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

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.

A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density

Statement

Assume the Axiom of Choice. Let $μ$ be a positive measure and let $ν$ be a signed measure on $(X, A)$ . Assume there is an increasing measurable exhaustion $(X_{n})$ with $⋃_{n} X_{n} = X$ , $μ (X_{n}) < + \infty$ , and $∣ ν ∣ (X_{n}) < + \infty$ for every $n$ . If $ν ≪ μ$ , then there exists a measurable real-valued function $f$ , unique up to $μ$ -almost-everywhere equality, such that $ν (E) = \int_{E} f d μ (E \in A) .$ If in addition $∣ ν ∣ (X) < + \infty$ , then $f \in L^{1} (μ)$ .

Facts & Assumptions

Given: AC, a positive measure $μ$ , a signed measure $ν$ with a common finite exhaustion $(X_{n})$ , and the hypothesis $ν ≪ μ$ .

[L1]

The Lebesgue decomposition theorem gives $ν = ν_{a} + ν_{s}$ with $ν_{a} ≪ μ$ , $ν_{s} ⊥ μ$ , and $ν_{a} (E) = \int_{E} f d μ$ for every measurable $E$ , with $f$ real-valued and $\int_{X_{n}} ∣ f ∣ d μ < \infty$ . (Every sigma-finite signed measure admits a Lebesgue decomposition relative to a sigma-finite positive measure)

[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)

[L3]

Real integrability means finiteness of both positive and negative integrals; the integral is linear on $L^{1}$ . (Integrable real and complex functions, and their integrals, The Lebesgue integral is linear on $L^{1} (μ)$ )

[L5]

Variation dominates $∣ ν (E) ∣$ by its partition definition. (The total variation |nu|(E) from countable measurable partitions)

[L4]

AC is assumed for the selections in [L1]. (The Axiom of Choice)

Proof

technique · direct

1.1L1choose

By [L1], write $ν = ν_{a} + ν_{s}$ , where $ν_{a} ≪ μ$ , $ν_{s} ⊥ μ$ , and $ν_{a} (E) = \int_{E} f d μ$ for every measurable $E$ , with $f$ real-valued and $\int_{X_{n}} ∣ f ∣ d μ < \infty$ .

2.1step 1.1L2algebra

Because $ν ≪ μ$ and $ν_{a} ≪ μ$ , on each $μ$ -null set $E$ the decomposition gives $0 = 0 + ν_{s} (E)$ , so $ν_{s} ≪ μ$ without subtracting infinite values. Step 1.1 already gives $ν_{s} ⊥ μ$ , so [L2] forces $ν_{s} = 0$ . Hence $ν = ν_{a}$ , and the displayed integral formula for $f$ represents $ν$ on every measurable set.

3.1step 2.1choosealgebra

Suppose $g$ is another measurable real-valued function with $ν (E) = \int_{E} g d μ$ for every measurable $E$ . Fix $n$ . The formula on $X_{n}$ gives $\int_{X_{n}} g d μ = ν (X_{n}) \in R$ , so both positive and negative integrals of $g$ there are finite. Thus $f, g \in L^{1} (X_{n})$ , and [L3] permits subtraction of their integrals over every measurable subset of $X_{n}$ . For each $m \geq 1$ , let $P_{n, m} : = X_{n} \cap {f - g \geq 1 / m}, N_{n, m} : = X_{n} \cap {g - f \geq 1 / m} .$ Applying the two representation formulas to $P_{n, m}$ gives $0 = \int_{P_{n, m}} (f - g) d μ \geq μ (P_{n, m}) / m,$ so $μ (P_{n, m}) = 0$ ; the same argument with $N_{n, m}$ gives $μ (N_{n, m}) = 0$ . Therefore $X_{n} \cap {f \neq g} = ⋃_{m \geq 1} P_{n, m} \cup ⋃_{m \geq 1} N_{n, m}$ is $μ$ -null. Since $X = ⋃_{n} X_{n}$ , the functions $f$ and $g$ agree $μ$ -almost everywhere on $X$ .

3.2L3step 2.1algebra

If $∣ ν ∣ (X) < \infty$ , use the representation on the measurable sets $P = {f \geq 0}$ and $N = {f < 0}$ . It gives $\int f^{+} d μ = ν (P)$ and $\int f^{-} d μ = - ν (N)$ . Both are finite: $∣ ν (E) ∣ \leq ∣ ν ∣ (X)$ for every $E$ , directly from the partition definition of variation. Hence $f \in L^{1} (μ)$ by [L3].

4.1step 2.1step 3.1step 3.2∎

Steps 2.1, 3.1, and 3.2 prove existence, almost-everywhere uniqueness, and the finite-measure integrability clause.

Depends on

Used by

Dependency tree · two levels

28 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 13.4 (standard reference, not scraped)
John K. Hunter, Measure Theory, Theorem 6.27 (standard reference, not scraped)