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Every finite signed or complex measure has a polar decomposition against its total variation

Statement

Assume the Axiom of Choice. Let $ν$ be a finite signed measure or a finite complex measure on $(X, A)$ . Then there exists a measurable function $h$ such that $ν (E) = \int_{E} h d ∣ ν ∣ (E \in A), ∣ h ∣ = 1 ∣ ν ∣ -almost everywhere.$ If $ν$ is signed, then $h$ may be chosen real-valued, and for a Hahn decomposition $X = P ⊔ N$ one may take $h = χ_{P} - χ_{N} ∣ ν ∣ -almost everywhere.$

Facts & Assumptions

Given: AC and a finite signed or finite complex measure $ν$ .

[L1]

For every measurable set $E$ , one has $∣ ν (E) ∣ \leq ∣ ν ∣ (E)$ , so $ν ≪ ∣ ν ∣$ ; finite signed measures therefore admit Radon-Nikodym densities with respect to $∣ ν ∣$ by the signed theorem, and finite complex measures do so by the complex corollary. (The total variation |nu|(E) from countable measurable partitions, The Radon-Nikodym derivative as an almost-everywhere equivalence class, A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density, A finite complex measure absolutely continuous with respect to a sigma-finite positive measure has an integrable complex density)

[L2]

For any complex L1 density $h$ on a measure space, the measure defined by its set integrals has total variation $\int_{E} ∣ h ∣$ . This also applies when $h$ is real-valued. (A complex L^1 density defines a complex measure whose total variation is |h| dmu)

[L3]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

[L4]

In the signed case, a Hahn decomposition $X = P ⊔ N$ exists, and on its positive and negative pieces the Jordan and total-variation formulas give $ν = ∣ ν ∣$ and $ν = - ∣ ν ∣$ , respectively (Hahn decomposition for signed measures, unique up to total-variation-null sets, Jordan decomposition of a signed measure into unique mutually singular positive parts, For a signed measure, total variation is nu-plus plus nu-minus, finite partitions suffice, and nu-plus and nu-minus are extremal).

[L5]

Total variation is a positive measure, finite for complex measures and for finite signed measures by their finite Hahn/Jordan masses. AC covers its partition selections and the RN/Hahn constructions. (The total variation of a signed or complex measure is a positive measure, Every complex measure has finite total variation, The Axiom of Choice)

[L6]

Finite L1 integrals may be subtracted and added linearly. (The Lebesgue integral is linear on $L^{1} (μ)$ )

Proof

technique · direct

1.1L1L5given

Because $∣ ν (E) ∣ \leq ∣ ν ∣ (E)$ for every measurable $E$ , the measure $ν$ is absolutely continuous with respect to $∣ ν ∣$ . By [L5], $∣ ν ∣$ is a finite positive measure, so the constant exhaustion is common and finite. Under AC, [L1] gives an L1 density $h = d ν / d ∣ ν ∣$ with $ν (E) = \int_{E} h d ∣ ν ∣ (E \in A) .$

2.1step 1.1L2L3L5L6algebra

Apply [L2] to this L1 function on the finite measure space with reference $∣ ν ∣$ . Its defined measure is exactly $ν$ by step 1.1. All integrals below are finite by step 1.1 and [L5], so [L6] licenses the displayed subtractions. Then $∣ ν ∣ (E) = \int_{E} ∣ h ∣ d ∣ ν ∣ (E \in A) .$ Let $A : = {∣ h ∣ > 1}$ . Using the displayed identity on $A$ gives $0 = \int_{A} ∣ h ∣ d ∣ ν ∣ - ∣ ν ∣ (A) = \int_{A} (∣ h ∣ - 1) d ∣ ν ∣ .$ Because $∣ h ∣ - 1 \geq 0$ on $A$ , [L3] yields $∣ ν ∣ (A) = 0$ . Now let $B : = {∣ h ∣ < 1}$ . Since $A$ is $∣ ν ∣$ -null, $0 = ∣ ν ∣ (B) - \int_{B} ∣ h ∣ d ∣ ν ∣ = \int_{B} (1 - ∣ h ∣) d ∣ ν ∣ .$ Again the integrand is nonnegative, so [L3] gives $∣ ν ∣ (B) = 0$ . Therefore $∣ h ∣ = 1$ $∣ ν ∣$ -almost everywhere.

3.1L4step 2.1algebra

If $ν$ is signed, let $X = P ⊔ N$ be a Hahn decomposition from [L4]. Then $χ_{P} - χ_{N}$ is real-valued and has modulus $1$ everywhere. For every measurable $E$ , additivity and the Jordan formulas give $ν (E) = ν (E \cap P) + ν (E \cap N) = ∣ ν ∣ (E \cap P) - ∣ ν ∣ (E \cap N) = \int_{E} (χ_{P} - χ_{N}) d ∣ ν ∣ .$ Hence the signed case may be represented by $h = χ_{P} - χ_{N}$ .

4.1step 1.1step 2.1step 3.1∎

Steps 1.1, 2.1, and 3.1 prove the general polar decomposition and the signed specialization.

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Sources

Sheldon Axler, Measure, Integration & Real Analysis, 9.41 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Exercise 13.7 (standard reference, not scraped)