Bourbaki

Exercise 20, § 8

Integration, V, p. 120

a) Let ν\nu be a positive measure on the unit circle U:∣z∣=1U : |z| = 1 in C\mathbf{C}, and let μ\mu be the complex measure j⋅νj \cdot \nu, where j:z↦zj : z \mapsto z is the canonical injection of UU into C\mathbf{C}, so that ∣μ∣=ν|\mu| = \nu. Let ρ\rho be the inverse image of ν\nu under the canonical local homeomorphism φ:t↦eit\varphi : t \mapsto e^{it} of R\mathbf{R} onto UU (§ 6, No. 6). Let AA be a ν\nu-measurable subset of UU; if μ(A)=∣μ(A)∣eiω\mu(A) = |\mu(A)|e^{i\omega}, show that ∣μ(A)∣≤∣∫ω−π/2ω+π/2ℜ(ei(θ−ω)) dρ(θ)∣|\mu(A)| \leq \left| \int_{\omega-\pi/2}^{\omega+\pi/2} \Re(e^{i(\theta-\omega)})\ d\rho(\theta) \right|.

b) Deduce from a) that, as AA runs over the set of μ\mu-measurable subsets of UU, one has

sup⁡A∣μ(A)∣≥sup⁡ω∣∫ω−π/2ω+π/2ℜ(ei(θ−ω)) dρ(θ)∣ \sup_A |\mu(A)| \geq \sup_\omega \left| \int_{\omega-\pi/2}^{\omega+\pi/2} \Re(e^{i(\theta-\omega)})\ d\rho(\theta) \right|
≥12π∫02πdω∫ω−π/2ω+π/2ℜ(ei(θ−ω)) dρ(θ) \geq \frac{1}{2\pi} \int_0^{2\pi} d\omega \int_{\omega-\pi/2}^{\omega+\pi/2} \Re(e^{i(\theta-\omega)})\ d\rho(\theta)
and conclude therefrom that
(∗)sup⁡A∣μ(A)∣≥1π∥μ∥ (*) \quad \sup_A |\mu(A)| \geq \frac{1}{\pi} \| \mu \|
(evaluate the last integral with the help of the Lebesgue–Fubini theorem).

c) Show that the inequality (*) is true for every bounded complex measure on a locally compact space X, where A runs over the set of μ\mu-measurable subsets of X. (Let μ=h⋅∣μ∣\mu = h \cdot |\mu|, where h takes its values in U. Consider the image measure ν=h(∣μ∣)\nu = h(|\mu|) on U and show that h(μ)=j⋅νh(\mu) = j \cdot \nu; then apply b) to h(μ)h(\mu).)

Solution

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Transcribed from the printed volume. Not translated.