The Borel-Cantelli Lemma

Lemma (The Borel-Cantelli Lemma): Let $(X, \mathcal A, \mu)$ be a measure space and let $(E_n)_{n=1}^{\infty}$ be a collection of measurable sets such that $\displaystyle{\sum_{n=1}^{\infty} \mu (E_n) < \infty}$. Then almost every $x \in X$ belongs to at most a finite number of sets in $(E_n)_{n=1}^{\infty}$.

• Proof: Since $\displaystyle{\sum_{n=1}^{\infty} \mu (E_n) < \infty}$, we have by countable additivity that:

(1)

\begin{align} \quad \mu \left ( \bigcup_{n=1}^{\infty} E_n \right ) \leq \sum_{n=1}^{\infty} \mu (E_n) < \infty \end{align}

• And in particular, for each $k \in \mathbb{N}$ we have that:

(2)

\begin{align} \quad \mu \left ( \bigcup_{n=k}^{\infty} E_n \right ) \leq \sum_{n=k}^{\infty} \mu (E_n) < \infty \end{align}

• Note that the set $\displaystyle{\bigcap_{k=1}^{\infty} \left ( \bigcup_{n=k}^{\infty} E_n \right )}$ is the set of all points $x \in X$ that must belong to an infinite number of sets in $(E_n)_{n=1}^{\infty}$. We see by the continuity of $\mu$ that:

(3)

\begin{align} \quad \mu \left ( \bigcap_{k=1}^{\infty} \left ( \bigcup_{n=k}^{\infty} E_n \right ) \right ) = \lim_{k \to \infty} \mu \left ( \bigcup_{n=k}^{\infty} E_n \right ) \leq \lim_{k \to \infty} \sum_{n=k}^{\infty} E_n = 0 \end{align}

• So almost every point $x \in X$ belongs to a finite number of sets in $(E_n)_{n=1}^{\infty}$. $\blacksquare$