Borel-Cantelli Lemma

Theorem
Let $(X, \Sigma, \mu)\ $ be a measure space, and $E_{n} \subseteq \Sigma$ be a countable collection of measurable sets.

If:
 * $\displaystyle \sum_{n=1}^{\infty}\mu \left({E_{n}}\right) < \infty$

then:
 * $\displaystyle \mu\left({\limsup_{n \to \infty} E_{n}}\right) = 0$

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

Thus, by the monoticity of the measure $\mu$:
 * $\displaystyle \mu \left({\limsup_{n \to \infty} E_n}\right) = \mu \left({\bigcap_{i=1}^\infty \bigcup_{j=i}^\infty E_j}\right) \le \mu \left({\bigcup_{j=i}^\infty E_j}\right)$

for each $i$.

By the subadditivity of $\mu$:
 * $\displaystyle \mu \left({\bigcup_{j=i}^\infty E_j}\right) \le \sum_{j=i}^\infty \mu \left({E_j}\right)$

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

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

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