proof of Borel-Cantelli 2

Let E denote the set of samples that are in Ai 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


where the superscript c means set complement. But for each k,

P(i=kAic) =i=kP(Aic)

Here we use the assumptionPlanetmathPlanetmath that the event Ai’s are independentPlanetmathPlanetmath. The inequality 1-ae-a and the assumption that the sum of P(Ai) diverges together imply that


Therefore Ec is a union of countableMathworldPlanetmath number of events, each of them has probability zero. So P(Ec)=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