Since the solution to this appears to be nowhere on the internet - a bit shocking given the ubiquity of the textbook and the (fairly) elementary nature of the problem in the context of Real Analysis - I'll post the whole thing. Hopefully no intractable errors.
Suppose $f$ is $\bar{\mathcal{M}}$-measurable and non-negative. Then there is an $\mathcal{M}$-measurable $g$ such that $f = g$ up to an $\mathcal{M}$-null set $N$. If $s = \sum_1^k a_j\cdot \chi_{A_j}$ is an arbitrary simple function, then $s^- = (\sum_1^k a_j\cdot \chi_{A_j \setminus N}) + 0\cdot \chi_N$ and $s^+ = (\sum_1^k a_j\cdot \chi_{A_j \setminus N}) + M \cdot \chi_N$ are simple functions, where $M$ is a bounding constant for $f(x)$.
As in Theorem $\textbf{2.10}$, write simple functions
$$\phi_n = \sum_{k = 0}^{2^{2n}-1} k2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_{F_n}, \hspace{1cm} \psi_n = \sum_{k = 0}^{2^{2n}-1} (k+1)2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_{F_n}$$
$$E_n^k = g^{-1}((k2^{-n}, (k+1)2^{-n}]), \hspace{1cm} F_n = g^{-1}((2^n, M))$$
Then these are simple functions converging pointwise to $g$ from above and below, respectively, in monotone fashion. Note that for $n$ large enough, $2^n > M$ so that for $n$ sufficiently large we can simply write
$$\phi_n^- = \sum_{k = 0}^{2^{2n}-1} k2^{-n}\cdot \chi_{E_n^k \setminus N}, \hspace{1cm} \psi_n^+ = \sum_{k = 0}^{2^{2n}-1} (k+1)2^{-n}\cdot \chi_{E_n^k} + M \cdot \chi_N$$
Since $f = g$ away from $N$, $\phi_n^- \leq f \leq \psi_n^+$ for all $n$, and on $N$ itself, $\phi_n^- \equiv 0 \leq f \leq M \equiv \psi_n^+$.
It remains to be shown that $\int(\psi_n^+ - \phi_n^-) d\mu \rightarrow 0$. For $n$ sufficiently large,
$$\int_X(\psi_n^+ - \phi_n^-) d\mu = \int_X(\psi_n^+ - \phi_n^-) d\bar{\mu} = \int_N(\psi_n^+ - \phi_n^-) d\bar{\mu} + \int_{N^c}(\psi_n^+ - \phi_n^-) d\bar{\mu} =$$
$$=\int_{N^c}(\psi_n^+ - \phi_n^-) d\bar{\mu} \leq \int_{N^c}|\psi_n^+ - f|+ |f - \phi_n^-| d\bar{\mu} \leq$$
$$ \leq \int_{N^c}|\frac{1}{2^n}|+ |\frac{1}{2^n}| d\bar{\mu} = \frac{1}{2^{n-1}} \cdot \bar{\mu}(N^c) = \frac{\bar{\mu}(X)}{2^{n-1}} = \frac{\mu(X)}{2^{n-1}} \rightarrow 0$$
Thus after reindexing $n$ if need be, the claim is proven for non-negative $f$. Taking positive and negative parts gives the general result.
Conversely, suppose such $\phi_n, \psi_n$ exist, and assume $f$ is non-negative. If $\psi_n = \sum_1^k a_i \cdot \chi_{A_i}$ with $a_i > M$ then we may simply adjust to the simple $\tilde{\psi_n}$ whose coefficients are the same as those of $\psi_n$ other than changing any such $a_i > M$ to $M$. Since $f$ is bounded by $M$, it is still the case that $f \leq \tilde{\psi_n}$. Construct monotone sequences of simple functions $\Phi_n = \max\lbrace \phi_1, \phi_2, \dots, \phi_n \rbrace, \Psi_n = \min \lbrace \tilde{\psi_1}, \tilde{\psi_2}, \dots, \tilde{\psi_n} \rbrace$. Then $0 \leq \Phi_n \leq f \leq \Psi_n \leq M$. Further,
$$|\Psi_n - \Phi_n| \leq |\psi_n - \phi_n| \implies \int(\Psi_n -\Phi_n) d\mu \leq \int (\psi_n - \phi_n) d\mu \rightarrow 0$$
Since $\Phi_n, \Psi_n$ are bounded above by $M$ they have finite integrals and are thus in $L^1(\mu)$. In particular so are their limits, which are $\mathcal{M}$-measurable, so that $\Phi = \lim \Phi_n, \Psi = \lim \Psi_n$ are contained in $L^1(\mu)$. For all $n$,
$$\int (\Psi - \Phi) d\mu \leq \int (\Psi_n - \Phi_n) d\mu \rightarrow 0$$
Thus $\Psi - \Phi = 0$ $\mu$-a.e., or said another way $\Psi$ and $\Phi$ differ only on an $\mathcal{M}$-null set. Since $\Phi \leq f \leq \Psi$, it follows that $f$ differs from either only on an $\mathcal{M}$-null set and so is $\bar{\mathcal{M}}$-measurable, since each is $\mathcal{M}$-measurable. Taking positive and negative parts gives the general result.
For the comment at the end of the exercise, the first equality was shown directly in each direction. For the second equality, taking $g \equiv M$ in the DCT for $\bar{\mu}$ it follows that since $\phi_n \rightarrow f$ $\bar{\mu}$-a.e. and $\phi_n \in L^1(\mu) \subset L^1(\bar{\mu})$ that $\int \phi_n d\mu = \int \phi_n d\bar{\mu} \rightarrow \int f d\bar{\mu}$.
I don't understand the sentence
So, we know convergence in $L^1$ is equivalent to: $\mu(E_n) < \int |f| < \infty.$
Anyway the main idea is correct. Convergence in $L^1$ implies that a subsequence $\{ \chi_{E_{n_k}} \}$ converges to $f$ pointwise a.e.
The second part of the proof is a bit confusing. You cannot define $f=\chi_E$ and anyway your set $E$ is not correct.
Since $\{ \chi_{E_{n_k}} \}$ converges to $f$ pointwise a.e., take a point $x$ such that $\chi_{E_{n_k}}(x)\to f(x)$. Since $\chi_{E_{n_k}}(x)$ only takes values $0$ and $1$ and the limit exists, necessarily $\chi_{E_{n_k}}(x)=0$ for all $k$ large, in which case $\chi_{E_{n_k}}(x)=0\to 0$, or $\chi_{E_{n_k}}(x)=1$ for all $k$ large in which case $\chi_{E_{n_k}}(x)=1\to 1$. Hence, $f(x)$ can only be $0$ or $1$, which shows that $f$ is the characteristic function of a set.
Best Answer
The question is:
Let $[f \neq g] =\{x \in X: f(x) \neq g(x)\}$. Note that $[f \neq g] \subseteq Z \subseteq N$. Since $N \in \mathcal{M}$ and $\mu(N)=0$, any subset $C$ of $N$ is in $\overline{\mathcal{M}}$ and $\bar{\mu}(C)=0$ . Since $[f \neq g] \subseteq N$, we have that $[f \neq g] \in \overline{\mathcal{M}}$ and $\bar{\mu}([f \neq g])=0$. So $f=g$ $\bar{\mu}-$almost everywhere.