Given a probability space $(\Omega,\mathscr{F},P)$ and a sequence $A_1,\cdots \in \mathscr{F}$, show that $P(\lim\sup A_n) = 1 $ if and only if $\sum_{n=1}^\infty P(A_n|A)$ diverges for all $A$ with $P(A)>0$. Hint: Show that $P(A_n \ i.o)<1 \iff \sum_{n=1}^\infty P(A_n|A)<\infty$ for some $A$ with $P(A)>0$.
I tried proving the hint, for $\Rightarrow$ we take $B_m=\bigcup_{k=m}^\infty A_k$ from which, since $P(A_n \ i.o)<1$ then $P(B_m) \downarrow P(A_n \ i.o) < 1$ and therefore $P(B_m) < 1$ for some $m$, taking $A=B_m^c = \bigcap_{k=m}^\infty A_k^c$ we find
$$\sum_{n=1}^\infty P(A_n|B_m^c)\le \frac{1}{P(B_m^c)}(P(A_1)+\cdots P(A_{m-1})) < \infty $$
I tried proving $\Leftarrow$ then: Since $\sum P(A_n|A)$ converges then $\sum P(A_n \cap A)$ also does and therefore $P(\lim\sup (A_n\cap A)) = 0$ from Borel Cantelli's first lemma, I still can't jump from this to $P(\lim \sup A_n)<1$ since the only restriction on $A$ is that it has positive probability.
Edit: I changed the question so the partial solution is actually correct
The solution to $\Leftarrow$ is: $P(\lim\sup(A_n \cap A)) = P(\lim\sup(A_n) \cap A) = 0$ implies that $P(\lim\sup A_n) = P(\lim\sup (A_n) \cap A^c) \le P(A^c) < 1$
Best Answer
Let $f=\sum I_{A_n}$, an extended real valued measurable function.
Then $f(\omega)=\infty$ iff $\omega \in \lim\sup A_n $.
If $P(\lim\sup A_n)=1$ then $f=\infty$ a.s. so $\int_A fdP=\infty$ for every set $A$ of positive probability. Hence, $\sum P(A \cap A_n)=\infty$ for every set $A$ of positive probability.
Conversely, if $\sum P(A \cap A_n)=\infty$ for every set $A$ of positive probability then $\int_A fdP=\infty$ for every set $A$ of positive probability. Taking $A=(f\leq n)$ we see that $P(f\leq n)=0$ for every $n$ so $f=\infty$ a.s., so $P(\lim\sup A_n)=1$.