Questions about ratio set in a dynamical system Given a dynamical system $(X, \Omega, \mu)$ ($\Omega$ the $\sigma$-algebra and $\mu$ a measure), we assume there is a group $G$ acting on the system in the sense that, for each group element $g$, $g$ corresponds to an invertible measurable mapping $T_g$ such that $g \cdot x = T_g x$. We also define $T_g\mu$ by, for each $A\in\Omega, T_g\mu(A) = \mu[T_g^{-1}(A)]$.
Then, we call the group action in $(X, \Omega, \mu, G)$ ergodic iff, for each $g\in G$, any $T_g$ invariant set ($T_g(A)=A$) will have $\mu$-measure zero; call the group action in $(X, \Omega, \mu, G)$ non-singular iff for each $g\in G$, $\mu$ is equivalent to $T_g\mu$; call the group action conservative iff for each $A\in\Omega_1$ with $\mu(A)>0$, we can find $h\in G$ such that $\mu[A\cap T_h^{-1}(A)]>0$. Now, given a ergodic, non-singular and conservative system $(X, \Omega, G, \mu)$, the definition of a ratio set of $(X, \Omega, \mu, G)$, denoted by $r_{\mu}(G)$, is defined below:
A positive real number $r$ is in $r_{\mu}(G)$ iff, for each $A\in\Omega_1$ with positive measure and $\epsilon>0$, we can find $g\in G$ such that $A\cap T_g^{-1}(A)\cap \{x\in X\,\vert\,\frac{d\,T_h\mu}{d\,\mu}(x)\in(r-\epsilon, r+\epsilon)\}$ has positive $\mu$ measure. Now we want to prove: given two equivalent measures $\mu_1, \mu_2$ defined on $\Omega$ such that both $(X, \Omega, \mu_1, G)$ and $(X, \Omega, \mu_2, G)$ are ergodic, non-singular and conservative, $r_{\mu_1}(G)=r_{\mu_2}(G)$ (i.e. the ratio set of $\mu_1$ and $\mu_2$ are the same).
Notice that, for each $A\in\Omega$ with positive measure, given $g\in G$, the following set will have positive measure (either $\mu_1$ or $\mu_2$).
$$
\{x\in A\,\vert\,\frac{d\,T_g\mu_1}{d\,\mu_1}(x)\in(\frac{T_g\mu_1(A)}{\mu_1(A)}-\epsilon, \frac{T_g\mu_1(A)}{\mu_1(A)}+\epsilon)\}
$$
because, otherwise, the integral $\int_A\frac{d\,T_g\mu_1}{d\,\mu_1}d\,\mu_1$ will not be $T_g\mu_1(A)$. The statement will also be true when we replace $\mu_1$ by $\mu_2$. Now it suffices to show that $r_{\mu_1}(G)\subseteq r_{\mu_2}(G)$. Given $r$ in the ratio set of $\mu_1$, if we fix an arbitrary $\epsilon$ and set $Q_A = A\cap T_h^{-1}(A)\cap \{x\in X\,\vert\,\frac{d\,T_h\mu}{d\,\mu}(x)\in(r-\epsilon, r+\epsilon)\}$ where $h$ is given by the definition of ratio set, then the following set will have positive measure:
$$
Q_{A, 1} = \{x\in Q_A\,\vert\,\frac{d\,\mu_1}{d\,\mu_2}(x)\in(\frac{\mu_1(Q_A)}{\mu_2(Q_A)}-\epsilon, \frac{\mu_1(Q_A)}{\mu_2(Q_A)}+\epsilon)\}
$$
I want to approach the problem by using $\frac{d\,T_h\mu_2}{d\,\mu_2}(x)=\frac{d\,T_h\mu_2}{d\,T_h\mu_1}(x)\cdot\frac{d\,T_h\mu_1}{d\,\mu_1}(x)\cdot\frac{d\,\mu_2}{d\,\mu_1}(x)$. From p.153 in Ergodic dynamics written by Jane Hawkins, we have that, for each $g\in G, \frac{d\,T_g\mu_1}{d\,T_g\mu_2}(x) = \frac{d,\mu_1}{d\,\mu_2}(T_g x)$. Now $Q_{A, 1}$ has positive measure and, for each element $x\in Q_{A, 1}, \frac{d\,T_g\mu_1}{d\,\mu_1}(x)\in (r-\epsilon, r+\epsilon)$. We want to find another set, say $B$ such that, for each $y\in B, \frac{d\,T_g\mu_2}{d\,T_g\mu_1}(y)$ is close to $\frac{\mu_2(Q_A)}{\mu_1(Q_A)}$ and this set need to have non-null intersection with $Q_{A, 1}$. I thought about $T_g^{-1}(Q_{A, 1})$ but did not know if $Q_{A, 1}$ and $T_g^{-1}(Q_{A, 1})$ will have non-null intersection. Any other hints will be appreciated
 A: In what follows I'll write $g$ for $T_g$, and i'll write $\alpha _i (g,x)$ for $\tfrac{dg^{-1}\mu _i}{d\mu _i} (x)$. Here is one way to argue that the ratio sets are the same: given $r>0$ in the ratio set of $\mu _2$ along with a positive measure set $A$ and an open neighborhood $U$ of $r$, find a neighborhood $V$ of $1$ with $V^2rV^{-2} \subseteq U$. The sets $Vs$, with $s>0$ rational, cover the positive reals, so we can find some $s>0$ such that the set $B$, of all $x\in A$ with $\tfrac{d\mu _1}{d\mu _2}(x)\in Vs$, has positive measure. Since $r$ is in the ratio set of $\mu _2$ we can find some $g\in G$ and a positive measure $C\subseteq B$ with $gC\subseteq B$ and $\alpha _2(g,x)\in VrV^{-1}$. Then each $x\in C$ satisfies $\alpha _1(g,x) = \tfrac{d\mu _1}{d\mu _2}(gx)\alpha _2(g,x)\tfrac{d\mu _1}{d\mu _2}(x)^{-1}\in V^2rV^{-2}\subseteq U$, so $r$ is in the ratio set of $\mu _1$.
I think it is nice to understand this from the broader perspective of cocycles of dynamical systems. Without going into any real details, i'll just mention some relevant definitions/jargon can be looked into more if you are interested in this perspective. The function $\alpha _i$ defined above is called the "Radon-Nikodym cocycle" of the system $(G,X,\mu _i)$. It satisfies the cocycle equation $\alpha _i (gh,x)=\alpha _i (g,hx )\alpha _i (h,x)$. The measures $\mu _1$ and $\mu _2$ being equivalent yields the equation $\tfrac{d\mu _1}{d\mu _2}(gx)\alpha _2(g,x)\tfrac{d\mu _1}{d\mu _2}(x)^{-1}=\alpha _1(g,x)$, so that $\alpha _1$ and $\alpha _2$ are cohomologous cocycles. The ratio set of the system $(G,X,\mu _i )$ coincides with what is sometimes called the ``essential range'' of the cocycle $\alpha _i$.
