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The following is an exercise from "A Course on Borel Sets" by S.M. Srivastava.

Exercise 4.7.10 Let $X$ be a Polish space and $B \subseteq X \times \mathbb R^n$ a Borel set with convex sections (i.e. $B_x=\{y\in\mathbb R ^n: (x,y)\in B\}$ are convex). Show that $B_X$ is Borel.

I am in particular interested in whether such a proof would generalize to having $\mathcal P^1(\mathbb R ^n)$ instead of $\mathbb R^n$, where $\mathcal P^1(\mathbb R ^n)$ denotes the space of finite first order probability measures on $\mathbb R ^n$.

Any help would be very much appreciated!

Many thanks in advance!

J.R.
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    I think the answr is in the affirmative and that it follows from the projection theorem ( see theorem here). The space you are interested is Polish with the Levy metric or the Prohorov metric) – Mittens Jun 12 '24 at 16:41
  • Thanks for the comment! The linked theorem only gives me universal measurability, not Borel measurability :. The above result holds for sections that are either -singletons, -compact, -$\sigma$-compact -open -nonmeager etc. Now I doubt that it holds for arbitrary Polish spaces (otherwise it would probably be a theorem in Srivastava), so I am hoping that the space of probability measures is similar enough to $\mathbb R ^n$ to make the proof valid there as well. – J.R. Jun 12 '24 at 17:55
  • The only similarity is that under the Prohorov-Levy metric for example, $\mathcal{P}_1(\mathbb{R}^n)$ is a complete metric space. It is not separable however. Universal measurability is perhaps the best thing you may get. – Mittens Jun 12 '24 at 18:22
  • What is $B_X$? In the second question, what exactly is being replaced? What is a “finite first order” probability measure and how can such be viewed as Borel? – Michael Jun 13 '24 at 05:17
  • @Michael $B_X$ is the projection of $B$ onto $X$. $\mathbb R ^n$ is being replaced by $\mathcal P ^1(\mathbb R^n)$. Finite first order means $\int |x| d\mu < \infty$. And being Borel can be defined in any topological space. – J.R. Jun 13 '24 at 05:56

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