ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-31

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The density $2 x$ on $[0, 1]$ is the Radon-Nikodym derivative of its density measure

Example

Let $ν$ be the measure on $(R, B (R))$ defined by $ν (E) : = \int_{E} 2 x χ_{[0, 1]} (x) d λ (x) .$ Then $ν ≪ λ$ and $\frac{d ν}{d λ} = 2 x χ_{[0, 1]} (x) λ -almost everywhere.$

Facts & Assumptions

Given: The measurable function $h (x) = 2 x χ_{[0, 1]} (x)$ .

[L1]

A nonnegative measurable function defines a measure by $E \mapsto \int_{E} h d λ$ . (The measure with density $f$ relative to $μ$ )

[L2]

Integrating against a density recovers the product integral. (Integrating against a density agrees with integrating the product)

[L3]

For a sigma-finite positive base and a signed measure satisfying the common finite-exhaustion hypothesis, the Radon--Nikodym derivative is the almost-everywhere class of a function whose measurable-set integrals recover the measure (The Radon-Nikodym derivative as an almost-everywhere equivalence class).

Verification

technique · direct

1.1

By [L1], $ν$ is a measure with density $h$ relative to $λ$ , so in particular $ν ≪ λ$ .

L1given

2.1

The measure $λ$ is sigma-finite, $ν$ is finite because $ν (R) = 1$ , and $[- n, n]$ is a common finite exhaustion. For every measurable $E$ , the definition of $ν$ gives $ν (E) = \int_{E} h d λ$ . Therefore [L3] identifies $h$ as a representative of $d ν / d λ$ .

step 1.1L2L3algebra∎

Depends on

Used by

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Dependency tree · two levels

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The density 2x2x on [0,1][0,1] is the Radon-Nikodym derivative of its density measure