Bourbaki

Exercise 20, § 2

Integration, VI, p. 66

Bourbaki marks this one as harder than the rest.

Let FF be a separable Fréchet space, F′F' its dual, μ\mu a positive measure on TT, mm a measure on TT with values in the weak dual Fs′F'_s, scalarly of base μ\mu. For mm to

³ According to the Feuille d’errata No 14, this exercise is to be suppressed.

have base μ\mu, it is necessary and sufficient that the following condition be fulfilled: for every ε>0\varepsilon > 0 and every compact subset KK of TT, there exists a compact subset K1⊂KK_1 \subset K such that μ(K−K1)≤ε\mu(K - K_1) \leq \varepsilon, and such that the image under mm of the set of μ\mu-measurable functions gg that are bounded, have support contained in K1K_1 and satisfy ∫∣g∣dμ≤1\int |g| d\mu \leq 1, is an equicontinuous subset of F′F'. (Recall that ∫gdm∈F′\int g dm \in F' for every μ\mu-measurable function gg that is bounded and has compact support; cf. Cor. 2 of Prop. 3. To show that the condition is necessary, use Prop. 13 of §1, No. 5 and Prop. 5 of §1, No. 2. To see that the condition is sufficient, consider first the case that TT is compact; applying Cor. 3 of Th. 1, No. 5, first show that there exist a partition of TT into a negligible set NN and a sequence (Kn)(K_n) of compact sets, and a measurable mapping gg of TT into Fs′F'_s, such that ∫fdm=∫gfdμ\int f dm = \int gf d\mu for every function f∈L∞(μ)f \in \mathscr{L}^\infty(\mu) that is zero on the complement of the union of a finite number of the sets KnK_n; to prove that m=g⋅μm = g \cdot \mu, use Exers. 16 b) and 20 a) of Ch. V, §5. Finally, to pass to the case that TT is any locally compact space, use Prop. 14 of Ch. IV, §5, No. 9.)

Solution

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