Bourbaki

Proposition 7

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

Integration, chapter V, § 1. Essential upper integral

— Let ff be an element of F+(T)\mathscr{F}_+(T). 1) If the function ff is not μ\mu-moderated, then μ∗(f)=+∞\mu^*(f) = +\infty. 2) If the function ff is μ\mu-moderated, then μ∗(f)=μ∗(f)\mu^*(f) = \mu^*(f). 3) If μ∗(f)<+∞\mu^*(f) < +\infty then there exists a μ\mu-moderated subset AA, the union of a sequence of compact subsets of TT, such that f=fφAf = f \varphi_A locally almost everywhere.

The first assertion follows immediately from Lemma 1 of Ch. IV, §5, No. 6. To establish the second, denote by AA a moderated subset such that f is zero outside A; A is the union of a negligible set A0A_0 and a sequence (An)n≥1(A_n)_{n \geq 1} of compact sets, which we may suppose to be increasing. The function ff is then almost everywhere equal to the upper envelope of the functions fφAnf \varphi_{A_n} (n≥1n \geq 1), therefore (Ch. IV, §1, No. 3, Th. 3 and §2, No. 3, Prop. 6)

μ∗(f)=lim⁡n→∞μ∗(fφAn)≤μ∙(f), \mu^*(f) = \lim_{n \to \infty} \mu^*(f \varphi_{A_n}) \leq \mu^\bullet(f),

whence the equality μ∗(f)=μ∙(f)\mu^*(f) = \mu^\bullet(f) by virtue of the formula (1). Finally, suppose that μ∙(f)<+∞\mu^\bullet(f) < +\infty; there exists an increasing sequence (An)(A_n) of compact sets such that

μ∙(f)=sup⁡nμ∗(fφAn). \mu^\bullet(f) = \sup_n \mu^*(f \varphi_{A_n}).

Set A=⋃nAnA = \bigcup_n A_n; the second member is equal to μ∗(fφA)\mu^*(f \varphi_A) (Ch. IV, §1, No. 3, Th. 3), that is, to μ∙(fφA)\mu^\bullet(f \varphi_A) (by Prop. 1, or by 2) above). Since μ∙(f)=μ∙(fφA)+μ∙(fφCA)\mu^\bullet(f) = \mu^\bullet(f \varphi_A) + \mu^\bullet(f \varphi_{\mathbf{C}_A}) (Prop. 2), we have μ∙(fφCA)=0\mu^\bullet(f \varphi_{\mathbf{C}_A}) = 0, from which 3) follows.

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.