Bourbaki

Proposition 7

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

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

We retain the notations of Prop. 6. If each μα\mu_\alpha is a left (resp. right) Haar measure on GαG_\alpha, then μ\mu is a left (resp. right) Haar measure on GG.

Suppose, for example, that the μα\mu_\alpha are left Haar measures. Let s∈Gs \in G. For every x∈Gx \in G,

(φα∘γ(s))(x)=φα(sx)=φα(s)φα(x)=(γ(φα(s))∘φα)(x); (\varphi_\alpha \circ \gamma(s))(x) = \varphi_\alpha(sx) = \varphi_\alpha(s)\varphi_\alpha(x) = (\gamma(\varphi_\alpha(s)) \circ \varphi_\alpha)(x);

therefore φα(γ(s)μ)=γ(φα(s))μα=μα\varphi_\alpha(\gamma(s)\mu) = \gamma(\varphi_\alpha(s))\mu_\alpha = \mu_\alpha. Therefore γ(s)μ=μ\gamma(s)\mu = \mu by Lemma 2 b), thus μ\mu is a left Haar measure.

We suppose henceforth the KαK_\alpha to be not only compact, but open in GG. The GαG_\alpha are then discrete and, for β≥α\beta \geq \alpha, Kα/KβK_\alpha/K_\beta is a compact and discrete group, hence is finite. The group GG is unimodular (Prop. 3).

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.