) are contained in N , hence are null sets by the completeness of μ. It follows that since f −^1 (E) is measurable, g−^1 (E) is measurable, therefore g is measurable. For b., suppose, again, that μ is complete, each fn is measurable, and fn → f μ-a.e. Set g(x) = lim fn(x) whenever the limit exists and g(x) = 0 otherwise. Then g is measurable by Proposition 2. and Exercise 3, so the hypotheses of a. are met and we conclude that f is measurable as well. Now suppose a., and let N be a subset of a null set. Then χN is identically 0 μ-a.e., so is measurable since 0 is measurable, hence, in particular, χ− N^1 (1) = N is measurable, from which it follows that μ is complete. Since b. implies a., the truth of b. likewise implies the completeness of μ.
- Prove Proposition 2.20. (See Proposition 0.20, where a special case is proved.)
2.20 Proposition. If f ∈ L+^ and
f < ∞, then {x : f (x) = ∞} is a null set and {x : f (x) > 0 } is σ-finite. Solution Let A = f −^1 ({∞}), and φn = nχA. Then φn ≤ f for all n, hence
φn ≤
∫ f <^ ∞^ for all^ n.^ But φn = nμ(A) → ∞ as n → ∞ unless μ(A) = 0, establishing the first claim. Let Bn = f −^1 ((1/n, ∞]) and B = f −^1 ((0, ∞]) =
n Bn.^ Then, for each^ n,^ χBn /n^ ≤^ f^ , hence μ(Bn)/n =
χBn /n ≤
f < ∞, so μ(Bn) < ∞. But B =
n Bn, hence^ B^ is^ σ-finite.
- Suppose {fn} ⊂ L+, fn → f pointwise, and
f = lim
fn < ∞. Then
E f^ = lim^
E fn^ for all E ∈ M. However, this need not be true if
f = lim
fn = ∞. Solution We establish first the following lemma: if a, b ∈ L+^ with a ≤ b and
a < ∞, then
(b − a) =
b −
a. Indeed, since b − a ∈ L+, by Theorem 2.15,
b =
(b − a) +
a, from which the claim immediately follows. Let gn ∈ L+^ be defined by gn(x) = inf k≥n fk(x);
then gn ≤ fn. Further, gn ↑ f , so by Theorem 2.14 (The Monotone Convergence Theorem), lim
∫ gn^ = f = lim
fn < ∞. Hence, for any E ∈ M, utilizing the above lemma, ∫
E
f − lim
E
gn = lim
E
(f − gn) ≤ lim
(f − gn) =
f − lim
gn = 0,
while, similarly,
lim
E
fn − lim
E
gn = lim
E
(fn − gn) ≤ lim
(fn − gn) = lim
fn − lim
gn = 0,
from which it follows that (^) ∫
E
f = lim
E
gn = lim
E
fn.
To show the necessity of the assumption
f < ∞, take μ = m and fn = χ(−∞,0) + χ[n,n+1]. Then fn → f = χ(−∞,0) and
f = lim
fn = ∞, but ∫
(0,∞)
f = 0 6 = 1 = lim
(0,∞)
fn.
a. By the uniform convergence of fn to f , there exists N such that |f − fn| < 1 for n ≥ N. It follows that |fN − fn| < 2 for n ≥ N , hence |fn| < |fN | + 2 = g. But g is integrable, as ∫ |g| =
|fN | + 2μ(X) < ∞,
so an application of Theorem 2.24 (The Dominated Convergence Theorem) yields the desired result. b. For X = [1, ∞) and μ = m, let fn(x) = χ1,n/x for n ≥ 2, and f (x) = 1/x. Then each fn ∈ L^1 (m) and fn → f uniformly, but f /∈ L^1 (m).
- (A generalized Dominated Convergence Theorem) If fn, gn, f, g ∈ L^1 , fn → f and gn → g a.e., |fn| ≤ gn, and
gn →
g, then
fn →
f. (Rework the proof of the dominated convergence theorem.) Solution Again, as in the proof of Thereom 2.24 (The Dominated Convergence Theorem), it suffices to assume all functions involved are real-valued, in which case we have gn + fn ≥ 0 a.e. and gn − fn ≥ 0 a.e. By Theorem 2.18 (Fatou’s Lemma), ∫ g +
f ≤ lim inf
(gn + fn) = lim
gn + lim inf
fn =
g + lim inf
fn,
∫ g −
f ≤ lim inf
(gn − fn) = lim
gn − lim sup
fn =
g − lim sup
fn,
where we have used the fact that lim inf(an + bn) = lim an + lim inf bn if {an} converges. Hence lim inf
fn ≥
f ≥ lim sup
fn, and the result follows.
