Bourbaki

Proposition 19

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

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

— For every subset A of X, μ∗(A)\mu^*(A) is the infimum of the outer measures of the open sets containing A.

The proposition is obvious if μ∗(A)=+∞\mu^*(A) = +\infty. In the contrary case, for every ε\varepsilon such that 0<ε<10 < \varepsilon < 1, there exists a function f∈J+f \in \mathscr{J}_+ such that φA≤f\varphi_A \leq f and μ∗(A)≤μ∗(f)≤μ∗(A)+ε\mu^*(A) \leq \mu^*(f) \leq \mu^*(A) + \varepsilon. Let G be the set of x∈Xx \in X such that f(x)>1−εf(x) > 1 - \varepsilon. Since ff is lower semi-continuous, G is open (GT, IV, §6, No. 2, Prop. 1) and contains A; on the other hand f≥(1−ε)φGf \geq (1 - \varepsilon)\varphi_G, whence

μ∗(G)≤11−εμ∗(f)≤11−ε(μ∗(A)+ε); \mu^*(G) \leq \frac{1}{1 - \varepsilon} \mu^*(f) \leq \frac{1}{1 - \varepsilon} (\mu^*(A) + \varepsilon);

since ε\varepsilon is arbitrary, we see that μ∗(G)\mu^*(G) differs as little as we please from μ∗(A)\mu^*(A), whence the proposition.

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.