For the rest of the section, we denote by $\Gamma$ the set of positive measures $\neq 0$ on G satisfying the conditions of Lemma 1, and for every $\alpha \in \Gamma$ we denote by $H_\alpha$ the closed subgroup of G that is the support of $\alpha$.

Replacing if necessary $\Phi$ by a finer filter, we can suppose that $s$ is a limit of $\xi \mapsto t_\xi$ with respect to $\Phi$; by Lemma 1, $\delta(t_\xi)\alpha(\xi) = \alpha(\xi)$ for every $\xi \in E$, and the conclusion follows from the continuity of the mapping $(u, \lambda) \mapsto \delta(u)\lambda$ on $G \times \mathscr{M}_+(G)$ (§ 3, No. 3, Prop. 13).

To prove Prop. 1 it suffices, by Lemma 1, to show that if a filter $\Psi$ on $\Gamma$ converges vaguely to a measure $\alpha_0 \neq 0$ and if $s$ belongs to the support of $\alpha_0$, then $\delta(s)\alpha_0 = \alpha_0$. Now, for every neighborhood $V$ of $s$ in $G$, there exists an $M \in \Psi$ such that, for every $\alpha \in M$, one has $V \cap H_\alpha \neq \varnothing$, by Lemma 2. For every neighborhood $V$ of $s$ and every $\alpha \in \Gamma$, let $t_{V,\alpha}$ be a point of $V \cap H_\alpha$ if $V \cap H_\alpha \neq \varnothing$, and any point of $H_\alpha$ in the contrary case; if $\Theta$ is the section filter of the filter of neighborhoods of $s$, and $\Phi$ is the product filter $\Theta \times \Psi$, then $s$ is, by the foregoing, a cluster point of $(V, \alpha) \mapsto t_{V,\alpha}$ with respect to $\Phi$. Since, on the other hand, the mapping $(V, \alpha) \mapsto \alpha$ has $\alpha_0$ as limit with respect to $\Phi$, the proposition follows from Lemma 3.

The set $\Gamma_\varphi$ is the intersection of $\Gamma$ with the hyperplane of $\mathscr{M}(G)$ formed by the $\alpha$ such that $\int \varphi(x)\,d\alpha(x) = 1$; since this hyperplane is vaguely closed in $\mathscr{M}(G)$ and does not contain 0, it follows from Prop. 1 that $\Gamma_\varphi$ is vaguely closed in $\mathscr{M}(G)$. It therefore suffices to show that for every compact subset $K$ of $G$, one has $\sup_{\alpha \in \Gamma_\varphi} \alpha(K) < +\infty$ (Ch. III, §1, No. 9, Prop. 15). Now, let $U$ be the open neighborhood of $e$ in $G$ defined by the inequality $\varphi(x) > \varphi(e)/2$; since $1 = \int \varphi(x)\,d\alpha(x) \geq \int_U \varphi(x)\,d\alpha(x)$ for $\alpha \in \Gamma_\varphi$, one sees that, on setting $c = 2/\varphi(e)$, one has $\alpha(U) \leq c$ for every $\alpha \in \Gamma_\varphi$. Let $V$ be a symmetric open neighborhood of $e$ in $G$ such that $V^2 \subset U$; let us show that $\alpha(Vx) \leq c$ for every $x \in G$ and every $\alpha \in \Gamma_\varphi$. Indeed, this relation is trivial if $Vx$ does not intersect the support $H_\alpha$ of $\alpha$; if, on the contrary, there exists an $h \in Vx \cap H_\alpha$, then $h = vx$ for some $v \in V$, whence

and since $\delta(h)\alpha = \alpha$, it follows that $\alpha(Vx) \leq \alpha(Uh) = \alpha(U) \leq c$. Now let $(x_i)_{1 \leq i \leq n}$ be a sequence of points of $K$ such that the $Vx_i$ form a covering of $K$; it follows from the foregoing that $\alpha(K) \leq \sum_{i=1}^n \alpha(Vx_i) \leq nc$ for every $\alpha \in \Gamma_\varphi$; Q.E.D.

Since the mapping $\alpha \mapsto \langle \varphi, \alpha \rangle$ is vaguely continuous, it suffices to observe that $\langle \varphi, \alpha \rangle \neq 0$ for every measure $\alpha \in \Gamma$, since $e$ belongs to the support $H_\alpha$ of $\alpha$ and $\varphi(e) > 0$.

