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Wednesday, May 29, 2013

1.16

This exercise is asking us to prove the Borel-Cantelli Lemma. In the measure theory settings, it states:
Suppose \lbrace E_n \rbrace_{n=1}^{\infty} is a countable family of subsets of \mathbb{R}^d, and (\dagger) \sum_{k=1}^{\infty} m(E_k) < \infty The set E = \bigcap_{n=1}^{\infty} \bigcup_{k \geq n} E_k has measure zero.
Part a.) We first need to show E is a measurable set. Recall that the measurable sets form a \sigma-algebra, \mathcal{M}, which is closed under countable unions and intersections.

Since each E_k \in \mathcal{M}, certainly A_n = (\bigcup_{k \geq n} E_k) \in \mathcal{M} \hspace{0.25cm} \forall n. It follows that since E = \bigcap_{n=1}^{\infty} A_n, that E \in \mathcal{M}.

Part b.) Assume to the contrary that m(E) = \delta > 0. Notice that if we define: S_N = \bigcap_{k=1}^N \bigcup_{n \geq k} E_n = \bigcup_{n \geq N} E_n Then, certainly, S_N \searrow E, and \forall N, \delta \leq m(S_N) = m(\bigcup_{n \geq N} E_n) \leq \sum_{n=N}^{\infty} m(E_n) Which certainly contradicts \dagger, completing the proof.

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