Proof of Wiener's local ergodic theorem So, I am trying to understand the proof of Wiener's local ergodic theorem. I've looked into quite a few books, for instance those by Krengel, Petersen and Einsiedler & Ward, but the phrasing is almost the same in every one of them. 
The theorem says that if $\{\tau_{\alpha} : \alpha \geq 0\}$ is a measurable and measure preserving semiflow on the probability space $(X, \mathcal{B}, \mu)$ and $f \in L^1(X,\mathcal{B}, \mu)$, then for almost every $x \in X$ we have: $$\lim_{\epsilon \rightarrow 0^+} \frac{1}{\epsilon}\int_{0}^{\epsilon} f(\tau_\alpha x)d\alpha = f(x).$$
The proof is starting by claiming that for almost all $x \in X$, $f(\tau_{\alpha}x)$ is a locally integrable function of $\alpha$. And that is, in fact, the part of the proof that I fail to understand. I'm inclined to believe that it is obvious but unfortunately I can't see why this holds.
Answers such as "Hint: use Fubini" are not what I'm looking for, I need a reasonably detailed explanation of what I'm missing.
 A: There is probably a better way to do this, but here is an argument using the Fubini and Fubini-Tonelli theorems.
By definition, the flow is measurable if the map $T:X \times [0,\infty) \to X$ given by $T(x,\alpha)=\tau_{\alpha}x$ is measurable (here, we consider $[0,\infty)$ with the Borel $\sigma$-algebra).
Thus the composition function $F=|f \circ T|: X \times [0, \infty) \to [0,\infty]$ is measurable. We want to prove that for almost every $x$ the slice function $F_x : [0,\infty) \to [0,\infty]$ defined by $F_x(\alpha) = F(x,\alpha) = |f(\tau_{\alpha}x)|$ is locally integrable.
Because the flow is measure-preserving, for any fixed $\alpha$ and any integrable function $g:X \to \mathbb{R}$ we have
$$
\int_X g \circ \tau_{\alpha} \; d \mu = \int_X g \; d\mu
$$
Thus for any fixed $\alpha \in [0,\infty)$, we have:
$$
\int_X |f(\tau_{\alpha}x)| \; d\mu = \int_X |f(x)| \; d\mu
$$
Integrating over any compact set $K \subset [0,\infty)$, this implies:
$$
\int_K \left( \int_X |f( \tau_{\alpha} x)| \; d\mu \right) d\alpha = \int_K \left(\int_X |f(x)| \; d\mu \right) d\alpha = \lambda(K) \int_X |f(x)| \; d\mu
$$
where $\lambda(K)$ is the measure of $K$. Since $f$ is integrable and $\lambda(K)$ is finite, we conclude that the iterated integral
$$
\int_K \left( \int_X F \; d\mu \right) d\alpha =\int_K \left( \int_X |f( \tau_{\alpha} x)| \; d\mu \right) d\alpha
$$
is finite for any compact set $K$. By the Fubini-Tonelli theorem, the function $F$ is integrable on $X \times K$. Now, since $F$ is integrable on $X\times K$, the Fubini theorem implies that almost every $x$-slice $F_x$ is integrable on $K$.
Thus we have proved that, for any fixed $K$, almost every function $F_x$ is integrable on $K$. This is not quite what we want. We want to prove that almost every function $F_x$ is integrable on all compact sets. However, we can cover $[0,\infty)$ with countably many compact sets $K_i$ in a way that any compact set intersects only finitely many of the $K_i$. A function is then locally integrable if and only if it is integrable on all of the $K_i$. By the preceding argument, for any fixed $i$ almost every function $F_x$ is integrable on $K_i$. But since there are countably many $K_i$ this implies that almost every function is integrable on all the $K_i$, and so almost every function $F_x$ is locally integrable.
