Is measure preserving action $\theta_t$ measure preserving with $t$ replaced by a stopping time?

Question

Suppose $\theta:\mathbb{Z}\times \Omega\to \Omega$ is a measure preserving action on a probability space. In particular $E[X] = E[X\circ \theta_t]$ for all $t$. Does the same hold if $t$ is replaced by a stopping time $\tau$ relative to $\{\mathcal{F}_t\}$ if $X$ is $\mathcal{F}_{\infty}$-measurable?

Answer 1

No, this is false if I understand the question correctly.

Take $\Omega = \{0,1\}^{\Bbb Z}$ with its product $\sigma$-algebra, and let $\mu$ be the infinite product of Bernoulli$(1/2)$. Let $\mathcal F_t$ be the filtration generated by coordinate maps $\pi_n(\omega)=\omega_n$ for $n \leq t$.

Let $\theta$ be the shift operator, that means that $$\theta(n,\omega) = (\omega_{n+k})_{k \in \Bbb Z}$$

Let $X(\omega)=\pi_1(\omega) = \omega_1$, the first positive coordinate of the infinite sequence $\omega$. Let $T(\omega)$ be the first $n \in \Bbb N$ such that $\omega_n=1$. Then clearly $E[X \circ \theta_T]=1$ whereas $E[X \circ \theta_t] = \frac{1}{2}$ for all fixed times $t$.