I'm having difficulties verifying a remark in Raymond Ryan's treatment of the Bochner Integral.
$\bf{\text{Remark:}}$ If $\mu$ is $\sigma$-finite, and $(f_{n})_{n=1}^{\infty}$ is a sequence of $\mu$-measurable functions which converges to $f$ almost everywhere, then $f$ is $\mu$-measurable.
$\bf{\text{Edit:}}$ At this point I am currently looking for an authoritative answer on what the hypothesis of this remark should be, given the comments below.
$\bf{\text{Background:}}$
Let $(\Omega,\Sigma,\mu)$ be a $\sigma$-finite measure space, and $X$ be a Banach space.
A function $f:\Omega\to X$ is simple if it assumes only finitely many values. That is, there exists subsets $E_{1}, … , E_{n}$ of $\Omega$ and scalars $x_{1}, … , x_{n}\in X$ such that $f = \sum_{i=1}^{n}\chi_{E_{i}}x_{i}$.
If the sets $E_{i}$ can be chosen from $\Sigma$, then $f$ is $\mu$-measurable simple.
A function $f:\Omega\to X$ is $\mu$-measurable if it is the limit of a sequence of $\mu$-measurable simple functions (almost everywhere).
A function $f:\Omega\to X$ is $\mu$-essentially separately valued if there exists $E\in \Sigma$ such that $\mu(\Omega\backslash E) = 0$ and $f(E)\subset Y$ for some separable subspace $Y$ of $X$.
These are the immediately preceding results (which may or may not be useful).
$\bf{\text{Lemma:}}$ Let $\mu$ be $\sigma$-finite. $f:\Omega\to X$ is $\mu$-measurable if and only if $\chi_{E}f$ is $\mu$-measurable for every $E\in \Sigma$, $\mu(E) < \infty$.
Proof (My previous question): Fact about measurable functions defined on $\sigma$-finite measure spaces.
$\bf{\text{(Pettis Measurability Theorem)}}$ Let $\mu$ be a $\sigma$-finite measure. The following are equivalent for $f:\Omega\to X$.
(i) $f$ is $\mu$-measurable.
(ii) $f$ is weakly $\mu$-measurable and $\mu$-essentially separately valued.
(iii) $f$ is Borel measurable and $\mu$-essentially separately valued.
Sorry for not having much of a start yet. My ideas consisted of trying to adapt the solution given as an answer to my previous question: Fact about measurable functions defined on $\sigma$-finite measure spaces.
but sadly went nowhere. I'm just looking for a hint not a full solution if possible.
$\bf{\text{Regarding My Issues Below:}}$
In Norbert's proof:
Case (1): The functions $f_{n}$ are each $\mu$-measurable simple as they can be written as $f_{n} = \chi_{\phi}$ and $\phi\in\Sigma$. Since $f_{n}\to\chi_{F}$ at every point outside of $F$, then by definition, $\chi_{F}$ is $\mu$-measurable. So I don't think a contradiction exists here.
Case (2): Before completeness is invoked, we have that since $f_{n}$ is $\mu$-measurable, $x^{*}\circ f_{n}$ is a $\mu$-measurable scalar valued function on $E$ for every $x^{*}\in X^{*}$. Therefore it is concluded that $x^{*}\circ f$ is a $\mu$-measurable scalar valued function on $E$ for every $x^{*}\in X^{*}$. I cannot manage to prove the details on this last step; it seems to actually require the remark which is to be proved?
$\bf{\text{Follow Up:}}$ I think I have found the heart of the matter, which I have posted as a separate question:
Contradiction achieved with the Pettis Measurability Theorem?
Best Answer
Let $X$ be a Banach space and $(\Omega,\Sigma,\mu)$ be a measure space. By $L_0(\Omega,\mu, X)$ we denote the linear space of $X$-valued measurable functions.
1) This is wrong counterexample, because anotherr definiton of measurability should ne used. If $(\Omega,\Sigma,\mu)$ is not complete the result is not true. Indeed the measure is not complete we have $E\in\Sigma$ and $F\subset E$ such that $F\notin \Sigma$ and $\mu(E)=0$. Let $f_n(\omega)=0$ for all $\omega\in\Omega$ and $n\in\mathbb{N}$, then $f_n\to\chi_F$ a.e, $\{f_n:n\in\mathbb{N}\}\subset L_0(\Omega,\mu,\mathbb{C})$ but $\chi_F\notin L_0(\Omega,\mu,\mathbb{C})$.
2) If $(\Omega,\Sigma,\mu)$ is complete we use Pettis measurability theorem. By assumption we have $E\in\Sigma$ such that $f(\omega)=\lim\limits_{n\to\infty} f_n(\omega)$ for all $\omega\in E$ and $\mu(\Omega\setminus E)=0$. Take arbitrary $x^*\in X^*$, then for all $\omega\in E$ we have $x^*(f(\omega))=\lim\limits_{n\to\infty} x^*(f_n(\omega))$. Since $f_n\in L_0(\Omega,\mu,X)$, then by Pettis measrability theorem $x^*\circ f_n$ is scalar valued measurable function. Since $x^*\circ f$ is limit on $E$ of measurable scalar valued functions $x^*\circ f_n$, then $x^*\circ f$ is scalar valued measurable on $E$ function. Since $\mu(\Omega\setminus E)=0$ and measure is complete then $x^*\circ f$ is scalar valued measurable on $\Omega$. Since $x^*$ is arbitrary $f$ is weakly $\mu$-measurable.
Since $f_n\in L_0(\Omega,\mu,X)$, then by Pettis measurability theorem $f_n$ is separably valued, i.e. there exist countable $S_n\subset X$, $E_n\in \Sigma$ such that $f_n(E_n)\subset \operatorname{cl}(S_n)$ and $\mu(\Omega\setminus E_n)=0$. Define $E_0=(\bigcap E_n)\cap E\in \Sigma$, then $\mu(\Omega\setminus E_0)=0$. Consider countable set $S=\bigcup_{n\in\mathbb{N}} S_n$, then $f_n(E_0)\subset \operatorname{cl}(S_n)\subset \operatorname{cl}(S)$ and $\bigcup_{n\in\mathbb{N}} f_n(E_0)\subset \operatorname{cl}(S)$. Recall $f(\omega)=\lim\limits_{n\to\infty} f_n(\omega)$ for all $\omega\in E$. Since $E_0\subset E$ we conclude that each point in $f(E_0)$ is limit of some sequence in $\bigcup_{n\in\mathbb{N}} f_n(E_0)$. In other words $f(E_0)\subset \operatorname{cl}(\bigcup_{n\in\mathbb{N}} f_n(E_0))$. Now we have $f(E_0)\subset \operatorname{cl}(\bigcup_{n\in\mathbb{N}}f_n(E_0))\subset$$\subset \operatorname{cl}(\operatorname{cl}(S))=\operatorname{cl}(S)$. Since $S$ is countable, $f(E_0)$ is separable. Since $\mu(\Omega\setminus E_0)=0$, then $f$ is separably valued.
Since $f$ is weakly measurable and separably valued by Pettis measurability theorem it is $\mu$ measurable, i.e. $f\in L_0(\Omega,\mu,X)$.