Generalisation of Dominated Convergence Theorem

Question

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 $\int|f_n-f| \rightarrow 0$

Answer 1

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.

Answer 2

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