Question: **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.

Answer: 
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* $j$...
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}