Show that $\int |f_n-f|\to 0$ 
Let $f$ be an integrable function and $(f_n)_{n=1}^\infty$ a sequence of non-negative measurable functions such that $f_n\to f$ and $\int_\Omega f_n\,d\mu\to \int_\Omega f\,d\mu$.
Show that
$
   \int_\Omega |f_n-f|d\mu\to 0.
  $

The naive idea of using $|\int f|\leq \int|f|$ fails because the $\leq$ sign points in the wrong directio.
This question is a follow-up to monotone convergence for decreasing functions, so perhabs we should use that?
I think that since $f_n\to f$ we can take $|f_n-f|$ decreasing, then
$$
\lim_{n\to\infty}\int_\Omega |f_n-f|d\mu = \int_\Omega \lim_{n\to\infty}|f_n-f|d\mu,
$$
and perhabs what's inside vanishes.
I think this is incorrect, because we didn't use the hypothesis $\int f_n \to \int f$.
Any ideas?
 A: The question asks to prove the result:

Let $f$ be an integrable function and $(f_n)_{n=1}^\infty$ a sequence of non-negative measurable functions such that $f_n\to f$ (pointwise) and $\int_\Omega f_n\,d\mu\to \int_\Omega f\,d\mu$. Then $ \int_\Omega |f_n-f|d\mu\to 0$.

We are going to present two ways proofs.
Proof $1$. Here is a proof using Dominated Convergence Theorem.
Since $\{f_n\}$ is a sequence of nonnegative measurable functions and $\{f_n\}$ converges to the function $f$ pointwise, we have that $f\geqslant 0$. So, since $f$ is integrable and $f\geqslant 0$, we have that $\int_\Omega f d\mu=\int_\Omega|f|d\mu<+\infty$
Now consider $f_n \wedge f$ defined by $(f_n \wedge f)(x)=\min\{f_n(x),f(x)\}$, for each $x \in \Omega$.
Since $\{f_n\}$ converges to $f$,
we have that $\{f_n \wedge f\}$ converges to $f$. But we know that, for all $n$, $\vert f_n \wedge f \vert =  f_n \wedge f \leqslant f $ and   $\int_{\Omega }  f d\mu< \infty $.  So we can apply Lebesgue Dominated Convergence Theorem and we have that
$$\lim_{n \to \infty}\int_{\Omega }  f_n \wedge f d\mu = \int_{\Omega }  f d\mu$$
To conclude the proof, note that
$$\vert f_n-f\vert = f_n+f-2(f_n\wedge f)$$
So
$$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } f_n d\mu +\int_{\Omega } f d\mu -2\int_{\Omega } (f_n\wedge f) d\mu $$
And since $\lim_{n \to \infty}\int_{\Omega }  f_n d\mu = \int_{\Omega }  f d\mu$, we have
$$  \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$

Proof $2$. Here is a proof using Fatou's lemma.
Since $\{f_n\}$ is a sequence of nonnegative measurable functions and $\{f_n\}$ converges to the function $f $ pointwise, we have that $f\geqslant 0$.
Note that $|f_n -f|\leqslant f_n +f$. So, for each $n$, the function $f_n +f - |f_n -f|$ is non-negative.
Note also that, since $\{f_n\}$ converges to $f $ pointwise, we have $|f_n-f|$ converge to $0$ pointwise, and so
$ 2f= \lim (f_n+f - |f_n-f|) = \lim\inf (f_n+f - |f_n-f|)$.
Since $\lim_{n \to \infty} \int_\Omega f_n d\mu = \int_\Omega f d\mu$, using Fatou's Lemma, we have:
\begin{align} 
2 \int_\Omega f d\mu 
&=\int_\Omega \lim\inf(f_n +f - |f_n -f|)d\mu \leqslant \lim\inf \int_\Omega (f_n +f - |f_n -f|)d\mu = \\ 
&=\lim\inf \left (\int_\Omega f_n d\mu +\int_\Omega f d\mu - \int_\Omega|f_n -f|d\mu \right) = \\
&= \left(\lim\inf\int_\Omega f_n d\mu\right) +\int_\Omega f d\mu - \left(\lim\sup\int_\Omega |f_n -f|d\mu\right)  = \\ 
&=2\int_\Omega f d\mu - \left(\lim\sup\int_\Omega|f_n -f|d\mu\right) 
\end{align}
So we have
$$2 \int_\Omega f d\mu \leqslant 2\int_\Omega f d\mu - \left(\lim\sup\int_\Omega |f_n -f|d\mu\right) $$
Since $f$ is integrable and $f \geqslant 0$, we know that $\int_\Omega f d\mu=\int_\Omega|f|d\mu<+\infty$, and so we get
$$\lim\sup\int_\Omega|f_n -f|d\mu \leqslant  0$$
So we can conclude that
$$\lim\int_\Omega|f_n -f|d\mu =  0$$
