Universal measurability of a kernel

Question

Let $X$ and $A$ be Borel topological spaces, that is it is they are homeomorphic to Borel subsets of a complete separable metric space. Let further $\pi$ be a universally measurable stochastic kernel on $X$ given $A$, and let $T$ be a Borel measurable stochastic kernel on $X$ given $X\times A$. Define $$ P(x,B) = \int_AT(x,a,B)\pi(x,\mathrm da) $$ for any universally measurable $B\subseteq X$. Is that true that $P(\cdot,B)$ is a universally measurable function for any universally measurable set $B$?

Answer 1

The following argument is taken directly from the appendix (page 376) to Michael Sharpe's General Theory of Markov Processes.

A bounded kernel $K$ from $(M,{\cal M})$ to $(E,{\cal E})$ extends automatically and uniquely to a kernel from $(M,{\cal M}^u)$ to $(E,{\cal E}^u)$. For $x\in M$, just extend $K(x,\cdot)$ in the only possible way to be a measure on $(E,{\cal E}^u)$. For $\mu$ a measure on $(M,{\cal M})$ and $f\in b{\cal E}^u$, choose $f_1,f_2\in b{\cal E}$ with $f_1\leq f\leq f_2$ and $\mu K(f_2-f_1)=0$. Then $\mu (Kf_2-Kf_1)=0$, hence $Kf\in {\cal M}^u$.