Borel-Cantelli Lemma

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $E_n \subseteq \Sigma$ be a countable collection of measurable sets.

If:


 * $\ds \sum_{n \mathop = 1}^\infty \map \mu {E_n} < \infty$

then:


 * $\ds \map \mu {\limsup_{n \mathop \to \infty} E_n} = 0$

where $\limsup$ denotes limit superior of sets.

Proof
By definition of limit superior:
 * $\ds \limsup_{n \mathop \to \infty} E_n = \bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j$

Thus, by Measure is Monotone and Intersection is Subset:
 * $(1): \quad \ds \map \mu {\limsup_{n \mathop \to \infty} E_n} = \map \mu {\bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j} \le \map \mu {\bigcup_{j \mathop = i}^\infty E_j}$

for all $i \in \N$.

By Measure is Subadditive:
 * $\ds \map \mu {\bigcup_{j \mathop = i}^\infty E_j} \le \sum_{j \mathop = i}^\infty \map \mu {E_j}$

However, by assumption $\ds \sum_{n \mathop = 1}^\infty \map \mu {E_n}$ converges.

By Tail of Convergent Series tends to Zero this implies:
 * $\ds \lim_{i \mathop \to \infty} \sum_{n \mathop = i}^\infty \map \mu {E_n} = 0$

Now $(1)$ implies, together with Lower and Upper Bounds for Sequences, that:
 * $\ds \map \mu {\limsup_{n \mathop \to \infty} E_n} \le 0$

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

Hence:


 * $\ds \map \mu {\limsup_{n \mathop \to \infty} E_n} = 0$

Borel-Cantelli Lemma in Probability
As each probability space $\struct {X, \Sigma, \Pr}$ 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.

Also see

 * Borel-Cantelli Lemma