Lemma 4

If the group G satisfies the condition (L), the mapping $\alpha \mapsto \alpha(\{e\})$ of $\Gamma_d$ into $\mathbf{R}_+^*$ is vaguely continuous.

Let us consider a measure $\alpha \in \Gamma_d$; let $V_0$ be a relatively compact open neighborhood of $e$ in $G$ such that $H_\alpha \cap V_0 = \{e\}$ and such that there exists no finite subgroup of $G$ contained in $V_0$ and not reduced to $e$. Let $V$ be a symmetric compact neighborhood of $e$ such that $V^3 \subset V_0$, and let $U$ be a symmetric neighborhood of $e$ such that $U^2 \subset V$. Let $\varphi$ (resp. $\psi$) be a function in $\mathscr{H}_+(G)$, with values in $[0, 1]$, equal to 1 on $V^3$ (resp. at the point $e$) and with support contained in $V_0$ (resp. in $U$). The set of measures $\beta \in \Gamma_d$ such that $|\beta(\varphi) - \alpha(\varphi)| \leq \varepsilon$ and $|\beta(\psi) - \alpha(\psi)| \leq \varepsilon$ is a neighborhood $W$ of $\alpha$. We propose to show that, provided $\varepsilon$ is taken to be sufficiently small, $H_\beta \cap V = \{e\}$ for every $\beta \in W$; it will then follow that $\beta(\psi) = \beta(\{e\})$, hence that $|\beta(\{e\}) - \alpha(\{e\})| \leq \varepsilon$, which will prove the lemma.

It will suffice to show that, for $\beta \in W$,

For, suppose that this point is established: then, for $x$ and $y$ in $V \cap H_\beta$, one has $xy^{-1} \in V^2 \cap H_\beta$; but, by virtue of (8), this implies $xy^{-1} \in V \cap H_\beta$; in other words, $V \cap H_\beta$ is a subgroup of $G$, which is obviously discrete and compact, hence finite; but then, by the choice of $V_0$, this implies that indeed $V \cap H_\beta = \{e\}$.

Let us argue by contradiction and so assume that there exists a point $z$ of $V^2 - V$ that belongs to $H_\beta$; by the choice of $U$ and $V$, we have $\psi(sz^{-1}) + \psi(s) \leq \varphi(s)$ in $G$, the relation $z \notin U^2$ implying $Uz \cap U = \varnothing$. Since

it follows that $2\beta(\psi) \leq \beta(\varphi) \leq \alpha(\varphi) + \varepsilon$; but we also have

and by construction $\alpha(\varphi) = \alpha(\psi) = \alpha(\{e\})$. We thus arrive at a contradiction by taking $\varepsilon < \alpha(\{e\})/3$. Q.E.D.

In graphic terms, a group $G$ satisfying the condition (L) is said to not have arbitrarily small finite subgroups. It can be shown that every Lie group satisfies the condition (L); but this condition is not characteristic of Lie groups; for example, the multiplicative group of $p$-adic integers congruent to 1 mod $p$ satisfies (L).*