Note: Here we consider the Lebesgue measure on $[0, 1]$.

Let $f_n: [0, 1] \to [0, 1]$ be a sequence of measurable functions.

We say a measurable subset $E$ of $[0, 1]$ is a condensation set of the sequence $f_n$ if there exists a subsequence $f_{n_k}$ and a function $f: [0, 1] \to [0, 1]$ (both depending on $E$) such that for almost every $x \in E$, we have $f_{n_k} (x) \to f(x)$.

Question:

Let $f_n: [0, 1] \to [0, 1]$ be a sequence of measurable functions.

Suppose there exists some $\varepsilon > 0$ such that any measurable subset of $[0, 1]$ of measure less than or equal to $\varepsilon$ is a condensation set of $f_n$.

Does it follow that there exists some measurable function $F$, and a subsequence of $f_n$ that converges pointwise a.e. to $F$?