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

Question

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.

Answer 1

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.

Answer 2

About Question 1: no, it is not true. Consider the subgroup $G < \mathrm{Homeo(\mathbb{R})}$ generated by $\{ x \mapsto 2x, x \mapsto x + a: a\in \mathbb{R} \}$. Then, Lebesgue measure is quasi-invariant (but not invariant) under the action of $G$,but no measures in the same measure class are invariant under $G$.

To show this we use that the measure is assumed to be $\sigma-$finite to apply the Radon-Nykodim theorem: we get $d \mu = f \,d\mathrm{Leb}$), by translation invariance we get that $f$ must be constant (Lebesgue-)a.e., and by invariance under $x \mapsto 2x$ we conclude $f \equiv 0$, which would mean $\mu$ is zero (and also not in the same measure class as Lebesgue measure!).

About Question 2: I interpret this to mean that there exists a measure class such that, for any $g \in G$ and any $\mu$ in this measure class, $g_*\mu$ belongs to the same measure class.