I'm reading "Ergodic Theory and Semisimple Groups" by Zimmer and at the very beginning of Chapter $2$ (pp. $8$) the author claims that

An action with quasi-invariant measure can be thought of as an action with an invariant measure class.

I interpreted this vague statement in the following way:

Every quasi-invariant measure is in the same measure class with an invariant measure.

Question1: is this statement true? I don't see how to prove this fact.

Question2: If question1 has a negative answer, how should such a statement be understood?

Here the author assumes the group $G$ be locally compact second countable, the action on a standard Borel space $S$ (i.e. Borel isomorphic to a Borel subset of a Polish space) be Borel (i.e. measurable). Moreover, a $\sigma$-finite measure $\mu$ is said to be quasi-invariant under the action of $G$ iff for all $A\subseteq S$, $g\in G$ we have $\mu(Ag)=0\iff\mu(A)=0$. It is invariant iff $\mu(Ag)=\mu(A)$ for all $A$, $g$. Finally, two measures are said to be in the same measure class iff they have the same null sets.

About my background: I have attended a basic measure theory course mostly focused on the real case. Whenever possible, a good reference that covers these topics is appreciated.

Thank you in advance for your help.

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    $\begingroup$ Thoughts: As far as I know there's a unique (up to constant multiplication) Borel measure that is also invariant, which is the Haar measure. I guess you need to show that your $\mu$ is equivalent to some Haar measure, though I'm not sure how to choose it properly. $\endgroup$ – Yanko Apr 18 at 15:12
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    $\begingroup$ A sentence with "can be thought" is not a mathematical statement. So I don't see any definite sense at "proving this fact". $\endgroup$ – YCor Apr 22 at 1:36
  • $\begingroup$ @YCor Ok, but then what does the author mean? Ad far as I'm concerned he wants to say something, otherwise it was sufficient don't write that phrase. I interpreted the above statement as "every quasi-invariant measure is in the same measure class with an invariant measure". Is this mathematical statement true? $\endgroup$ – LBJFS Apr 22 at 8:32
  • $\begingroup$ I also edit the question to make it clear $\endgroup$ – LBJFS Apr 22 at 8:34
  • $\begingroup$ OK, I posted an answer. I actually can't make any sense of your sentence "Every quasi-invariant measure is in the same measure class with an invariant measure." $\endgroup$ – YCor Apr 22 at 9:30

Well, in general, let $G$ is be group acting on a set $X$ with an equivalence relation $\sim $ which preserved by $G$ in the sense that $x\sim y$ implies $gx\sim gy$ for all $x,y$. So $G$ naturally acts on $X/\sim$; let $p:X\to X/\sim$ be the projection.

Then clearly for $x\in X$, we have $gx\sim x$ for all $g\in G$ if and only if $p(x)\in X/\sim$ is a fixed point of the $G$-action on $X/\sim$.

Here $X$ is the set of measures and $\sim$ identifies measures with the same null subsets. So we interpret the quasi-invariance $g\mu\sim \mu$ for all $g\in G$, as the class of $\mu$ being a fixed point for the $G$-action on the quotient set (the set of class of measures). All this is trivial, but it is a useful point of view.

  • $\begingroup$ I upvoted your answer, but I want to think more about it before I accept it. Anyway, thank you very much. $\endgroup$ – LBJFS Apr 23 at 8:56

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