Proposition 13

The mapping $(s, \mu) \mapsto \gamma(s)\mu$ of $G \times \mathscr{M}_+(X)$ into $\mathscr{M}_+(X)$ is continuous when the set $\mathscr{M}_+(X)$ of positive measures on $X$ is equipped with the vague topology.

Since $\gamma(s)\mu = \gamma(ss_0^{-1})\gamma(s_0)\mu$, it follows from Remark 1 that it suffices to prove the continuity of the mapping under consideration at a point of the form $(e, \mu_0)$ with $\mu_0 \in \mathscr{M}_+(X)$. Given a function $f \in \mathscr{K}(X)$ and a number $\varepsilon > 0$, it is thus a matter of showing that there exist a neighborhood U of e in G and a neighborhood W of $\mu_0$ in $\mathscr{M}_+(X)$ such that the relations $s \in U, \mu \in W$ imply

Let V be a compact neighborhood of the support K of $f$ in X, and let $\varphi \in \mathscr{K}_+(X)$ be such that $\varphi(x) = 1$ on V; there exists a neighborhood $W_0$ of $\mu_0$ in $\mathscr{M}_+(X)$ such that $a = \sup_{\mu \in W_0} \mu(V)$ is finite: it suffices to take for $W_0$ the set of $\mu \in \mathscr{M}_+(X)$ such that $|\langle \varphi, \mu - \mu_0 \rangle| \leq 1$. Since the mapping $(s, x) \mapsto sx$ is continuous, there is, on the other hand, a compact neighborhood $U_0$ of e in G such that $sK \subset V$ for all $s \in U_0$; the function $(s, x) \mapsto f(sx)$ is then uniformly continuous in $U_0 \times V$ and so there is a neighborhood $U \subset U_0$ of e such that $|f(sx) - f(x)| \leq \varepsilon / 2a$ for all $s \in U$ and $x \in V$. For $s \in U$ and $\mu \in W_0$, we therefore have

if $W \subset W_0$ is the neighborhood of $\mu_0$ in $\mathscr{M}_+(X)$ formed by the measures $\mu \in W_0$ such that $\left| \int f(x)\, d\mu(x) - \int f(x)\, d\mu_0(x) \right| \leq \varepsilon / 2$, U and W meet the requirements.