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Let $X$ be a separable metric space, $\mathcal{B}(X)$ its Borel algebra and let $P:X \times \mathcal{B}(X)\rightarrow \mathbb{R}$ be a function such that $P(x, \cdot)$ are probability measures for every $x\in X$ and $P(A,\cdot)$ are Borel functions for all $A\in\mathcal{B}(X)$. My question is

Let $B: X^2 \times\mathcal{B}(X^2)\rightarrow\mathbb{R}$ be a function given by $$B(x,y,\cdot)=P(x,\cdot)\otimes P(y,\cdot)\;\;\;(x,y\in X),$$ where $\otimes$ denotes the product of measures. Is the function $B(\cdot,A)$ measurable with respect to $\mathcal{B}(X^2)$, for $A\in\mathcal{B}(X^2)$?

It seems to me that it is true by the classical induction, but I am not sure...

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up vote 1 down vote accepted

For a product set $B_1\times B_2$, the map $(x,y)\mapsto B(x,y, B_1\times B_2)=P(x,B_1)P(y,B_2)$ is measurable. The collection of such product sets $B_1\times B_2$ for $B_1,B_2\in{\cal B}(X)$ is a $\pi$-system that generates ${\cal B}(X\times X)={\cal B}(X)\times{\cal B}(X)$. Finally, an application of Dynkin's $\pi-\lambda$ theorem shows that $(x,y)\mapsto B(x,y, A)$ is measurable for any $A\in {\cal B}(X\times X)$.

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Thanks! I can not believe it could be so easy:) – dawid Feb 2 '13 at 1:04

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