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 $\displaystyle \mu \left({\limsup_{n \to \infty} E_n}\right) = 0$.

$\blacksquare$

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.

Source of Name

This entry was named for Émile Borel and Francesco Paolo Cantelli.