[Math] Monotone convergence theorem assuming convergence in measure

Question

I have heard that the monotone convergence theorem holds, if the hypothesis of almost everywhere convergence is replaced by convergence in measure.
I concur; if $f_n$ converges in measure then there exists a subsequence of $f_n$ which converges to $f$ a.e. so that we can apply MCT for that subsequence but I couldn't see why the conclusion holds for the original sequence $f_n$.

Answer 1

Let $f_n: X \to [0, \infty]$ be $\mu$-measurable functions, such that $f_n \leq f_{n+1}$ for each $n \in \mathbb N$ and assume that the sequence $\{ f_n \}_{n=1}^\infty$ converges to a function $f\colon X \to [0, \infty]$ in measure. By a theorem of F. Riesz we know, that that there exists a subsequence $\{ f_{n_j} \}_{j=1}^\infty$ of $\{ f_n \}_{n=1}^\infty$, such that $$ f_{n_j} \to f \quad \text{$\mu$-a.e.}$$ On the other hand, since $f_n \leq f_{n+1}$, we know that the limit $g := \lim_{n \to \infty} f_n$ exists, so we have $f = g$ $\mu$-a.e., i.e. $$f_n \to f \quad \text{$\mu$-a.e.}$$ Now the conditions for the monotone convergence theorem are fullfilled, and we deduce that $$ \lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu \; . $$