Bourbaki

Exercise 20, § 2

Integration, VI

Bourbaki marks this one as harder than the rest.

Let FF be a Fréchet space of countable type, 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, scalarwise of basis μ\mu. For mm to be of basis 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 that the image by mm of the set of μ\mu-measurable functions gg, bounded, with support contained in K1K_1 and verifying ∫∣g∣dμ≤1\int |g| d\mu \leq 1, is an equicontinuous subset of F′F'. (One will recall that ∫gdm∈F′\int g d\mathbf{m} \in F' for every bounded μ\mu-measurable function gg with 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 where TT is compact; applying cor. 3 of th. 1, No. 5, first show that there is a partition of TT into a negligible set NN and a sequence (Kn)(K_n) of compact sets, and a measurable mapping gg from TT into Fs′F'_s, such that ∫fdm=∫gfdμ\int f d\mathbf{m} = \int g f d\mu for every function f∈L∞(μ)f \in \mathscr{L}^\infty(\mu) null in the complement of a union of a finite number of sets KnK_n; to prove that m=g.μ\mathbf{m} = g . \mu, use exerc. 18 b) and 22 a) of chap. V, § 5. Finally, to pass to the case where TT is an arbitrary locally compact space, use prop. 4 of chap. V, § 1, No. 4.)

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.