We say a sequence of continuous functions $f_n: [0, 1] \to \mathbb R$ is equicontinuous on average if for every $x \in [0, 1]$ and $\varepsilon > 0$ there exists some $\delta > 0$ such that $\limsup_{N \to \infty} \frac{1}{N} \sum_{n = 0} ^{N-1} |f_n (x) - f_n (y)| < \varepsilon$ whenever $|x - y| < \delta$.

Suppose $f_n$ are continuous functions that are equicontinuous on average and uniformly bounded - that is, $\sup_n \sup_x |f_n(x)| < \infty$.

Question: Does there exist a subsequence $f_{n_k}$ that converges in measure to some continuous $f$?

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    $\begingroup$ What do you mean by convergence in measure for a sequence of continuous functions? $\endgroup$ May 25, 2021 at 13:18
  • $\begingroup$ @DieterKadelka The usual notion of convergence in measure of functions with respect to the Lebesgue measure. $\endgroup$
    – Nate River
    May 25, 2021 at 13:19
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    $\begingroup$ I guess $f_n(x) = \arctan(n(x-a_n))$ with $a_n$ sampled uniformly from $[\tfrac13, \tfrac23]$ is a counterexample (almost surely), but I did not check the details. $\endgroup$ May 25, 2021 at 14:24

1 Answer 1


$\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$The answer is no. To show this, let us use a slightly modified suggestion in a comment by Mateusz Kwaśnicki, as follows. Let \begin{equation} f_n:=g_{a_n,1/n}, \end{equation} where \begin{equation} g_{a,h}(x):=1(a<x\le a+h)\frac{x-a}h+1(x>a+h) \end{equation} and $(a_n)$ is any sequence uniformly distributed on $[1/4,3/4]$, so that \begin{equation} \frac1N\sum_1^N 1(a<a_n\le a+h)\to\frac h{3/4-1/4}=2h \end{equation} (as $N\to\infty$) for any $a$ and $h$ such that $1/4\le a\le a+h\le3/4$. Note that $f_n$ is continuous and $0\le f_n\le1$, for each $n$.

Take now any $x$ and $y$ such that $0\le x<y\le1$. Then \begin{align*} \sum_{n=1}^N |f_n(x)-f_n(y)|=\sum_{n=1}^N (f_n(y)-f_n(x))\le S_1+S_2+S_2, \end{align*} where \begin{equation} S_1:=\sum_{n=1}^N 1(a_n\le x\le a_n+1/n)=\sum_{n=1}^N 1(x-1/n\le a_n\le x)=o(N), \end{equation} since $(a_n)$ is uniformly distributed on $[1/4,3/4]$; similarly, \begin{equation} S_2:=\sum_{n=1}^N 1(a_n\le y\le a_n+1/n)=o(N); \end{equation} and \begin{equation} S_3:=\sum_{n=1}^N 1(x\le a_n, y\ge a_n+1/n)\le\sum_{n=1}^N 1(x\le a_n\le y)\lesssim 2N(y-x). \end{equation} Thus, \begin{align*} \limsup_N\frac1N\sum_{n=1}^N |f_n(x)-f_n(y)|\le2|y-x| \end{align*} for any $x,y$ in $[0,1]$.

On the other hand, the limit $f$ of any subsequence $f_{n_k}$ that converges in measure is of the form $f(x)=1(x>a)$ for some $a\in[1/4,3/4]$ and almost all $x\in[0,1]$. So, the limit cannot be continuous.


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