TITLE: Radon-Nikodym derivative as a measurable function in a product space
QUESTION [6 upvotes]: Let $X$ be a Polish space with the probability measure $P$ and the Borel sigma-algebra.
Suppose that $X$ is also a group such that $(x,y)\mapsto xy^{-1}$ is Borel measurable and the probability $P$ is left and right quasiinvariant.
Let $P_x$ denote the probability measure $P_x(A)=P(xA)$ for every $x$ in $X$. Obviously, for each $x$ in $X$, the Radon-Nikodym derivative $dP_x/dP$ is Borel measurable.
I am trying to show that there is a measurable function $\Phi:X \times X\rightarrow[0,\infty )$ such that for every $x$ in $X$,  $\Phi(x,y)=(dP_x/dP)(y)$  for a.e. $y$ (notice that $\Phi$ needs to be measurable with respect to the Borel sigma-algebra of the product space).
I can show that there is a measurable function $\Phi$ such that for $P$ - almost every $x$ in $X$, $\Phi(x,y)=(dP_x/dP)(y)$  for  a.e. $y$,  by taking the derivative $dm/dP\times P$, where $m=(P\times P)\circ S$ and $S:X \times X\rightarrow X\times X$ is the function $S(x,y)=(x,x^{-1}y)$. But I need the equality for every $x$ in $X$.
Any suggestions?

REPLY [4 votes]: Just to add to Theo's excellent answer, I will quote in full
Exercise 6.10.72 from  page 63 of V. Bogachev's Measure Theory (volume 2).
A long hint is also given there.

Let $(X,{\cal A},\mu)$ be a probability space, ${\cal A}$ a countably generated
  $\sigma$-algebra, $(T,{\cal B})$ a measurable space, and let $\mu_t$, where $t\in T$, be 
  a family of bounded measures on $\cal A$ absolutely continuous with respect to $\mu$
  such that for every $A\in {\cal A}$,
  the function $t\mapsto \mu_t(A)$ is measurable with respect to $\cal B$. Prove that
  one can find an ${\cal A}\otimes{\cal B}$-measurable function $f$ on $X\times T$ such
  that for every $t\in T$, the function $x\mapsto f(x,t)$ is the Radon-Nikodym density
  of the measure $\mu_t$ with respect to $\mu$.