A “compactness theorem” for measurable functions

Question

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 $F$?

Answer 1

Counterexample. For each $n$ let $k_n$ be the characteristic function of $[0,\frac{1}{2^n}] \cup [\frac{2}{2^n},\frac{3}{2^n}] \cup \cdots$. Next observe that there are only countably many subsets of $[0,1]$ of the form: a finite union of open intervals with rational endpoints, whose total length is $\frac{1}{10}$. Enumerate these sets as $(A_n)$, and for each $n$ let $f_n$ be the function which is constantly $1$ on $A_n$ and equals $k_n$ on $[0,1]\setminus A_n$.

For any subset $E$ of $[0,1]$ whose measure is at most $\frac{1}{10}$, we can find a subsequence $(A_{n_k})$ of $(A_n)$ such that $m(E\setminus A_{n_k}) \to 0$. Then $(f_{n_k}) \to 1$ pointwise a.e. on $E$.

Now suppose that some subsequence $(f_{n_j})$ converges pointwise a.e. on $[0,1]$ to some function $F$. By Egoroff, we can assume that $(f_{n_j}) \to F$ uniformly on $A$, for some measurable $A \subseteq [0,1]$ with $m(A) > \frac{9}{10}$. But this is impossible, because for any $n_0$ the set $B$ on which $f_{n_0}$ is constantly zero has measure at least $\frac{4}{10}$ (namely, $\frac{1}{2}$ on which $k_{n_0}$ is zero, minus at most $\frac{1}{10}$ on $A_{n_0}$), hence has positive measure intersection with $A$, and no subsequence of $(f_n)$ converges to zero on a set of positive measure (thus showing that for sufficiently large $n$, $f_n$ is not constantly zero, and hence has uniform distance $1$ to $f_{n_0}$, on this intersection).