Bourbaki

Lemma 6

Tag 09SN, permanent. Cite it as [Bourbaki, Tag 09SN].

Integration, IX, p. 90, § 6. Promeasures and measures on a locally convex space

Let μ∈M1\mu \in \mathscr{M}^1. There exists a sequence of measures μn∈D\mu_n \in \mathscr{D} such that μ(f)=lim⁡n→∞μn(f)\mu(f) = \lim_{n \to \infty} \mu_n(f) for all f∈Cf \in \mathscr{C} and W(μ)=lim⁡n→∞W(μn)W(\mu) = \lim_{n \to \infty} W(\mu_n).

Let I=[0,1]I = [0, 1]. The space M1\mathscr{M}^1 of bounded measures on T=]0,1[T = ]0, 1[ will be identified with the subspace of M(I)\mathscr{M}(I) formed by the measures that place no weight at 00. (3) We equip M(I)\mathscr{M}(I) with the vague topology. The mapping

(3) That is, the measures on II that are concentrated on T=I−{0}T = I - \{0\}.

t↦εtt \mapsto \varepsilon_t of II into M(I)\mathscr{M}(I) is continuous (Ch. III, §1, No. 9, Prop. 13); since D is dense in I, the closure D‾\overline{\mathscr{D}} of D\mathscr{D} contains all of the point measures. Let A be the set of measures ν∈D\nu \in \mathscr{D} such that ∥ν∥≤∥μ∥\| \nu \| \leq \| \mu \|; the measure μ\mu is in the closure of A (Ch. III, §2, No. 4, Cor. 1 of Th. 1). The set A is relatively compact in M(I)\mathscr{M}(I) (Ch. III, §1, No. 9, Prop. 15) and the compact subsets of M(I)\mathscr{M}(I) are metrizable (TVS, III, §3, No. 4, Cor. 2 of Prop. 6,(4) and GT, X, §3, No. 3, Th. 1). Therefore there exists a sequence of measures μn∈A\mu_n \in A converging to μ\mu in M(I)\mathscr{M}(I). Since C\mathscr{C} is identified with the subspace of continuous functions on I zero at the origin, we have μ(f)=lim⁡n→∞μn(f)\mu(f) = \lim_{n \to \infty} \mu_n(f) for all f∈Cf \in \mathscr{C}. Moreover, since C(I)⊗C(I)\mathscr{C}(I) \otimes \mathscr{C}(I) is dense in the normed space C(I×I)\mathscr{C}(I \times I) (Ch. III, §4, No. 1, Lemma 1), the relations lim⁡n→∞μn=μ\lim_{n \to \infty} \mu_n = \mu and ∥μn∥≤∥μ∥\| \mu_n \| \leq \| \mu \| imply that lim⁡n→∞(μn⊗μn)=μ⊗μ\lim_{n \to \infty} (\mu_n \otimes \mu_n) = \mu \otimes \mu (Ch. III, §1, No. 10, Prop. 17); since the measures μn\mu_n and μ\mu place no weight at 0, we have

W(μn)=∫I∫Iinf⁡(t,t′)dμn(t)dμn(t′), W(\mu_n) = \int_I \int_I \inf(t, t') d\mu_n(t) d\mu_n(t'),
W(μ)=∫I∫Iinf⁡(t,t′)dμ(t)dμ(t′), W(\mu) = \int_I \int_I \inf(t, t') d\mu(t) d\mu(t'),

whence lim⁡n→∞W(μn)=W(μ)\lim_{n \to \infty} W(\mu_n) = W(\mu).

It remains to prove that the Fourier transform of w′w' is equal to e−W/2e^{-W/2}. Let μ∈M1\mu \in \mathscr{M}^1; choose measures μn∈D\mu_n \in \mathscr{D} as in Lemma 6. The measure w′w' is bounded, and ∣ei⟨f,μn⟩∣=1|e^{i \langle f, \mu_n \rangle}| = 1 for all nn; Lemma 5 and Lebesgue's convergence theorem (Ch. IV, §4, No. 3, Th. 2) then imply

∫Cei⟨f,μ⟩dw′(f)=lim⁡n→∞∫Cei⟨f,μn⟩dw′(f)=lim⁡n→∞e−W(μn)/2=e−W(μ)/2. \int_{\mathscr{C}} e^{i \langle f, \mu \rangle} dw'(f) = \lim_{n \to \infty} \int_{\mathscr{C}} e^{i \langle f, \mu_n \rangle} dw'(f) = \lim_{n \to \infty} e^{-W(\mu_n)/2} = e^{-W(\mu)/2}.

Q.E.D.

The measure ww on C\mathscr{C} whose Fourier transform is equal to e−W/2e^{-W/2} is called the Wiener measure on C\mathscr{C}.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.