Generalisation of Dominated Convergence Theorem Wikipedia claims, if $\sigma$-finite the Dominated convergence theorem is still true when pointwise convergence is replaced by convergence in measure, does anyone know where to find a proof of this? Many thanks!
Statement of the theorem:
Let $\mu$ be $\sigma$-finite, $|f_n|\leq g$ and $f_n\rightarrow f$ in measure, then we must have
$\int f_n \rightarrow \int f$ and $\int|f_n-f| \rightarrow 0$
 A: I don't know why this question is reopened. However,  given none of the proofs really pleasing me, I'll present one.
So just by interchanging the integral signs, we see that:
$$\int |f-f_n| =  \int_{0}^{\infty} \underbrace{ \mu\left( |f_n-f|>t\right) }_{=:A_n(t)}dt \text{ (1) }$$
And under the condition of boudedness of $f_n$, we have :
$$0 \le A_n(t) \le \mu( 2g >t) := G(t)$$
Note that $$\int_{0}^{\infty} G(t)= \int 2g < \infty$$
Thus the DCT conditions for the first integrals are fulfilled,  so :
$$ \lim_{n \rightarrow + \infty} \int |f-f_n| = \lim_{n \rightarrow + \infty} \int_{0}^{\infty}  \mu\left( |f_n-f|>t\right) = \int_{0}^{\infty} \lim_{n \rightarrow + \infty}  \mu\left( |f_n-f|>t\right) = 0 $$
A: I think there is a point of reviving this post because I think this is also a nice proof pointed out to me when I posted this question by Chris Janjigian. 
I have seen this question posted numerous times on this site, so I think there is a point of writing this out.
Let us consider the sequence $\int |f_n-f|$. then consider the following, take a subsequence $\int |f_{n_j}-f|$
For $f_{n_j}$, there must exist a sub-subsequence $f_{n_{j_k}}$ such that $f_{n_{j_k}}$ converges to $f(x)$ almost everywhere. (Since $f_n$, hence $f_{n_j}$ converges to $f(x)$ in measure)
It must also be the case $|f_{n_{j_k}}-f| \leq 2g$, we now apply dominated convergence to see that $\int|f_{n_{j_k}}-f| \rightarrow 0$
What we have shown is that, for every subsequence of $\int |f_n-f|$, we have a further subsequence, which converges to 0. Now using the lemma:
If for every subsequence of $x_n$, there exists a sub-subsequence which converges to 0, then $x_n$ converges to 0.
We are done.
The proof of the last lemma can be found Sufficient condition for convergence of a real sequence
A: 
Let $(X,\mathcal B,\mu)$ be a measure space, $\{f_n\}$ a sequence of functions which converges to $f$ in measure, and for almost every $x$ and all $n$, $|f_n(x)|\leqslant g(x)$, where $g$ is integrable.
Then $\lVert f_n-f\rVert_{L^1}\to 0$.

Let $A_k:=\{g\gt 1/k\}$; then $A:=\bigcup_k A_k =\{g\neq 0\}$ and $X\setminus A\subset\bigcap_n\{f_n=0\}\cap\{f=0\}$.  We have for each $k$,
$$\int_X|f_n(x)-f(x)|d\mu\leqslant 2\int_{X\setminus A_k}|g(x)|\mathrm d\mu(x)+\int_{A_k}|f_n(x)-f(x)|\mathrm d\mu(x).$$
If $\lVert f_n-f\rVert_{L^1}$ doesn't converge to $0$, we can find a $\delta>0$ and a subsequence $\{f_{n'}\}$ such that $\lVert f_{n'}-f\rVert_{L^1}\geqslant 2\delta$. We fix $k$ such that $2\int_{X\setminus A_k}|g(x)|\mathrm d\mu(x)\leqslant\delta$ (such a $k$ exists by the dominate convergence theorem, since $\lim_{k\to\infty}\int_{X\setminus A_k}|g(x)|\mathrm d\mu(x)= \int_{X\setminus A}|g(x)|\mathrm d\mu(x)$ ). Then
$$\delta\leqslant \int_{A_k}|f_{n'}(x)-f(x)|\mathrm d\mu(x).$$
Now, as $A_k$ has a finite measure, we can extract a subsequence $\{f_{n''}\}$ of $\{f_{n'}\}$ which converges almost everywhere on $A_k$. Applying the classical dominated convergence theorem to this sequence, we get a contradiction.
