# The Borel-Cantelli Lemma

/Today we're chatting about the

**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 $$\mu\left(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k \right)=0.$$

*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

*Train of Thought: Our goal for the proof is to make $|c_nf_n(x)|\leq\frac{1}{n}$ for $n$ large enough and for a.e. $x\in X$, for then we can bound it above by $\epsilon$ for any $\epsilon>0$. So we ask ourselves, "Is there a set (of positive measure) of $x\in X$ where $|f_n(x)|\leq k_n$ for some positive constant $k_n$?" If so, then we can write $$|c_nf_n(x)|\leq c_nk_n \overset{\text{want}}{=}1/n.$$ The last equality is what we want, and this behooves us to choose $c_n=1/nk_n$. But does such a $k_n$ even exist? The answer is yes (!), and this is where Borel-Cantelli comes in.*