Borel-Cantelli Lemma

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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 $i \in \N$.


By Measure is Subadditive:

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

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.


Sources