probability theory - $P(limsup A_n)=1 $ if $forall A in mathfrak{F}$ s.t. $sum_{n=1}^{infty} P(A cap A_n) = infty$

Let $\mathfrak{F} = (A_n)_{n \in \mathbb{N}}$. Prove that $P(\limsup A_n)=1$ if $\forall A \in \mathfrak{F}$ s.t. $P(A) > 0, \sum_{n=1}^{\infty} P(A \cap A_n) = \infty$. (Side question 1 Is second Borel-Cantelli out because we don't know if the $A_n$'s are independent?)

Suppose $\forall A \in \mathfrak{F}$ s.t. $P(A) > 0, \sum_{n=1}^{\infty} P(A \cap A_n) = \infty$, but $P(\limsup A_n)<1$. My prof gave us this convenient hint: Show $\exists M > 0$ s.t. $P(\bigcap_{j=M}^{\infty} A_j^{c}) > 0$.

$1 > P(\limsup A_n)$

$1 - P(\limsup A_n) > 0$

$P(\liminf A_n^{c}) > 0$

By definition of $\liminf A_n^{c}$, $\exists M > 0 $ s.t. $P(\bigcap_{j=M}^{\infty} A_j^{c}) > 0$. (Side questions 2 and 3 Isn't this =1? ...Wait, is that what we're trying to prove? Haha)

$\to 1 - P(\bigcup_{j=M}^{\infty} A_j) > 0$

$\to 1 > P(\bigcup_{j=M}^{\infty} A_j)$

Main question: Now I guess we have to come up with some A s.t.

$\to 1 > P(\bigcup_{j=M}^{\infty} A_j) \geq \sum_{n=1}^{\infty} P(A \cap A_n)$, or is there something else to do?

Answer

Show that the series $\sum\limits_n P(A \cap A_n)$ converges for $A=\bigcap\limits_{n=M}^{\infty} A_n^{c}$.