Let $E_n$ be sequence of arbitrary events.
$P(E_n\ \ \text{i.o.}) = 1$ if and only if $\sum_n P(A\cap E_n) =\infty$ for all $A$ with $P(A)>0$.
Question
It is a very short proof, but couldn't make sure if it has a logic flaw. I appreciate if one can verify.
Thanks in advance.
Attempt
If $P(E_n\ \ \text{i.o.}) = 1$, first note that $A\cap\{E_n \text{ i.o.}\}=\{A\cap E_n \text{ i.o.}\}$ hence
\begin{equation}
0 < P(A) = P(A\cap \{E_n \text{ i.o.}\}) = P(\{A\cap E_n \text{ i.o.}\})
\end{equation}
By Borel-Cantelli Lemma, $\sum_n P(A\cap E_n)$ is not finite. (Contradiction otherwise.)
If $P(E_n\ \ \text{i.o.}) < 1$, note that $\cup_{n>N} E_n$ decreases to $\{E_n \text{ i.o.}\}$.
Fix any $N$ s.t. $P(\cup_{n>N} E_n) < 1$ and let $A= \Omega - \cup_{n>N} E_n$. Now $P(A)>0$ but,
\begin{equation}
\sum_n P(A \cap E_n) = E\sum_n \mathbb{1}\{A\cap E_n\} \leq N < \infty
\end{equation}
Question
Is there a direct argument by assuming $\sum_n P(A\cap E_n)=\infty$? I believe not, however if there is, it will be surely instructive.
Note$\{E_n \text{ i.o.}\}$ is $\limsup E_n$.
Note $\mathbb{1}\{A\}$ is the indicator function on the set $A$.