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The Borel-Cantelli Lemma: Let $\{E_k\}_{k=1}^\infty$ be a countable collection of measurable sets for which $\sum_{k=1}^\infty m(E_k) < \infty$. Then almost all $x \in \mathbb R$ belong to at most finitely many of the $E_k$'s.

Here is the proof from Royden's book. For each n, the the countable subadditivity of measure, $$m(\bigcup_{k=n}^\infty E_k) \leq \sum_{k=n}^\infty m(E_k) < \infty.$$

Hence, by the countinuity of measure, $$m(\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)) = \lim_{n \to \infty} m(\bigcup_{k=n}^\infty E_k) \leq \lim_{n \to \infty} \sum_{k=n}^\infty m(E_k) = 0.$$

Therefor almost all $x \in \mathbb R$ fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$ and therefore belong to at most finitely many $E_k$'s.

I don't understand the last sentence. How can we conclude that if almost all $x \in \mathbb R$ fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$ then they belong to at most finitely many $E_k$'s. If $x \in \bigcap_{n=1}^\infty (\bigcup_{k=2n}^\infty E_k)$, it may fail to belong to $\bigcap_{n=1}^\infty (\bigcup_{k=n}^\infty E_k)$, but it still belongs to infinitely many $E_k$'s.

BCLC
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TH000
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1 Answers1

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Suppose that $x \notin \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$. Applying de Morgan, we find that $$ x \in \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty \overline{E_k}. $$ Therefore for some $n \ge 1, x \in \overline{E_k}$ for $k \ge n$ or $$ x \in \bigcap_{k=n}^\infty \overline{E_k}. $$ Applying de Morgan again, $$ x \notin \bigcup_{k=n}^\infty E_k. $$ So $x$ can only belong to $E_1,\ldots,E_{n-1}$, a finite number of sets.

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Yuval Filmus
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  • Thank you for the great answer, but can you tell me where my logic goes wrong? – TH000 May 04 '15 at 03:06
  • Try using your argument to construct a counterexample. Then see what goes wrong when you apply my proof to your purported counterexample. – Yuval Filmus May 04 '15 at 03:07
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    I think I get it. If $x \notin \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$, it is impossible that $x \in \bigcap_{n=1}^\infty \bigcup_{k=2n}^\infty E_k$. – TH000 May 04 '15 at 03:16