Given an ergodic and non-singular dynamic system (definition provided [here](https://web.williams.edu/Mathematics/csilva/NonsingularET_Apr.pdf)) $(X, \mathcal{B}, \mu\_1, T)$ where $(X, \mathcal{B}, \mu\_1)$ is a measure space and $T$ is a fixed transformation, we then will have $\mu\_1$ equivalent to $T\mu$ where $T\mu\_1$ is defined by $T\mu\_1(A)=\mu\_1[T^{-1}(A)]$ for each $A\in \mathcal{B}$. Now suppose $\mu\_2$ is another measure that is equivalent to $\mu\_1$ and makes $(X, \mathcal{B}, \mu\_2, T)$ ergodic and non-singular. Hence, we will have $T\mu\_2\sim\mu\_2\sim\mu\_1\sim T\mu\_1$. Do we have:


$$
\frac{d\,\mu\_1}{d\,\mu\_2}\circ T = \frac{d\,T\mu\_1}{d\,T\mu\_2}
$$


$\mu\_1$-almost everywhere? Here the "$\mu\_1$-almost everywhere" can be replaced by "$\mu\_2$-almost everywhere" or any one of those four measures. By definition of $T\mu\_2$, for each $A\in\mathcal{B}$, we have:


$$
\int\_X\chi\_A(x)d\,T\mu\_2 = \int\_X\chi\_A(Tx)d\,\mu\_2
$$


I tried to show the following equation holds for each $A\in\mathcal{B}$ but cannot:


$$
\int\_A\frac{d\,\mu\_1}{d\,\mu\_2}(x)d\,T\mu\_2 = \int\_A\frac{d\,\mu\_1}{d\,\mu\_2}\circ T(x)d\,\mu\_2 = \int\_A\frac{d\,T\mu\_1}{d\,T\mu\_2}(x)d\,\mu\_2
$$


Any hints or counter-examples will be appreciated



In general is $\frac{d\,\mu_1}{d\,\mu_2}\circ T = \frac{d\,T\mu_1}{d\,T\mu_2}$?
I am fairly sure this is not the case (unless I am missing something in the definition of $T$)


Consider the special case where $T$ is invertible with inverse $T\_{\rm inv}$. Then, it holds for any test function $f$


\begin{aligned}
\int f \,dT\mu\_1 &= \int f \circ T \,d\mu\_1 \\
&= \int f \circ T \,\frac{d\mu\_1}{d\mu\_2} \,d\mu\_2 \\
&= \int f \circ T \, \frac{d\mu\_1}{d\mu\_2} \,d(T\_{\rm inv} (T \mu\_2))\\
&= \int f \,\frac{d\mu\_1}{d\mu\_2} \circ T\_{\rm inv} \,dT\mu\_2
\end{aligned}
and thus it actually holds
$$
\frac{dT\mu\_1}{dT\mu\_2} = \frac{d\mu\_1}{d\mu\_2} \circ T\_{\rm inv}.
$$


If I understand correctly, the following simple example satisfies your assumptions and shows that your claim does not hold: Take $X = \{1, 2, 3\}$, $\mu\_1 = (1/6, 2/6, 3/6)$ (meaning $\mu\_1 = \frac{1}{6}\delta\_1 + \frac{2}{6} \delta\_2 + \frac{3}{6} \delta\_3$), $\mu\_2 = (2/6, 3/6, 1/6)$ and $T(1) = 2, T(2) = 3, T(3) = 1$, and one quickly finds that $T$ is invertible but also that $\frac{dT\mu\_1}{dT\mu\_2} = \frac{d\mu\_1}{d\mu\_2} \circ T\_{\rm inv} \neq \frac{d\mu\_1}{d\mu\_2} \circ T$.


