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.

The Lebesgue decomposition of a sigma-finite signed measure is unique

Statement

Let $μ$ be a positive measure and let $ν$ be a signed measure satisfying the common finite-exhaustion hypothesis of Every sigma-finite signed measure admits a Lebesgue decomposition relative to a sigma-finite positive measure. If $ν = ν_{a} + ν_{s} = {\tilde{ν}}_{a} + {\tilde{ν}}_{s},$ with $ν_{a}, {\tilde{ν}}_{a} ≪ μ$ and $ν_{s}, {\tilde{ν}}_{s} ⊥ μ$ , then $ν_{a} = {\tilde{ν}}_{a}, ν_{s} = {\tilde{ν}}_{s} .$

Facts & Assumptions

Given: Two Lebesgue decompositions of the same signed measure $ν$ relative to a positive measure $μ$ .

[L1]

A measure concentrated on a $μ$ -null set is singular with respect to $μ$ . (A positive, signed, or complex measure concentrated on a measurable set)

[L2]

A signed or complex measure that is both absolutely continuous and singular with respect to $μ$ is zero. (A signed or complex measure that is both absolutely continuous and singular with respect to the same positive measure is zero)

Proof

technique · direct

1.1

Subtract the two decompositions to obtain $ν_{a} - {\tilde{ν}}_{a} = {\tilde{ν}}_{s} - ν_{s} .$ The left-hand side is absolutely continuous with respect to $μ$ , because differences of absolutely continuous measures are again absolutely continuous.

givenalgebra

1.2

Choose $μ$ -null sets $N$ and $\tilde{N}$ on which $ν_{s}$ and ${\tilde{ν}}_{s}$ are concentrated. Then ${\tilde{ν}}_{s} - ν_{s}$ is concentrated on $N \cup \tilde{N}$ , which is still $μ$ -null, so [L1] makes the right-hand side singular with respect to $μ$ .

givenL1choose

2.1

The common difference in steps 1.1 and 1.2 is therefore both absolutely continuous and singular with respect to $μ$ , so [L2] forces $ν_{a} - {\tilde{ν}}_{a} = 0$ . Substituting back into the decomposition identity gives $ν_{s} - {\tilde{ν}}_{s} = 0$ as well.

step 1.1step 1.2L2∎

Depends on

Used by

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, Theorem 13.5 (standard reference, not scraped)
Sheldon Axler, Measure, Integration & Real Analysis, 9.35 (standard reference, not scraped)