Proving this statement without the independence condition (Borel 0-1 Law) 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.
 A: 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.
