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A finite complex measure absolutely continuous with respect to a sigma-finite positive measure has an integrable complex density

Statement

Assume the Axiom of Choice. Let $μ$ be a positive measure and let $ν$ be a finite complex measure on $(X, A)$ . If $ν ≪ μ$ and $μ$ is sigma-finite, then there exists a complex function $h \in L^{1} (μ)$ , unique up to $μ$ -almost-everywhere equality, such that $ν (E) = \int_{E} h d μ (E \in A) .$

Facts & Assumptions

Given: AC, a sigma-finite positive measure $μ$ and a finite complex measure $ν$ with $ν ≪ μ$ .

[L1]

The real and imaginary parts of $ν$ are finite signed measures. (The real and imaginary parts of a complex measure are finite signed measures, and nu = Re nu + i Im nu)

[L2]

Absolute continuity means vanishing on every measurable reference-null set. Thus $ν ≪ μ$ implies both real and imaginary parts vanish there too. (Absolute continuity of a signed or complex measure with respect to a positive measure)

[L3]

A signed measure satisfying a common finite exhaustion with $μ$ and absolutely continuous with respect to $μ$ has an almost-everywhere unique density; if its total variation is finite, the density is integrable (A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density).

[L4]

A complex function is integrable exactly when its real and imaginary parts are integrable, using $∣ u ∣, ∣ v ∣ \leq ∣ u + i v ∣ \leq ∣ u ∣ + ∣ v ∣$ . (Integrable real and complex functions, and their integrals)

[L5]

A complex measure has finite total variation (Every complex measure has finite total variation).

[L6]

AC supplies the hypotheses of the signed RN existence theorem and its underlying Hahn/Jordan selections. (The Axiom of Choice)

Proof

technique · direct

1.1L1L2L3L5L6choosealgebra

By [L1] and [L2], the signed measures $Re ν$ and $Im ν$ are absolutely continuous with respect to $μ$ . By [L5], partition sums for their total variations are bounded by those of $∣ ν ∣$ , so both have finite total variation. Choose an increasing exhaustion $(X_{n})$ with $μ (X_{n}) < \infty$ ; it is then common to $μ$ and both signed measures. Under AC as in [L6], applying [L3], choose real-valued $u, v \in L^{1} (μ)$ such that $Re ν (E) = \int_{E} u d μ, Im ν (E) = \int_{E} v d μ (E \in A) .$

2.1step 1.1L3L4algebra∎

Put $h : = u + i v$ . Then [L4] gives $h \in L^{1} (μ)$ , and for every measurable $E$ one has $ν (E) = Re ν (E) + i Im ν (E) = \int_{E} u d μ + i \int_{E} v d μ = \int_{E} h d μ .$ If $h^{'}$ is another such density, then its real and imaginary parts give alternative signed densities for $Re ν$ and $Im ν$ , so the uniqueness part of [L3] makes $h^{'} = h$ $μ$ -almost everywhere.

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John K. Hunter, Measure Theory, Theorem 6.30 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Exercise 13.3 (standard reference, not scraped)