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Let $X$ and $Y$ be uncountable Polish spaces, $\Delta(Y)$ be the space of Borel probability measures on $Y$ endowed with the Borel $\sigma$-algebra induced by the variation distance, and let $g:X\to \Delta(Y)$ be Borel measurable. Is $g(X)$ separable? I'm also interested in the weaker question of whether we can find for every Borel probability measure on $X$ a null set $N$ such that $g(X\setminus N)$ is separable.

Separable subspaces of $\Delta(Y)$ have a nice structure; they correspond exactly to subspaces of $L_1(\mu)$ for some Borel probability measure $\mu$ on $Y$. But the topology on all of $\Delta(Y)$ is extremely fine. For example, every set of Dirac measures is closed. My vague intuition is that it is so fine that functions with a nonseparable range cannot be Borel measurable.

Michael Greinecker
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    Your problem seems in a way a bit similar to the one in my first MO-question. I have not thought anything of the possible details but maybe a solution could be obtained along the same lines. – TaQ Jul 08 '23 at 20:51
  • Thank you, that is very useful! Actually, there should be a simpler solutions since the domain here is Polish. I'll try to figure out the details. – Michael Greinecker Jul 08 '23 at 21:10
  • @TaQ The answer there shows that for some null set $N$, the image $g(X\setminus N)$ is separable. The other answer would follow from the problem in Dudley's textbook, but I'm not sure how that problem works. There might be an implicit assumption that if $T$ is uncountable, then $2^T>2^\mathbb{N}=\mathfrak{c}$. This need not be the case. – Michael Greinecker Jul 08 '23 at 22:44
  • Dudley has Problem 9 on page 387 where it is asked to show that under the continuum hypothesis the range of a Borel measurable map from a separable metric space to a metric space is separable. So, indeed, it seems that without that kind of additional assumption (or something alike) problem 10 on page 99 cannot be established. – TaQ Jul 10 '23 at 00:53
  • @TaQ Thank you. Curiously, the second edition of the book has the same problem but is missing the assumption of CH. – Michael Greinecker Jul 10 '23 at 05:39
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    This MO-question and this MO-answer are related to the gap in the above mentioned problem of Dudley. – TaQ Jul 13 '23 at 09:00

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