Bourbaki

Proposition 5

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

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

— Let (Gι)ι∈I(G_\iota)_{\iota \in I} be a family of locally compact groups. For every ι∈I\iota \in I let μι\mu_\iota be a left (resp. right) Haar measure on GιG_\iota. Assume that there exists a finite subset JJ of II such that, for every ι∈I−J\iota \in I - J, GιG_\iota is compact and μι(Gι)=1\mu_\iota(G_\iota) = 1. Then the product measure ⨂ι∈Iμι\bigotimes_{\iota \in I} \mu_\iota is a left (resp. right) Haar measure on G=∏ι∈IGιG = \prod_{\iota \in I} G_\iota. If x=(xι)∈Gx = (x_\iota) \in G then

ΔG(x)=∏ι∈IΔGι(xι). \Delta_G(x) = \prod_{\iota \in I} \Delta_{G_\iota}(x_\iota).
For every finite subset JJ of II, ⨂ι∈Jμι\bigotimes_{\iota \in J} \mu_\iota is a left (resp. right) Haar measure on ∏ι∈JGι\prod_{\iota \in J} G_\iota, as follows immediately from the definitions. Therefore ⨂ι∈Iμι\bigotimes_{\iota \in I} \mu_\iota is a left (resp. right) Haar measure on GG (Ch. III, §4, No. 6, Prop. 9). On the other hand, if the μι\mu_\iota are left Haar measures then
δ(x)(⨂ι∈Iμι)=⨂ι∈Iδ(xι)μι=⨂ι∈I(ΔGι(xι)μι)=(∏ι∈IΔGι(xι))⨂ι∈Iμι, \delta(x) \left( \bigotimes_{\iota \in I} \mu_\iota \right) = \bigotimes_{\iota \in I} \delta(x_\iota) \mu_\iota = \bigotimes_{\iota \in I} (\Delta_{G_\iota}(x_\iota) \mu_\iota) = \left( \prod_{\iota \in I} \Delta_{G_\iota}(x_\iota) \right) \bigotimes_{\iota \in I} \mu_\iota,
whence ΔG(x)=∏ι∈IΔGι(xι)\Delta_G(x) = \prod_{\iota \in I} \Delta_{G_\iota}(x_\iota).

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

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Cites

Transcribed from the printed volume. Not translated.