Bourbaki

Proposition 5

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

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

a) Let H be an increasing directed set of functions ⩾0\geqslant 0 that are lower semi-continuous on every compact subset of T. Then

μ∙(sup⁡h∈Hh)=sup⁡h∈Hμ∙(h). \mu^\bullet \left( \sup_{h \in H} h \right) = \sup_{h \in H} \mu^\bullet(h).

b) Let H be a decreasing directed set of functions ⩾0\geqslant 0 that are upper semi-continuous on every compact subset of T. If there exists in H a function h0h_0 such that μ∙(h0)<+∞\mu^\bullet(h_0) < +\infty, then

μ∙(inf⁡h∈Hh)=inf⁡h∈Hμ∙(h). \mu^\bullet \left( \inf_{h \in H} h \right) = \inf_{h \in H} \mu^\bullet(h).

For every compact set K⊂TK \subset T, we have in case a)

μ∙(sup⁡h∈HhφK)=μK∙(sup⁡h∈HhK)=sup⁡h∈HμK∙(hK)=sup⁡h∈Hμ∙(hφK), \mu^\bullet \left( \sup_{h \in H} h \varphi_K \right) = \mu_K^\bullet \left( \sup_{h \in H} h_K \right) = \sup_{h \in H} \mu_K^\bullet(h_K) = \sup_{h \in H} \mu^\bullet(h \varphi_K),

and in case b)

μ∙(inf⁡h∈HhφK)=μK∙(inf⁡h∈HhK)=inf⁡h∈HμK∙(hK)=inf⁡h∈Hμ∙(hφK), \mu^\bullet \left( \inf_{h \in H} h \varphi_K \right) = \mu_K^\bullet \left( \inf_{h \in H} h_K \right) = \inf_{h \in H} \mu_K^\bullet(h_K) = \inf_{h \in H} \mu^\bullet(h \varphi_K),

by Prop. 2 of No. 2, and Prop. 8 of Ch. V, §1, No. 2. Case a) follows at once, by passage to the supremum with respect to K (No. 2, Prop. 2). To treat case b), denote by ε\varepsilon a number >0> 0, and choose a compact set K such that μ∙(h0φK)⩾μ∙(h0)−ε\mu^\bullet(h_0 \varphi_K) \geqslant \mu^\bullet(h_0) - \varepsilon. We then have (No. 5, Prop. 4) μ∙(h0φCK)⩽ε\mu^\bullet(h_0 \varphi_{C_K}) \leqslant \varepsilon; for every function h∈Hh \in H that is ⩽h0\leqslant h_0, we therefore have μ∙(hφCK)⩽ε\mu^\bullet(h \varphi_{C_K}) \leqslant \varepsilon, and finally μ∙(hφK)⩾μ∙(h)−ε\mu^\bullet(h \varphi_K) \geqslant \mu^\bullet(h) - \varepsilon by Prop. 4 of No. 5. Therefore

μ∙(inf⁡h∈Hh)⩾μ∙(inf⁡h∈HhφK)=inf⁡h∈H,h⩽h0μ∙(hφK)⩾inf⁡h∈H,h⩽h0μ∙(h)−ε. \mu^\bullet \left( \inf_{h \in H} h \right) \geqslant \mu^\bullet \left( \inf_{h \in H} h \varphi_K \right) = \inf_{h \in H, h \leqslant h_0} \mu^\bullet(h \varphi_K) \geqslant \inf_{h \in H, h \leqslant h_0} \mu^\bullet(h) - \varepsilon.

Consequently the left side of (6) is ⩾\geqslant the right side; the reverse inequality being obvious, the proposition is established.

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.