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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 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?

  • 0
    Are you aware that your hypotheses imply that $X$ is a locally compact second countable group and that $P$ is in the Haar class by a theorem of Weil and Mackey? (Essentially the point is that you can embed $X$ into the unitary group of $L^2(X)$ and it inherits a locally compact group topology from the unitary group of $L^2(X)$.) If you're willing to accept this, your question becomes obvious. However, you might as well be trying to prove that result, so I'm asking for a clarification on your ultimate goal.2011-08-17
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    I am familiar with that theorem. In fact, I am trying to show that this embedding exactly is Borel measurable (as part of the proof of this theorem). In order to do so, I am trying to show that such a function Φ exists.2011-08-17
  • 0
    Okay, I'll write something later today or tomorrow if nobody else does. However, the proof is somewhat technical and I'd probably refer you to [Fisher—Witte Morris—Whyte](http://nyjm.albany.edu/j/2004/10-15.html) anyway, so I can as well point you to that reference now. What you ask is a special case of their Proposition 2.22.2011-08-18
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    Thank you, I believe that lemma 2.7 is the one I need.2011-08-18
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    Yes, exactly, that's the main point. But note that it is *not* quite obvious that your $\Phi$ actually gives a Borel map $X \to \mathbf{F}(X,[0,\infty))$ (in their notation) doing what you want; you can't just shove $\Phi$ into lemma 2.6, apply 2.7 and get the thing you're looking for entirely for free! You need an argument similar to the one in 2.22 but you already have the main ingredients.2011-08-18
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    If memory serves, the result in the exercise from Bogachev's book that is cited by Byron Schmuland goes back to Doob, as an application of the martingle convergence theorem.2011-08-20

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