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First are relevant definitions from textbook $\textbf{Analysis III}$ by Amann.

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 $\mu$-simple if $f(X)$ is finite, $f^{-1}(e) \in \mathcal{A}$ for every $e \in E,$ and $\mu\left(f^{-1}(E \backslash\{0\})\right)<\infty .$ We denote by $\mathcal{S}(X, \mu, E)$ the set of all $\mu$-simple functions.

A function $f \in E^{X}$ is said to be $\mu$-measurable if there is a sequence $\left(f_{j}\right)$ in $\mathcal{S}(X, \mu, E)$ such that $f_{j} \rightarrow f$ $\mu$-almost everywhere as $j \rightarrow \infty$. We set $$\mathcal{L}_{0}(X, \mu, E):=\left\{f \in E^{X} \mid f \text { is } \mu \text {-measurable}\right\}$$

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 $\mu$-measurable if $\mathcal{A}$ contains $f^{-1}(-\infty), f^{-1}(\infty)$, and $f^{-1}(O)$ for every open subset $O$ of $\mathbb{R} .$ We denote the set of all $\mu$-measurable $\overline{\mathbb{R}}$-valued functions on $X$ by $\mathcal{L}_{0}(X, \mu, \overline{\mathbb{R}})$

The authors continue to present propositions:

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My questions:

From the authors' definitions, we have

Def: $f \in \mathcal{L}_{0} (X, \mu, E)$ if and only if there is a sequence $\left(f_{j}\right)$ in $\mathcal{S}(X, \mu, E)$ such that $f_{j} \rightarrow f$ $\mu$-almost everywhere as $j \rightarrow \infty$.

and

Thm: $f \in \mathcal{L}_{0} (X, \mu, \overline{\mathbb{R}}^{+})$ if and only of there is an increasing sequence $\left(f_{j}\right)$ in $\mathcal{S}\left(X, \mu, \mathbb{R}^{+}\right)$ such that $f_{j} \to f$ for $j \to\infty$.

In light of Def, I fell it's very intuitive for the following statement (St) to be correct.

St: $f \in \mathcal{L}_{0} (X, \mu, \overline{\mathbb{R}}^{+})$ if and only of there is a sequence $\left(f_{j}\right)$ in $\mathcal{S}\left(X, \mu, \mathbb{R}^{+}\right)$ such that $f_{j} \to f$ $\mu$-almost everywhere for $j \to\infty$.

Clearly, Thm $\implies$ St. As such, I would like to ask if St $\implies$ Thm. If it's not the case, could you please give an intuitive explanation why it's not correct that St $\implies$ Thm?

Thank you so much for your clarification!

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I've figured a proof and posted it as an answer here to peacefully close this question.


$\textbf{Theorem}$ If there is a sequence $\left(f_{n}\right)$ of $\mu$-simple functions $f_n: X \to \mathbb{R}^{+}$ that converges to $f: X \to \overline{\mathbb{R}}^{+}$ $\mu$-almost everywhere, then there is an increasing sequence $\left(g_{m}\right)$ of $\mu$-simple functions $g_m: X \to \mathbb{R}^{+}$ that converges to $f: X \to \overline{\mathbb{R}}^{+}$.

$\textbf{My attempt}$

First, we need the following lemma.

If there is a sequence $\left(f_{n}\right)$ of $\mu$-simple functions $f_n: X \to \mathbb{R}^{+}$ that converges to $f: X \to \overline{\mathbb{R}}^{+}$ $\mu$-almost everywhere, then $\mathcal{A}$ contains $f^{-1}(+\infty)$ and $f^{-1}(O)$ for every open subset $O$ of $\mathbb{R}$.

For all $(m, k) \in \mathbb{N} \times\left\{0, \ldots, 2^{m}-1\right\}$, we define $$\begin{cases} b_{m,k} &=k m 2^{-m} \\ B_{m,k} &=f^{-1} ([k m 2^{-m}, (k+1)m 2^{-m})) \\ B_m &= f^{-1} ([m, +\infty)) \end{cases}$$

It follows from our Lemma that $B_{m,k}$ and $B_{m}$ belong to $\mathcal A$. Then we define a sequence $(g_m)$ by $$g_{m}=\mathbb{1}_{B_{m}} + \sum_{k=0}^{2^{m}-1} b_{m,k} \mathbb{1}_{B_{m,k}}$$

It's not hard to verify that $(g_m)$ is an increasing sequence of $\mu$-simple functions and converges to $f$.

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