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?