Bourbaki

Remark 2

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

Integration, chapter IX, § 1. Premeasures and measures on a topological space

Let us show that the measures introduced in Example 2 of No. 3 are measures with compact support in TT. Let μ\mu be a positive measure on TT whose support is a compact set KK, and let ν\nu be the measure defined by μK\mu_K (in the sense of No. 3). Let f∈F+(T)f \in \mathscr{F}_+(T); then

ν∙(f)=μK∙(fK)(No. 3, formula (3)). \nu^\bullet(f) = \mu_K^\bullet(f_K) \quad \text{(No. 3, formula (3))}.

The encumbrance μ∙\mu^\bullet being concentrated on KK, we also have

μ∙(f)=μ∙(fφK)=μ∙((fK)0)=μK∙(fK), \mu^\bullet(f) = \mu^\bullet(f \varphi_K) = \mu^\bullet((f_K)^0) = \mu_K^\bullet(f_K),

whence μ∙=ν∙\mu^\bullet = \nu^\bullet, and finally μ=ν\mu = \nu. Conversely, if KK is a compact set in TT and λ\lambda a measure on KK, and if μ\mu is the measure on TT defined by λ\lambda, then μ∙(CK)=0\mu^\bullet(\mathbf{C}K) = 0 (No. 3, formula (3)); consequently, the support of μ\mu is contained in KK, hence is compact.

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.