# A variation of Borel Cantelli Lemma

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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What does $P(A_ni. o.) = 0$ mean? Also, if this is homework, please tag it as such. –  mixedmath May 29 '11 at 23:12
Probably $P(\limsup A_n)$. (infinitely often). Further continuity from above/below of the measure. –  Jonas Teuwen May 29 '11 at 23:12
@epsilon: Care to accept an answer? –  Did Aug 15 '11 at 13:00
How do you prove this if the complement is switched, i.e. if we know $$\sum_{n=1}^{\infty}{P(A_n\cap A_{n+1}^c})<\infty$$ –  nelson meier Oct 13 '12 at 18:35
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.