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I am trying to prove this statement, "if $P(S_n) \to 1$ as $n \to \infty$, prove that there exists subsequence $\{n_k\}$ such that $P(\cap_{n_k}S_{n_k}) > 0$".

As $lim_{n \to \infty} P(S_n) = 1 \ne 0 \Rightarrow \sum_{n=1}^{\infty} P(S_n) = \infty$.

If sequence $\{S_n\}$ is independent, then I can use Borel 0-1 law to get $P(S_n \text{ i.o}) = 1 \Rightarrow P(\{s : s \in S_{n_k} \}) = 1$ for k = 1,2, ... $\Rightarrow P(\cap_{n_k} S_{n_k}) = 1$.

On the other hand, how do I prove this statement without the independence condition ? Because I can not use Borel 0-1 law otherwise.

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I don't think the Borel-Cantelli lemma is suitable for this exercise. To solve it, you will only need to use the basic definitions of probability theory.

Three ingredients are involved. (1) Since $P\left(S_{n}\right)\to1$, we know that $P\left(S_{n}^{c}\right)\to0$. (2) By De Morgan's law, $P\left(\bigcap_{n=1}^{\infty}S_{n}\right)=1-P\left(\bigcup_{n=1}^{\infty}S_{n}^{c}\right)$. (3) By countable subadditivity, $P\left(\bigcup_{n=1}^{\infty}S_{n}^{c}\right)\leq\sum_{n=1}^{\infty}P\left(S_{n}^{c}\right)$. It's probably a good idea to show the result using these hints.

I'll show a slightly more general result, namely that for any $\delta>0$ there is subsequence $\delta$-dependent subsequence $n_{k}$ such that $P\left(\bigcap_{k=1}^{\infty}S_{n_{k}}\right)\geq1-\delta$: Since $P\left(S_{n}^{c}\right)\to0$, there is a subsequence $S_{n_{k}}$ satisfying $P\left(S_{n_{k}}^{c}\right)<\delta2^{-k}$, which implies $\sum_{k=1}^{\infty}P\left(S_{n_{k}}^{c}\right)<\delta$. By De Morgan's law and subadditivity, $P\left(\bigcap_{n=1}^{\infty}S_{n}\right)\geq1-\delta$ as claimed.

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