Bourbaki

Exercise 8, § 5

Integration, IX, p. 115

Bourbaki marks this one as harder than the rest.

Let T be a complete separable metric space, and d its metric. For every closed subset F of T and every real number a>0a > 0, denote by FaF^a the set of x∈Tx \in T such that d(x,F)<ad(x, F) < a. Given two bounded positive measures λ\lambda and μ\mu on T, denote by D(λ,μ\lambda, \mu) the infimum of the set of real numbers a>0a > 0 satisfying the inequalities λ∙(F)≤μ∙(Fa)+a\lambda^\bullet(F) \leq \mu^\bullet(F^a) + a, μ∙(F)≤λ∙(Fa)+a\mu^\bullet(F) \leq \lambda^\bullet(F^a) + a for every closed subset F of T.

a) Show that D is a metric on M+b\mathscr{M}_+^b and that D(εx,εy)=d(x,y)D(\varepsilon_x, \varepsilon_y) = d(x, y) for two points x and y of T such that d(x,y)<1d(x, y) < 1.

b) Let f be a Borel mapping of T into T (cf. TG, IX, §6, No. 3 or GT, IX, §6, Exer. 16) and λ\lambda a bounded positive measure on T. Show that there exists a bounded positive measure μ\mu on T such that μ∙(A)=λ∙(f−1(A))\mu^\bullet(A) = \lambda^\bullet(f^{-1}(A)) for every Borel subset A of T. Let a>0a > 0, and let H be the set of x∈Tx \in T such that d(x,f(x))≥ad(x, f(x)) \geq a; show that H is Borel in T and that D(λ,μ)≤sup⁡(a,λ(H))D(\lambda, \mu) \leq \sup(a, \lambda(H)) (for every closed subset F of T, one has F∩(T−H)⊂f−1(Fa)F \cap (T - H) \subset f^{-1}(F^a) and λ∙(F)≤λ∙(F∩(T−H))+λ∙(H)\lambda^\bullet(F) \leq \lambda^\bullet(F \cap (T - H)) + \lambda^\bullet(H)).

c) Let μ\mu and μn\mu_n (for n≥1n \geq 1) be bounded positive measures on T such that lim⁡n→∞D(μn,μ)=0\lim_{n \to \infty} D(\mu_n, \mu) = 0. Show that the sequence (μn)(\mu_n) converges tightly to μ\mu (show that lim⁡n→∞μn∙(T)=μ∙(T)\lim_{n \to \infty} \mu_n^\bullet(T) = \mu^\bullet(T) and μ∙(F)≥lim sup⁡n→∞μn∙(F)\mu^\bullet(F) \geq \limsup_{n \to \infty} \mu_n^\bullet(F) for every closed subset F of T; deduce from this that μ∙(f)≤lim inf⁡n→∞μn∙(f)\mu^\bullet(f) \leq \liminf_{n \to \infty} \mu_n^\bullet(f) for every lower semi-continuous function f≥0f \geq 0 by the method of Lemma 3 of §2, No. 6).

d) Conversely, show that for every sequence of bounded positive measures μn\mu_n tending tightly to μ\mu in M+b\mathscr{M}_+^b, one has lim⁡D(μn,μ)=0\lim D(\mu_n, \mu) = 0. One can proceed as follows: let ε>0\varepsilon > 0, and let K be a compact subset of T such that μ∙(T−K)<ε\mu^\bullet(T - K) < \varepsilon and sup⁡nμn∙(T−K)<ε\sup_{n} \mu_n^\bullet(T - K) < \varepsilon; construct a finite family (Bi)1≤i≤p(B_i)_{1 \leq i \leq p} of sets with μ\mu-negligible boundary, of diameter ≤ε/2\leq \varepsilon/2, pairwise disjoint, whose union contains K (cf. Exer. 1). Let f be a mapping of T into T, constant on each of the sets B1,…,BpB_1, \ldots, B_p and C(B1∪⋯∪Bp)\mathscr{C}(B_1 \cup \cdots \cup B_p) and such that f(Bi)⊂Bif(B_i) \subset B_i for 1≤i≤p1 \leq i \leq p. Define the measures πn\pi_n and π\pi by πn∙(A)=μn∙(f−1(A))\pi_n^\bullet(A) = \mu_n^\bullet(f^{-1}(A)) and π∙(A)=μ∙(f−1(A))\pi^\bullet(A) = \mu^\bullet(f^{-1}(A)) for every Borel subset A of T; deduce from b) the relations D(πn,μn)≤εD(\pi_n, \mu_n) \leq \varepsilon, D(π,μ)≤εD(\pi, \mu) \leq \varepsilon and show that lim⁡n→∞D(πn,π)=0\lim_{n \to \infty} D(\pi_n, \pi) = 0.

e) The metric D on M+b\mathscr{M}_+^b is compatible with the tight topology (observe that M+b\mathscr{M}_+^b is metrizable for the tight topology).

Solution

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