# limit of probability

Assume $\{A_n\}$ is a sequence of events satisfying $\mathop\lim\limits_{n\to\infty}P\{A_n\}=1$, $\{B_n\}$ is a sequence of events satisfying $\mathop\lim\limits_{n\to\infty}P\{B_n\}=1$. Do we have $\mathop\lim\limits_{n\to\infty}P\{A_n\cap B_n\}=1$? If it holds, how to prove it? Thank you.

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$P(A_n \cup B_n) = P(A_n) + P(B_n) - P(A_n \cap B_n)$ by the inclusion-exclusion principle. Does $P(A_n \cup B_n) \rightarrow 1$? If so, what does that say about $\lim_n P(A_n \cap B_n)$? –  A Blumenthal Jan 20 '13 at 23:43
Using the same method Mercy gave, we can prove that $P(A_n\cup B_n)\to 1$. –  Zhou Heng Jan 20 '13 at 23:56

We have $$1 \ge P(A_n\cap B_n)=P(A_n)+P(B_n)-P(A_n\cup B_n) \ge P(A_n)+P(B_n)-1\ \forall n.$$ Passing to the limit we get $$\lim_nP(A_n\cap B_n)=1.$$