TITLE: Proving the Borel-Cantelli Lemma
QUESTION [6 upvotes]: Let $\{{E_k}\}^{\infty}_{k=1}$ be a countable family of measurable subsets of $\mathbb{R}^d$ and that
  \begin{equation} % Equation (1)
\sum^{\infty}_{k=1}m(E_k)<\infty
\end{equation}
  Let
  \begin{align*}
E&=\{x\in \mathbb{R}^d:x\in E_k, \text{ for infinitely many $k$ }\} \\
&= \underset{k\rightarrow \infty}{\lim \sup}(E_k).\\
\end{align*}
(a) Show that $E$ is measurable
(b) Prove $m(E)=0.$

My Proof Attempt:
Proof. Let the assumptions be as above. We will prove part (a) by showing that
\begin{equation*}
E=\cap^{\infty}_{n=1}\cup_{k\geq n}E_k. 
\end{equation*}
Hence, E would be measurable, since for every
fixed $n$, $\cup_{k\geq n}E_k$ is measurable since it is a countable union of measurable sets.
Then $\cap^{\infty}_{n=1}\cup_{k\geq n}E_k$ is the countable intersection of measurable sets.
From here, we shall denote $\cup_{k\geq n}E_k$ as $S_n$.
 Let $x\in \cap^{\infty}_{n=1}S_n$.  Then $x\in S_n$ for every
$n\in \mathbb{N}$. Hence, $x$ must be in $E_k$ for infinitely many $k$, otherwise there would exist
an $N\in \mathbb{N}$ such that $x\notin S_N$. Leaving $x$ out of the intersection.
Thus, $\cap^{\infty}_{n=1}S_n\subset E$.
Conversely, let $x\in E.$ 
Then $x\in E_k$ for infinitely many $k$. Therefore, $\forall n\in \mathbb{N}$,
 $x\in S_n$. Otherwise, $\exists N\in \mathbb{N}$ such that $x\notin S_N$.
Which would imply that $x\in E_k$ for only $k$ up to $N$, i.e. finitely many. A contradiction.
Therefore, $x\in \cap^{\infty}_{n=1}S_n$. Hence, they contain one
another and equality holds. This proves part (1).
Now for part (b). Fix $\epsilon>0$. We need to show that there exists $N\in \mathbb{N}$ such that
\begin{equation*}
m(S_N)\leq \epsilon
\end{equation*}
Then since $E\subset S_N$, monotonicity of the measure would imply that $m(E)\leq \epsilon$.
Hence, proving our desired conclusion as we let $\epsilon \rightarrow 0$.
Since $\sum^{\infty}_{k=1}m(E_k)<\infty$, there exists $N\in \mathbb{N}$ such that
\begin{equation*}
\left| \sum^{\infty}_{k=N}m(E_k)\right |\leq \epsilon
\end{equation*}
By definition, 
\begin{equation*}
m(S_N)=m(\cup_{k\geq N}E_k)=\sum^{\infty}_{k=N}m(E_k)
\end{equation*}
Thus, $m(S_N)\leq \epsilon$. This completes our proof.

Any corrections of the proof, or comments on style of the proof are welcome and appreciated. Thank you all for your time.

REPLY [4 votes]: Your proof is fine, modulo the comment in the other answer. For another approach, which I think is the way Rudin does it, note that $x\in E$ if and only if the series $\sum^{\infty}_{k=1}1_{E_k}(x)$ diverges. 
Set $s_n(x)=\sum^{n}_{k=1}1_{E_k}(x).$ Then, $s_n(x)\to s(x)=\sum^{\infty}_{k=1}1_{E_k}(x)$ and the monotone converge theorem gives $\sum^{n}_{k=1}m(E_k)\to \sum^{\infty}_{k=1}m(E_k)<\infty.$ Thus, $s\in L^1(m)$, so the series converges almost everywhere $m$. That is, the set on which it diverges, namely $E$, has Lebesgue measure zero and so $m(E)=0.$
Remark: since we proved that $m(E)=0,$ we get part $(a)$ for free.