False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-31

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FALSE: absolutely continuous measures always have bounded Radon-Nikodym derivatives

Statement

False claim: If $ν ≪ μ$ , then $d ν / d μ$ is bounded.

Facts & Assumptions

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

[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).

Refutation

technique · direct

1.1

The function $x^{- 1 / 2} χ_{(0, 1]}$ defines a finite measure by [L1], and [L2] makes it absolutely continuous with respect to $λ$ . Because $λ$ is sigma-finite and the measure is finite, the exhaustion $[- n, n]$ verifies the remaining Radon--Nikodym hypotheses. Moreover $\int_{0}^{1} x^{- 1 / 2} d λ = 2.$

L1L2givenalgebra

2.1

The same function is a representative of $d ν / d λ$ , and it is unbounded near $0$ . Therefore absolute continuity alone does not force boundedness of the derivative.

step 1.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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, Chapter 13 (standard reference, not scraped)