Suppose $f \in L^1(X)$ and let $\phi_s :X \to X$ be a one parameter measure preserving flow that is also measurable on $\mathbb{R} \times X$. Show that for almost every $x \in X$ the function $f(\phi_s(x))$ is integrable on $[0,1]$ with respect to the Lebesgue measure.
This seems like a straightforward application of Fubini's theorem but I'm not sure how to apply it. The intuition is that if $f$ is integrable on all of $X$ then it should also be integrable on the entire orbit of x along the flow for a.e. $x \in X$. The only issue is that the orbit of $x$ might create a path of measure zero in $X$ that $f$ blows up on. How do we know that this happens infrequently enough?
This is a quick consequence of the fact that $\phi_s$ is measure-preserving. In particular, for any integrable function $f$, we have $$\int f\circ\phi_s\,d\mu = \int f\,d\mu.\tag{1}$$ Therefore, $$ \int_0^1\int |f|\circ \phi_s\,d\mu\,ds = \int_0^1\int |f|\,d\mu\,ds = \int |f|\,d\mu < \infty. $$ The claim follows by Fubini's theorem.
To prove $(1)$, observe $(1)$ holds for indicator functions of measurable sets $E\subset X$ by definition, then simple functions by linearity, non-negative functions by monotone convergence, and then integrable functions by decomposing $f = f^+-f^-$.