# Borel-Cantelli Lemma Proof Verification

Exercise 16 (Stein): The Borel-Cantelli Lemma: Suppose $$\{E_k\}_{k=1}^\infty$$ is a countable family of measurable subsets of $$\mathbb{R}^d$$ and that $$\begin{equation*} \sum_{k=1}^\infty m(E_k) < \infty. \end{equation*}$$ Let \begin{align*} E = \{x \in \mathbb{R}^d : x \in E_k, \text{ for infinitely many } k \}\\ = \lim_{k\rightarrow\infty} \sup (E_k). \end{align*} Show that $$E$$ is measurable.

Remark: Now, after looking at some resources I now understand that the proof of this extremely simple once one remembers that the set of measurable sets is a $$\sigma$$-algebra, however I was curious if the proof I have worked out is also true. Any suggestions on writing proofs in general is always welcome of course!

Let $$\epsilon > 0$$ be arbitrary. First, recall that the countable union of measurable sets is itself measurable, and notice that $$\begin{equation*} m\biggr(\bigcup_{k = j}^\infty E_k \biggr) \leq \sum_{k=j}^\infty m(E_k) < \sum_{k=1}^\infty m(E_k) < \infty, \end{equation*}$$ for all $$j \in \mathbb{N}$$. Now consider the following decreasing sequence $$\begin{equation*} \bigcup_{k=1}^\infty E_k \supset \bigcup_{k=2}^\infty E_k \supset \dots \supset \bigcup_{k=N}^\infty E_k \supset \dots \supset E. \end{equation*}$$ From Corollary 3.3, since the sequence decreases to $$E$$ and $$\displaystyle m\biggr(\bigcup_{k=1}^\infty E_k\biggr) < \infty$$, it follows that $$\begin{equation*} m(E) = \lim_{N \rightarrow \infty}\biggr(\bigcup_{k = N}^\infty E_k\biggr). \end{equation*}$$ Thus it follows that there exists an $$N \in \mathbb{N}$$ such that for $$\epsilon' = \frac{\epsilon}{2}$$, $$\begin{equation*} m\biggr(\bigcup_{j=N}^\infty E_j \setminus E \biggr) < \epsilon'. \end{equation*}$$ Moreover, notice that since each $$\bigcup_{j=k}^\infty E_j$$ is measurable for every $$k \in \mathbb{N}$$, it follows that there exists an $$\mathcal{O}_N$$ for $$N$$ such that for $$\epsilon''=\frac{\epsilon}{2}$$ $$\begin{equation*} m\biggr(\mathcal{O}_n \setminus \bigcup_{j=N}^\infty E_j\biggr) < \epsilon''. \end{equation*}$$ Therefore, we can conclude that then there exists $$\mathcal{O}_N$$ such that $$\begin{equation*} m(\mathcal{O}_N - E) = m\biggr(\mathcal{O}_n \setminus \bigcup_{j=N}^\infty E_j\biggr) + m\biggr(\bigcup_{j=N}^\infty E_j \setminus E \biggr) < \epsilon' + \epsilon'' = \epsilon. \end{equation*}$$ Since $$\epsilon$$ is arbitrary, it follows that $$E$$ is a measurable set.

Remark 2: The second part of the Borel-Cantelli Lemma ($$m(E) = 0$$) is clear as if this were not true the hypothesis of the finiteness of the sum of the measure of the members of the family would not hold.

• What do you mean by $\sum_{k=1}^\infty E_k<\infty$? Jul 27, 2020 at 18:39
• @AlonsoDelfín oops that was a typo, I edited the above post. Meant $\sum_{k=1}^\infty m(E_k) < \infty$. Jul 27, 2020 at 18:42

We have $$E= \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$$ (why?) Hence, $$E$$ is trivially measurable by the axioms of $$\sigma$$-algebra. The assumption $$\sum_k m(E_k) < \infty$$ is not necessary.
Notice that $$E=\bigcap_n\bigcup_{m\geq n}E_m$$. Measurability follows immediately.
If you meant to say "prove that $$E$$ has $$\mu$$ measure zero, then that follows from monotone convergence since $$\int_X\sum_{k}\mathbb{1}_{E_k}\,d\mu\stackrel{MCT}{=}\sum_k\int_X\mathbb{1}_{E_k}\,d\mu=\sum_k\mu(E_k)<\infty$$ Notice that $$E=\Big\{\sum_k\mathbb{1}_{E_k}=\infty\Big\}= \bigcap_n\bigcup_{k\geq n}E_k$$. Since $$f=\sum_n\mathbb{1}_{E_k}$$ is integrable $$\mu(E)=\mu(|f|=\infty)=0$$
• I guess indeed that the OP meant to prove that $m(E) = 0$ (given the name Borel-Cantelli and the assumption on the series). (+1) Jul 27, 2020 at 20:40