Bourbaki

Exercise 13, § 4

Integration, IV, p. 124

Bourbaki marks this one as harder than the rest.

a) Let XX be a locally compact space, and let Γ\Gamma be a base for the topology of XX consisting of relatively compact sets. Let U\mathscr{U} be a uniform structure compatible with the topology of XX, and let S\mathscr{S} be a fundamental system of entourages for this structure. Let M↦λ(M)M \mapsto \lambda(M) be a finite numerical function ⩾0\geqslant 0 defined on Γ\Gamma. For every compact set K⊂XK \subset X and every entourage V∈SV \in \mathscr{S}, let αV(K)\alpha_V(K) be the infimum of the numbers ∑iλ(Ui)\sum_i \lambda(U_i) for all finite coverings (Ui)(U_i) of KK formed by sets in Γ\Gamma that are small of order VV; suppose that as VV runs over S\mathscr{S}, the supremum α(K)\alpha(K) of the numbers αV(K)\alpha_V(K) is finite. Show that there exists a measure μ\mu on XX (and only one) such that μ(K)=α(K)\mu(K) = \alpha(K) for every compact set KK (make use of Th. 5).

b) Let Ψ\Psi be the set of Borel subsets of XX, α\alpha a mapping of Ψ\Psi into [0,+∞][0,+\infty] satisfying the following conditions: (i) If B1,B2B_1, B_2 are two disjoint Borel subsets of XX, then α(B1∪B2)=α(B1)+α(B2)\alpha(B_1 \cup B_2) = \alpha(B_1) + \alpha(B_2). (ii) If BB is a compact subset of XX then α(B)<+∞\alpha(B) < +\infty. (iii) If BB is a Borel subset of XX then α(B)=inf⁡α(U)\alpha(B) = \inf \alpha(U), where UU runs over the set of open subsets of XX containing BB.

Then, there exists one and only one positive measure μ\mu on XX such that α(B)=μ∗(B)\alpha(B) = \mu^*(B) for every Borel subset BB of XX that can be covered by a sequence of compact sets.

c) For every B∈ΦB \in \Phi, set α(B)=0\alpha(B) = 0 if BB can be covered by a sequence of compact sets, and α(B)=+∞\alpha(B) = +\infty in the contrary case. Then, the conditions (i), (ii), (iii) of b) are satisfied. One has μ=0\mu = 0, therefore α(B)≠μ∗(B)\alpha(B) \neq \mu^*(B) if BB cannot be covered by a sequence of compact sets.

Solution

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Transcribed from the printed volume. Not translated.