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$\bf{\text{(Pettis Measurability Theorem)}}$ 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?

roo
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  • To anyone who was trying to read this question before now, you probably hate me for all the edits. A thousand apologies for not thoroughly correcting it before posting! – roo Sep 12 '13 at 16:40
  • The problem is that definition of Pettis measurability for the case $X=\mathbb{C}$ is NOT the same as standard definition of measurability. Hence paragraph 1) of my answer to your previous question is incorrect. I suggest you to delete this question and continue discussion within your previous question. Together we will find the truth :-) – Norbert Sep 12 '13 at 17:31
  • Thank you for your alliance in this quest. But I don't want to delete this question as it contains the very contradiction that I am trying to resolve, which is that the observation you make in your comment contradicts the Pettis measurability theorem. – roo Sep 12 '13 at 18:09
  • I already confirmed that solution in 1) is wrong. So do you have other contradictions? – Norbert Sep 12 '13 at 18:35
  • The contradiction is written in this question. But I have resolved it I think. I will post back shortly explaining more thoroughly. Te upshot is that I think completeness I required for the theorem that was being used. With this added condition everything suddenly works out fine. – roo Sep 12 '13 at 18:42

1 Answers1

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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.

roo
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