Exercise 14, § 1
Integration, VII, p. 80
Bourbaki marks this one as harder than the rest.
Let be a locally compact space, a uniform structure on for which there exists an entourage such that is compact for all (GT, II, §4, Exer. 9). On the other hand, let be a uniformly equicontinuous group of homeomorphisms of (for the uniform structure ), such that there exists an whose orbit under is dense. Show that there exists on a positive measure invariant under , and that this measure is unique up to a constant factor. One may proceed as follows:
As for the existence of the invariant measure, follow the method of No. 2, Th. 1; one proves first that if is a compact subset of and is an open neighborhood of , there exists a finite number of elements such that .
As for uniqueness, observe first that the entourages of that are invariant under every homeomorphism of , for , form a fundamental system of entourages . For every relatively compact set and every relatively compact set with nonempty interior, let be the smallest number of elements of a covering of by sets of the form , where ; if is a third relatively compact set with nonempty interior, then . Let be a compact subset of , a relatively compact open set, a symmetric open entourage contained in such that , and a symmetric closed entourage contained in . For every symmetric entourage such that and , show that, for every positive measure on invariant under ,
(1)
For this, one proves that if is a covering of formed by sets, every belongs to at least sets of the form , and that every belongs to at most sets of the form (note that if , then is contained in ).
Let be an ultrafilter finer than the section filter of ; let be a nonempty, relatively compact open set in , and set for every symmetric entourage and every relatively compact subset of . Let ; for every compact subset of , set , where runs over the set of relatively compact open neighborhoods of . Deduce from (1) and (2) that
Conclude from this that if are two compact subsets of , and , are two relatively compact open sets, then
and finally that .
Solution
No solution has been written yet.