TITLE: Lebesgue Dominated Convergence Theorem with Convergence in Measure
QUESTION [7 upvotes]: I have an issue with the solution to the following problem. I now want to prove that the Lebesgue Dominated Convergence Theorem still works when the condition $\{ f_{n} \}$ converges to $f$ a.e. is replaced by $\{ f_{n} \}$ converges to $f$ in measure. 
Relax, I know a lot of people have asked this question already. But what I do not understand is that the solutions I found online insist to work with a sub-subsequence instead of just a subsequence. For instance, the following is an expressed answer I summarise from someone:
We can take a subsequence $\{f_{n_{k}} \}$ converges to $f$ in measure. Then, by a proved proposition, there is a sub-subsequence $\{f_{n_{k_{l}}} \}$ converges to $f$ a.e.. Then it reduces to the usual LDCT conditions and we can have the conclusion. 
But how about I simplify a little bit? Can't I just take a subsequence $\{f_{n_{k}} \}$ which converges to $f$ a.e. and have the conclusion? I fail to see the need to take a subsequence that converges to $f$ in measure first. Is that really necessary?

REPLY [8 votes]: If $f_n$ converges to $f$ in measure then there is a subsequence $f_{n_k}$ that converges to $f$ almost everywhere, and so we could apply Lebsegue dominated convergence to that subsequence, which is not good enough. 
So the proof does seem to need a sub-sub-sequence as follows: 

Suppose $|f_n(x)|\leq g(x)$ for all $x \in X$, $\int g < \infty$, and $f_n$ converges to $f$ in measure.  Then 
$$ \lim_{n\rightarrow\infty} \int f_n = \int f $$
Proof: Suppose not (we reach a contradiction). Then there is an $\epsilon>0$ and a subsequence $f_{n_k}$ such that 
$$\left|\int f_{n_k} - \int f\right| \geq \epsilon \quad \forall k \in \{1, 2, 3, ...\} \quad  \mbox{ (Eq. 1)}$$
Now $f_{n_k}$ converges to $f$ in measure so there is a sub-subsequence $f_{n_{k[m]}}$ that converges to $f$ almost everywhere.  So the usual Lebsegue dominated convergence theorem applies to that sub-subsequence to ensure
$$ \lim_{m\rightarrow\infty}\int f_{n_{k[m]}} = \int f$$
which contradicts (Eq. 1).  $\Box$