Strong measurability of $L^1$ representation Let $G$ be a locally compact group and $X$ be a Banach space. Consider a uniformly bounded strongly continuous representation $\pi \colon G \to B(X)$. Let $f \in L^1(G)$ and $x \in X$. Consider the map $G \to X$, $s \mapsto f(s)\pi(s)x$.
Why this map is strongly measurable?
I have no doubt on this result but I never seen a complete proof. What is precisely the used result? 
 A: Pettis's Theorem says that a function $G \to X$ is strongly measurable if and only if it's weakly measurable and almost everywhere separably valued.  
Assume $G$ is second countable*, and, thus, separable.  $\pi$ is strongly continuous so $\pi(G)x$ is separable.  Therefore, its span is separable so $\{f(s) \pi(s) x \, \mid \, s \in G_{1}\}$ is separable.  Here I've taken a subset $G_{1} \subseteq G$ such that $G \setminus G_{1}$ is null with respect to Haar measure and $f$ is finite on $G_{1}$.  That proves the map $s \mapsto f(s) \pi(s) x$ is separably-valued.  
It remains to show that the map in question is weakly measurable.  In other words, we must show that if $F$ is a bounded linear functional on $X$, then
$$s \mapsto F(f(s)\pi(s)x)$$
is a measurable function on $G$.  By linearity, we can write
$$F(f(s) \pi(s) x) = f(s) F(\pi(s) x),$$
which is a product of two functions on $G$, one of which we know is measurable.  I claim that $s \mapsto F(\pi(s)x)$ is measurable.  Indeed, $s \mapsto \pi(s)x$ is continuous and $y \mapsto F(y)$ is continuous.  It follows that $s \mapsto F(\pi(s)x)$ is continuous and, therefore, measurable.  We showed $s \mapsto F(f(s)\pi(s)x)$ is a product of measurable functions.  Therefore, it is measurable itself.  
The previous paragraph proves $s \mapsto f(s) \pi(s)x$ is weakly measurable. We already proved that map is almost everywhere separably valued.  Invoking Pettis's Theorem, we see that our map is strongly measurable.    
*Are there interesting examples of locally compact topological groups that aren't second countable?  Would $X$ be separable in that case?  This seems like a technicality to me.
