Borel-Cantelli Lemma

Theorem

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

Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-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:

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

$\blacksquare$