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A generalized version of the second Borel-Cantelli lemma says

Theorem 5.3.2. Second Borel-Cantelli lemma, II. Let $\mathcal F_n, n \ge 0$ be a filtration with $F_0 = \{\emptyset, \Omega\}$ and $A_n , n \ge 1$ a sequence of events with $A_n ∈ \mathcal F_n$ . Then $$ \{A_n \,i.o.\} = \left\{\sum_{n \ge 1} P (A_n |\mathcal F_{n−1}) =\infty \right\}. $$

An exercise of this lemma is

Exercise 5.3.6. Show $\sum_{n \ge 2} \mathbb P (A_n | \cap_{m=1}^{n−1} A_m^c ) = \infty$ implies $P (∩_{m \ge 1} A_m^c ) = 0$.

I think we can actually prove that $\cap_{m \ge 1} A_m^c = \emptyset$. My proof is like this: Let $B_n = \cup_{m = 1}^n A_n$. Let $\mathcal F_n = \sigma(A_1,\dots,A_n)$, which forms a filtration. It is easy to verify that $$ \mathbb P (A_n | B_{n-1}^c ) = \mathbb P (A_n | \mathcal F_{n-1})(\omega) $$ when $\omega \in B_{n-1}^c$.

Assume there exist $\omega \in \cap_{m \ge 1} A_m^c$. It follows from the previous lemma that $$ \omega \in \left\{\sum_{n \ge 1} P (A_n |\mathcal F_{n−1}) =\infty \right\} = \{A_n \,i.o.\}. $$ In other words, $\omega \in \cap_{m \ge 1} \cup_{n \ge m} A_n$, which contradicts $\omega \in \cap_{m \ge 1} A_m^c$.

Is this the right proof?

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+1 for a well-explained and well-formatted question! – Zev Chonoles Feb 23 '13 at 17:39
up vote 1 down vote accepted

The set equality in the lemma you mention only holds a.s., so it's not possible to prove that $\cap_{n \ge 1} A_n^C = \emptyset$. The best you can prove is that it has probability $0$. Apart from that, your proof looks correct.

To fix the proof, you just need to conclude from the lemma that since, as you correctly point out, $$\bigcap_{n\ge 1} A_n^C\subseteq \left\{\sum_{n \ge 1} P (A_n |\mathcal F_{n−1}) =\infty \right\},$$ then, up to a set of probability $0$, $\{A_n \ i.o.\}$ contains the set $\cap_{n\ge 1} A_n^C$. As these two sets are disjoint, this implies that $P(\cap_{n\ge 1} A_n^C)=0$.

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