Bourbaki

Proposition 4

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

Integration, chapter VII, § 1. Construction of a Haar measure

*Let G be a locally compact group, Γ\Gamma a topological group, and γ↦uγ\gamma \mapsto u_\gamma a homomorphism of Γ\Gamma into the group G\mathscr{G} of automorphism of G, such that (γ,x)↦uγ(x)(\gamma, x) \mapsto u_\gamma(x) is a continuous mapping of Γ×G\Gamma \times G into G. Then, the mapping γ↦mod⁡(uγ)\gamma \mapsto \operatorname{mod}(u_\gamma) is a continuous representation of Γ\Gamma in R+∗\mathbf{R}_+^*.

This mapping is obviously a representation (algebraic) of Γ\Gamma in R+∗\mathbf{R}_+^*; it will suffice to prove its continuity. Let f∈K(G)f \in \mathscr{K}(G) and let S be its support.*

Let γ0∈Γ\gamma_0 \in \Gamma and let UU be a relatively compact neighborhood of uγ0−1(S)u_{\gamma_0}^{-1}(S). The mapping γ↦uγ\gamma \mapsto u_\gamma is a continuous mapping of Γ\Gamma into G\mathscr{G} equipped with the topology of compact convergence (GT, X, §3, No. 4, Th. 3); therefore uγ−1(S)⊂Uu_\gamma^{-1}(S) \subset U for γ\gamma sufficiently near γ0\gamma_0. Lemma 1 of No. 1 then proves that ∫f(uγ(x))dμ(x)\int f(u_\gamma(x)) d\mu(x) (where μ\mu denotes a left Haar measure of GG) depends continuously on γ\gamma; whence the proposition.

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

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Cites

  • X, §3, No. 4, Th. 3 not yet on this site
  • Lemma 1
Transcribed from the printed volume. Not translated.