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 strictly less than $\frac{1}{10}$, 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.