How to prove that if $\sum_{n=1}^{\infty} \int_{|\Omega}|f_{n}|d\mu<\infty$ then $\sum_{n=1}^{\infty} f_{n}(x)$ converges $\mu-a.e.$ Let $(\Omega, \mathcal{A}, \mu)$ be a measure space and let $f_{n}\in \mathcal{L}^{1}(\Omega,\mathcal{A},\mu)$ or $n\in \mathbb{N}$. 
Prove the following:
If $\sum_{n=1}^{\infty} \int_{\Omega}|f_{n}|d\mu<\infty$ then  $\sum_{n=1}^{\infty} f_{n}(x)$ converges for $\mu$-almost every $x\in\Omega$, i.e. for every $x\in \Omega\backslash N$, with $N\in\mathcal{A}$ and $\mu(N)=0$.
I thought if I can prove that if $\int_{\Omega}|f_{n}|d\mu\leq f_{n}(x)$ implicates that $\mu(x)=0$ then I'm done, is this the right way?
 A: What we know is that
$$ \sum_{n=1}^\infty \int_\Omega |f_n|d \mu<\infty. $$
We can apply the monotone convergence theorem to pass the series inside the integral since the partial sum $S_m(x) = \sum_{n=1}^m |f_n(x)|$ is increasing for all $x$. We get:
\begin{align}
\sum_{n=1}^\infty \int_\Omega |f_n|d \mu& = \lim_{m\to\infty} \sum_{n=1}^m \int_\Omega |f_n|d\mu = \lim_{m\to\infty} \int_\Omega \sum_{n=1}^m |f_n|d\mu \\
&\overset{\text{monotone thm}}{=} \int_\Omega \lim_{m\to\infty} \sum_{n=1}^m |f_n|d\mu =
 \int_\Omega \sum_{n=1}^\infty |f_n|d\mu <\infty.
\end{align}
Since the integral of the function $\sum_{n=1}^\infty |f_n(x)|$ is finite then we conclude that the function is finite $d\mu$-a.e. For all $x$ for which $\sum_{n=1}^\infty |f_n(x)|$ converges, then $\sum_{n=1}^\infty f_n(x)$ absolutely converges, so that it converges (elementary fact about convergence of series). This means that $\sum_{n=1}^\infty f_n(x)$ converges $d\mu$-a.e.
A: This can be easily proved by contradiction. Assume there is some non null set $K$ for which $\sum_n f_n(x)$ does not converge. Then we have surely that $\sum_n |f_n(x)|=\infty$ for all $x\in K$. Now by MCT we have:
\begin{align}
\infty=\int_K \sum_{n=1}^\infty |f_n(x)|d\mu \stackrel{MCT}{=} \sum_{n=1}^\infty \int_K |f_n(x)|d\mu \leq \sum_{n=1}^\infty\int_\Omega |f_n(x)|d\mu<\infty
\end{align}
A contradiction.
