If a sequence of monotone functions converges in measure, does it also converge almost everywhere? Let $\{f_n\}$ be a sequence of monotone functions from $\Bbb R$ to $\Bbb R$ such that $f_n$ converges in measure to some function $f$. Is it true that $f_n$ converges to $f$ a.e.?
I am sure it has a sub-sequence that converges to $f$ a.e., and intuitively it seems that it must be true, but I am not able to prove it rigorously. Could you please give me some hint?
 A: By dividing $\{f_n\}$ into two subseqences, the increasing functions and the decreasing ones, and working with each subsequence separately, we may assume without loss of generality that $\{f_n\}$ is a sequence of increasing functions.
Note that $\{{f_n}_{|[-1,1]}\}$ is a sequence of functions defined on a finite measure space converging in measure to $f_{|[-1,1]}$, so a standard result tells us that there is a subsequence, say $\{{f^{(1)}_n}_{|[-1,1]}\}$, of $\{{f_n}_{|[-1,1]}\}$, that converges a.e. to $f$ on $[-1,1]$. Furthermore there is a subsequence, say $\{{f^{(2)}_n}_{|[-2,2]}\}$, of $\{{f^{(1)}_n}_{|[-2,2]}\}$ converging a.e. to $f$ on $[-2,2]$. If we continue producing subsequences of subsequences in this fashion we see that the "diagonal" subsequence $\{f^{(n)}_n\}$ converges pointwise to $f$ on a co-null set $X\subset \mathbb{R}$. Since the $f^{(n)}_n$ functions are increasing we get that $f_{|X}$ is increasing. This also means that $f_{|X}$ is continuous on a co-countable set $A\subset X$. Then $A$ is also co-null.
We claim that $f_n \rightarrow f$ pointwise on $A$. Indeed, fix $x'\in A$ and $\epsilon>0$. Choose $a, b\in A$ such that $x'\in (a, b)$ and $f(b)-f(a)<\epsilon$. Pick $\delta$ such that $x'\in (a+\delta, b-\delta)$. Since $f_n \rightarrow f$ in measure there exists $N\in \mathbb{N}$ such that $\mu(|f-f_n|\geq \epsilon)<\delta$ for $n\geq N$. 
For such $n$ we must have $|f(x')-f_n(x')|<2\epsilon$. Indeed, suppose for contradiction that either $f_n(x')\leq f(x')-2\epsilon$ or $f(x')+2\epsilon\leq f_n(x')$. In the first case we get that for $x\in (a, x')\cap A$: $f_n(x)\leq f_n(x')\leq f(x')-2\epsilon\leq f(x)-\epsilon$ where the last inequality follows from the fact that $f$ does not vary by more than $\epsilon$ on $(a,b)\cap A$. But this means in particular that $(a, x')\cap A\subset \{|f-f_n|\geq \epsilon\}$ which is impossible since the LHS has measure greater than $\delta$ and the RHS measure less than $\delta$. Similarly, if $f(x')+2\epsilon\leq f_n(x')$ were to hold we would get that $(x', b)\cap A\subset \{|f-f_n|\geq \epsilon\}$ which again is impossible. Hence $|f(x')-f_n(x')|<2\epsilon$.
