Bourbaki

Lemma 2

Tag 09M0, permanent. Cite it as [Bourbaki, Tag 09M0].

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

Let X be a locally compact space and for every measure α∈M+(X)−{0}\alpha \in \mathscr{M}_+(X) - \{0\}, let SαS_\alpha be the support of α\alpha. Let Φ\Phi be a filter on M+(X)−{0}\mathscr{M}_+(X) - \{0\} that converges vaguely to a measure α0≠0\alpha_0 \neq 0. Then, for every neighborhood V of a point s of the support of α0\alpha_0, there exists a set M∈ΦM \in \Phi such that, for every α∈M\alpha \in M, one has V∩Sα≠∅V \cap S_\alpha \neq \varnothing. For, if φ∈K+(X)\varphi \in \mathscr{K}_+(X) is a function with support contained in V and such that ∫φ(x)dα0(x)>0\int \varphi(x) d\alpha_0(x) > 0, by definition there exists a set M∈ΦM \in \Phi such that ∫φ(x)dα(x)>0\int \varphi(x) d\alpha(x) > 0 for all α∈M\alpha \in M, which implies V∩Sα≠∅V \cap S_\alpha \neq \varnothing.

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

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Cites

Nothing.

Transcribed from the printed volume. Not translated.