Theorem 2
Tag 09RJ, permanent. Cite it as [Bourbaki, Tag 09RJ].
Integration, chapter IX, § 5. Measures on completely regular spaces
— Let T be a locally compact space, or a Polish space, and let H be a relatively compact subset of ; then H satisfies Prokhorov’s condition.
We may restrict ourselves to the case that H is closed, hence compact. The total masses of the measures are obviously bounded, because the mapping is continuous, and everything comes down to proving the assertion b) of Def. 2.
Suppose first that T is locally compact. Let be a number > 0. Let us associate to every measure a compact set in T such that , then a relatively compact open neighborhood of . The function being upper semi-continuous on (No. 3, Prop. 6), the set of measures such that is a neighborhood of in H. Therefore there exists a finite subset of H such that the sets () cover H. Denoting by K the compact set , we have for all .
Suppose next that T is Polish. We do not restrict the generality by assuming that T is the intersection of a decreasing sequence of open subsets of a compact space X (GT, IX, §6, No. 1, Cor. 1 of Th. 1). Let be the injection of T into , and let be the set of measures of the form for ; since is compact in , it follows that there exists a compact set such that for every measure , by the preceding result applied to the locally compact space . Therefore also , and finally for every measure . Now set ; the set K is compact and is contained in T, and, for every measure , we have . Prokhorov’s condition is thus verified.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.