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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&={xin mathbb{R}^d:xin E_k, text{ for infinitely many $k$ }} \
&= underset{krightarrow 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_{kgeq n}E_k.
end{equation*}
Hence, E would be measurable, since for every
fixed $n$, $cup_{kgeq n}E_k$ is measurable since it is a countable union of measurable sets.
Then $cap^{infty}_{n=1}cup_{kgeq n}E_k$ is the countable intersection of measurable sets.

From here, we shall denote $cup_{kgeq n}E_k$ as $S_n$.
Let $xin cap^{infty}_{n=1}S_n$. Then $xin S_n$ for every
$nin mathbb{N}$. Hence, $x$ must be in $E_k$ for infinitely many $k$, otherwise there would exist
an $Nin mathbb{N}$ such that $xnotin S_N$. Leaving $x$ out of the intersection.
Thus, $cap^{infty}_{n=1}S_nsubset E$.

Conversely, let $xin E.$
Then $xin E_k$ for infinitely many $k$. Therefore, $forall nin mathbb{N}$,
$xin S_n$. Otherwise, $exists Nin mathbb{N}$ such that $xnotin S_N$.
Which would imply that $xin E_k$ for only $k$ up to $N$, i.e. finitely many. A contradiction.
Therefore, $xin cap^{infty}_{n=1}S_n$. 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 $Nin mathbb{N}$ such that
begin{equation*}
m(S_N)leq epsilon
end{equation*}
Then since $Esubset S_N$, 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 $Nin mathbb{N}$ 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_{kgeq 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.

The proof is almost perfect, only in the end it is not necessary true that $m(cup_{kgeq N}E_k)=sum_{k=N}^infty m(E_k)$ since the sets $E_k$ might not be pairwise disjoint. So the measure of the union is only at most the sum of measures. But in our case it doesn't change anything for the proof, since we anyway get $m(S_N)leqsum_{k=N}^infty m(E_k)leqepsilon$. Still it is important to remember the correct properties of measure.

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 $xin E$ 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, $sin L^1(m)$, 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.