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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