For every nonnegative integer $n$, define $A_n:\mathbb R\to\mathbb R$ by $A_n(t)=k2^{-n}$ for every $t$ such that $k\leqslant2^nt\lt k+1$, for some integer $k$ (in other words, $2^nA_n(t)$ is the integer part of $2^nt$).
Let $\phi_n=A_n(f)$ and $\psi_n=-A_n(-f)$. Then $\phi_n$ and $\psi_n$ are step functions, $\phi_n\leqslant f\leqslant \psi_n$, and $\psi_n-\phi_n\leqslant2^{-n}$ for every $n$. Furthermore, the sequence $(\phi_n)_n$ is nondecreasing and the sequence $(\psi_n)_n$ is nonincreasing.
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 $\int \phi_n d\mu = \int \phi_n d\bar{\mu} \rightarrow \int f d\bar{\mu}$.
Best Answer
If $f$ is bounded and $M-$ measurable there exist simple functions $g_n$ increasing to $f^{+}$ and simple functions $h_n$ decreasing to $f^{-}$. Hence $\phi_n \equiv g_n-h_n \leq f$ and $\int (f-\phi_n) \to 0$. By applying this to $-f$ we can find the $\psi_n$'s. Once you get $\int (\psi_n-\phi_n) \to 0$ you can go to a subseqeunce to get the inequality $\int (\psi_n-\phi_n) \leq \frac 1 n$ for all $n$.
Now suppose $f$ is bounded and $\overset {-}M$ measurable. There exists an $M-$ measurable function $g$ such that $f=g$ almost everywhere. Suppose $f(x)=g(x)$ for $x \notin E$ where $E$ has measure $0$. Suppose $-C<f(x)<C$ for all $x$. Now choose $\phi_n$ and $\psi_n$ as above with $f$ changed to $g$ and then define $\Phi_n=\phi_n-I_E C,\Psi_n=\psi_n+I_E C$. Then $\Phi_n$ and $\Psi_n$ are simple functions satisfying the required properties.