# Borel Cantelli-Type problem from Billingsley's Probability and Measure

Problem 4.11(d) from Billingsley's Probability and Measure book states: Show that $$P\left(\limsup A_n\right)=1$$ if and only if $$\sum_n P(A\cap A_n)$$ diverges for each $$A$$ of positive probability.

Notably, there is no assumption that the $$A_i's$$ are independent. I'm not sure how to proceed here! I assume there is a nice trick. Any hints are most welcome.

It seems to be simpler to show the equivalence of the negation of the assertions, that is, we will prove that $$\mathbb P\left(\limsup_{n\to\infty}A_n\right)<1\Leftrightarrow \mbox{ there exists }B \mbox{ of positive probability such that }\sum_n\mathbb P\left(B\cap A_n\right)<\infty.$$ Let us show $$\Leftarrow$$. By the classical Borel-Cantelli lemma applied to $$(B\cap A_n)_n$$, we derive that $$\mathbb P\left(\limsup_n (B\cap A_n)\right)=0$$. By elementary operations on sets, we derive that $$\mathbb P(A\cap B)=0$$, where $$A=\limsup_nA_n$$. As a consequence, $$\mathbb P(A)=\mathbb P(A\cap B^c)\leqslant \mathbb P(B^c)$$. Since $$\mathbb P(B)>0$$, $$\mathbb P(B^c)<1$$ hence $$\mathbb P(A)<1$$.
It remains to show $$\Rightarrow$$. By assumption, $$\mathbb P(\liminf_n A_n^c)=\mathbb P\left(\left(\limsup_n A_n\right)^c\right)>0$$. By definition of $$\liminf$$, there exists a $$k_0$$ such that $$\mathbb P\left(\bigcap_{k\geqslant k_0}A_k^c\right)>0$$. Let $$B=\bigcap_{k\geqslant k_0}A_k^c$$. For $$n\geqslant k_0$$, $$B\cap A_n$$ is empty hence the series $$\sum_n\mathbb P\left(B\cap A_n\right)$$ converges.