Bourbaki

Exercise 2, § 4

Integration, IX, p. 113

Bourbaki marks this one as harder than the rest.

Hypotheses and notations are those of Th. 1 of No. 2, for which we propose to indicate a new proof.

a) Let KK be a compact subset of TT; for every i∈Ii \in I, set Ki=pi(T)K_i = p_i(T), and denote by qiq_i the mapping of KK onto KiK_i that coincides on KK with pip_i; for i≤ji \leq j, denote by qijq_{ij} the mapping of KjK_j into KiK_i that coincides on KjK_j with pijp_{ij}. Deduce from the preceding exercise the existence of a greatest positive measure πK\pi^K on KK such that qi(πK)=(μi)Kiq_i(\pi^K) = (\mu_i)_{K_i} for all i∈Ii \in I.

b) If KK and LL are compact subsets of TT such that K⊂LK \subset L, show that (πL)K≥πK(\pi^L)_K \geq \pi^K; deduce from this that the set of measures of the form iK(πK)i^K(\pi^K) (where KK runs over the set of compact subsets of TT, and iKi^K is the canonical injection of KK into TT) admits a supremum μ\mu in M+(T)\mathscr{M}_+(T).

c) Show that pi(μ)=μip_i(\mu) = \mu_i for every i∈Ii \in I (it suffices to observe that pi(μ)≤μip_i(\mu) \leq \mu_i and that the measures pi(μ)p_i(\mu) and μi\mu_i have the same total mass by the hypothesis (P)).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.