The concept of infinitely often and the Borel Cantelli lemma

In order for an event $\omega$ to belong to infinitely many of the $A_n$, we must have the following: if we take any $n\ge 1$, $\omega$ must belong to at least one of the sets $A_n,A_{n+1},A_{n+2},\ldots)$ for all $n\ge 1$. Thus we can write

Next, note the equivalence:

Thus $\eqref{a154401}$ is equivalent to

Next, observe that, given a sequence of sets $B_m$,

In particular, taking $B_n=\bigcup_{m=n}^\infty A_m$, we see that $\eqref{a154501}$ is equivalent to

DEFINITION. Let $A_{1}, \ldots, A_{n}, \ldots$ be events in a probability space. Then the event $\left\{\omega \in \Omega | \omega \text { belongs to infinitely many of the } A_{n}\right\}=\bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} A_{m}$ is called " $A_{n}$ infinitely often", abbreviated " $A_{n}$ i.o.".

The Borell Cantelli lemma states that if $\sum_{n=1}^\infty P(A_n)<+\infty$ then the probability that an event takes place infinitely often is zero. The proof of this statement is based on the fact that if a series converges then its residual sum tends to zero, i.e., $\lim_{n\to \infty}\sum_{m\ge k}P(A_m)=0$. Now, observe that $\sum_{m\ge n}P(A_m)\ge P(\cup_{m\ge n}A_m)=P(B_n)$ where $B_n=\cup_{m\ge n}A_m$ . Thus, $\lim_n P(B_n)=0$. Since the elements $B_m$ constitute a monotonic sequence, we have $P(\cap_{n}B_n)=\lim_{n}P(B_n)$. We therefore conclude that $P(\cap_n B_n)=0$. Thus the probability of an event taking place infinitely often is zero.