folland Exercise 2.24 folland Exercise 2.24. Let $(X, \mathcal{M}, \mu)$ be a measure space with $\mu(X)<\infty,$ and let $(X, \overline{\mathcal{M}}, \bar{\mu})$ be its completion. Suppose $f: X \rightarrow \mathbb{R}$ is bounded. Then $f$ is $\overline{\mathcal{M}}$ -measurable (and hence in $\left.L^{1}(\bar{\mu})\right)$ iff there exist sequences $\left\{\phi_{n}\right\}$ and $\left\{\psi_{n}\right\}$ of $\mathcal{M}$ -measurable simple functions such that $\phi_{n} \leq f \leq \psi_{n}$ and $\int\left(\psi_{n}-\phi_{n}\right) d \mu< \frac{1}{n} .$ In this case, $\lim \int \phi_{n} d \mu=\lim \int \psi_{n} d \mu=\int f d \bar{\mu}$ .
also show that for case $\mu(X)=\infty$ or $f$ is $un$bounded this is not true .
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, $\phi_n^- \equiv 0 \leq f \leq M \equiv \psi_n^+$.
 A: $f: X \rightarrow \mathbb{R}$ is bounded, say $|f| \leq M, \quad$  Suppose $f$ is $\bar{m}$ -measurable. Let $g_{+}(x)=f(x)+M$ and $g_{-}(x)=-f(x)+M .$ Then $g_{\pm} \in L^{+}(\bar{\mu})$ so there are increasing sequences $\left\{\phi_{j}^{+}\right\}$ and $\left\{\phi^- _{j}\right\}$ of simple functions in $L^{+}(\bar{\mu}) , \phi_{j}^{+} \rightarrow g_{+}$ and $\phi_{j}^{-} \rightarrow g_{-} b y$ Theorem 2.10. By the Monotone Convergence Theorem, $\int \mathrm{\phi}_{j}^{\pm} d \bar{\mu} \rightarrow \int g_{\pm} d \bar{\mu}$. Given $n \in \mathbb{N},$ since $g_{\pm} \in \mathrm{L}^{\prime}$
$\exists j \in \mathbb{N} , 0 \leq \int g_{+} d \bar{\mu}-\int \phi_{j}^{+} d \bar{\mu}<\frac{1}{2 n}, \quad 0 \leq \int g_{-} d \bar{\mu}-\int \phi_{j}^{-} d \bar{\mu} .<\frac{1}{2 n} . \quad$ Suppose the standard representation of
$\phi_{j}^{+}$ is $\sum_{k=1}^{m^{k}} a_{k}^{\pm} \chi_{A_{k}^{+}}$ where $A_{k}^{\pm} \in \bar{m} ;$ for each $k$, choose $E_{k}^{\pm} \in M$ 3 $E_{k}^{\pm} \subset A_{k}^{\pm}$ and $\bar{\mu}\left(A_{k}^{\pm}-E_{k}^{\pm}\right)=0$
(which we can do by Theorem( 1.9)). Then $0 \leq \sum a_{k}^{\pm} x_{E_{k}^{\pm}} \leq \phi_{j}^{\pm}+M . \quad$ Let $\varphi_{n}=\sum q_{k}^{+} x_{E_{k}^{+}}-N$
and $\psi_{n}=M-\sum a_{k}^{-} X_{E_{k}}^{-} .$ Then $\varphi_{n}$ and $\Psi_{n}$ are simple functions which are $m$ -measurable, $\varphi_{n} \leq f \leqslant \Psi_{n}$ and $\int\left(\psi_{n}-\varphi_{n}\right) d \mu=\int\left(\psi_{n}-f\right) d \bar{\mu}+\int\left(f-\varphi_{n}\right) d \bar{\mu}=\int\left(g_{-}-\rho_{j}^{-}\right) d \bar{\mu}+\int\left(g_{+}-\rho_{j}^{+}\right) d \bar{\mu}<\frac{1}{2 n}+\frac{1}{2 n}=\frac{1}{n}$
Conversely. Let $g=\sup \varphi_{n}$ and $h=\inf \psi_{n}$ gand $h$ are $m$ -measurable, $g \leq f \leq h$ and $(\forall n) \quad \int(h-g) d \mu \leq \int\left(\psi_{n}-\varphi_{n}\right) d \mu<\frac{1}{n} \quad$ So $\quad h-g \in L^{+}(\mu) \quad$ and $\quad \int(h-g) d \mu=0,$ so by
Prop. $2.16, \quad g=h \quad$ a.e. $(\mu) . \quad$ Let $E=\{x: g(x)<h(x)\} .$ Then $E \in m$ and $\mu E=0 .$ Let $F=\{x: g(x)<f(x)\}$
Then $F \subset E,$ so $F \in \bar{m}$ and $\bar{\mu} F=0 .$ since $g$ is also $\bar{m}$ -measurable and we now have $f=g$ a.e. $(\bar{\mu})$
$f$ is $\bar{m}$ -measurable by Prop. $2.11(a)$ Moreover, since $\int f d \bar{\mu} \leq \int \psi_{n} d \bar{\mu}=\int \psi_{n} d \mu<\int \varphi_{n} d \mu+\frac{1}{n}=\int \varphi_{n} d \bar{\mu}+\frac{1}{n} \leq \int f d \bar{\mu}+\frac{1}{n}$
$\lim \int \varphi_{n} d \mu=\lim \int \psi_{n} d \mu=\int f d \bar{\mu}$
