I am trying to show that two measures are equivalent. This is the same as showing that the Radon-Nikodym derivative is bounded from above and below. I'll break it down so that things are easy to follow. If you need more information, please do ask!
Setup: Consider the probability space $(X,\rho$) and the map $R:X\to X$. Let $Y\subset X$ with $\rho(Y)>0$.
Assumption 1: There exists a measure $\mu_Y$ on $Y$ such that $\mu_Y$ is equivalent to $\rho|_Y$, where $\rho|_Y$ denotes the restriction of the measure $\rho$ to $Y$ (suitably normalised).
Assumption 2: There exist;
(i) A probability space $(Z=Y\times K,\nu)$ where $\nu=\mu_Y\times \kappa$ for some suitably normalised measure $\kappa$ on some space $K$.
(ii) An ergodic invariant function $F:Z\to Z$ and a measurable surjection $s:Z\to X$ such that $s\circ F=R\circ s$.
Assumption 3: Define the measure $\mu=s_*\nu:=\nu(s^{-1}(\cdot))$ on $X$. It is assumed that $R$ is ergodic and invariant with respect to $\mu$. Moreover, we assume that $R^{-1}A=A$ implies $\rho(A)\in\{0,1\}$, but invariance of R with respect to $\rho$ fails.
Question: Can we say that $\mu$ and $\rho$ are equivalent?
My "attempt": I think a similar sort of argument for $\frac{d\mu}{d\rho}=\frac{ds_*\nu}{d\rho}$ to the following could work.
If we define the measure $m=\rho|_Y\times \kappa$, note that $$\frac{1}{C}\le \frac{d\nu}{dm}= \frac{d\mu_Y}{d\rho|_Y}\le C$$ for some $C>0$, so that $\nu$ and $m$ are equivalent.
So far however I am not sure how to get the result. Any help would be very much appreciated! Thanks!