Remark 1

— Under the hypotheses of Prop. 4, the mapping $\mu \mapsto \mu *^\beta f$ of $\mathscr{M}_+(G)$ into $\mathscr{C}(X)$ is continuous when $\mathscr{M}_+(G)$ is equipped with the vague topology, for every $f \in \mathscr{K}(X)$. For, let $K$ be a compact subset of $X$, $S$ the (compact) support of $f$; since $G$ operates properly in $X$, the set of $s \in G$ for which there exists an $x \in K$ with $s^{-1}x \in S$ is a compact subset $L$ of $G$ (GT, III, § 4, No. 5, Th. 1). Let $\varepsilon$ be a number $> 0$, $\varphi$ a function in $\mathscr{K}_+(G)$ equal to 1 on the compact set $L$, $\mu_0$ an element of $\mathscr{M}_+(G)$; the set $W_0$ of measures $\mu \in \mathscr{M}_+(G)$ such that

is a neighborhood of $\mu_0$ in $\mathscr{M}_+(G)$. On the other hand, the function $(s, x) \mapsto f(s^{-1}x)\chi(s^{-1}, x)$ is uniformly continuous on $L \times K$, therefore there exists a finite number of points $x_i \in K$ ($1 \leq i \leq n$) such that for every $x \in K$, there is an $i$ for which for all $s \in L$. Since $\mu(L) \leq \int \varphi(s)\, d\mu_0(s) + \varepsilon$ for all $\mu \in W_0$, also for every $x$ satisfying the preceding inequality and every $\mu \in W_0$. Now let $W$ be the neighborhood of $\mu_0$ in $\mathscr{M}_+(G)$ formed by the measures $\mu \in W_0$ such that for $1 \leq i \leq n$. It is clear that for every measure $\mu \in W$ and every $x \in K$, and since $\varepsilon$ is arbitrary, this proves our assertion.