Proposition 6
Tag 08OM, permanent. Cite it as [Bourbaki, Tag 08OM].
Integration, chapter III, § 4. Products of measures
When and are equipped with the topology of strictly compact convergence (§ 1, No. 10), the bilinear mapping of into is hypocontinuous for the set of vaguely bounded subsets of and (TVS, III, § 5, No. 3).
Let be two compact sets, a compact subset of , and a vaguely bounded and closed subset of ; it is known that is vaguely compact (§ 1, No. 9, Prop. 15), hence also compact for the topology of strictly compact convergence (§ 1, No. 10, Prop. 17). On the other hand, the Banach space is isometric to (No. 1, Lemma 1); the mapping of into , such that is the function defined by , is separately continuous by virtue of § 3, No. 4, Props. 8 and 9. Since is barreled, it follows that the mapping is hypocontinuous relative to the vaguely bounded subsets of (TVS, III, § 5, No. 3, Prop. 6); the restriction of this mapping to is therefore continuous (loc. cit., Prop. 4). The image of under this mapping is consequently a compact subset of the Banach space . Now, is none other than the set of functions as runs over and runs over ; by virtue of formula (3) of No. 1, the conditions and therefore imply . In view of the definition of the topology of strictly compact convergence, this proves the proposition (TVS, III, § 5, No. 3, Def. 2).
The conclusion of Prop. 6 is no longer valid when the topology of strictly compact convergence is replaced by the vague topology (Exer. 2 c)). However, if B (resp. C) is a vaguely bounded subset of (resp. ), then the image of under the mapping is vaguely bounded in and therefore the restriction of this mapping to is vaguely continuous, by virtue of Prop. 6, of §1, No. 10, Prop. 17, and of Prop. 4 of TVS, III, §5, No. 3 (cf. Exer. 3).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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