Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and let $X$ be a compact metric space. Let $F \colon \Omega \times X \to X$ and $\bar{F} \colon \Omega \times X \to X$ be measurable functions such that for each $\omega$, $F(\omega,\,\cdot\,) \colon X \to X$ is a homeomorphism with inverse $\bar{F}(\omega,\,\cdot\,) \colon X \to X$.


Suppose we have a probability measure $\rho$ on $X$ such that
$$ \rho(A) \ = \ \mathbb{P} \otimes \rho(F^{-1}(A)) \ = \ \mathbb{P} \otimes \rho(\bar{F}^{-1}(A)) $$
for all $A \in \mathcal{B}(X)$. Does it follow that $\mathbb{P}$-almost every $\omega \in \Omega$ has the property that for all $A \in \mathcal{B}(X)$,


$\hspace{50mm} \rho(A) \ = \ \rho(\{x \in X : F(\omega,x) \in A\}) \ $?


(Note: It is sufficient to show that for each $A$, the above equality holds for $\mathbb{P}$-almost all $\omega$; the fact that the $\mathbb{P}$-full measure set can then be chosen independently of $A$ follows by standard results.)



If a probability measure is stationary in both forward time and reverse time, does this imply that the measure is incompressible?
Not at all. Let me reformulate your question in terms of the group $G$ of homeomorphisms of $X$ (and replacing your $\mathbb P$ with $\mu$ - which is more conventional in this context). Essentially you are saying that there is a measure $\mu$ on $G$ and a measure $\rho$ on $X$ which is stationary with respect to both $\mu$ and its reflection $\check\mu$ (the image of $\mu$ under the group inversion), and you are asking whether the measure $\rho$ is then necessarily invariant with respect to $\mu$-a.e. homeomorphism. I am going to give a counterexample in which the space $\Omega$ is finite, so that one does not have to worry about the measurability issues.


Let $X$ be the boundary of the free group $F=\langle a,b\rangle$ (the space of infinite irreducible words in the alphabet $\Omega=\{a,b,a^{-1},b^{-1}\}$ which consists of the generators and their inverses), and let $\mu$ be the uniform distribution on $\Omega$, so that $\mu=\check\mu$. Then the natural "uniform" measure on $X$ is $\mu$-stationary (and therefore $\check\mu$-stationary as well), but not invariant with respect to the action of $F$ on $X$.


