Given a measurable subset $A$ of $[0, 1]$, a sequence of functions $f\_n: [0, 1] \to \mathbb R$ is said to be *equi-Lebesgue continuous* on $A$ if for every $x \in A$, and $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $0 < r < \delta$, we have


$$\frac{1}{2r} \int\_{B\_r (x)} \lvert f\_n (x) - f\_n (y)\rvert \, dy < \varepsilon$$


for all $n \in \mathbb N$.


Let $f\_n: [0, 1] \to \mathbb R$ be a sequence of functions equibounded in $L^\infty$, that is, $\sup\_{n \in \mathbb N} \lVert f\_n \rVert\_{L^\infty} < \infty$. Suppose further that there exists a subset $E$ of $[0, 1]$ of measure $1$ such that $f\_n$ are equi-Lebesgue continuous on $E$.


**Question:** Does there exist a subsequence $f\_{n\_k}$ of $f$ converging a.e.?



Arzelà–Ascoli for equi-Lebesgue continuous functions
$\newcommand\ep\varepsilon\newcommand\ze\zeta\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$The answer is yes.


Indeed, take any real $\be>0$. Let
\begin{equation\*}
\al:=\be/2,\quad\ep:=\be^2/48,\quad\ze:=\eta:=\be/4. 
\end{equation\*}
Write $B\_x(r):=[0,1]\cup(x-r,x+r)$ instead of $B\_r(x)$.


Without loss of generality (wlog), $|f\_n|\le M$ on $E$ for some real $M>0$ and all $n$.


By the regularity of the Lebesgue measure, there is a compact subset $K\_\al$ of $E$ such that
\begin{equation\*}
 |E\setminus K\_\al|=|[0,1]\setminus K\_\al|\le\al, \tag{0}\label{0}
\end{equation\*}
where $|A|$ denotes the Lebesgue measure of a subset $A$ of $\R$.


By the main condition in the OP,
\begin{equation\*}
 \forall x\in E\ \exists \de\_{x,\ep}\in(0,\infty)\ \forall r\in[0,3\de\_{x,\ep}]\ \forall n\ 
\end{equation\*}
\begin{equation\*}
 \int\_{B\_x(r)}|f\_n(y)-f\_n(x)|\,dy\le\ep|B\_x(r)|. \tag{1}\label{1}
\end{equation\*}


Since $K\_\al$ is compact, there is a finite set $G\_{\al,\ep}\subset K\_\al$ such that
\begin{equation\*}
 K\_\al\subseteq\bigcup\_{x\in G\_{\al,\ep}}B\_x(\de\_{x,\ep}). 
\end{equation\*}
Moreover, by the [Vitali covering lemma](https://en.wikipedia.org/wiki/Vitali_covering_lemma#Finite_version),
there is a finite set $F\_{\al,\ep}\subseteq G\_{\al,\ep}$ such that
the balls $B\_x(\de\_{x,\ep})$ for $x\in F\_{\al,\ep}$ are **pairwise disjoint** and
\begin{equation\*}
 K\_\al\subseteq\bigcup\_{x\in F\_{\al,\ep}}B\_x(3\de\_{x,\ep}). \tag{1.5}\label{1.5}
\end{equation\*}


By \eqref{1} and Markov's inequality,
\begin{equation\*}
 |A\_{n,r,x,\eta}|\le\frac\ep\eta\,|B\_x(r)|
\end{equation\*}
for all natural $n$, all $x\in F\_{\al,\ep}$, and all $r\in[0,3\de\_{x,\ep}]$, where
\begin{equation\*}
 A\_{n,r,x,\eta}:=\{y\in B\_x(r)\colon|f\_n(y)-f\_n(x)|\ge\eta\}. 
\end{equation\*}
So, recalling that the balls $B\_x(\de\_{x,\ep})$ for $x\in F\_{\al,\ep}$ are pairwise disjoint, for
\begin{equation\*}
 A\_{n,\ep,\eta}:=\bigcup\_{x\in F\_{\al,\ep}}A\_{n,3\de\_{x,\ep},x,\eta}
\end{equation\*}
we have
\begin{equation\*}
 |A\_{n,\ep,\eta}|\le\sum\_{x\in F\_{\al,\ep}}\frac\ep\eta\,|B\_x(3\de\_{x,\ep})|
 \le3\frac\ep\eta\,\sum\_{x\in F\_{\al,\ep}}|B\_x(\de\_{x,\ep})|\le3\frac\ep\eta. \tag{2}\label{2}
\end{equation\*}


Recalling that $|f\_n|\le M$ on $E$ for all $n$ and $F\_{\al,\ep}\subset E$, and passing to a subsequence if needed, wlog we have
$f\_n(x)\to g(x)\ \forall x\in F\_{\al,\ep}$
(as $n\to\infty$), where $g$ is some real-valued function on $F\_{\al,\ep}$, so that for some natural $n\_{\al,\ep,\ze}$ we have
\begin{equation\*}
 n\ge n\_{\al,\ep,\ze}\implies\forall x\in F\_{\al,\ep}\ |f\_n(x)-g(x)|\le\ze. 
\end{equation\*}
So, if $m,n\ge n\_{\al,\ep,\ze}$ and
$y\in B\_x(3\de\_{x,\ep})\setminus A\_{m,\ep,\eta}\setminus A\_{n,\ep,\eta}$ for some $x\in F\_{\al,\ep}$, then
\begin{equation\*}
 |f\_m(y)-f\_n(y)|\le|f\_m(y)-f\_m(x)|+|f\_m(x)-g(x)|+|g(x)-f\_n(x)|+|f\_n(x)-f\_n(y)|
 \le\eta+\ze+\ze+\eta, 
\end{equation\*}
whence, in view of \eqref{1.5},
\begin{equation\*}
 |f\_m(y)-f\_n(y)|\le2\eta+2\ze=\be
\end{equation\*}
if $m,n\ge n\_{\al,\ep,\ze}$ and
$y\in K\_\al\setminus A\_{m,\ep,\eta}\setminus A\_{n,\ep,\eta}$.


So,  

\begin{equation\*}
 |\{x\in[0,1]\colon |f\_m(y)-f\_n(y)|>\be\}|\le|[0,1]\setminus K\_\al|
 +|A\_{m,\ep,\eta}|+|A\_{n,\ep,\eta}|
 \le\al+2\times3\frac\ep\eta=\be
\end{equation\*}
if $m,n\ge N\_\be:=n\_{\al,\ep,\ze}=n\_{\be/2,\be^2/48,\be/2}$.
So, the sequence $(f\_n)$ is Cauchy convergent in measure, and hence convergent in measure. So, a subsequence of $(f\_n)$ is convergent almost everywhere, as claimed.




---


An almost the same proof will work for the corresponding general statement for functions $f\_n$ on $[0,1]^d$ for any natural $d$ and, even more generally, for any complete separable metric space $S$ with a finite [doubling Borel measure](https://en.wikipedia.org/wiki/Doubling_space#Definition) $\mu$ over $S$, so that $\mu(B\_x(3r))\le C\mu(B\_x(r))$ for some real $C>0$, all $x\in S$, and all real $r>0$, where $B\_x(r)$ is, of course, the ball in $S$ of radius $r$ centered at $x$.




---


Also, the main condition in the OP can be relaxed to the following:


\begin{equation}
 \forall x\in E\ \forall\ep>0\ \exists\de>0\ \forall n
\end{equation}
\begin{equation}
 \int\_{B\_x(\de)} |f\_n (x)-f\_n (y)|\,dy<\ep|B\_x(\de)|. 
\end{equation}


