The setting for DCT is some fixed measure space $(X,M,\mu)$
Proof. $f$ is measurable (perhaps after redefinition on a null set) by Propositions 2.11 and 2.12, and since $\left |f \right | \leq g$ a.e. we have $f\in L^1$...
where 2.12 says that given any measure space $(X,M,\mu)$ and its natural completion $(X,\overline{M},\overline{\mu})$, then for all $f$ that $\overline{M}$-measurable there exists a $g$ that is $M$- measurable such that $f=g$ $\overline{\mu}$-a.e.
and 2.11 says that Given any complete measure $\mu$ on $(X,M)$, and $\{f_n\}$ is $M$-measurable for all $n$ and $f_n\to f$ a.e. then $f$ is measurable.
I dont understand how these two proposition makes $f\in L^1$, since $\mu$ is not a complete measure. Does the logic goes like this? Take $\{h_n\}$ be sequence of $\overline{M}$-measurable functions such that $h_n=f_n$ $\overline{\mu}$-a.e. and hence $h_n\to f$ $\overline{\mu}$-a.e. and since $\overline{\mu}$ is complete then $f$ is $M$-measurable?