An exercise about Lebesgue integrable functions & convergence Let $(\Omega ,F,\mu)$ a be a measure space & $\{f_n\}$ a  sequence of nonnegative Lebesgue integrable functions.
If $\{f_n\}$ converges in measure  to function $f $ & the integrals converges to $\int_\Omega f < \infty $, prove that  $\int_\Omega |f_n -f|\,d\mu$ converges to  $0$, when  $n $ goes to  infinity.
It seems easy, but I didn't success.
Help me please. 
 A: Since $\{f_n\}$ is a  sequence of nonnegative Lebesgue integrable functions and $\{f_n\}$ converges in measure  to function $f $, we have that $\mu([f<0])=0$. So $f\geqslant 0$ a.e.. So, without loss of generality, we can assume that $f\geqslant 0$. 
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 for any $\varepsilon>0$, 
$$ [\vert (f_n \wedge f)-f \vert\geqslant \varepsilon ]= [f-(f_n \wedge f)\geqslant \varepsilon ] = [f-f_n \geqslant \varepsilon ] \subseteq [\vert f_n -f \vert \geqslant \varepsilon ] $$
we have that $\{f_n \wedge f\}$ converges in measure  to function $f$. 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 the Dominated Convergence Theorem (for convergence in measure) and we have that 
$$\lim_{n \to \infty}\int_{\Omega }  f_n \wedge f d\mu = \int_{\Omega }  f d\mu$$
To conclute 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 so, since $\lim_{n \to \infty}\int_{\Omega }  f_n d\mu = \int_{\Omega }  f d\mu$ we have 
$$  \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$
Remark: Note that it was NOT assumed that $\mu(\Omega)<+\infty$.
