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Let $(E,\|.\|)$ be a (not necessarily separable) Banach space and $\Sigma_E$ the Borel $\sigma$-algebra (w.r.t. the norm topology). Let $(\Omega,\Sigma_\Omega)$ be a measurable space (which we can assume to be standard Borel if required later on).

A $\Sigma_\Omega$-$\Sigma_E$-measurable map $X: \Omega \to E$ is said to be "strongly measurable" iff there exists a closed separable sub-Banach space $F \subseteq E$ such that for the image we have the inclusion: $X(\Omega) \subseteq F$.

It is now clear that for such a strongly measurable map $X$ there exists a countable subset $S \subseteq \Sigma_E$ that separates the points of the image $X(\Omega)$. The latter means that for every two distinct points $y_1,y_2 \in X(\Omega)$, $y_1 \neq y_2$, there exists an $A \in S$ with either $y_1 \in A$ and $y_2 \notin A$, or, $y_1 \notin A$ and $y_2 \in A$. For brevity, lets call this condition "countably separated image".

My question is now about the reverse: Is it true that a $\Sigma_\Omega$-$\Sigma_E$-measurable map $X: \Omega \to E$ with countably separated image is always strongly measurable?

Added comment: Since it was pointed out in the comments that there are already counter examples for general measurable spaces $(\Omega,\Sigma_\Omega)$, I now want to insist that $(\Omega,\Sigma_\Omega)$ is a standard Borel space.

Comment about the answer: The provided reference in the answer shows that the extra condition for $X$ to have countably separated image is not needed in the case, where $(\Omega,\Sigma_\Omega)$ is a standard Borel space. In that case we directly get the equivalence:

"$X$ is strongly measurable iff $X$ is measurable."

Edit 1: Clarification what was meant with "countably separated image".

Edit 2: Comment about the assumption of $(\Omega,\Sigma_\Omega)$ being standard Borel.

Edit 3: Comment about the answer.

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    $\begingroup$ Not sure about standard Borel $(\Omega,\Sigma_\Omega)$, but without that condition there are easy counterexamples: the identity map on $\ell^\infty$ is measurable but not strongly measurable, but $\ell^\infty$ is countably separated in your sense because its predual $\ell^1$ is separable. Another counterexample is $(\Omega,\Sigma_\Omega)=([0,1],\mathcal{P}([0,1]))$ and the map $X:[0,1]\to\ell^\infty([0,1])$ (the sequence space $\ell^\infty([0,1])$, not the function space) defined by $X(t)=e_t$. Again, it is measurable but not strongly measurable. The image is countably separated as $[0,1]$ is. $\endgroup$
    – David Gao
    Commented Aug 30 at 13:11
  • $\begingroup$ @DavidGao: Do I understand correctly that you claim that $\ell^\infty$ is a non-separable Banach space, but whose Borel $\sigma$-algebra is countably separated (in the sense as defined above)? $\endgroup$
    – Packo
    Commented Aug 30 at 17:15
  • $\begingroup$ Yes, that is the claim. Just choose a countable dense subset $\{\varphi_n\}_n$ of $\ell^1$ and a countable basis $\{O_m\}_m$ of topology on $\mathbb{C}$. Then the Borel sets $A_{m, n}$, indexed by $m, n$, defined by $A_{m, n} = \{x \in \ell^\infty: \varphi_n(x) \in O_m\}$, separate points. $\endgroup$
    – David Gao
    Commented Aug 30 at 18:50
  • $\begingroup$ (In case of $\ell^\infty$, you can also just define $A_{m, n}$ as the set of elements of $\ell^\infty$ whose $n$-th coordinates lie within $O_m$. Though the argument using a dense subset of $\ell^1$ has the advantage of being applicable to any Banach space with a separable predual.) $\endgroup$
    – David Gao
    Commented Aug 30 at 18:59
  • $\begingroup$ Apart from separable Banach spaces and Banach spaces with separable preduals, there are also examples of countably separated Banach spaces like $B(E, F)$ with separable Banach space $E$ and countably separated Banach space $F$; $L^p([0, 1]; E)$ with $1 \leq p \leq \infty$ and countably separated Banach space $E$; $C_b(K; E)$ with separable topological space $K$ and countably separated Banach space $E$; and any Banach subspace of the above. It’s really not that hard a condition to satisfy. $\endgroup$
    – David Gao
    Commented Aug 30 at 19:17

1 Answer 1

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If $(\Omega,\Sigma_\Omega)$ is standard Borel, $E$ metrizable, and $X:\Omega_\Sigma\to E$ measurable, then $X(\Omega)$ is separable. This is Proposition 1.11 in "Probability Distributions on Banach Spaces" (1987) by Vakhania, Tarieladze, and Chobanyan. So, the answer is yes.

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  • $\begingroup$ I’ve wondered for a long time why I’ve never seen an example of a measurable but not strongly measurable function defined on a standard Borel space. This is very nice to learn about. +1 $\endgroup$
    – David Gao
    Commented Aug 31 at 17:25
  • $\begingroup$ @MichaelGreinecker: Thank you very much for the reference to the more general statement. I presume that just saying that the set $X(\Omega)$ is "separable" means that $X(\Omega)$ is a separable topological space in the subspace topology of $E$. For the final step of my question for the case that $E$ is a Banach space, one then needs to take a countable dense subset $C \subseteq X(\Omega)$, which existence comes from the reference, and put $F:=\overline{\mathrm{span}}(C)$, the closure of the span of $C$ in $E$, which then itself is separable sub-Banach space of $E$. $\endgroup$
    – Packo
    Commented Sep 1 at 5:28
  • $\begingroup$ @Packo Yes, exactly. $\endgroup$
    – Michael Greinecker
    Commented Sep 1 at 8:06
  • $\begingroup$ As I do not have the Vakhania et al. book at hand, I ask whether you have personally checked the proof of Proposition 1.11, or is there possibly some missing assumption like CH or the weaker Luzin's hypothesis? Namely, in view of the discussion in the second paragraph in this MO-question of mine, I suspect that it might be so. $\endgroup$
    – TaQ
    Commented Sep 1 at 22:24
  • $\begingroup$ @TaQ I did check the proof. The proof has two steps: First, a discrete uncountable subset is constructed. Every subset of that set is closed, so there must be as many Borel sets as the power set of that set. Under CH, we are done; there are more than continuum many Borel sets in the standard Borel space. The second step is to construct a Borel function of the standard Borel space onto itself whose range has the same cardinality as the discrete set. Since uncountable analytic sets have the cardinality of the continuum, the proof is done. $\endgroup$
    – Michael Greinecker
    Commented Sep 2 at 5:53

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