Problem Related To Borel Cantelli Lemma and independence

{An} is a sequence of events which may NOT be independent. Also A is any event such that P(A) greater than zero. Again the question says nothing about whether {An} is an independent sequence and also it says nothing about whether A or{An} are independent (we assume no independence).

we are asked to prove P[An i.o.]=1 iff ∑ P(A ∩ An) = ∞. (Sum goes fron n=1 to infinity).

Here is what I have so far. I think if we assume that ∑ P(A ∩ An) = ∞, That means P[(A ∩ An) i.o.]=1, that is the lim sup (A ∩ An) =1. If I am not wrong,lim sup (A ∩ An) is a subset of lim sup (An).Therefore P(lim sup (An)) must be greater than or equal to P (lim sup (A ∩ An)) and we have proved that P[An i.o.]= P( lim sup An ) = 1.

If we assume P[An i.o.]= P( lim sup An ) = 1, how can we prove this implies ∑ P(A ∩ An) = ∞ ? Once again please remember we cannot use independence.

can someone enlighten me on this one please :)

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This can't be true in general, since the assumption of independence (or at least pairwise independence) is required; see en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma, the section "Converse result". –  Shai Covo Nov 30 '10 at 1:30

This is not correct as stated. Suppose $B$ is an event with probability $1/2$, and let $A$ and all the $A_n$ be equal to $B$. Then $P(A_n \text{i.o.}) = P(B) = 1/2$ but $\sum_{n=1}^\infty P(A \cap A_n) = \sum_{n=1}^\infty \frac{1}{2} = \infty$.

Your sentence beginning "I think if we assume" is not true in general. It is the so-called second Borel-Cantelli lemma, which requires independence.

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Hmm. I see i dont understand B-C lemma after all. Thanks to Nate and Shai for your kind response. –  useroo7 Dec 2 '10 at 3:01
I hope you will be kind enough to help me understand the following. Is it true that P ([An i.o.]∩ A) = P [( A ∩ An) i.o.]. If so, how do we prove this? or is it trivial? And assuming that we know that P[An i.o.] =1 , can we establish that the sum {P(A ∩ An)} diverges (meaning = infinity) ? –  useroo7 Dec 2 '10 at 3:09
$[An\cap A \ i.o] = lim\ sup\ (An\cap A) \subset (lim\ sup\ An)\cap lim\ sup\ A = [An \ i.o]\cap A$. So $P([An\cap A \ i.o])\leq P([An \ i.o]\cap A)$ –  Rodolfo Apr 19 '12 at 21:11