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