Egorov's theorem application Problem
Let $\{f_k\}_{k \in \mathbb N}$ be a sequence of measurable functions defined on $E \subset \mathbb R^n$ with $E$ measurable and $|E|<\infty$, such that $f_k \to 0$ a.e.. Show that there exists a subsequence $\{f_{k_j}\}_{j \in \mathbb N}$ such that $\sum_{j \in \mathbb N} |f_{k_j}|<\infty$ a.e.
I thought of using Egorov's theorem, so for each $\epsilon_n=\dfrac{1}{n}$, there exists a closed subset $F_n \subset E$ with $|E \setminus F_n|<\epsilon_n$ and such that $f_n \rightrightarrows 0$ on $F_n$. For each $j$, I can pick $n_j$ with $|f_{n_j}|<\dfrac{1}{2^j}$ in $F_j$ and I can also pick $n_1<n_2<...<n_j<...$
It is easy to see that complement of the set $F=\bigcup_{j \in \mathbb N} F_j$ has measure zero. The problem is that I cannot affirm $|f_{n_j}|<\dfrac{1}{2^j}$ in all $F$ but just in $F_{n_j}$. I can assure this on the intersection $\bigcap_{j \in \mathbb N} F_j$ but the complement of this set is not of measure zero, so the series is not convergent almost everywhere.
Any hints to solve this problem would be greatly appreciated.
 A: Construct $F_n$ as you did, but then let $F_n' = F_1 \cup \dots \cup F_n$.  Then we again have $|E \setminus F_n'| < \epsilon_n$ and $f_n \rightrightarrows 0$ on $F_n'$.  Moreover, $F_1' \subset F_2' \subset \dots$ which will be useful later.  Choose the $n_j$s such that $|f_{n_j}| < 2^{-j}$ on $F_{j}'$ instead.
Now let's set $F = \bigcup_n F_n'$ instead.
Hint 2: If you do that, then for any fixed $x \in F$, you cannot guarantee $|f_{n_j}(x)| < \frac{1}{2^j}$ for every $j$, but you can guarantee it for all but finitely many $j$...
A: Convergence a.e. on a finite measure space implies convergence in measure. So we may choose $k_1$ such that $k\geqslant k_1$ implies $\mu\left(\left\{x:|f_k(x)|> 2^{-k}\mu(E)^{-1}\right\}\right)<2^{-k}\mu(E)^{-1}$, and inductively $k_j\geqslant k_{j-1}$ such that $k\geqslant k_j$ implies $\mu\left(\left\{x:|f_k(x)|>2^{-k_j}\mu(E)^{-1}\right\}\right)<2^{-k_j}\mu(E)^{-1}$ and $k_j\stackrel{j\to\infty}\longrightarrow\infty$. We may assume WLOG that $2^{-k_1}\mu(E)^{-1}<1$. Hence
\begin{align}
\sum_{j=1}^\infty |f_{k_j}| &\leqslant \sum_{j=1}^\infty 2^{-k_j}\mu(E)^{-1}\mu(E)+2^{-k_j}\mu(E)^{-1}\\
&\leqslant \mu(E)^{-1}\sum_{j=1}^\infty 2^{-k_j}\\
&\leqslant \mu(E)^{-1}\\
&<\infty.
\end{align}
