TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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The total variation of an absolutely continuous signed or complex measure has density the absolute value of the Radon-Nikodym derivative

Statement

Assume the Axiom of Choice. Let $μ$ be a sigma-finite positive measure.

If $ν$ is a finite signed measure with $ν ≪ μ$ and Radon-Nikodym derivative $h = d ν / d μ$ , then $∣ ν ∣ (E) = \int_{E} ∣ h ∣ d μ (E \in A) .$
If $ν$ is a finite complex measure with $ν ≪ μ$ and Radon-Nikodym derivative $h = d ν / d μ$ , then $∣ ν ∣ (E) = \int_{E} ∣ h ∣ d μ (E \in A) .$

Facts & Assumptions

Given: The Axiom of Choice, a sigma-finite positive measure $μ$ and an absolutely continuous finite signed or finite complex measure $ν$ .

[L1]

A real $L^{1}$ density defines a finite signed measure whose total variation is the integral of its absolute value. (A real L^1 density defines a finite signed measure with its canonical Hahn and Jordan data)

[L2]

A complex $L^{1}$ density defines a complex measure whose total variation is the integral of its modulus. (A complex L^1 density defines a complex measure whose total variation is |h| dmu)

[L3]

Under the Axiom of Choice, finite absolutely continuous signed and complex measures admit integrable representatives of their Radon-Nikodym derivatives (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, The Axiom of Choice).

Proof

technique · direct

1.1L1L3

In the finite signed case, [L3] gives an integrable real-valued representative $h$ of $d ν / d μ$ , and [L1] applied to that density yields exactly $∣ ν ∣ (E) = \int_{E} ∣ h ∣ d μ (E \in A) .$

1.2L2L3

In the finite complex case, [L3] gives an integrable complex representative $h$ of $d ν / d μ$ , and [L2] applied to that density yields the same formula $∣ ν ∣ (E) = \int_{E} ∣ h ∣ d μ (E \in A) .$

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the signed and complex clauses.

Depends on

Used by

Dependency tree · two levels

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Sources

Richard F. Bass, Real Analysis for Graduate Students, Exercise 13.4 (standard reference, not scraped)
John K. Hunter, Measure Theory, §6.9 (standard reference, not scraped)