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?
Best Answer
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 $P(\cap_{n\ge 1} A_n^C)=0$.