Suppose that $f_n \colon \Bbb{R}\to [0,1]$ for $n\in \Bbb{N}$ and that each one of these functions is nondecreasing. Prove that there exists a function $g\colon \Bbb{R} \to [0,1]$, a countable set $A\subset \Bbb{R}$ and a subsequence $f_{n_k}$ such that $\lim\limits_{k\to\infty} f_{n_k}(x)=g(x)$ for all $x\in \Bbb{R}\setminus A$.

I'm reviewing real analysis and came across this problem from a homework assignment: http://www.math.washington.edu/~toro/Courses/14-15/524/1-hw.pdf

(number 5, part 2)

My question is, what does nondecreasing or countable set have to do with anything? Why can't we just fix an $x$, then $f_n(x)$ is just a sequence in $[0,1]$, which is compact, hence sequentially compact, and note that it has a convergent subsequence. Then $g$ is just the function that maps to these points. In part 1 of the problem we prove that nondecreasing functions have countably many discontinuities which is very routine, so I assume the convergence doesn't happen on those points. But what goes wrong?

  • $\begingroup$ The set of all possible discontinuities for $\{f_n\}$ is a countable union of countable sets, so it is countable, and maybe $A$ will contain it. But the problem with your construction is that you choose for one $x$ a convergent subsequence $f_{n_k}$, then for another $x$ you would choose a further subsequence, but all $x$ will be uncountably many. $\endgroup$
    – Svetoslav
    Feb 13, 2016 at 23:46
  • $\begingroup$ @Svetoslav Why not just let $A=\emptyset$? $\endgroup$
    – user223391
    Feb 13, 2016 at 23:50
  • $\begingroup$ @Svetoslav We're dealing with pointwise convergence, right? Each $x$ is independent of the others. $\endgroup$
    – user223391
    Feb 13, 2016 at 23:52
  • $\begingroup$ Well, maybe you can let $A$ to be empty, I just mentioned it to avoid possible difficulties around the discontinuity points. But this is not the problem. The problem is how you construct the subsequences. $\endgroup$
    – Svetoslav
    Feb 13, 2016 at 23:54
  • $\begingroup$ Yes, pointwise convergence. $\endgroup$
    – Svetoslav
    Feb 13, 2016 at 23:55

1 Answer 1


The problem with what you outline in the second paragraph is that the subsequence may depend on $x$, but you need a single sequence that works for all $x\notin A$.

To do this, choose a subsequence such that $f_{n_k}(x)$ converges for all $x$ from a countable dense subset of $[0,1]$ (use a diagonal procedure: first find a subsequence $n_k^{(1)}$ so that you have convergence at $x_1$, then a subsequence of this sequence, $n_k^{(2)}$ that gives you convergence at $x_2$ etc., and finally let $n_k=n_k^{(k)}$).

Then $g(x)=\lim f_{n_k}(x)$ is defined on a countable dense set $B$, and this function is increasing. Therefore, it has a unique increasing extension to all of $[0,1]$. Then $f_{n_k}(x)\to g(x)$ at all points of continuity of $g$. To see this, let $\epsilon>0$ be given. Pick $a,b\in B$, $a<x<b$, such that $g(x)-g(a)$ and $g(b)-g(x)$ are both $<\epsilon$. For large $k$, we have that $f_{n_k}(t)$ almost agrees with $g(t)$ at $t=a,b$. Since $f_{n_k}$ is increasing, this means that $f_{n_k}(x)$ is close to $g(x)$, too.

(A more abstract view of your statement is obtained by considering the measures generated by the $f_n$; then your claim becomes a routine consequence of the Banach-Alaoglu theorem.)

  • $\begingroup$ Of course, thank you! I thought about this after I posted the question but I'm glad you confirmed. $\endgroup$
    – user223391
    Feb 14, 2016 at 1:50

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