Lemma 1

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

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

— If the encumbrance $p$ is concentrated on $A \subset T$ , then $p(f) = p(f \varphi_A)$ for every $f \in \mathscr{F}_+(T)$ .

For, set $T - A = B$ , so that $p(\varphi_B) = 0$ ; then

f \varphi_B \leq (+\infty) \cdot \varphi_B = \sup_{n \in \mathbf{N}} n \varphi_B,

therefore $p(f \varphi_B) = 0$ by properties $a), b), d)$ of Def. 1. It follows from c) that $p(f) \leq p(f \varphi_A) + p(f \varphi_B) = p(f \varphi_A)$ , and finally $p(f) = p(f \varphi_A)$ by a).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Proposition 14, § 1

Cites

Def. 1