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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]

A representative of a Radon-Nikodym derivative recovers the measurable-set values of the measure: for signed measures by the Radon-Nikodym theorem, and for finite complex measures by the complex corollary. (The Radon-Nikodym derivative as an almost-everywhere equivalence class, A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density, A finite complex measure absolutely continuous with respect to a sigma-finite positive measure has an integrable complex density)

[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

1.1

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

L1given

2.1

If $g = \sum_{j = 1}^{m} c_{j} χ_{E_{j}}$ is canonical disjoint and each $∣ ν ∣ (E_{j}) < + \infty$ , then [L3] and step 1.1 give $\int g d ν = \sum_{j = 1}^{m} c_{j} ν (E_{j}) = \sum_{j = 1}^{m} c_{j} \int_{E_{j}} h d μ .$ If, in addition, the Lebesgue integral of $g h$ is defined, then [L4] identifies the same finite sum with $\int g h d μ$ .

L3L4step 1.1algebra∎

Depends on

Used by

Radon-Nikodym derivatives add almost everywhere Theorem

Dependency tree · two levels

21 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)