If a probability measure is stationary in both forward time and reverse time, does this imply that the measure is incompressible?

Question

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.)

Answer 1

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 $X$.