$P(\limsup A_n)=1 \Leftrightarrow \sum_{n=1}^{\infty} P(A \cap A_n) = \infty\; \forall A, P(A)>0$

Question

Let $\{A_n\}$ be a sequence of independent events. How to prove that $$P(\limsup A_n)=1 \Leftrightarrow \sum_{n=1}^{\infty} P(A \cap A_n) = \infty\; \forall A, P(A)>0?$$

As the $A_n$ are independent I guess I should be using Borel-Cantelli ($\sum_{n=1}^{\infty} P(A_n)=\infty \Rightarrow P(\limsup A_n)=1$) somewhere, but I'm not sure how to get from $P(A \cap A_n)$ to $P(A_n)$.

Answer 1

Note that $\omega \in \lim \sup A_n$ iff $\sum_n I_{A_n}(\omega)=\infty$. Hence, $P(\lim \sup A_n)=1$ implies $\sum_n I_{A_n}=\infty$ almost surely. This implies that $\int_A \sum_n I_{A_n}dP=\infty$ whenver $P(A)>0$. Hence $\sum P(A\cap A_n)= \infty$. The converse follows by Borel Cantelli Lemma: just take $A=\Omega$.

Answer 2

Since $\mathbb P(A^c) <1$ we have $$\mathbb P(\liminf (A_n\cap A)^c) = \mathbb P(\liminf A_n^c \cup A^c) \leq \underbrace{\mathbb P(\liminf A_n^c)}_{0} + \mathbb P(A^c) <1. $$ Equivalently, $\mathbb P(\limsup A_n\cap A) >0$ and $\sum \mathbb P(A_n\cap A)=\infty$ by Borel-Cantelli.