Remark 3

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

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

For every Borel subset $A$ of $T$ (in particular for $A = T$ ) and every positive measure $\mu$ on $T$ , the number $\mu^*(A)$ is the supremum of the measures $\mu^*(K)$ of the compact subsets of $A$ . Indeed, for every compact subset $K$ of $A$ , one has $\mu^*(K) \leq \mu^*(A)$ ; on the other hand, if $\mathfrak{K}$ is the set of compact subsets of $T$ , then

\mu^*(A) = \sup_{K \in \mathfrak{K}} \mu_K^*(A \cap K) = \sup_{K \in \mathfrak{K}} \sup_{L \in \mathfrak{K}, L \subset A \cap K} \mu_K^*(L) \leq \sup_{L \in \mathfrak{K}, L \subset A} \mu^*(L)

by Cor. 1 of Th. 4 of Ch. IV, §4, No. 6.

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

Cited by

Theorem 2, § 3
Proposition 3, § 4
Theorem 1 (Prokhorov), § 4

Cites

IV, §4, No. 6