Given $\{A_n\}$, with $P\left(A_n\right)\to 1$, there exists a subsequence $\{n_k\}_{k\ge 1} \;$ such that $\; P\left(\bigcap_k A_{n_k}\right) > 0$

Question

This is problem 4.21a is Resnick's Probability Path.

Suppose $\{A_n\}$ is a sequence of events. If $P(A_n)\to 1$ as $n \to \infty$, prove that there exists a subsequence $\{n_k\}$ tending to infinity such that $P(\bigcap_k A_{n_k})>0$. (Hint: Use Borel-Cantelli.)

Having searched a bit on here for related problems, I've gotten the impression that what most people mean by Borel-Cantelli is something a little more expansive than what Resnick means, which is that for some sequence of (not necessarily independent) events $\{A_n\}$, if $\sum P(A_n) < \infty$, then $P(\limsup_n A_n) = 0$. Since the $\{A_n\}$ in question are not independent, I can't use any results on sequences whose sums don't converge.

Clearly, $P(A_n)\to 1$ implies that $\sum P(A_n) = \infty$, and so I can't directly apply Borel-Cantelli. Moreover, even though $P(A_n^c)\to 0$, this does not imply that $\sum P(A_n^c)<\infty$, so I can't do anything with the complements. I'm not sure what else to do: I mean, if I could somehow determine that $P(\limsup_n A_n) = 1$, I could show that $\sum P(A_n) = \infty$ ... but then I could just reread the first sentence of this paragraph.

Could anyone offer any suggestions?

Answer 1

I see no need to use Borel-Cantelli here. Observe that $$P\left(\bigcap_{k=1}^{\infty} A_{n_k}\right) > 0 \quad \Leftrightarrow \quad P\left(\bigcup_{k=1}^{\infty} A_{n_k}^{c}\right) < 1$$ Now because $P(A_{n}^{c}) \to 0$, we can choose a subsequence $n_{k}$ such that $$P(A_{n_k}^{c}) < 3^{-(k+1)}$$

Answer 2

Take any $A_i$ s.t. $P(A_i)=a>0$. Take any series of positive numbers $b_i$ s.t. $\sum_1^{\infty} b_i=b<a$. Let $n_0=i$ and for each $i\geq 1$ pick $n_i>n_{i-1}$ s.t. $P(A^c_{n_i})<1-b_i$. Then $P(\bigcap_k A_{n_k})> a-b>0$