# 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:

$\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({E_n}\right) < \infty$

then:

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

where $\limsup$ denotes limit superior of sets.

## Proof

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

Thus, by Measure is Monotone and Intersection Subset:

$(1): \quad \displaystyle \mu \left({\limsup_{n \to \infty} E_n}\right) = \mu \left({\bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j}\right) \le \mu \left({\bigcup_{j \mathop = i}^\infty E_j}\right)$

for all $i \in \N$.

$\displaystyle \mu \left({\bigcup_{j \mathop = i}^\infty E_j}\right) \le \sum_{j \mathop = i}^\infty \mu \left({E_j}\right)$

However, by assumption $\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({E_n}\right)$ converges, and by Tail of Convergent Series this implies:

$\displaystyle \lim_{i \to \infty} \sum_{n \mathop = i}^\infty \mu \left({E_n}\right) = 0$

Now $(1)$ implies, together with Lower and Upper Bounds for Sequences, that:

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

But as $\mu$ is a measure, the converse inequality also holds.

Hence:

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

$\blacksquare$

## 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.