TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Radon-Nikodym derivatives add almost everywhere

Statement

Let $μ$ be a sigma-finite positive measure. Let $ν_{1}, ν_{2}$ be either two finite signed measures or two finite complex measures, with $ν_{1} ≪ μ$ and $ν_{2} ≪ μ$ . Then $\frac{d (ν_{1} + ν_{2})}{d μ} = \frac{d ν_{1}}{d μ} + \frac{d ν_{2}}{d μ} μ -almost everywhere.$

Facts & Assumptions

Given: Measures $ν_{1}, ν_{2}$ absolutely continuous with respect to $μ$ .

[L1]

A representative of $d ν / d μ$ recovers the measurable-set values of $ν$ . (Integrating against a Radon-Nikodym derivative recovers integration against the measure)

[L2]

The representing density is unique up to $μ$ -almost-everywhere equality for signed measures under the Radon--Nikodym hypotheses and for finite complex measures (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).

Proof

technique · direct

1.1

Choose representatives $h_{j}$ of $d ν_{j} / d μ$ for $j = 1, 2$ . For every measurable set $E$ , [L1] gives $(ν_{1} + ν_{2}) (E) = ν_{1} (E) + ν_{2} (E) = \int_{E} h_{1} d μ + \int_{E} h_{2} d μ = \int_{E} (h_{1} + h_{2}) d μ .$

L1choosealgebra

2.1

The function $h_{1} + h_{2}$ is therefore a density for $ν_{1} + ν_{2}$ , so [L2] yields $\frac{d (ν_{1} + ν_{2})}{d μ} = h_{1} + h_{2} = \frac{d ν_{1}}{d μ} + \frac{d ν_{2}}{d μ} μ -almost everywhere.$

step 1.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Richard F. Bass, Real Analysis for Graduate Students, Exercise 13.8 (standard reference, not scraped)