How to show measurability of a function implies existence of bounding simple functions If $(X,\mathscr{M},\mu)$ is a measure space with $\mu(X) < \infty$, and $(X,\overline{\mathscr{M}},\overline{\mu})$ is its completion and $f\colon X \to \mathbb{R}$ is bounded. Then $f$ is $\overline{\mathscr{M}}$-measurable (and hence in $L^1(\overline{\mu}))$ iff there exist sequences $\{\phi_n\}$ and $\{\psi_n\}$ of $\mathscr{M}$-measurable simple functions such that $\phi_n \le f \le \psi_n$ and $\int (\psi_n - \phi_n)d \mu < n^{-1}$. In this case, $\lim \int \phi_n d \mu = \lim \int \psi_n d \mu = \int f d \bar{\mu}$.
I am able to prove everything except the part that $f$ is $\overline{\mathscr{M}}$-measurable $\implies$ there exist sequences $\{\phi_n\}$ and $\{\psi_n\}$ of $\mathscr{M}$-measurable simple functions such that $\phi_n \le f \le \psi_n$ and $\int (\psi_n - \phi_n)d \mu < n^{-1}$.
I know that $f$ is $\overline{\mathscr{M}}$-measurable $\implies$ there exists an $\mathscr{M}$-measurable function $g$ s.t. $f=g$ $\overline{\mu}$-almost everywhere but I'm not sure where to proceed after that
Thank you!

A set $N$ is $(\mathcal M,\mu)$-negligible if we can find $N'\in\mathcal M$ such that $\mu(N')=0$ and $N\subset N'$. 
Recall that 
$$\overline{\mathcal M}^{\mu}=\{B\cup N,B\in\mathcal M,N\mbox{ is }(\mathcal M,\mu)-\mbox{negligible}\}.$$
It can indeed be shown that the latter collection is a $\sigma$-algebra, the smallest containing both $\mathcal M$-measurable sets and $(\mathcal M,\mu)$-negligible ones. 
First, as $f$ is $\overline{\mathcal M}^{\mu}$-measurable, it can be approximated pointwise by simple functions, that is, linear combinations of elements of  $\overline{\mathcal M}^{\mu}$. So if we deal with the case $f=\chi_S$, where $S\in\overline{\mathcal M}^{\mu}$, we write it as $S=B\cup N$, and we notice that 
$$\chi_B\leqslant \chi_S\leqslant \chi_{B\cup N'},$$
where $N'$ is as in the definition of negligible.
