Bourbaki

Proposition 4

Tag 09N9, permanent. Cite it as [Bourbaki, Tag 09N9].

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

— Let ff be a positive function and (gn)n≥1(g_n)_{n \geq 1} a sequence of μ\mu-measurable positive functions on T. Setting g=∑n≥1gng = \sum_{n \geq 1} g_n, one has

μ∙(fg)=∑n≥1μ∙(fgn). \mu^\bullet(fg) = \sum_{n \geq 1} \mu^\bullet(fg_n).

Set hn=∑i=1ngih_n = \sum_{i=1}^n g_i for all n≥1n \geq 1. For every compact subset K of T,

μK∙((fhn)K)=∑i=1nμK∙((fgi)K) \mu_K^\bullet((fh_n)_K) = \sum_{i=1}^n \mu_K^\bullet((f g_i)_K)

by Prop. 2 of Ch. V, §1, No. 1 applied to the compact space K. Passing to the limit with respect to the increasing directed set of compact subsets of T, one obtains

μ∙(fhn)=∑i=1nμ∙(fgi). \mu^\bullet(fh_n) = \sum_{i=1}^n \mu^\bullet(f g_i).

Now, fgfg is the limit of the increasing sequence (fhn)n≥1(fh_n)_{n \geq 1}, whence μ∙(fg)=lim⁡n→∞μ∙(fhn)\mu^\bullet(fg) = \lim_{n \to \infty} \mu^\bullet(fh_n); the preceding formula then immediately implies (4).

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.