Radon-Nikodym derivative as a measurable function in a product space. - MathOverflow

Radon-Nikodym derivative as a measurable function in a product space.

2011-08-16T12:23:13Z

Let $X$ be a Polish space with the probability measure $P$ and the Borel sigma algebra. Suppose that $X$ is also a group 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 traying to show that there is a measureable 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 eqaulity for every x in X.

Any suggestions?

Answer by Marc Palm for Radon-Nikodym derivative as a measurable function in a product space.

2011-08-16T14:53:05Z

Here is a rough sketch of things, I am pretty sure that they are true:

Consider a topological Hausdorff group $G$ with a quasi invariant measure, then the group is necessary locally compact. Now, since you assume implicitely that the transform of a measurable set by multiplication is measurbale, the action will be continuous, as measurable group isomorphisms are continuous. Hence we may assume that $X$ is a topological group. Now since your action is transitive, there is only one orbit and your quasi invariant measure is necesary continuous against the Haar measure $\mu$. The function $d P_x / d P$ can writen as the product of two measurable functions $d P_x / d \mu$ and $d \mu / d P$ by the chain rule of the Radon Nykodym derivatives, hence is measurable.

Note $\mu_x = \mu$: So in fact, your function $\Phi(x,y) = \frac{\lambda(xy)}{\lambda(y)}$ for $\lambda =d P /d \mu$.