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 $M$.

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 $E$?

*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.

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