30 1. Measure theory Proof. The first claim is trivial, so we focus on the second. We deal with an easy case when all of the En are compact. By repeated use of Lemma 1.2.5 and Exercise 1.2.4, we have m( N n=1 En) = N n=1 m(En). Using monotonicity, we conclude that m( n=1 En) N n=1 m(En). (We can use m instead of m∗ throughout this argument, thanks to Lemma 1.2.13). Sending N we obtain m( n=1 En) n=1 m(En). On the other hand, from countable subadditivity one has m( n=1 En) n=1 m(En), and the claim follows. Next, we handle the case when the En are bounded but not necessarily compact. We use the ε/2n trick. Let ε 0. Applying Exercise 1.2.7, we know that each En is the union of a compact set Kn and a set of outer measure at most ε/2n. Thus m(En) m(Kn) + ε/2n and hence n=1 m(En) ( n=1 m(Kn)) + ε. Finally, from the compact case of this lemma we already know that m( n=1 Kn) = n=1 m(Kn) while from monotonicity m( n=1 Kn) m( n=1 En). Putting all of this together we see that n=1 m(En) m( n=1 En) + ε
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