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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)
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Radon-Nikodym derivatives add almost everywhere

Statement

Assume the Axiom of Choice. Let μ be a sigma-finite positive measure. Let ν1,ν2 be either two finite signed measures or two finite complex measures, with ν1≪μ and ν2≪μ. Then d(ν1+ν2)dμ=dν1dμ+dν2dμμ-almost everywhere.

Facts & Assumptions

Given: AC and finite measures ν1,ν2 absolutely continuous with respect to μ.

[L2]

The integral is linear on L1. (The Lebesgue integral is linear on L1(μ))

[L3]

The Jordan parts of a finite-valued signed measure have finite totals (their totals are finite measure values on Hahn pieces); their sum is the total variation. (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)

Proof

technique · direct
1.1L1L3givenchoose

The sum ν1+ν2 is finite-valued and countably additive, and vanishes on every μ-null set, so is a finite signed or finite complex measure of the same type and is absolutely continuous. In the signed case [L3] gives finite total variation for both summands and the sum. Any increasing finite-measure exhaustion for μ is therefore common to these measures. Under AC apply [L1] to choose h1,h2∈L1(μ) representing ν1,ν2, and to obtain the almost-everywhere unique density of their sum.

2.1L1L2step 1.1algebra∎

By [L2], h1+h2 is L1, and for every measurable E, (ν1+ν2)(E)=∫Eh1 dμ+∫Eh2 dμ=∫E(h1+h2) dμ. Thus it is a density of the sum. Uniqueness in [L1] gives the asserted equality of RN classes.

Depends on

Used by

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Sources