TITLE: Universal measurability of a kernel
QUESTION [5 upvotes]: 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 $B$?

REPLY [5 votes]: 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 $Kf\in {\cal M}^u$.