TITLE: Generalisation of Dominated Convergence Theorem
QUESTION [27 upvotes]: 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 $\int|f_n-f| \rightarrow 0$

REPLY [6 votes]: 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 :
$$ \lim_{n \rightarrow + \infty} \int |f-f_n| = \lim_{n \rightarrow + \infty} \int_{0}^{\infty}  \mu\left( |f_n-f|>t\right) = \int_{0}^{\infty} \lim_{n \rightarrow + \infty}  \mu\left( |f_n-f|>t\right) = 0 $$