Bourbaki

Theorem 1

Tag 08P7, permanent. Cite it as [Bourbaki, Tag 08P7].

Integration, chapter IV, § 1. Upper integral of a positive function

Let HH be a nonempty set of functions in J+\mathscr{J}_+, directed for the relation ≤\leq. For every positive measure μ\mu on XX,

μ∗(sup⁡g∈Hg)=sup⁡g∈Hμ∗(g)=lim⁡g∈Hμ∗(g). \mu^*\left( \sup_{g \in H} g \right) = \sup_{g \in H} \mu^*(g) = \lim_{g \in H} \mu^*(g).

Let f=sup⁡g∈Hgf = \sup_{g \in H} g. We shall first prove the theorem for the special case that the functions g∈Hg \in H and their upper envelope ff belong to H+\mathscr{H}_+. It then follows from Dini’s theorem (GT, X, §4, No. 1, Th. 1) that the section filter of HH converges uniformly to ff on every compact subset of XX, and in particular on the support K of ff. Since 0≤g≤f0 \leq g \leq f for every function g∈Hg \in \mathbf{H}, the support of every function in H\mathbf{H} is contained in KK; but by definition μ\mu is continuous on the vector space K(X,K;C)\mathscr{K}(X, K; \mathbf{C}) of continuous functions with support contained in KK, for the topology of uniform convergence; whence the relation (2) in this case.

Let us pass to the general case. It is clear that μ∗(g)≤μ∗(f)\mu^*(g) \leq \mu^*(f) for every function g∈Hg \in \mathbf{H}. By Def. 1, it all comes down to showing that, for every function ψ∈K+\psi \in \mathscr{K}_+ such that ψ≤f\psi \leq f,

μ(ψ)≤sup⁡g∈Hμ∗(g). \mu(\psi) \leq \sup_{g \in \mathbf{H}} \mu^*(g).

For every function g∈Hg \in \mathbf{H}, let Φg\Phi_g be the set of functions φ∈K+\varphi \in \mathscr{K}_+ such that φ≤g\varphi \leq g, and let Φ\Phi be the union of the sets Φg\Phi_g as gg runs over H\mathbf{H}; since H\mathbf{H} is directed, so is Φ\Phi, and f=sup⁡φ∈Φφf = \sup_{\varphi \in \Phi} \varphi. Since ψ≤f\psi \leq f, ψ\psi is the upper envelope of the set of functions inf⁡(ψ,φ)\inf(\psi, \varphi) as φ\varphi runs over Φ\Phi; but since ψ\psi and the functions inf⁡(ψ,φ)\inf(\psi, \varphi) belong to K+\mathscr{K}_+, the first part of the proof shows that μ(ψ)=sup⁡φ∈Φμ(inf⁡(ψ,φ))\mu(\psi) = \sup_{\varphi \in \Phi} \mu(\inf(\psi, \varphi)). Now, each φ∈Φ\varphi \in \Phi belongs to a set Φg\Phi_g, therefore

μ(inf⁡(ψ,φ))≤μ(φ)≤μ∗(g)≤sup⁡g∈Hμ∗(g), \mu(\inf(\psi, \varphi)) \leq \mu(\varphi) \leq \mu^*(g) \leq \sup_{g \in \mathbf{H}} \mu^*(g),

from which it follows at once that μ(ψ)≤sup⁡g∈Hμ∗(g)\mu(\psi) \leq \sup_{g \in \mathbf{H}} \mu^*(g). We have thus proved that μ∗(f)=sup⁡g∈Hμ∗(g)\mu^*(f) = \sup_{g \in \mathbf{H}} \mu^*(g); the relation μ∗(f)=lim⁡g∈Hμ∗(g)\mu^*(f) = \lim_{g \in \mathbf{H}} \mu^*(g) is then a consequence of the monotone limit theorem (GT, IV, §5, No. 2, Th. 2).

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.