Borel-Cantelli Lemma

Theorem
Given a measure space $$(X, \Sigma, \mu)$$ and a countable collection of measurable sets $$E_{n} \subseteq \Sigma$$, if $$ \sum_{n=1}^{\infty}\mu \Big(E_{n}\Big) < \infty$$ then $$ \mu \Big(\limsup_{n \to \infty} E_{n}\Big) = 0.$$

Proof
By definition, $$\limsup_{n\to \infty}E_{n} = \bigcap_{i=1}^{\infty}\bigcup_{j=i}^{\infty}E_{j}$$.

Thus, by the monoticity of the measure $$\mu$$, $$\mu \Big(\limsup_{n \to \infty} E_{n}\Big) = \mu \Big(\bigcap_{i=1}^{\infty}\bigcup_{j=i}^{\infty}E_{j}\Big) \leq \mu \Big( \bigcup_{j=i}^{\infty}E_{j}\Big)$$ for each $$i$$.

By the subadditivity of $$\mu$$, $$\mu \Big( \bigcup_{j=i}^{\infty}E_{j}\Big) \leq \sum_{j=i}^{\infty}\mu\Big(E_{j}\Big)$$.

However, by assumption $$\sum_{n=1}^{\infty}\mu\Big(E_{n}\Big)$$ converges, and the tails of a convergent series themselves converge to zero.

Hence by selecting an appropriate $$i$$, $$\sum_{j=i}^{\infty}\mu\Big(E_{j}\Big)$$ can be made arbitrarily small, so $$\mu \Big(\limsup_{n \to \infty} E_{n}\Big) = 0$$.

Borel-Cantelli Lemma in Probability
As each probability space $$(X, \Sigma, \Pr)$$ is a measure space, the result carries over to probability theory.

Hence, given any sequence of countable events $$E_{n}$$ the sum of whose probabilities is finite, the probability that infinitely many of the events occur is zero.