1.2. Lebesgue measure 27 (vi) If E1,E2,E3,... ⊂ Rd are a sequence of Lebesgue measurable sets, then the union ∞ n=1 En is Lebesgue measurable. (vii) If E1,E2,E3,... ⊂ Rd are a sequence of Lebesgue measurable sets, then the intersection ∞ n=1 En is Lebesgue measurable. Proof. Claim (i) is obvious from definition, as are Claims (iii) and (iv). To prove Claim (vi), we use the ε/2n trick. Let ε 0 be arbitrary. By hypothesis, each En is contained in an open set Un whose difference Un\En has Lebesgue outer measure at most ε/2n. By countable subaddi- tivity, this implies that ∞ n=1 En is contained in ∞ n=1 Un, and the differ- ence ( ∞ n=1 Un)\( ∞ n=1 En) has Lebesgue outer measure at most ε. The set ∞ n=1 Un, being a union of open sets, is itself open, and the claim follows. Now we establish Claim (ii). Every closed set E is the countable union of closed and bounded sets (by intersecting E with, say, the closed balls B(0,n) of radius n for n = 1, 2, 3,...), so by (vi), it suffices to verify the claim when E is closed and bounded, hence compact by the Heine-Borel theorem. Note that the boundedness of E implies that m∗(E) is finite. Let ε 0. By outer regularity (Lemma 1.2.12), we can find an open set U containing E such that m∗(U) ≤ m∗(E) + ε. It suffices to show that m∗(U\E) ≤ ε. The set U\E is open, and so by Lemma 1.2.11 is the countable union ∞ n=1 Qn of almost disjoint closed cubes. By Lemma 1.2.9, m∗(U\E) = ∑∞ n=1 |Qn|. So it will suffice to show that ∑N n=1 |Qn| ≤ ε for every finite N. The set N n=1 Qn is a finite union of closed cubes and is thus closed. It is disjoint from the compact set E, so by Exercise 1.2.4 followed by Lemma 1.2.5 one has m∗(E ∪ N n=1 Qn) = m∗(E) + m∗( N n=1 Qn). By monotonicity, the left-hand side is at most m∗(U), which is in turn at most m∗(E) + ε. Since m∗(E) is finite, we may cancel it and conclude that m∗( N n=1 Qn) ≤ ε, as required. Next, we establish Claim (v). If E is Lebesgue measurable, then for every n we can find an open set Un containing E such that m∗(Un\E) ≤ 1/n. Letting Fn be the complement of Un, we conclude that the comple- ment Rd\E of E contains all of the Fn, and that m∗((Rd\E)\Fn) ≤ 1/n. If we let F := ∞ n=1 Fn, then Rd\E contains F , and from monotonicity m∗((Rd\E)\F ) = 0, thus Rd\E is the union of F and a set of Lebesgue outer measure zero. But F is in turn the union of countably many closed sets Fn. The claim now follows from (ii), (iii), (iv).
Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2011 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.