I came across the following problem in an old notebook.
Let $f_n:\mathbb{R}\to [0,1]$ be a sequence of non-decreasing functions, that is, $f_n(x)\le f_n(y$ whenever $x\le y$ for all $n$. Show that there exists a subsequence $f_{n_k}$, a function $g$ and an atmost countable subset $A\subset \mathbb{R}$ such that $f_{n_k}(x)\to g(x)$ for all $x\in \mathbb{R}\setminus A$.
Since each $f_n$ is non-decreasing the point of discontinuities of $f_n$, say $D_n$, is at most countable. The candidate for set $A$ in the question is probably $\cup D_n$. Outside this set $A$ then I have to find a subsequence which is convergent. For any fixed $x$ we can obtain a convergent subsequence of $f_n$ which converges on $x$. Therefore, the strategy would be to get a countable dense subset of $A^c$, and using a diagonalization argument obtain a subsequence which will converge on this countable dense subset of $A^c$. My trouble is to show that this same subsequence would work for all other points in $A^c$. Any help along this line or an alternate proof will be much appreciated.