Bourbaki

Exercise 9, § 7

Integration, IV, p. 147

Bourbaki marks this one as harder than the rest.

Let Y be the topological space whose underlying set is the product I×{−1,0,1}I \times \{-1, 0, 1\} in R2\mathbf{R}^2, where I=[0,1]I = [0, 1]; for every a∈Ia \in I and every ε>0\varepsilon > 0, let Ua,εU_{a, \varepsilon} be the set of (x,y)∈Y(x, y) \in Y such that ∣x−a∣≤ε|x - a| \leq \varepsilon and (x,y)(x, y) is distinct from (a,1)(a, 1) and (a,−1)(a, -1); the sets {(x,y)}\{(x, y)\} for y≠0y \neq 0 and x∈Ix \in I together with the sets Ua,εU_{a, \varepsilon} form a base for a topology on Y for which Y is a non-metrizable compact space (cf. GT, IX, §2, Exer. 13 d)). Denote by X the topological sum space of Y and a set reduced to a point ω\omega. Let H\mathscr{H} be the set of continuous finite numerical functions hh on X such that

h(x,0)=12(h(x,1)+h(x,−1)) h(x,0) = \frac{1}{2}(h(x,1) + h(x,-1))

for all x∈Ix \in I, and h(ω)=∫01h(x,0) dxh(\omega) = \int_0^1 h(x,0)\, dx. Show that ChH(X)\mathrm{Ch}_{\mathscr{H}}(X) consists of the points (x,y)(x,y) such that y≠0y \neq 0 and x∈Ix \in I, but that there exists no positive measure μ\mu of mass 1 on X such that h(ω)=μ(h)h(\omega) = \mu(h) for all h∈Hh \in \mathscr{H}, and μ(X−ChH(X))=0\mu(X - \mathrm{Ch}_{\mathscr{H}}(X)) = 0.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.