CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 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.

Equivalent sigma-finite positive measures have reciprocal Radon-Nikodym derivatives almost everywhere

Statement

Let $μ$ and $ν$ be equivalent sigma-finite positive measures on the same measurable space. Then $\frac{d ν}{d μ} \frac{d μ}{d ν} = 1 ν -almost everywhere,$ and therefore also $μ$ -almost everywhere.

Facts & Assumptions

Given: Sigma-finite positive measures $μ$ and $ν$ with $μ ≪ ν ≪ μ$ .

[L1]

Under one exhaustion finite for the outer and intermediate positive measures and the variation of the inner measure, the chain rule gives $d η / d λ = (d η / d κ) (d κ / d λ)$ almost everywhere along $η ≪ κ ≪ λ$ (Radon-Nikodym derivatives satisfy the chain rule along nu << mu << lambda).

[L2]

The constant function $1$ represents $d μ / d μ$ because $μ (E) = \int_{E} 1 d μ$ for every measurable set, and the representing density is unique up to almost-everywhere equality. (A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density)

Proof

technique · direct

1.1

Choose increasing finite-measure exhaustions $(A_{n})$ for $μ$ and $(B_{n})$ for $ν$ , and put $X_{n} : = A_{n} \cap B_{n}$ . After replacing both exhaustions by finite unions, $(X_{n})$ is increasing, covers $X$ , and is finite for both measures. Apply [L1] to the chain $μ ≪ ν ≪ μ$ on this common exhaustion. Then $\frac{d μ}{d μ} = \frac{d μ}{d ν} \frac{d ν}{d μ} μ -almost everywhere.$ By [L2], $d μ / d μ = 1$ almost everywhere, so $\frac{d μ}{d ν} \frac{d ν}{d μ} = 1 μ -almost everywhere.$

L1L2construct

2.1

Interchanging the roles of $μ$ and $ν$ gives $\frac{d ν}{d μ} \frac{d μ}{d ν} = 1 ν -almost everywhere.$ Because $μ$ and $ν$ have the same null sets, the two almost-everywhere conclusions are equivalent.

step 1.1L1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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 13.5 and Exercise 13.6 (standard reference, not scraped)