*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 $F$?



A “compactness theorem” for measurable functions
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).


