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

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An absolutely continuous finite measure can have an unbounded Radon-Nikodym derivative

Statement refuted

Every absolutely continuous finite measure has a bounded Radon-Nikodym derivative.

Facts & Assumptions

Given: The measure $ν (E) = \int_{E} x^{- 1 / 2} χ_{(0, 1]} (x) d λ (x)$ on $R$ .

[L1]

A nonnegative measurable density defines a positive measure; when that measure is absolutely continuous with respect to a sigma-finite base and satisfies the common finite-exhaustion hypothesis, a density recovering all measurable-set values represents its Radon--Nikodym derivative (The measure with density $f$ relative to $μ$ , The Radon-Nikodym derivative as an almost-everywhere equivalence class).

[L2]

The integral over a null set vanishes (A nonnegative integral over a null set vanishes).

Counterexample

technique · direct

1.1

The density $x^{- 1 / 2} χ_{(0, 1]}$ is integrable on $(0, 1]$ , since $\int_{0}^{1} x^{- 1 / 2} d λ = 2.$ Therefore [L1] makes $ν$ a finite measure, and [L2] shows that it vanishes on every Lebesgue-null set. Thus $ν ≪ λ$ ; the exhaustion $[- n, n]$ verifies the remaining Radon--Nikodym hypotheses.

L1L2givenalgebra

2.1

The same density represents $d ν / d λ$ , but it is unbounded near $0$ . Hence a finite absolutely continuous measure need not have a bounded derivative.

step 1.1L1∎

Depends on

Used by

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