Proposition 7

Let $N \subset \Gamma^0$ be the subspace of normalized Haar measures on the discrete subgroups of G, and for every relatively compact open neighborhood U of e in G, let $N_U$ be the subset of N formed by the $\alpha$ such that $H_\alpha \cap U = \{e\}$. Then: a) $N_U$ is compact. b) The interiors of the sets $N_U$ in N form a covering of N, as U runs over the set of relatively compact open neighborhoods of e in G. c) For a subset M of N to be relatively compact in N, it is necessary and sufficient that there exist a relatively compact open neighborhood U of e in G such that $M \subset N_U$.

Since $D_U$ is the image of $N_U$ under the canonical continuous mapping $\Gamma \to \Sigma$, the assertion (iii) of Th. 1 will result at once from Prop. 7 a).

To prove Prop. 7, we observe that $N_U$ can be defined as the subset of $\Gamma^0$ formed by the $\alpha$ such that both

Now, if $A$ is compact (resp. open and relatively compact) in $G$, then the mapping $\alpha \mapsto \alpha(A)$ of $\mathscr{M}_+(G)$ into $\mathbf{R}$ is upper (resp. lower) semi-continuous for the vague topology (Ch. IV, §4, No. 4, Cor. 3 of Prop. 5 and loc. cit., §1, No. 1, Prop. 4); we thus see that $N_U$ is a closed subset of $\Gamma^0$. Moreover, let $\varphi \in \mathscr{K}_+(G)$ be a function such that $\varphi(e) = 1$ and $\varphi(x) = 0$ on $G - U$; it is clear that $\int \varphi(x) d\alpha(x) = 1$ for all $\alpha \in N_U$; Prop. 2 of No. 1 therefore shows that $N_U$ is a compact set, which proves a). On the other hand let $V$ be a relatively compact open neighborhood of $e$ in $G$ such that $\overline{V} \subset U$, and let $\varphi \in \mathscr{K}_+(G)$, with support contained in $U$ and such that $\varphi(x) = 1$ on $V$. Then $\alpha(\varphi) = 1$ for $\alpha \in N_U$, therefore there exists a neighborhood $W$ of $\alpha$ in $N$ such that $\beta(\varphi) < 2$ for $\beta \in W$; it is then clear that $W \subset N_V$, therefore $N_V$ is a neighborhood of $N_U$. Since the $N_U$ cover $N$, this proves b). Finally, every compact subset $M$ of $N$ is contained in a finite union of sets $N_{U_i}$ ($1 \leq i \leq n$), and since $\bigcup_i N_{U_i} \subset N_U$, where $U = \bigcap_i U_i$, this proves c).