Bourbaki

Proposition 3

Tag 096G, permanent. Cite it as [Bourbaki, Tag 096G].

Integration, chapter V, § 7. Integration with respect to an induced measure

— Let X be a locally compact subspace of T, and λ\lambda a complex measure on X. The following properties are equivalent: a) the canonical injection i:X→Ti : X \to T is λ\lambda-proper; b) for every compact subset K of T, K∩XK \cap X is essentially λ\lambda-integrable; c) every point t∈Tt \in T admits a neighborhood V such that V∩XV \cap X is essentially λ\lambda-integrable; d) there exists a measure θ\theta on T such that θX=λ\theta_X = \lambda. If these equivalent conditions are satisfied, we have, with notations as in d),

(i(λ))X=λandi(λ)=i(θX)=φX⋅θ. (i(\lambda))_X = \lambda \quad \text{and} \quad i(\lambda) = i(\theta_X) = \varphi_X \cdot \theta .

The injection ii being continuous, the equivalence of the properties a), b) and c) follows from Prop. 1 of §6, and the remark that follows it, applied to the positive measure ∣λ∣|\lambda|. If λ\lambda is induced on X by a measure θ\theta on T, then ∣λ∣=∣θ∣X|\lambda| = |\theta|_X (formula (6)), consequently ∣λ∣(K∩X)=∣θ∣(K∩X)≤∣θ∣(K)<+∞|\lambda|(K \cap X) = |\theta|(K \cap X) \leq |\theta|(K) < +\infty (Prop. 1) for every compact subset K of T, so that d) implies b). Suppose that a) is satisfied, and let us show that (i(λ))X=λ(i(\lambda))_X = \lambda, which will imply d). Let g be an element of H(X;C)\mathscr{H}(X; \mathbf{C}); denoting by g′g' the extension by 0 of g to T we have, by the definition of induced measure and then by Prop. 7 of §6, No. 4,

∫g d(i(λ))X=∫g′ d(i(λ))=∫(g′∘i) dλ=∫g dλ. \int g \, d(i(\lambda))_X = \int g' \, d(i(\lambda)) = \int (g' \circ i) \, d\lambda = \int g \, d\lambda .

This completes the proof of the equivalence of the four properties. If λ=θX\lambda = \theta_X and g∈H(T;C)g \in \mathscr{H}(T; \mathbf{C}), then

∫g d(i(θX))=∫(g∘i) d(θX)=∫gφX dθ, \int g \, d(i(\theta_X)) = \int (g \circ i) \, d(\theta_X) = \int g \varphi_X \, d\theta ,

because gφXg \varphi_X is the extension by 0 of g∘ig \circ i to T. This proves the second formula of (7).

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

Cited by

Cites

Transcribed from the printed volume. Not translated.