Bourbaki

Lemma 1

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

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

Let α\alpha be a positive measure ≠0\neq 0 on G, S its support; the following two conditions are equivalent: a) S is a closed subgroup of G and the measure induced by α\alpha on S is a right Haar measure on S. b) δ(s)α=α\delta(s)\alpha = \alpha for every s∈Ss \in S. Moreover, when these conditions are satisfied, the set of t∈Gt \in G such that δ(t)α=α\delta(t)\alpha = \alpha is equal to S. It is clear that a) implies b); conversely, the relation b) implies that Sx=SSx = S for every x∈Sx \in S; in other words, the relations x∈Sx \in S and y∈Sy \in S imply that y∈Sxy \in Sx, or again that yx−1∈Syx^{-1} \in S, and since S is nonempty, S is a closed subgroup of G. The set of t∈Gt \in G such that St=SSt = S is then equal to S itself, whence the last assertion.

For the rest of the section, we denote by Γ\Gamma the set of positive measures ≠0\neq 0 on G satisfying the conditions of Lemma 1, and for every α∈Γ\alpha \in \Gamma we denote by HαH_\alpha the closed subgroup of G that is the support of α\alpha.

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.