Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $E := [0, \infty]$ endowed with order topology.
$f \in E^{X}$ is called $\mu$-simple if $f = \sum_{k=1}^n e_k 1_{A_k}$ where $e_k \in E \setminus \{\infty\}$ and $(A_k)_{k=1}^n$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A$. Let $\mathcal S (X, \mu, E)$ be the space of such $\mu$-simple functions.
$f \in E^{X}$ is called $\mathcal A$-measurable if $f^{-1} (O) \in \mathcal A$ for every open subset $O$ of $E$.
$f \in E^{X}$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n) \subset \mathcal S (X, \mu, E)$.
I'm trying to prove below characterization for non-complete measure space.
Theorem: $f \in E^X$ is $\mu$-measurable if and only if there is an increasing sequence $(f_n) \subset \mathcal S (X, \mu, E)$ such that $f_n \to f$ $\mu$-a.e.
Could you please have a check on my proof?
My attempt: We need the following result.
Lemma 1: Let $(X, \mathcal A, \mu)$ be complete. If $f$ is $\mu$-measurable, then $f$ is $\mathcal A$-measurable.
Remark: The original proof of Lemma 1 works when $E$ is a Banach space. However, if we look closer at the proof (for direction $\implies$), we see that it only uses the metric structure of $E$. On the other hand, the order topology of $[0, \infty]$ is metrizable and separable. As such, the original proof is applicable here.
The reverse direction $\impliedby$ is trivial. Let's prove the other one $\implies$. By $\sigma$-finiteness of $\mu$, it suffices to consider the case $\mu$ is finite. Fix a $\mu$-measurable function $f \in E^X$.
- $(X, \mathcal A, \mu)$ is complete.
For $n, k \in \mathbb{N}$, set $$ A_{nk}:=\begin{cases} [k 2^{-n} \leq f<(k+1) 2^{-n}] & \text {if} \quad k=0, \ldots, n 2^{n}-1, \\ [f \geq n] & \text {if} \quad k=n 2^{n}. \end{cases} $$ The sets $A_{nk}$ are obviously disjoint for $k=0, \ldots, n 2^{n}$ and by Lemma 1 they lie in $\mathcal{A}$. Since $\mu(X)<\infty$, each $A_{nk}$ has finite measure. Then, $$ f_{n}:=\sum_{k=0}^{n 2^{n}} k 2^{-n} 1_{A_{nk}} \in \mathcal S (X, \mu, E) \quad \forall n \in \mathbb{N}. $$
Further one verifies that $0 \leq f_{n} \leq f_{n+1}$ for $n \in \mathbb{N}$. Now suppose $x \in X$. If $f(x)=\infty$, we have $f_{n}(x)=n$, so $\lim_n f_{n}(x)=f(x)$. On the other hand, if $f(x)<\infty$, then $f_{n}(x) \leq f(x)<f_{n}(x)+2^{-n}$ for $n>f(x)$, so $\lim_n f_{n}(x)=f(x)$ in this case as well. This shows $(f_n)$ converges pointwise to $f$ everywhere.
- $(X, \mathcal A, \mu)$ is non-complete.
Let $(X, \mathcal A', \mu')$ be the completion of $(X, \mathcal A, \mu)$. Clearly, $f$ is $\mu'$-measurable. By 1., there is an increasing sequence $(g_n) \subset \mathcal S (X, \mu', E)$ such that $g_n \to f$ everywhere. Assume $g_n = \sum_{i=1}^{\varphi (n)} e_{ni} 1_{A_{ni}}$ with $(A_{ni})_{i=1}^{\varphi (n)}$ a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A'$. By construction of $\mathcal A'$, there exist $B_{ni} \in \mathcal A$ and $C_{ni} \in \mathcal Z$ such that $A_{ni} = B_{ni} \cup C_{ni}$ and $\mu(B_{ni}) = \mu'(A_{ni}) < \infty$. Let $f_n := \sum_{i=1}^{\varphi (n)} e_{ni} 1_{B_{ni}}$. Then $(f_n) \subset \mathcal S (X, \mu, E)$.
Let's prove that $f_n \to f$ $\mu$-a.e. Notice that $\{x \in X \mid f_n (x) \neq g_n (x)\} \subset C_n := \bigcup_{i=1}^{\varphi (n)} C_{ni}$. Let $N := \bigcup_n C_n$. Then $N$ is a $\mu'$-null set. It follows that $f_n \to f$ on $A := N^c$. There exist $B \in \mathcal A$ and $C \in \mathcal Z$ such that $A = B \cup C$ and $\mu(B) = \mu'(A) = 1$. It follows that $f_n \to f$ on $B$. This completes the proof.
