5

Let G be a topological group and X a measure space. Let $G \times X \rightarrow X$ be a measurable group action, $\mu$ a $\sigma$-finite measure on $X$, and $g\mu$ (for any $g \in G$) the measure $g\mu(A)= \mu(g^{-1}A)$. Assume that $g\mu$ is always equivalent to $\mu$, that is, $\mu(A)=0$ iff $g\mu(A)=0$ for every measurable $A \subset X$.

Define $$\phi : G \times X \rightarrow \mathbb{R}, (g,x) \mapsto \frac{dg\mu}{d\mu}(x)$$ where $\frac{dg\mu}{d\mu}$ is the Radon-Nikodym derivative.

Why is $\phi$ measurable?

dessin d'enfant terrible
  • 543
  • What do you mean by $\phi$ being measurable? The group $G$ is not given any measure. (If $G$ were locally compact, one could take the/a Haar measure.) Maybe you mean that $\phi$ is measurable in the variable $x$? Then this is just the statement of Radon-Nikodym theorem. Same question for the notion of a "measurable group action", where $G$ is merely a topological group. – Moishe Kohan Jul 12 '14 at 12:58
  • Of course, maybe you are asking the same question as in http://math.stackexchange.com/questions/58018/radon-nikodym-derivative-as-a-measurable-function-in-a-product-space?rq=1. Then you should just award your bounty to this user: http://math.stackexchange.com/users/5363/t-b – Moishe Kohan Jul 12 '14 at 13:04
  • @studiosus I mean that G has the sigma algebra of the borel sets, and $G\times X$ has the product sigma algebra. – dessin d'enfant terrible Jul 12 '14 at 14:52
  • @studiosus Thank you. That question is very similar. But whats proved is different from the statement that interests me. For instance, the proof you linked to assumes X is a standard Borel space. – dessin d'enfant terrible Jul 16 '14 at 00:55
  • Why do you think this statement is true or provable in this utter generality? While it seems to be reasonable at first glance, the Radon-Nikodym derivative is subject to a lot of constraints which it must fulfill simultaneously. It is not even clear how to make $\phi$ into a well-defined function on $G\times X$ without using the axiom of choice: a priori it is only clear that you obtain a map from $G$ to the measurable function classes $X \to \Bbb R$, but how do you turn this into a function on $G \times X$? Reading the answers in the in the thread linked by @studiosus (to be continued) – user160629 Jul 17 '14 at 08:49
  • (continued) it seems quite unlikely that the answerers didn't consider the possibility of doing some general gymnastics before deciding that it is necessary to do some rather hard work. – user160629 Jul 17 '14 at 08:50
  • @user160629 I don't see the well-definedness issue. Specifying the class of $f(g,-)$ for all $g$ specifies the class of $f$ (by Fubini's theorem). In any case, this would also be a problem for the proof to which studiosus linked. You may be right, however, that this is not true in greater generality. The reason I thought it was true is that it is stated on page 325 of the book "Kazhdan's Property (T)" by Bachir Bekka. – dessin d'enfant terrible Jul 17 '14 at 18:45
  • @user160629 Or maybe you would object that there is no "class of $f$" since $G\times X$ has no specified measure. In that case you may prefer the phrasing in the linked post: There exists measurable $\rho (g,x)$ such that $\rho (g,-)$ is always a derivative of $g\mu$ with respect to $\mu$. – dessin d'enfant terrible Jul 17 '14 at 22:53

0 Answers0