Continuity(?) of induced group representation in the isometries of $L^p$ Let $G$ be a topological group which acts on a measure space $(X,\mu)$ by measure preserving transformations. It's well known that this induces a representation $\pi$ of $G$ in $O(L^p (X,\mu))$, the group of linear isometries on $L^p (X,\mu)$, for any $1\leq p<\infty$. 
My question is, under what circumstances do we have the following continuity property? Namely for each $f\in L^p (X,\mu)$, $g\mapsto \pi (g) f$ is continuous?
Edit: another, fancier way of saying this is that $\pi : G \rightarrow O(L^p (X,\mu))$ is continuous provided that $O(L^p (X,\mu))$ is endowed with the strong operator topology.
 A: The answer is easier than expected. 
It suffices to check that for any $g_0 \in G$, and any $\varepsilon >0$, there exists a neighbourhood $U$ of $g_0$ such that for any $g\in U$, $\Vert \pi(g)f - \pi(g_0)f \Vert_p <\varepsilon$. Since $G$ is a topological group, it actually suffices to pick $g_0 = e$. 
The induced representation $\pi$ is given by $\pi (g) f(x) := f(g^{-1} \cdot x)$ for $f$ a point realization of an element of $L^p (X,\mu)$. It follows that for any indicator function of a measurable set $\chi_A$, $\pi(g) \chi_A = \chi_{gA}$. Compute that if $\mu(A) - \mu(A\cap gA) < (\varepsilon / 2)^p$, then $\Vert \chi_A - \chi_{gA}\Vert_p < \varepsilon$. Therefore it suffices to pick $U$ to be a small enough neighborhood around $e$ that $\mu(A) - \mu(A\cap gA) < (\varepsilon / 2)^p$, which we can do by assumption. In other words, as $g\rightarrow e$ in the topology of $G$, we have that $\pi(g)\chi_{A} \rightarrow \chi_A$ in $L^p$. 
The same argument works for any simple function via the triangle inequality, and then by the density of simple functions, it also holds for any $f\in L^p(X,\mu)$ that $\pi(g)f\rightarrow f$ in $L^p$ as $g\rightarrow e$, so we're done.
