Let $G$ be a locally compact topological group, $X, \mu$ a probability space, and $G\times X \rightarrow X$ a measurable group action which preserves $\mu$ (i.e. $\mu (gA)=\mu(A)$) . Does it follow that for any $f(x)\in L^2(X)$, the map $$G \rightarrow L^2(X), g \mapsto f(gx)$$ is continuous?
I'm sure it suffices to prove that $\mu (A \Delta g_nA) \rightarrow 0$ whenever $g_n \rightarrow e$ (where $\Delta$ is the symmetric difference) but I don't even see how to prove this.
One would like to believe this to be true in the stated generality and it is often asserted either without proof or accompanied with some fervent hand-waving involving the name Fubini. Detailed and correct proofs are quite hard to find.
One good source is Lemme A.1.1 of Babillot et al, Theoremes ergodiques pour les actions de groupes. They do assume in the proof that $X$ is a standard Borel space, that $\mu$ is $\sigma$-finite and $G$-invariant and that $G$ is second countable. They prove that the representation of $G$ on $L^p(X,\mu)$ is strongly continuous for $1 \leq p < \infty$.