Let $(X, \mathcal A, \mu)$ be a measure space and $(E, | \cdot |)$ a Banach space. Let $f \in E^X$.
$f$ is called $\mu$-simple if $f = \sum_{k=1}^n e_k 1_{A_k}$ where $0 \neq e_k \in E$ 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.
$f$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n)$ in $\mathcal S (X, \mu, E)$.
Several months ago, I posted a proof of below theorem
Theorem 1: Let $(X, \mathcal A, \mu)$ be complete and $\sigma$-finite, $(f_n)$ a sequence of $\mu$-measurable functions, and $f \in E^X$ the $\mu$-a.e. limit of $(f_n)$. Then $f$ is also $\mu$-measurable.
The proof of Theorem 1 relies on both assumptions completeness and $\sigma$-finiteness. Now I want to prove Theorem 2 which generalizes Theorem 1 by removing the requirements completeness.
Theorem 2: Let $(X, \mathcal A, \mu)$ be $\sigma$-finite, $(f_n)$ a sequence of $\mu$-measurable functions, and $f \in E^X$ the $\mu$-a.e. limit of $(f_n)$. Then $f$ is also $\mu$-measurable.
Could you have a check on my attempt?
My attempt: The proof relies on Lemma 2.
Lemma 2: Let $(X, \mathcal A', \mu')$ be the completion of $(X, \mathcal A, \mu)$. A function $f \in E^X$ is $\mu$-measurable if and only if $f$ is $\mu'$-measurable.
Let $(f_n)$ be a sequence of $\mu$-measurable functions and $f \in E^X$ the $\mu$-a.e. limit of $(f_n)$. Let $(X, \mathcal A', \mu')$ be the completion of $(X, \mathcal A, \mu)$. Then $(X, \mathcal A', \mu')$ is complete and $\sigma$-finite. Because $\mathcal A \subset \mathcal A'$ and $\mu' \restriction \mathcal A = \mu$, we get
- $(f_n)$ is a sequence of $\mu'$-measurable functions,
- and $f \in E^X$ the $\mu'$-a.e. limit of $(f_n)$.
By Theorem 1, $f$ is $\mu'$-measurable. The claim then follows from Lemma 2. This completes the proof.
I will use a social argument to disprove your Lemma 2.
First, notice that being $\mu'$-measurable is just the same as being measurable in the completed $\sigma$=algebra, as you have stated in another post.
If Lemma 2 was true, that would kind of make the whole $\mu$-measurable concept a little useless as it would simply be the same as measurable over the completed $\sigma$-algebra. If Lemma 2 is true, I would reinforce my comment that this is a very horrible definition of $\mu$-measurable. :-)