In Durett's book, the second Borel Cantelli lemma is as follows:

Let $\{F_n\}$ be a filtration, and $A_n\in F_n$ be a sequence of events. Then, $\{A_n \text{ i.o.}\}=\{\omega:\sum_{n=1}^\infty P(A_n|F_{n-1})=\infty\}$ a.s..

But the more common version is: Let $A_n$ be a sequence of events. If $\sum_n P(A_n)=\infty$, and $A_n$ are independent, then $P(A_n \text{ i.o.})=1$.

How can Durett's version imply the common version? It seems that the independence of $A_n$ cannot imply $A_n$ is independent with $F_{n-1}$.

Thank you for your help!


Assume the $A_i$ are independent.

Take $F_n := \sigma (A_1 , \ldots , A_n)$. Then $A_n$ is independent from $F_{n-1}$. Thus $P(A_n \vert F_{n-1}) = P(A_n )$.

Now if $\sum_n P (A_n ) = \infty$ then

$$P( A_n \text{ i.o.}) = P\left( \sum_n P(A_n \vert F_{n-1}) = \infty\right) = P \left( \sum_n P(A_n) = \infty \right) = P (\Omega)= 1$$

But (for completeness) there is more: The common Borel-Cantelli usually includes also that for arbitrary $A_n$, if $\sum_n P(A_n ) < \infty$, then $P(A_n \text{ i.o })=0$.

By monotone convergence theorem:

$$E[\sum_n P (A_n \vert F_{n-1} )]= \sum_n E[P(A_n \vert F_{n-1})] = \sum_{n} P(A_n) <\infty$$ This implies that $P(\sum_n P(A_n \vert F_{n-1}) = \infty ) = 0$, thus $P(A_n \text{ i-o.}) = 0$.


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