In Folland, Theorem 2.32 states
If $f_n \to f$ in $L^1$ then there is a subsequence $\{f_{n_j}\}$ such that $f_{n_j} \to f$ almost everywhere.
The proof given just says to "combine propositions 2.29 and 2.30" which say
(2.29) If $f_n \to f$ in $L^1$ then $f_n \to f$ in measure.
and
(2.30) Suppose $\{f_n\}$ is Cauchy in measure then there is a measureable $f$ such that $f_n \to f$ in measure and there is a subsequence $f_{n_j}$ that converges to $f$ almost everywhere. Moreover if $f_n \to g$ in measure then $f =g$ almost everywhere.
I'm unsure how how this is implied? I think part of this is because I don't fully understand 2.30. I know that given $f_n \to f$ in L^1 then by 2.29 we know that $f_n \to f$ in measure. 2.30, however, deals with a sequence that is Cauchy in measure. However, in general convergence in measure does not imply Cauchy in measure, correct?
Convergence in measure implies Cauchy in measure.
Let our measure be $\mu$. if $f_n \to f$ in measure, then given $\epsilon > 0$, we have $\mu(\{x : |f_n(x) - f(x)| > \epsilon/2\})\to 0$ as $n \to \infty$. But as $|f_n(x) - f_m(x)| \leq |f_n(x) - f(x)| + |f(x) - f_m(x)|$, we have $$ \{x: |f_n(x) - f_m(x)| > \epsilon\} \subseteq \{x: |f_n(x) - f(x)| > \epsilon/2\} \cup \{x : |f(x) - f_m(x)| > \epsilon/2\}.$$ Monotonicity of measure then gives that $\mu(\{x: |f_n(x) - f_m(x)| > \epsilon\})$ is arbitrarily small as $n,m \to \infty$. Concretely: given $\delta > 0$, there exists $N$ such that if $n \geq N$, then $\mu(\{x : |f_n(x) - f(x)|> \epsilon/2\}) < \delta/2$. For this $N$, if $n,m \geq N$, then also $\mu(\{x: |f_n(x) - f_m(x)| > \epsilon\}) < \delta$.