Borel-Cantelli Lemma Proof Verification

Question

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.

Answer 1

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.

Answer 2

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 $$\mu(E)=\mu(|f|=\infty)=0$$