Let $(X, \mathcal{A}, \mu)$ be a complete, $\sigma$-finite measure space and $(E,|\cdot|)$ a Banach space.
We say $f \in E^{X}$ is $\boldsymbol{\mu}\textbf{-simple}$ if $f(X)$ is finite, $f^{-1}(e) \in \mathcal{A}$ for every $e \in E,$ and $\mu\left(f^{-1}(E \backslash\{0\})\right)<\infty$.
Suppose $f_n, f \in E^{X}$ for $n \in \mathbb{N} .$ Then $(f_n)_{n \in \mathbb N}$ converges to $f$ $\boldsymbol{\mu}\textbf{-almost everywhere}$ if and only if there is a $\mu$-null set $N$ such that $f_{n}(x) \rightarrow f(x)$ for all $x \in N^{c}$.
In the theory of integration, it is useful to consider not only real-valued functions but also maps into the extended number line $\overline{\mathbb{R}}$. Such maps are called $\overline{\mathbb{R}}$-valued functions.
An $\overline{\mathbb{R}}$-valued function $f: X \rightarrow \overline{\mathbb{R}}$ is said to be $\boldsymbol{\mu}\textbf{-measurable}$ if $\mathcal{A}$ contains $f^{-1}(-\infty), f^{-1}(\infty)$, and $f^{-1}(O)$ for every open subset $O$ of $\mathbb{R}$.
After dicussing with @Thorgott, I came up with the following theorem. I've tried but to no avail. Could you please leave me some hints to finish it?
Theorem $f: X \to \overline{\mathbb{R}}$ is $\mu$-measurable if and only if there is a sequence of $\mu$-simple functions $f_n: X \to \mathbb R$ such that $f_n \to f$ $\mu$-almost everywhere.
From @copper.hat's hints and my textbook, I've figured out a proof. I would be great if someone helps verify it. Thank you so much!
$\textbf{My attempt}$
$\Longrightarrow$
(i) We consider first the case $\mu(X)<\infty$. Let $(a_k)_{k \in \mathbb N}$ be an enumeration of $\mathbb Q$ and $A_{k,n} = f^{-1} [ \mathbb B (a_k, 1/(n+1))]$. Let $A_{+} = f^{-1}(+\infty)$ and $A_{-} = f^{-1}(-\infty)$. Then $\{A_{k,n},A_{+},A_{-}\} \subseteq \mathcal A$ for all $(k,n) \in \mathbb N^2$. The continuity of $\mu$ from above and the assumption $\mu(X)<\infty$ implies there are $m_n$ and $B_n \in \mathcal A$ such that $$B^c_n =A_+ \cup A_- \cup \bigcup_{k=0}^{m_n} A_{k,n} \quad \text{and} \quad \mu(B_n) < \frac{1}{2^{n+1}}$$
Now define $\varphi_{n} \in {\mathbb R}^{X}$ by $$\varphi_{n}(x) = \begin{cases} {a_{0}} & {\text {if} \quad x \in A_{0,n}} \\ {a_{k}} & {\text {if} \quad x \in A_{k, n} \setminus \bigcup_{p=0}^{k-1} A_{p, n} \quad \text {for} \quad 1 \le p \leq m_{n}} \\ n & {\text {if}} \quad x \in A_{+} \\ -n & {\text {if}} \quad x \in A_{-} \\ {0} & {\text {otherwise}} \end{cases}$$
Clearly, $\varphi_{n}$ is $\mu$-simple and $\|\varphi_{n}(x) - f(x)\| < 1/(n+1)$ for all $x \in B_n^c$. Define a decreasing sequence $(C_n)_{n \in \mathbb N}$ by $C_n = \bigcup_{p=0}^{\infty} B_{n+p}$. Then $C_n^c \subseteq B_n^c$ and $\mu(C_n) \le \sum_{p=0}^\infty \mu(B_{n+p}) < 1/2^n$. It therefore follows from the continuity of $\mu$ from above that $C = \bigcap_{n=0}^\infty C_{n}$ is $\mu$-null. We now set $$\psi_{n}(x) = \begin{cases} {\varphi_{n}(x)} & {\text {if} \quad x \in C_{n}^{c}} \\ n & {\text {if}} \quad x \in A_{+} \\ -n & {\text {if}} \quad x \in A_{-} \\ {0} & {\text {otherwise}}\end{cases}$$
Clearly, $\psi_{n}$ is $\mu$-simple. For $x \in C^c$, there exists $n \in \mathbb N$ such that $x \in C_n^c$. Then $x \in C_{n+p}^c$ for all $p \in \mathbb N$. So $\|\psi_{n+p}(x) - f(x)\| =\|\varphi_{n+p}(x) - f(x)\| < 1/(n+p+1)$ for all $p \in \mathbb N$. Hence $\psi_{n} (x) \to f(x)$ for all $x \in A_+ \cup A_- \cup C^c$.
