Bourbaki

Proposition 4

Tag 09M4, permanent. Cite it as [Bourbaki, Tag 09M4].

Integration, chapter VIII, § 5. The space of closed subgroups

— Let Γ0\Gamma^0 be the set of measures α∈Γ\alpha \in \Gamma such that HαH_\alpha is unimodular, and for every α∈Γ0\alpha \in \Gamma^0 set μα=μ/α\mu_\alpha = \mu / \alpha; then the mapping α↦∥μα∥\alpha \mapsto \| \mu_\alpha \| of Γ0\Gamma^0 into R‾\overline{\mathbf{R}} is lower semi-continuous for the vague topology.

For every α∈Γ0\alpha \in \Gamma^0 and every function f∈K(G)f \in \mathscr{K}(G), set

fα(x˙)=∫Hαf(xs)dα(s)=(f∗α)(x), f_\alpha(\dot{x}) = \int_{H_\alpha} f(xs) d\alpha(s) = (f * \alpha)(x) ,

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

∥μα∥=sup⁡f∈K+(G),f≠0μα(fα)/∥fα∥=sup⁡f∈K+(G),f≠0μ(f)/∥fα∥, \| \mu_\alpha \| = \sup_{f \in \mathscr{K}_+(G), f \neq 0} \mu_\alpha(f_\alpha) / \| f_\alpha \| = \sup_{f \in \mathscr{K}_+(G), f \neq 0} \mu(f) / \| f_\alpha \|,

where one has set

∥fα∥=sup⁡x˙∈Qα∣fα(x˙)∣=sup⁡x∈G∣(f∗α)(x)∣. \|f_\alpha\| = \sup_{\dot{x} \in Q_\alpha} |f_\alpha(\dot{x})| = \sup_{x \in G} |(f * \alpha)(x)| .

To establish the proposition, it will suffice to show that, given f∈H+(G)f \in \mathscr{H}_+(G), the mapping α↦∥fα∥\alpha \mapsto \|f_\alpha\| is vaguely continuous. Now, let K be the support of ff; the function f∗αf * \alpha has its support contained in KHαKH_\alpha and is invariant on the right under HαH_\alpha; consequently

∥fα∥=sup⁡x∈K∣(f∗α)(x)∣. \|f_\alpha\| = \sup_{x \in K} |(f * \alpha)(x)| .

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

Recall that if, for a measure α∈Γ0\alpha \in \Gamma^0, ∥μα∥\|\mu_\alpha\| is finite, then G is necessarily unimodular (Ch. VII, §2, No. 6, Cor. 3 of Th. 3).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.