Contradiction achieved with the Pettis Measurability Theorem?

$\bf{\text{(Pettis Measurability Theorem)}}$ Let $(\Omega,\Sigma,\mu)$ 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.


$\bf{\text{Relevant Definitions:}}$

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 $X$.


$\bf{\text{My Question:}}$

Since the scalar field is one-dimensional and thus separable, every scalar function on $\Omega$ is $\mu$-essentially separately valued.

Therefore by the Pettis Measurability Theorem, Borel measurability and $\mu$-measurability are equivalent for scalar valued functions. But the example $\chi_{F}$, for any $F\notin \Sigma$, $\mu(F)= 0$ seems to contradict this fact for incomplete spaces, as the function $\chi_{F}$ is not Borel measurable when $F\notin \Sigma$, but is $\mu$-measurable as the almost everywhere limit of the sequence $f_{n} = 0$.

I am led to suspect, unless I have an error in my reasoning above, that completeness of $\mu$ may be needed for the Pettis Measurability Theorem somewhere?

Solutions Collecting From Web of "Contradiction achieved with the Pettis Measurability Theorem?"

Completeness of $\mu$ is required for this version of the Pettis measurability theorem.

In the proof of the direction $(ii)\Rightarrow (iii)$, as in the definition of $\mu$-essentially separately valued, there is a set $E\in\Sigma$, $\mu(\Omega \backslash \Sigma) = 0$ such that $f(E)\subset Y$ for a separable subspace $Y$ of $X$. There is a reduction to the case where $f(\Omega)\subset Y$, as changing the values of $f$ on a $\mu$-null set does not harm weakly $\mu$-measurability or Borel measurability.

This last claim is only true for Borel measurability if $\mu$ is complete. And thus the reduction is not valid without that assumption, as the counter example in my question (supplied by Norbert) shows.