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Let $X$ and $Y$ be standard Borel spaces and let $\mathcal P(X)$ denote the space of probability measures over $X$ endowed with the topology of weak convergence. Consider a map $f:X\times \mathcal P(Y)\to \mathcal P(X\times Y)$ given by $$ f(x,p)(A):=(\delta_x\otimes p)(A) = p(A_x) $$ where $\delta_x$ is the Dirac measure concentrated at $x\in X$ and $A_x = \{y:(x,y)\in A\}$ is the $x$-section of the set $A$. I wonder whether $f$ is Borel-measurable.

My attempt to the proof is as follows: I need to show that $f_A:X\times \mathcal P(Y) \to\Bbb R$ is Borel-measurable for any measurable rectangle $A = B\times C$, where $f_A(x,p):= f(x,p)(A)$. I have $$ f_A(x,p) = 1_B(x)p(C) $$ which is a Borel map of $(x,p)$. Hence, so is $f$. Please tell me, whether the proof is correct, and whether there is a simpler proof of this argument.

SBF
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    I think you need to use a monotone class argument to show it is enough to prove this for rectangles. This is mostly a special case of the general result that products of kernels are kernels. – Michael Greinecker Mar 24 '14 at 19:11
  • @Michael: how would you suggest reducing my problems to the product of kernels - that's what I thought of, but didn't manage to do. Do you mean that the first kernel is $x\mapsto \delta x$ and the second one is $p \mapsto p$? – SBF Mar 24 '14 at 20:31
  • Exactly. And there is of course a natural correspondence between kernels and measure valued maps. – Michael Greinecker Mar 24 '14 at 20:42

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