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).
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