Given a dynamic 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 groupa 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 the p153 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