Bourbaki

Proposition 9

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

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

Let G be a locally compact group, V an open subset of G, and μ\mu a nonzero positive measure on V having the following property: if U is an open subset of V and if s∈Gs \in G is such that sU⊂VsU \subset V, then the image of the measure μU\mu_U induced by μ\mu on U, under the homeomorphism x↦sxx \mapsto sx of U onto sU, is μsU\mu_{sU}. Then, there exists one and only one left Haar measure α\alpha on G that induces μ\mu on V.

For every s∈Gs \in G, let μs\mu_s be the image of μ\mu under the homeomorphism x↦sxx \mapsto sx of V onto sV. The restriction of μs\mu_s to V∩sVV \cap sV is the image of μs−1V∩V\mu_{s^{-1}V \cap V} under the restriction of x↦sxx \mapsto sx to s−1V∩Vs^{-1}V \cap V; by hypothesis, this image is μV∩sV\mu_{V \cap sV}. By translation, one concludes from this that μs\mu_s and μt\mu_t have the same restriction to sV∩tVsV \cap tV for any s,ts, t. Therefore, by Prop. 1 of Ch. III, §2, No. 1, there exists a measure α\alpha on G that induces μs\mu_s on sV for every ss. It is clear that α\alpha is the unique left Haar measure on G inducing μ\mu on V.

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.