A question from a past qualifying exam at my university reads: Let $f_n$ be non-decreasing functions on $X:=(-\infty,0]$ such that $f_n\rightarrow 0$ in (Lebesgue) measure as $n\rightarrow \infty$. Proof or counterexample: Necessarily $f_n\rightarrow 0$ almost everywhere on $X$ w.r.t Lebesgue measure.
Proof: We have that $f_n\leq f_{n+1}\leq 0$. Thus $|f_n|\geq |f_{n+1}|\geq 0$ for all $n$. Now for each $k\in \mathbb{N}$, pick $n_k$ large enough so that $n_{k+1}>n_k$ and $m(\{x\in X:|f_{n_k}(x)|>1/k \})<1/k$. We may do this because $f_n\rightarrow f$ in measure. Next, define $$ E_k=(\{x\in X:|f_{n_k}(x)|>1/k \})$$ Because $|f_n|\geq|f_{n+1}|$ for all $n$, then $E_{k+1}\subset E_{k}$. Suppose that $|f_n(x)|$ does not converge to $0$ for some $x\in X$ The measure of each $E_k$ is finite, so we may use the continuity of integration and conclude that $m(\cap_1^\infty E_k)=\lim(E_k)=0$. Now, for $x\in X$, if $f_n(x)$ does not converge to $0$, then it will not any of the $E_k$. This set has measure zero. I was wanting $\cap_1^\infty E_k$ to be the set where $f_n$ does not converge to 0 pointwise, but I don't think this is the case.
Next, I was thinking about looking at the sets $ E_k=\cup_{m=k}^\infty\{x\in X:|f_{m}(x)|>1/k \}$, but I'm not sure how to control the measure of such sets. I know that in general, convergence in measure does not imply pointwise convergence almost every, but I feel like it is true in this special case.
Edit: I think I misinterpreted the question as saying that the sequence of functions is non-decreasing. Rather, each $f_n$ is non-decreasing.