3
$\begingroup$

Be $(\Omega, \mathcal{A}, P)$ and $E$ a probability space and a Banach space respectively.

This paper of G.A. Edgar contains a proof that, for a function $X: \Omega \rightarrow E$, being weakly measurable is equivalent to being a $\big(\mathcal{A}, {\mathcal{Ba}}(E,\sigma(E, E^{\,\vee}))\big)$-measurable function. ${\mathcal{Ba}}(E,\sigma(E, E^{\,\vee}))$ is the Baire $\sigma$-algebra w.r.t. the weak topology on $E$.

If $X:\Omega\to E$ is a strongly/Bochner measurable function this is equivalent to being weakly measurable and separably valued or equivalent to being $(\mathcal{A}, \mathcal{B}(E))$-measurable and separably valued.

Question: If $X: \Omega \rightarrow E$ is a strongly/Bochner measurable function, does there exist a characterization of it solely as a random element i.e. an $(\mathcal{A}, \mathcal{E})$-measurable function for some $\sigma$-algebra $\mathcal{E}$ ? (Looks like the property 'separably valued' needs to be encoded into such a $\sigma$-algebra)

$\endgroup$
3
  • $\begingroup$ @user95282 Yep. Clarified the text. $\endgroup$
    – M1011
    Commented Feb 7, 2022 at 18:46
  • $\begingroup$ Not a formal proof, but a heuristics why it is very unlikely that such a characterization exists: Measurability is an assumption about preimages, while the separability assumption you require is on the image. $\endgroup$ Commented Feb 7, 2022 at 19:30
  • $\begingroup$ Further to Martin Väth's comment: If $\mathcal{A}$ contains all subsets of $\Omega$ then all functions to $E$ are $(\mathcal{A},\mathcal{E})$ measurable, no matter what $\mathcal{E}$ is. $\endgroup$
    – user95282
    Commented Feb 8, 2022 at 12:09

0

You must log in to answer this question.