(ii) We next consider the case $\mu(X)=\infty$. Because $\mu$ is $\sigma$-finite, there is a sequence $(A_k)_{k \in \mathbb N}$ of pairwise disjoint subsets in $\mathcal{A}$ such that $\bigcup_{k=0}^\infty A_{k}=X$ and $\mu (A_{k}) < \infty$. As in (i), for each $A_k$, there is a sequence $(\psi^k_{n})_{n \in \mathbb N}$ of $\mu$-simple functions and a $\mu$-null set $C_k$ such that $\psi^k_{n} (x) \to f(x)$ for all $x \in A_k \setminus C_k$. Moreover, $C=\bigcup_{k=0}^\infty C_{k}$ is $\mu$-null. We define a sequence $(\psi_{n})_{n \in \mathbb N}$ by $$\psi_{n}(x) = \begin{cases} {\psi^k_{n}(x)} & {\text {if} \quad x \in \bigcup_{k=0}^n A_{k}} \\ {0} & {\text {otherwise}}\end{cases}$$
Clearly, $(\psi_{n})_{n \in \mathbb N}$ is a sequence of $\mu$-simple functions such that $\psi_{n} (x) \to f(x)$ for all $x \in \bigcap_{k=0}^\infty C_k^c$.
$\Longleftarrow$
Assume there exist a sequence $(\psi_{n})_{n \in \mathbb N}$ of $\mu$-simple functions and a $\mu$-null set $N$ such that $\psi_{n} (x) \to f(x)$ for all $x \in N^c$.
Let $O$ be open in $\mathbb R$. We define a sequence $(O_k)_{k \in \mathbb N^*}$ by $O_{k} = \{y \in O \mid d(y, O^{c})>1 / k \}$. Then $O_{k}$ is open and $\overline{O}_{k} \subseteq O$. Let $x \in N^{c}$. We have $x \in O \iff \exists k \in \mathbb{N}^{*}: x \in O_k$. Therefore, $f(x) \in O$ if and only if there exists $(k,m_k) \in \mathbb{N}^{*} \times \mathbb{N}$ such that $\forall n \ge m_k: \varphi_{n}(x) \in O_{k}$. Consequently, $x \in f^{-1}(O)$ if and only if there exists $(k,m_k) \in \mathbb{N}^{*} \times \mathbb{N}$ such that $\forall n \ge m_k: x \in \varphi^{-1}_{n}(O_{k})$. As a result, $$f^{-1}(O) \cap N^{c} = \left ( \bigcup_{(k,m_k) \in \mathbb{N}^{*} \times \mathbb{N}} \bigcap_{n \ge m_k} \varphi_{n}^{-1} (O_{k}) \right ) \cap N^{c} = \bigcup_{(k,m_k) \in \mathbb{N}^{*} \times \mathbb{N}} \bigcap_{n \ge m_k} \left ( \varphi_{n}^{-1} (O_{k}) \cap N^{c}\right )$$
Because $\varphi_{n}$ is $\mu$-simple, $\varphi_{n}^{-1}(O_{k}) \in \mathcal{A}$ for all $(n,k) \in \mathbb{N} \times \mathbb{N}^{*}$. Hence $f^{-1}(O) \cap N^{c} \in \mathcal{A}$. Furthermore, the completeness of $\mu$ implies $f^{-1}(O) \cap N$ is a $\mu$-null set. Altogether, we obtain $$f^{-1}(O)=\left(f^{-1}(O) \cap N\right) \cup\left(f^{-1}(O) \cap N^{c}\right) \in \mathcal{A}$$
Let $x \in N^c$. We have $f(x) = +\infty \iff \forall M \in \mathbb N, \exists N \in \mathbb N,\forall n \ge N: \varphi_n(x) \ge M$. Consequently, $x \in f^{-1}(+\infty) \iff \forall M \in \mathbb N, \exists N \in \mathbb N,\forall n \ge N: x \in \varphi^{-1}_n ([M, \infty))$. As a result, $$\begin{aligned} f^{-1}(+\infty) \cap N^c &= \left( \bigcap_{M=0}^\infty \bigcup_{N=0}^\infty \bigcap_{n=N}^\infty \varphi^{-1}_n ([M, \infty)) \right) \cap N^c \\ &= \bigcap_{M=0}^\infty \bigcup_{N=0}^\infty \bigcap_{n=N}^\infty \left( \varphi^{-1}_n ([M, \infty)) \cap N^c \right) \end{aligned}$$
Because $\varphi_{n}$ is $\mu$-simple, $\varphi_{n}^{-1}([M, \infty)) \in \mathcal{A}$ for all $(n,M) \in \mathbb{N} \times \mathbb{N}$. Hence $f^{-1}(+\infty) \cap N^{c} \in \mathcal{A}$. Furthermore, the completeness of $\mu$ implies $f^{-1}(+\infty) \cap N$ is a $\mu$-null set. Altogether, we obtain $$f^{-1}(+\infty)=\left(f^{-1}(+\infty) \cap N\right) \cup\left(f^{-1}(+\infty) \cap N^{c}\right) \in \mathcal{A}$$
With similar reasoning, we have $f^{-1}(-\infty) \in \mathcal{A}$.