Integration, IV, p. 147
Bourbaki marks this one as harder than the rest.
The general notations and hypotheses being those of Exer. 6, set S=SˇH(X). A function f∈C(S;R) is said to be H-resolvent if there exists a function g∈Hc such that f=g∣S (this function is then unique).
a) Show that the following conditions on a function f∈C(S;R) are equivalent: (i) f is H-resolvent. (ii) For every x∈X, HxS(f)=HxS(f). (iii) For every x∈X and every pair of measures μ1,μ2 in MxS, one has ∫fdμ1=∫fdμ2. (Make use of Exer. 6 b) and c).) b) Show that the following properties are equivalent: (i) Every function f∈C(S;R) is H-resolvent. (ii) For every x∈X and every function f∈C(S;R), HxS(f)=HxS(f). (iii) The set MxS reduces to a single element. (iv) The set Hc is lattice-ordered. (v) For every function u∈EH (Exer. 6 g)), there exists a greatest lower bound hu of u in Hc. (To show that (iv) implies (i), apply Exer. 7 d) to Hc, proving that for every point s∈S, the linear form h↦h(s) is latticial in Hc; then use Stone’s theorem and Exer. 6 c). To see that (i) implies (v), note that Hc⊂EH and consider the unique function hu∈Hc such that hu∣S=u∣S. To see that (v) implies (iv), note that if h∈Hc then −h−=inf(h,0)∈EH, and apply (v) to u=−h−.)
c) If the conditions of b) are satisfied and if γx is the unique element of MxS, show that the mapping x↦γx of X into M+(X) (equipped with the vague topology) is continuous; from this, deduce that one then has ChH(X)=SˇH(X) (note that γx=εx in ChH(X)).
d) In order that every function f∈C(S;R) be the restriction to S of a function h∈H, it is necessary and sufficient that H be closed in C(X;R) and be lattice-ordered.