Prove that $\lim_{n \to \infty}\int_{A}f_{n}d\mu=\int_{A}fd\mu.$ 
Let $(f_{n})_{n=1}^{\infty}$ be a sequence of measurables functions non-negatives such that  $\displaystyle \lim_{n\to \infty}f_{n}(x)=f(x)$ almost everywhere and $$\lim_{n \to \infty} \int_{X}f_{n}d\mu =\int_{X}fd\mu<+\infty. $$
Prove that for all measurable set $E\subseteq X$, $$\lim_{n \to \infty}\int_{E}f_{n}d\mu=\int_{E}fd\mu.$$

My approach: Let $E\subseteq X$ measurable set, so we have
$$\lim_{n \to \infty}\int_{E}f_{n}d\mu\overbrace{=}^{definition}\lim_{n\to \infty}\int_{X} f_{n}\cdot \chi_{E}d\mu\overbrace{=}^{de: \chi_{E}}\lim_{n \to \infty} \int_{E}f_{n}d\mu\overbrace{=}^{hypothesis?}\int_{E}fd\mu $$

Is my proof correct?
I don't know how can I use the hypothesis $f_{n}\overset{a.e.x}{\to}f$
How can I conclude in my prove?
 A: Let $(X,\mathcal{F},\mu)$ be a measure space. Let $(f_{n})$ be a
sequence of real-valued integrable functions defined on $X$. Let
$f:X\rightarrow\mathbb{R}$ be an integrable function. Suppose that
$f_{n}\rightarrow f$ a.e., then $\lim_{n}\left(||f_{n}||-||f||-||f_{n}-f||\right)=0$,
where $||f||=\int|f|d\mu$ etc...
We go to prove the above assertion. Observe that $|f_{n}-f|\leq|f_{n}|+|f|$,
so $|f_{n}|-|f|-|f_{n}-f|\geq-2|f|.$ On the other hand, $|f_{n}|\leq|f_{n}-f|+|f|$
and hence $|f_{n}|-|f|-|f_{n}-f|\leq0\leq2|f|$. Combining, we have
$\Biggl||f_{n}|-|f|-|f_{n}-f|\Biggr|\leq2|f|$. Clearly, $|f_{n}|-|f|-|f_{n}-f|\rightarrow0$
a.e.. By Dominated Convergence Theorem, we have that $\int\left(|f_{n}|-|f|-|f_{n}-f|\right)d\mu\rightarrow 0$.
That is, $||f_{n}||-||f||-||f_{n}-f||\rightarrow0$.
Go back to our question. Since $f_{n}$ and $f$ are non-negative,
$||f_{n}||=\int f_{n}$ and $||f||=\int f$. It is given that $||f_{n}||\rightarrow||f||$.
Therefore, $||f_{n}-f||=\left(||f_{n}||-||f||\right)-\left(||f_{n}||-||f||-||f_{n}-f||\right)\rightarrow0$.
Finally, if $E\subseteq X$ is measurable, we have $\int_{E}|f_{n}-f|\leq||f_{n}-f||\rightarrow0$.
Therefore $\int_{E}f_{n}\rightarrow\int_{E}f$.
