Using a complete measure space, show the existence of sequences of functions Fron Folland's Real Analysis

Let $(X,\mathcal{M},\mathcal{u})$ be a measure space with $\mathcal{u}(X) < \infty$, and let $(X,\mathcal{\overline{M}},\mathcal{\overline{u}})$ be its completion.
Suppose, $f:\:X \rightarrow \mathbb{R}$ is bounded. Then,
$f$ is $\mathcal{\overline{M}}$-measurable $\iff$ $\exists$ sequences $\{\phi_n\}$ and $\{\psi_n\}$ of $\mathcal{M}$-measurable simple functions such that $\phi_n \le f \le \psi_n$ and $\int(\psi_n - \phi_n)\,\mathrm{d}u < n^{-1}$. In this case, $\lim \int \phi_n\,\mathrm{d}u = \lim\int\psi_n\,\mathrm{d}u = \int f\,\mathrm{d}\overline{u}$


My work thus far:
$\implies$ Suppose $f$ is $\mathcal{\overline{M}}$-measurable. Then by properties of a complete measure space there exists a function $g$ that is $\mathcal{M}$-measureable such that $f = g \:\:\: \overline{u}\:a.e.$
Since $g$ is measurable, there exists a seq $\{ \phi_n\}$ of simple functions such that $0 \le \phi_1 \le \phi_2 \le \cdots \le g$
Here is where I get a little hazy. I want a function that converges to g from above. I feel like $\{\psi_n\}$ where $\psi_n = \sum_{j=0}^\infty \phi_j - \sum_{i=0}^n \phi_i$ will do the trick, but I'm not confident. Plus I am also wondering if I am missing something about the fact that $f = g$ a.e.
Any thoughts on how to proceed (or where to begin if I'm way off) are appreciated. Thanks!
P.S. there is a similar question here If a function is measurable with respect to the completion then it is equal to some measurable (with respect to the measure space) function a.e. but I have been unable to use it to make any headway.
 A: Hopefully it is okay to answer this here as I figure it out.
$\Longrightarrow $ Suppose $ f $ is $ \mathcal{\overline{M}} $ - measurable. Then by properties of a complete measure space there exists a function $g$ that is $\mathcal{M}$-measureable such that $f = g \:\:\:\:u -a.e.$
For every $n \in \mathbb{N} $ partion $\mathbb{R_+}$ into $(0,n] \cup (n,\infty)$. Partion $(0,n]$ into:
$$I_{n,k} = \left(\frac{k}{2n},\frac{k+1}{2n}\right], \:\:\: k = 0,1,2,...,2n^2-1$$
and let 
$$\mathcal{M_{n,k}} = g^{-1}(I_{n,k}) = \left\{x \in X \:|\: \frac{k}{2n} \lt |g(x)| \le   \frac{k+1}{2n} \right\} \quad and \quad \mathcal{M_{n,2n^2}} = g^{-1}\left( (n,\infty)\right)$$
where $\mathcal{M_{n,k}} \in \mathcal{M}$ since $g$ is measurable. Then if $\phi_n = \sum \frac{k}{2n}\chi_\mathcal{M_{n,k}}$ and $\psi_n = \sum \frac{k+1}{2n}\chi_\mathcal{M_{n,k}}$
$$|\phi_n| \le |g(x)| \le |\psi_n| \quad\quad |\psi_n - \phi_n| < \frac{1}{n}, \forall n$$
If everything up till now is correct, my question is do I use the fact that $f$ is bounded to use Dominated Convergence and show that $\int|\psi_n - \phi_n| \lt \frac{1}{n}$?
