# The Borel-Cantelli Lemma

#### Borel-Cantelli Lemma: Let $(X,\Sigma,\mu)$ be a measure space with $\mu(X)< \infty$ and suppose $\{E_n\}_{n=1}^\infty \subset\Sigma$ is a collection of measurable sets such that $\displaystyle{\sum_{n=1}^\infty \mu(E_n)< \infty}$. Then $$\mu\left(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k \right)=0.$$

When I first came across this lemma, I struggled to understand what it meant "in English." What does $\mu(\cap_n\cup_k E_k)=0$ really signify?? There's a pretty simple explanation if $(X,\Sigma,\mu)$ is a probability space, but how are we to understand the result in the context of general measure spaces?

The first step towards answering this question is to recognize that $\mu(\cap_n\cup_k E_k)=0$ is the same as saying if $A=\{x\in X:\text{there exists infinitely many$n$such that$x\in E_n$}\}$ then $\mu(A)=0$. And this is equivalent to the statement

##### almost every $x\in X$ lives in at most finitely many $E_n$.

So in other words, for almost every $x\in X$, there is some finite indexing set $A_x=\{n_1,\ldots,n_m\}\subset\mathbb{N}$ such that  $$x\in\bigcap_{i\in A_x} E_i,$$  which is to say that for almost each $x\in X$, we have a picture like this:

To make things a little more concrete, let's look at an example to see the Borel-Cantelli Lemma in action.

## Example

Suppose $(X,\Sigma,\mu)$ is a measure space with $\mu(X)< \infty$ and suppose $\{f_n:X\to\mathbb{C}\}$ is a sequence of measurable functions. Show there exists a sequence $\{c_n\}$ of positive numbers such that $\displaystyle\lim_{n\to\infty} c_nf_n(x)=0$ for almost every $x\in X$.

## Proof

We start by claiming that for $n$ large enough, there is $k_n\in\mathbb{R}$ such that $$\mu(\{x\in X:|f_n(x)|>k_n\})< \frac{1}{2^n}.$$ Indeed notice that for a fixed $n$, the sets $E_k=\{x\in X:|f_n(x)|>k\}$ for $k=1,2,3,\ldots$ satisfy $E_1\subset E_2\subset E_3\subset\cdots$. Further we have $$\bigcap_{k=1}^\infty E_k=\varnothing$$ since the $f_n$ map into $\mathbb{C}$ (i.e. there is no $x\in X$ nor $n\in\mathbb{N}$ for which $f_n(x)=\infty$). Thus, since $\mu(E_i)<\mu(X)<\infty$, we may use continuity from above to conclude $$0=\mu(\varnothing)=\mu\left(\bigcap_{k=1}^\infty E_k\right)=\lim_{k\to\infty}\mu(E_k).$$ Hence there is a $k=k_n$ large enough so that $$E_n=\{x\in X:|f_n(x)|>k_n\} \qquad \text{satisfies} \qquad \mu(E_n)< \frac{1}{2^n}.$$ Now consider the collection $\{E_n\}_{n=1}^\infty$ and notice that $\displaystyle{\sum_{n=1}^\infty \mu(E_n)=1<\infty}$. Thus by the Borel-Cantelli Lemma, almost every $x\in X$ lives in at most finitely many $E_n$. In other words, fix $x\in X$ (off of a set of measure zero). Then there is an finite indexing set $A_x=\{n_1,\ldots,n_m\}\subset\mathbb{N}$ (with, say, $n_i< n_{i+1}$) such that $$x\in\bigcap_{i=\in A_x}E_i=E_{n_m}$$ where the equality holds since the $E_i$ are nested. Since $x\in E_{n_m}$ but $x\not\in E_n$ for any $n>n_m$, it follows that $|f_n(x)|\leq k_n$ for all $n>n_m$. This prompts us to choose $$c_n=\frac{1}{nk_n}.$$ Indeed fix $\epsilon >0$. Then for all $n>\text{max}\{n_m,\frac{1}{\epsilon}\}$, we have $$|c_nf_n(x)|< \frac{1}{nk_n}\cdot k_n=\frac{1}{n}< \epsilon$$ which implies $c_nf_n\to 0$ almost everywhere in $X$.

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