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

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The Radon-Nikodym derivative is integrable exactly when the absolutely continuous part is finite

Statement

Let $μ$ be a sigma-finite positive measure and let $ν_{a}$ be the absolutely continuous part of a signed measure under the common finite-exhaustion hypothesis relative to $μ$ . Then $\frac{d ν_{a}}{d μ} \in L^{1} (μ) ⟺ ν_{a} is finite.$

Facts & Assumptions

Given: The absolutely continuous part $ν_{a}$ of a measure relative to $μ$ .

[L1]

Integrability means finiteness of the integral of the absolute value. (Integrable real and complex functions, and their integrals)

[L2]

A representative of $d ν_{a} / d μ$ recovers the measurable-set values of $ν_{a}$ . (The Radon-Nikodym derivative as an almost-everywhere equivalence class)

[L3]

For an absolutely continuous finite signed or finite complex measure, the total variation has density $∣ d ν_{a} / d μ ∣$ . (The total variation of an absolutely continuous signed or complex measure has density the absolute value of the Radon-Nikodym derivative)

[L4]

For a finite signed measure, $∣ ν_{a} ∣ (X) = ν_{a}^{+} (X) + ν_{a}^{-} (X)$ (For a signed measure, total variation is nu-plus plus nu-minus, finite partitions suffice, and nu-plus and nu-minus are extremal).

[L5]

Total variation is the supremum of absolute-value sums over measurable partitions (The total variation |nu|(E) from countable measurable partitions).

Proof

technique · direct

1.1

Let $f$ be a representative of $d ν_{a} / d μ$ . If $f \in L^{1} (μ)$ , then for every countable measurable partition $(E_{j})$ of $X$ , [L2] gives $\sum_{j} ∣ ν_{a} (E_{j}) ∣ \leq \sum_{j} \int_{E_{j}} ∣ f ∣ d μ = \int_{X} ∣ f ∣ d μ < + \infty .$ Taking the supremum over partitions in [L5] yields $∣ ν_{a} ∣ (X) < + \infty$ , so $ν_{a}$ is finite.

L1L2L5givenalgebra

1.2

Conversely, assume $ν_{a}$ is finite. By [L4], this is equivalent to $∣ ν_{a} ∣ (X) < + \infty$ .

L4given

2.1

Since $ν_{a}$ is finite and absolutely continuous, [L3] gives $∣ ν_{a} ∣ (X) = \int_{X} ∣ \frac{d ν_{a}}{d μ} ∣ d μ < + \infty .$ Hence [L1] shows that $d ν_{a} / d μ \in L^{1} (μ)$ .

L1L3step 1.2

3.1

Step 1.1 proves $d ν_{a} / d μ \in L^{1} (μ) \Rightarrow ν_{a}$ finite, and steps 1.2-2.1 prove the converse.

step 1.1step 1.2step 2.1∎

Depends on

Used by

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Sources

Richard F. Bass, Real Analysis for Graduate Students, Chapter 13 (standard reference, not scraped)