Bourbaki

Exercise 5, § 7

Integration, IV, p. 144

Bourbaki marks this one as harder than the rest.

The notations being those of Exer. 4 e), let A′\mathscr{A}' denote the dual of the normed space A\mathscr{A}, equipped with the order structure deduced from that of A\mathscr{A} (Ch. II, §2). On the other hand, denote by C the convex cone with vertex 0 generated by X×{1}X \times \{1\} in the space E×RE \times \mathbf{R}.

a) Show that the following properties are equivalent: (i) Every point x∈Xx \in X is the barycenter of a unique maximal measure βx\beta_x of mass 1 on XX. (ii) The vector space FF generated by CC is a lattice for the order structure having CC as the set of elements ⩾0\geqslant 0 (in other words, XX is a simplex (TVS, II, §2, Exer. 41)). (iii) For every function f∈Pf \in \mathscr{P}, S(f)S(f) is both concave and convex. (iv) If μ\mu and ν\nu are two maximal measures such that μ(h)=ν(h)\mu(h) = \nu(h) for all h∈Ah \in \mathscr{A}, then μ=ν\mu = \nu. (v) The vector space A′\mathscr{A}' is a Riesz space. (vi) If ff and gg are two functions in P\mathscr{P}, then S(f+g)=S(f)+S(g)S(f + g) = S(f) + S(g). (First prove that (i) implies the following property: (vii) The mapping x↦βxx \mapsto \beta_x may be extended in only one way to a bijective linear mapping of FF onto the subspace of M(X)\mathscr{M}(X) generated by the maximal measures, and this mapping transforms CC into the cone of maximal measures. Then deduce (ii) from (vii), using Exer. 2 e). To deduce (iii) from (ii), use the decomposition lemma in the Riesz space FF and Exer. 4 f). To deduce (iv) from (iii), use the fact that if g∈C(X;R)g \in \mathscr{C}(X; \mathbf{R}) is both concave and convex, then the set of h∈Ah \in \mathscr{A} such that h(x)>g(x)h(x) > g(x) for all x∈Xx \in X is a decreasing directed set (TVS, II, §5, Prop. 6) and has gg for its lower envelope; then apply Exer. 4 e). To deduce (v) from (iv), observe that A′\mathscr{A}' may be identified with the subspace of M(X)\mathscr{M}(X) generated by the maximal measures, with the help of Exer. 4 e). To deduce (vi) from (v), apply the decomposition lemma in the Riesz space A′\mathscr{A}' and Exer. 4 f). Finally, to deduce (i) from (vi), consider the mapping f↦S(f)(x)f \mapsto S(f)(x) for f∈Pf \in \mathscr{P} and x∈Xx \in X, and use Exer. 4 e).) b) When the conditions of a) are satisfied, show that for every function f∈C(X;R)f \in \mathscr{C}(X; \mathbf{R}), the mapping x↦βx(f)x \mapsto \beta_x(f) is in the closure, for the topology of uniform convergence in XX, of the set of bounded functions on XX that are the difference of two upper semi-continuous functions (make use of the fact that P\mathscr{P} is total in C(X;R)\mathscr{C}(X; \mathbf{R})); consequently, this mapping is measurable for every measure on XX. For every positive measure μ\mu on XX, the unique maximal measure ν≻μ\nu \succ \mu is given by ν=∫βxdμ(x)\nu = \int \beta_x d\mu(x) (cf. Ch. V). c) When the conditions of a) are satisfied and XX is in addition metrizable, show that for every x∈Xx \in X, βx\beta_x is the only positive measure μ\mu of mass 1 and barycenter xx such that μ(X−L)=0\mu(X - L) = 0, where LL is the set ChA(X)\mathrm{Ch}_{\mathscr{A}}(X) of extremal points (argue as in Th. 1). d) When the conditions of a) are satisfied, show that the following properties are equivalent: α)\alpha) The set L=ChA(X)L = \mathrm{Ch}_{\mathscr{A}}(X) is closed, in other words SˇA(X)=ChA(X)\check{S}_{\mathscr{A}}(X) = \mathrm{Ch}_{\mathscr{A}}(X). β)\beta) For every function f∈Pf \in \mathscr{P}, the restriction of S(f)S(f) to L‾\overline{L} is continuous. γ)\gamma) For every function f∈Pf \in \mathscr{P}, the function S(f)S(f) is continuous on XX. δ)\delta) The mapping x↦βxx \mapsto \beta_x of XX into M+(X)\mathscr{M}_+(X) (equipped with the vague topology) is continuous. (To show that α\alpha) implies β)\beta), observe that βx=εx\beta_x = \varepsilon_x in LL. To prove that β)\beta) implies γ)\gamma), use the fact that S(f)S(f) is the lower envelope of a decreasing directed set of functions in A\mathscr{A} (TVS, II, §5, Prop. 6), Dini’s theorem and Exer. 3 d) applied to the functions in A\mathscr{A}. To prove that γ)\gamma) implies δ)\delta), use Exer. 4 e). Finally, to see that δ)\delta) implies α)\alpha), use Exer. 4 a).)

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