Equivalence in law of two processes I am having some trouble with showing the following. I will explain the setup below.
Let us suppose we have two probability measure preserving dynamical systems $(X_1,\Sigma_1,\mu_1,T_1)$ and $(X_2,\Sigma_2,\mu_2,T_2)$. 
Suppose there exists a measurable map $\pi:X_2\to X_1$ such that $\pi$ is onto, $\pi\circ T_2=T_1\circ \pi$, and $\mu_1=\pi_\star\mu_2$. That is, for $A\in \Sigma_1$, we have $\mu_1(A)=\mu_2(\pi^{-1}(A))$.
If $f\in L^1(X_1)$, then it is immediate that $g=f\circ \pi\in L^1(X_2)$. I want to show that the two processes $\{f\circ T_1^k :k\ge 0\}$ and $\{g\circ T_2^k :k\ge 0\}$ are equivalent in distribution. For this, it suffices to show that their finite dimensional distributions coincide. However I am having trouble with this. Can anyone help?
 A: We officially would have to check that for all $n\geq 1$ and all Borel subset $B$ of $\mathbb R^n$, $$\mu_1\left( \left(f\circ T_1^k\right)_{k=1}^n\in B\right)=\mu_2\left( \left(g\circ T_2^k\right)_{k=1}^n\in B\right).$$
But actually, it suffices to check the previous equality when $B$ is a product of Borel sets, say $B=B_1\times\dots\times B_n$. Then 
$$
\mu_2\left( \left(g\circ T_2^k\right)_{k=1}^n\in B\right)= 
\mu_2\left( \bigcap_{k=1}^n \{g\circ T_2^k \in B_k\}\right)=\mu_2\left( \bigcap_{k=1}^n \{f\circ\pi\circ T_2^k \in B_k\}\right).
$$
Now write $ \{f\circ\pi\circ T_2^k \in B_k\}$ as $$(f\circ \pi)^{-1}\{x\in X_2,T_2^kx\in B_k\}= \pi^{-1}\left( f^{-1}\{x\in X_2,T_2^kx\in B_k\}\right)$$
hence letting $A_k:=\{x\in X_2,f\circ T_2^kx\in B_k\}$, we derive that 
$$
\mu_2\left( \left(g\circ T_2^k\right)_{k=1}^n\in B\right)=\mu_2\left( \bigcap_{k=1}^n \pi^{-1}\left(A_k\right)\right)=\mu_2\left(  \pi^{-1}\left(\bigcap_{k=1}^nA_k\right)\right)
$$
and the result, by going back to the equality $\mu_1=\pi_\star\mu_2$.