For every $\alpha \in \Gamma^0$ and every function $f \in \mathscr{K}(G)$, set

where the convolution product is taken relative to the right Haar measure $\mu$ and where one makes use of the fact that $\dot{\alpha} = \alpha$ (§ 4, No. 4, formula (11)). We know (Ch. VII, §2, No. 1, Prop. 2) that the mapping $f \mapsto f_\alpha$ of $\mathscr{K}_+(G)$ into $\mathscr{K}_+(Q_\alpha)$ is surjective; therefore, by (2),

where one has set

To establish the proposition, it will suffice to show that, given $f \in \mathscr{H}_+(G)$, the mapping $\alpha \mapsto \|f_\alpha\|$ is vaguely continuous. Now, let K be the support of $f$; the function $f * \alpha$ has its support contained in $KH_\alpha$ and is invariant on the right under $H_\alpha$; consequently

The conclusion therefore follows from the fact that the mapping $\alpha \mapsto f * \alpha$ of $\mathscr{M}_+(G)$ equipped with the vague topology, into $\mathscr{C}(G)$ equipped with the topology of compact convergence, is continuous (§ 4, No. 2, Remark 1).

Recall that if, for a measure $\alpha \in \Gamma^0$, $\|\mu_\alpha\|$ is finite, then G is necessarily unimodular (Ch. VII, §2, No. 6, Cor. 3 of Th. 3).

For every measure $\alpha \in \Gamma^0(G)$, recall (Ch. VII, §2, No. 3, Prop. 5) that the function

is defined $\mu_\alpha$-almost everywhere on $Q_\alpha$, is $\mu_\alpha$-integrable, and

In view of Prop. 4, it suffices to prove that, in $\Gamma^0(g)$, $\alpha \mapsto \|\mu_\alpha\|$ is upper semi-continuous. Fix a measure $\alpha \in \Gamma^0(g)$, and let K be a compact subset of G. There exists on $Q_\alpha$ a continuous function with compact support, taking its values in $[0, 1]$, equal to 1 on the compact set $\pi_\alpha(K)$; since the mapping $f \mapsto f_\alpha$ of $\mathscr{H}_+(G)$ into $\mathscr{H}_+(Q_\alpha)$ is surjective (Ch. VII, §2, No. 1, Prop. 2), one sees that there exists a function $f \in \mathscr{H}_+(G)$ such that

Since $\beta \mapsto f * \beta$ is a continuous mapping of $\mathscr{M}_+(G)$, equipped with the vague topology, into $\mathscr{C}(G)$ equipped with the topology of compact convergence (§ 4, No. 2, Remark 1), one sees that for every $\varepsilon > 0$, the set $U_\varepsilon$ of $\beta \in \Gamma^0(G)$ such that

is an open neighborhood of $\alpha$ in $\Gamma^0(g)$; for every $\beta \in U_\varepsilon$, we then have, by virtue of the formula (2), Given a number $\varepsilon > 0$, let us choose a function $h \in \mathscr{K}_+(G)$ such that $\int_G |g(x) - h(x)|\, d\mu(x) \leq \varepsilon$, and let us take $K = \mathrm{Supp}(h)$ in the foregoing. For every $\beta \in \Gamma^0(g)$, by hypothesis $g_\beta(\dot{x}) \geq 1$ almost everywhere (for $\mu_\beta$) in $Q_\beta$, therefore by virtue of (4); since $h$ is zero outside $K$, and $a fortiori$ outside $KH_\beta$, it follows that combining this result with (5), one sees that when $\beta \in U_\varepsilon$, which completes the proof.

For, let $g \in \mathscr{K}_+(G)$ be a function such that $g(x) \geq 1/c$ for $x \in KV$. For every $x \in K$,

for $\alpha$ satisfying the conditions of the statement; since, moreover, $\pi_\alpha(K) = Q_\alpha$, one has $\alpha \in \Gamma^0(g)$, whence the corollary.

For $a \in A$ and $\alpha \in N_A$,

and since $\pi_\alpha(A) = Q_\alpha$, one has $N_A \subset \Gamma^0(\varphi_A)$, and the corollary therefore follows from Prop. 5.

(i) The set $\Sigma^0$ of unimodular closed subgroups of G is closed in $\Sigma$ (hence is compact).

(ii) If G is generated by a compact neighborhood of e, then the set $\Sigma_c^0$ of unimodular closed subgroups H of G such that the quotient space $G/H$ is compact, is open in $\Sigma^0$ (hence is locally compact).

(iii) For every relatively compact open neighborhood U of e in G, the set $D_U$ of discrete subgroups H of G such that $H \cap U = \{e\}$ is closed in $\Sigma^0$ (hence is compact).

It follows from Prop. 3 of No. 1 that $\Sigma$ is homeomorphic to $\Gamma_\varphi$, hence is compact by Prop. 2 of No. 1. Moreover, it was noted at the beginning of No. 2 that the set $\Gamma^0$ of measures $\alpha \in \Gamma$ such that $H_\alpha$ is unimodular is closed in $\Gamma$; since $\Gamma^0$ is stable under the homotheties with ratio $> 0$, the image $\Sigma^0$ of $\Gamma^0$ in $\Sigma$ is a closed subset of $\Sigma$, which proves (i).

