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

$$E^c=\bigcup _{k=1}^{\mathrm{\infty }}\bigcap _{i=k}^{\mathrm{\infty }}{A}_i^c$$

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

$P({\cap }_{i=k}{A}_i^c)$ $={\displaystyle \prod _{i=k}^{\mathrm{\infty }}}P(A_i^c)$

$={\displaystyle \prod _{i=k}^{\mathrm{\infty }}}(1-P(A_i))$

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

$$P({\cap }_{i=k}{A}_i^c)\le \exp (-\sum _{i=k}^{\mathrm{\infty }}P(A_i))=0$$

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
Canonical name	ProofOfBorelCantelli2
Date of creation	2013-03-22 14:29:35
Last modified on	2013-03-22 14:29:35
Owner	kshum (5987)
Last modified by	kshum (5987)
Numerical id	4
Author	kshum (5987)
Entry type	Proof
Classification	msc 60A99