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](https://link.springer.com/book/10.1007/978-3-030-59242-4) 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



Questions about ratio set in a dynamical system
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 $\alpha \_i$.


