Exercise 2, § 5
Integration, IX, p. 113
Bourbaki marks this one as harder than the rest.
Let be a completely regular space and a filter on that converges tightly to a bounded measure . A Borel subset of is said to be a convergence set (for ) if .
a) If the disjoint sets and are convergence sets, then so is .
b) Let be an open or closed set. For to be a convergence set, it is necessary and sufficient that . If is a convergence set, then so is .
c) Let be an open or closed convergence set in , and a convergence set such that is also a convergence set. Then and are convergence sets, as are their complements.
d) The clan generated by the open convergence sets consists of convergence sets.
e) Let be a subset of whose boundary is -negligible. Then is a convergence set.
f) Suppose that is locally compact or Polish. Show that every compact subset of is contained in a compact convergence set (in case is locally compact, use Exercise 1 b). If is Polish, use the same exercise to construct a sequence of finite coverings of by open subsets of , with -negligible boundary and of diameter ; show that is compact and conclude that it is a convergence set by b)).
g) Extend the result of f) to the case of a convergent sequence of measures on a complete metric space.
Solution
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