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If $P(A_n) \rightarrow 0$ and $\sum_{n=1}^{\infty}{P(A_n^c\cap A_{n+1}})<\infty$ then $P(A_n \text{ i.o.})=0$.

How to prove this? Thanks.

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    How do you prove this if the complement is switched, i.e. if we know \sum_{n=1}^{\infty}{P(A_n\cap A_{n+1}^c})<\infty 2012-10-13

1 Answers 1

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Hint: $\lim \sup A_n \subseteq A_N \cup \bigcup_{n=N}^\infty (A^c_n \cap A_{n+1})$. Estimate the probability of this.

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    Do you think the RHS will go to 0 if we take probability in both side?2016-10-07