# Every quasi-invariant measures is in an invariant measure class (Zimmer)

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.

• 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. – Yanko Apr 18 at 15:12
• A sentence with "can be thought" is not a mathematical statement. So I don't see any definite sense at "proving this fact". – YCor Apr 22 at 1:36
• @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? – LBJFS Apr 22 at 8:32
• I also edit the question to make it clear – LBJFS Apr 22 at 8:34
• 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." – 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.