Showing convergence and integrability from $\sum_{n=0}^\infty \int_X |f_n| \, d \mu \lt \infty$ I have seen this question here but I didnt find a comprehensible answer so ill try my luck.
Let $(f_n)$ be a sequence of almost everywhere integrable functions defined on X with $$\sum_{n=0}^\infty \int_X |f_n| \, d \mu \lt \infty$$
Show  that $$f(x) := \sum_{n=0}^\infty f_n(x)$$
$1)$ converges for almost all $x$ 
$2)$ the limit function $f$ is integrable
$3) \int_X f \, d\mu = \sum_{n=0}^\infty \int_Xf_n \, d\mu$
My try:
$1)$ Define $g_N := \sum_{n=0}^N|f_n|.$ It follows that $g_N$ is a monotone increasing function so I can use the monotone convergence theorem. 
It follows that : $$\lim_{N \rightarrow \infty}\int_X g_N \, d \mu  = \int_X \lim_{N \rightarrow \infty} g_N \, d \mu $$
$\Longrightarrow \sum_{n=0}^\infty|f_n| \lt \infty$ so the sum converges for almost all $x$
$2)$ Since all $f_n $ are measurable and we showed that $f = \lim f_n$ is finite it follows that $f$ is integrable
$3)$ I would say it strictly follows from dominated convergence theorem with $1)$ and   $2)$.
Any help here is appreciated thanks

For number two, note that 
\begin{align*}
|f(x)|\leq\sum_{n=1}^{\infty}|f_{n}(x)|,
\end{align*}
so $\displaystyle\int|f(x)|dx\leq\int\displaystyle\sum_{n=1}^{\infty}|f_{n}(x)|dx=\sum_{n=1}^{\infty}\int|f_{n}(x)|dx<\infty$.
For number three, the dominated integrable function can be taken as $g(x):=\displaystyle\sum_{n=1}^{\infty}|f_{n}(x)|$.
