TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)

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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 on the same measurable space. No choice or sigma-finiteness assumption is needed for uniqueness. 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]

Singularity provides a measurable $μ$ -null carrier; concentration means vanishing on every measurable subset of its complement. (Mutual singularity for signed or complex measures, A positive, signed, or complex measure concentrated on a measurable set)

[L2]

An absolutely continuous measure vanishes on every measurable $μ$ -null set. (Absolute continuity of a signed or complex measure with respect to a positive measure)

Proof

technique · direct

1.1L1L2givenchoose

Choose measurable $μ$ -null carriers $N_{1}, N_{2}$ for $ν_{s}, {\tilde{ν}}_{s}$ , respectively. These are just two existential witnesses, so no choice axiom is needed. Let $N = N_{1} \cup N_{2}$ , again measurable and $μ$ -null. Both singular parts vanish on every measurable subset of $X ∖ N$ , and both absolutely continuous parts vanish on every measurable subset of $N$ .

2.1step 1.1givenalgebra

For every measurable $E$ , evaluate each given decomposition on $E ∖ N$ . It gives $ν (E ∖ N) = ν_{a} (E ∖ N) = {\tilde{ν}}_{a} (E ∖ N)$ . As both absolutely continuous parts vanish on $E \cap N$ , additivity yields $ν_{a} (E) = ν (E ∖ N) = {\tilde{ν}}_{a} (E)$ . No difference of infinite values occurs.

3.1step 1.1givenalgebra∎

Similarly, evaluating on $E \cap N$ gives $ν (E \cap N) = ν_{s} (E \cap N) = {\tilde{ν}}_{s} (E \cap N)$ . Both singular parts vanish on $E ∖ N$ , so $ν_{s} (E) = ν (E \cap N) = {\tilde{ν}}_{s} (E)$ . Since $E$ was arbitrary, both components are unique.

Depends on

Used by

Dependency tree · two levels

6 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)