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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?

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Concerning your last statement (I must have overlooked it first), you won't need equality for all $x \in X$, but only for a.a. $x \in X$, in order to show measurablity, so you argument is slicker than mine. Sets of measure $0$ are always $P$ measurable, by definition, and the completion wrt. to any quasiinvariant measure on a transitive $G$ space will coincide, so the completion do not depend upon the quasiinvariant measure chosen in the first place, and you should enrich your $\sigma$ algebra by these sets. –  plusepsilon.de Aug 17 '11 at 7:58
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Asked and answered at MathSE: math.stackexchange.com/questions/58018/… –  Byron Schmuland Aug 30 '11 at 15:27
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1 Answer

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 $\lambda =d P /d \mu$.

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See also: mathoverflow.net/questions/64116/… –  plusepsilon.de Aug 16 '11 at 15:02
    
Btw, $\Phi$ is a cocycle and the data $P$, $\lambda$ or $\phi$ are all equivalent, and can be recovered from each other under some mild assumptions. And assuming right quasi invariance, or left quasi invariance, should be sufficient to get quasi invariance from both sides. –  plusepsilon.de Aug 16 '11 at 15:09
    
Could you please be a little more thorough in detailes? I didn't understand why we can assume that X is a topological group. All we know is that X is a measurable group (the group action is borel measurable) and that the measure is left (hence also right) quasiinvariant. Now, even if i accept that X is a topological group, why are P and the haar measure equivalent and why does Φ(x,y)=λ(xy)/λ(y) for λ=dP/dμ? –  BBB Aug 16 '11 at 15:49
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One quick way to topologize $X$ is to use the quasi-invariant measure to produce an action of $X$ on $L^2(X)$ by unitaries. This embeds $X$ into the unitary group on a Hilbert space, which is a topological group if endowed with the strong operator topology. –  Dima Shlyakhtenko Aug 16 '11 at 17:51
    
Yes, but how do you show that the embedding is measurable? This is exactly what I am now trying to prove, and in order to do so I need the function mentioned above to be measurable. If you know another way to show that the embedding is measureable, I will be happy to see it. –  BBB Aug 16 '11 at 19:21
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