**Definitions:**


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 $f$?
> 
> 
> 



Almost Arzela Ascoli
$\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$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.


