Let $f\_n: [0, 1] \to \mathbb R$ be a sequence of functions.


Given a measurable subset $E$ of $[0, 1]$, we say that the sequence $f\_n$ is *equicontinuous on $E$* if for every $x \in E$, and $\varepsilon > 0$ there exists a $\delta > 0$ and all $n \in \mathbb N$ we have $|f\_n(x) - f\_n(y)| < \varepsilon$ for all $y \in [0, 1]$ such that $|y - x| < \delta$.


We say a sequence of functions $f\_n: [0, 1] \to \mathbb R$ has uniformly bounded variation if there exists some $M > 0$ such that each $f\_n$ has total variation less than or equal to $M$.



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> **Question:** Given any sequence of functions $f\_n: [0, 1] \to \mathbb R$ of uniformly bounded variation, is it true that there exists a measurable subset $E$ of $[0, 1]$ with Lebesgue measure $1$ and a subsequence $f\_{n\_k}$ of lower density $1$ such that $f\_{n\_k}$ is equicontinuous on $E$?
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*Note:* We define the *lower density* of a subsequence $f\_{n\_k}$ to be the number


$$\liminf\_n\frac {\#\{k \, | \, k \in \mathbb N, \, n\_k < n\}}{n}$$


where $\#$ denotes the cardinality of a finite set.



Is every sequence of functions with uniformly bounded variation almost equicontinuous?
Consider the pairs $(n,k)\in\mathbb{N}^2$ with $0\leq k<2^n$, they form a sequence in lexicographic order. Consider now the functions $f\_{n,k}$ which are defined as $0$ in $[0,\frac{k}{2^n}]$, $1$ in $[\frac{k+1}{2^n},1]$, and $2^nx-k$ in $[\frac{k}{2^n},\frac{k+1}{2^n}]$, and suppose there is a set $E\subseteq [0,1]$ and a subsequence of $f\_{n,k}$ as in the question.


If we define $A\_n=\bigcup\{(\frac{k}{2^n},\frac{k+1}{2^n});(n,k)\text{ is in the subsequence}\}$, then the fact that the subsequence has density $1$ implies that when $n$ tends to $\infty$, the measure of $A\_n$ tends to $1$. This implies that there is a point $x\in E$ which is in infinitely many of the $A\_n$: for example, take a subsequence $A\_{n\_i}$ such that $\sum\_i(1-m(A\_{n\_i}))<1$; then $m(\cap\_k A\_{n\_i})>0$), so $E\cap(\cap\_k A\_{n\_i})\neq\emptyset$.


Suppose then that $x\in E\cap(\cap\_i A\_{n\_i})$ and we are given $\varepsilon<\frac{1}{2}$; there is no $\delta$ that satisfies the definition of equicontinuity at $x$: indeed, take $i$ such that $2^{-n\_i}<\delta$, then there is some interval $(\frac{k}{2^{n\_i}},\frac{k+1}{2^{n\_i}})$ contained in $(x-\delta,x+\delta)$ and such that $(n\_i,k)$ is in the subsequence. So the image of $(x-\delta,x+\delta)$ under the function $f\_{n\_i,k}$ is all $[0,1]$, thus it contains values at distance $>\varepsilon$ of $f\_{n\_i,k}(x)$.


