Let $(X, \mathcal B, \mu)$ be a Borel probability space. A measurable flow (to my understanding) is an automorphism $\mathbb R \to \mathrm{Aut}(X)$, i.e. a collection of measure-preserving maps $T_t : X \to X$ such that $T_{t + s} = T_t \circ T_s$ for every $t, s \in \mathbb R$. A flow is ergodic if, whenever $A \in \mathcal B$ satisfies $T_t(A) = A$ for every $t \in \mathbb R$, we have $\mu(A) = 0$ or $1$.
I am trying to prove that the first of the following two statements implies the second:
- The flow $T_t$ is ergodic.
- There exists a time $t_0 \in \mathbb R$ such that the discrete-time map $\hat T : X \to X$ defined by $\hat T(x) = T_{t_0}(x)$ is ergodic, i.e. whenever $A \in \mathcal B$ satisfies $\hat T^{-1}(A) = A$, we have $\mu(A) = 0$ or $1$.
This does not seem to be a trivial problem: given a set $A \in \mathcal B$ with $0 < \mu(A) < 1$, the first statement does not a priori preclude the existence of a time $t_0$ such that $T_{t_0}(A) = A$. In fact, it is a known result that an ergodic flow can have at most countably many non-ergodic times $t_0$. However, without using this result (which I'm currently working on proving independently), it is not clear to me why every time $t_0$ cannot be non-ergodic.
So, does Statement 1 imply Statement 2?