Property (ii) will be a consequence of the following proposition:

With the notations of Prop. 5 of No. 2, we have, for $g \in \mathscr{K}_+(G)$,

For, if K is the support of $g$, the relation $\int g(xs)\, d\alpha(s) \geq 1$ for all $x \in G$ implies $KH_\alpha = G$, the integral obviously being zero on the complement of $KH_\alpha$, therefore $G/H_\alpha = \pi_\alpha(K)$ is compact. Given a measure $\alpha \in \Gamma_c^0$, it will therefore suffice to define a function $g \in \mathscr{K}_+(G)$ such that $\Gamma^0(g)$ is a neighborhood of $\alpha$ in $\Gamma^0$. Since $G/H_\alpha$ is compact and the canonical mapping $f \mapsto f_\alpha$ of $\mathscr{K}_+(G)$ into $\mathscr{K}_+(G/H_\alpha)$ is surjective (Ch. VII, §2, No. 2), there exists a function $g \in \mathscr{K}_+(G)$ such that $\int g(xs)\, d\alpha(s) = 2$ for every $x \in G$. Let K be the (compact) support of $g$, L a symmetric compact neighborhood of e in G that generates G; the mapping $\beta \mapsto g * \beta$ of $\mathscr{M}_+(G)$ into $\mathscr{C}(G)$ being vaguely continuous (§ 4, No. 2, Remark 1), there exists a neighborhood W of $\alpha$ in $\Gamma^0$ such that

for all $\beta \in W$ and $x \in LK$. The first member of (7) being equal to zero outside $KH_\beta$, the relation $\beta \in W$ implies

from which one deduces, by induction on $n$, that $L^n K \subset KH_\beta$ for every integer $n > 0$; since L generates G, we therefore have $G = KH_\beta$ for every measure $\beta \in W$, which proves that $W \subset \Gamma_c^0$. On the other hand, the first member of (7) being invariant on the right under $H_\beta$, the inequality (7) is also valid for $x \in LKH_\beta = G$; therefore $W \subset \Gamma^0(g)$, which proves the proposition.

Finally, (iii) will be a consequence of the following proposition:

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).

We have $D_c = D \cap \Sigma_c^0$, therefore (iii) is a consequence of (i), and Th. 1 (ii) of No. 3.

With the notations of No. 3, Prop. 7, it suffices, for proving (i) and (ii), to prove that:

Now, if $\Gamma_d$ is the set of Haar measures on the discrete subgroups of $G$, then $D$ is canonically homeomorphic to the space of orbits of the group of homotheties in $\Gamma_d$ with ratio $> 0$ (GT, I, §5, No. 2, Prop. 4). It therefore suffices to prove that the canonical mapping $\alpha \mapsto (\alpha(\{e\}), \alpha/\alpha(\{e\}))$ of $\Gamma_d$ onto $\mathbf{R}_+^* \times \mathbf{N}$ is a homeomorphism, which will result from the following lemma:

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).*

(i) There exists an open neighborhood $U$ of $e$ in $G$ such that $H_\alpha \cap U = \{e\}$ for all $\alpha \in A$.

(ii) There exists a constant $k$ such that $\mu_\alpha(G/H_\alpha) \leq k$ for all $\alpha \in A$.

If $A \subset N_c$ is relatively compact in $N_c$, it is a fortiori so in $N$, and the necessity of the conditions (i) and (ii) therefore follows from No. 3, Props. 6 and 7 (without assuming $G$ to be abelian). Conversely, suppose that $A \subset N_c$ satisfies these conditions; if $\overline{A}$ is the closure of $A$ in $N$, then $\overline{A}$ is compact by virtue of No. 3, Prop. 7; moreover, since $\alpha \mapsto \| \mu_\alpha \|$ is lower semi-continuous on $\Gamma^0$ for the vague topology (No. 2, Prop. 4), the condition (ii) implies that one also has $\| \mu_\alpha \| \leq k$ for all $\alpha \in \overline{A}$. Now, since $G$ is abelian, $\mu_\alpha = \mu/\alpha$ is a Haar measure on the group $G/H_\alpha$, and $G/H_\alpha$ is therefore compact for every $\alpha \in \overline{A}$ (Ch. VII, §1, No. 2, Prop. 2). This means that $\overline{A} \subset N_c$, thus $A$ is relatively compact in $N_c$.

Let $D_c$ be the set of discrete subgroups $H$ of $G$ such that $G/H$ is compact, and, for every $H \in D_c$, let $v(H)$ be the total mass $\mu_\alpha(G/H)$, where $\mu_\alpha$ is the quotient measure of $\mu$ by the normalized Haar measure $\alpha$ of $H$. For a subset $A$ of the space $D_c$ to be relatively compact in $D_c$, it is necessary and sufficient that it satisfy the following two conditions:

(i) There exists an open neighborhood $U$ of $e$ in $G$ such that $H \cap U = \{e\}$ for all $H \in A$.

(ii) There exists a constant $k$ such that $v(H) \leq k$ for all $H \in A$.

Taking into account Prop. 9, this follows at once from the fact that $D_c$ is the image of $N_c$ under the canonical bijection of $N$ onto $D$, and the fact that, under the hypotheses made, this bijection is a homeomorphism (No. 4, Prop. 8).