# proof of Borel-Cantelli 2

Let $E$ denote the set of samples that are in $A_{i}$ infinitely often. We want to show that the complement of $E$ has probability zero.

As in the proof of Borel-Cantelli 1, we know that

 $E^{c}=\bigcup_{k=1}^{\infty}\bigcap_{i=k}^{\infty}A_{i}^{c}$

where the superscript ${}^{c}$ means set complement. But for each $k$,

 $\displaystyle P(\cap_{i=k}A_{i}^{c})$ $\displaystyle=\prod_{i=k}^{\infty}P(A_{i}^{c})$ $\displaystyle=\prod_{i=k}^{\infty}(1-P(A_{i}))$

Here we use the assumption that the event $A_{i}$’s are independent. The inequality $1-a\leq e^{-a}$ and the assumption that the sum of $P(A_{i})$ diverges together imply that

 $P(\cap_{i=k}A_{i}^{c})\leq\exp(-\sum_{i=k}^{\infty}P(A_{i}))=0$

Therefore $E^{c}$ is a union of countable number of events, each of them has probability zero. So $P(E^{c})=0$.

Title proof of Borel-Cantelli 2 ProofOfBorelCantelli2 2013-03-22 14:29:35 2013-03-22 14:29:35 kshum (5987) kshum (5987) 4 kshum (5987) Proof msc 60A99