TheoremStatement: Literature-sourcedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)

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Integrating against a Radon-Nikodym derivative recovers integration against the measure

Statement

Let $μ$ be a sigma-finite positive measure and let $ν$ be a signed measure or a finite complex measure with $ν ≪ μ$ . If $h$ is a representative of $d ν / d μ$ , then $ν (E) = \int_{E} h d μ (E \in A) .$ More generally, if $g = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ is the canonical disjoint representation of a simple measurable function and $∣ ν ∣ (E_{j}) < + \infty$ for every $j$ , then $\int g d ν = \sum_{j = 1}^{m} c_{j} \int_{E_{j}} h d μ .$ In particular, whenever the Lebesgue integral of $g h$ is defined, one has $\int g d ν = \int g h d μ .$

Facts & Assumptions

Given: A representative $h$ of $d ν / d μ$ .

[L1]

By definition, a representative of a Radon-Nikodym derivative satisfies the measurable-set integral formula (The Radon-Nikodym derivative as an almost-everywhere equivalence class). Its existence is not needed here because $h$ is supplied.

[L3]

For a simple function in canonical disjoint form with each $∣ ν ∣ (E_{j}) < + \infty$ , the simple integral against $ν$ is $\int g d ν = \sum_{j = 1}^{m} c_{j} ν (E_{j}) .$ (The simple integral against a signed or complex measure)

[L4]

The Lebesgue integral is linear on $L^{1} (μ)$ . (The Lebesgue integral is linear on $L^{1} (μ)$ )

Proof

technique · direct

1.1L1given

The measurable-set identity $ν (E) = \int_{E} h d μ$ is exactly [L1].

2.1L1givenalgebra

For each nonzero level set $E_{j}$ , the hypothesis $∣ ν ∣ (E_{j}) < + \infty$ makes $h 1_{E_{j}}$ integrable. In the signed case, apply step 1.1 to $E_{j} \cap {h \geq 0}$ and $E_{j} \cap {h < 0}$ : the positive and negative integrals are finite because the corresponding values of $ν$ have absolute value at most $∣ ν ∣ (E_{j})$ . In the finite complex case, $h \in L^{1} (μ)$ already by the definition of its derivative.

3.1L3L4step 1.1step 2.1algebra∎

If $g = \sum_{j = 1}^{m} c_{j} χ_{E_{j}}$ is canonical disjoint, then [L3] and step 1.1 give $\int g d ν = \sum_{j = 1}^{m} c_{j} ν (E_{j}) = \sum_{j = 1}^{m} c_{j} \int_{E_{j}} h d μ .$ By step 2.1, every $h 1_{E_{j}}$ is in $L^{1} (μ)$ , so [L4] identifies this finite sum with $\int g h d μ$ .

Depends on

Used by

Two common diagonalizations differ by a bimeasurable base isomorphism and a measurable field of unitaries Lemma

Dependency tree · two levels

15 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)