1.2. Lebesgue measure 31 for every ε 0, while from countable subadditivity we have m( ∞ n=1 En) ≤ ∞ n=1 m(En). The claim follows. Finally, we handle the case when the En are not assumed to be bounded or closed. Here, the basic idea is to decompose each En as a countable disjoint union of bounded Lebesgue measurable sets. First, decompose Rd as the countable disjoint union Rd = ∞ m=1 Am of bounded measurable sets Am for instance, one could take the annuli Am := {x ∈ Rd : m − 1 ≤ |x| m}. Then each En is the countable disjoint union of the bounded measurable sets En ∩ Am for m = 1, 2,..., and thus m(En) = ∞ m=1 m(En ∩ Am) by the previous arguments. In a similar vein, ∞ n=1 En is the countable disjoint union of the bounded measurable sets En ∩ Am for n, m = 1, 2,..., and thus m( ∞ n=1 En) = ∞ n=1 ∞ m=1 m(En ∩ Am), and the claim follows. From Lemma 1.2.15 one of course can conclude finite additivity, m(E1 ∪ . . . ∪ Ek) = m(E1) + . . . + m(Ek), whenever E1,...,Ek ⊂ Rd are Lebesgue measurable sets. We also have another important result: Exercise 1.2.11 (Monotone convergence theorem for measurable sets). (i) (Upward monotone convergence) Let E1 ⊂ E2 ⊂ . . . ⊂ Rn be a countable non-decreasing sequence of Lebesgue measurable sets. Show that m( ∞ n=1 En) = limn→∞ m(En). (Hint: Express ∞ n=1 En as the countable union of the lacunae En\ n−1 n =1 En .) (ii) (Downward monotone convergence) Let Rd ⊃ E1 ⊃ E2 ⊃ . . . be a countable non-increasing sequence of Lebesgue measurable sets. If at least one of the m(En) is finite, show that m( ∞ n=1 En) = limn→∞ m(En). (iii) Give a counterexample to show that in the hypothesis that at least one of the m(En) is finite in the downward monotone convergence theorem cannot be dropped.

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