Am reading the proof of the Pettis measurability theorem on p.56 of Vector Measures by Diestel & Uhl.
At some point we have $\mu$-measurabe function $f:X\to\mathbb{R}$ defined on a measure space $(X,\Sigma,\mu)$ and taking real values. Here $\mu$-measurable means that there exist a sequence of simple functions converging almost everywhere to $f$. We are considering a set $E:=\{x\in X : f(x) \in B \}$ with $B$ Borel. Diestel & Uhl write
If $\mu$ is complete, then $E$ belongs to $\Sigma$. In any case, there is a set $E'\in \Sigma$ with $\mu(E\triangle E')=0$.
I understand the claim when $\mu$ is complete, but I can't show that such $E'$ exist. Any help is very appreciated.
I believe the correct statement is:
where $\mu^*$ is the outer measure induced by $\mu$.
Let $(h_n)$ be be a sequence of simple function converging to $f$ almost everywhere. So, there is $N\in \Sigma$ such that $\mu(N)=0$ and $h_n \to f$ pointwise in $X \setminus N$. We know that it $\mu$ is not complete, then $f$ may not be measurable.
Let $g$ be a function defined as $g(x)=f(x)$ for $x \in X \setminus N$ and $g(x)=0$ for $x \in N$. It is easy the see the $g$ is a measurable function and that $f=g$ a.e..
Now, le $B \subseteq \Bbb R$ be a Borel set. Let $E:=\{x\in X : f(x) \in B \}$ and $E':=\{x\in X : g(x) \in B \}$. Since $f$ may not be measurable, $E$ may not be measurable. However, $g$ is measurable, so, $E'$ is measurable, that is $E'\in \Sigma$.
Now, $E\triangle E' \subseteq \{x \in X: f(x)\neq g(x)\} \subseteq N$ and $\mu(N)=0$.
So we can conclude that $E\triangle E'$ is a subset of a null set. So $\mu^*(E\triangle E')=0$).
However, we can not conclude that $\mu(E\triangle E')=0$, because $E\triangle E'$ may not be a measurable. that is, it may be the case that $E\triangle E' \notin \Sigma$.