Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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Luzin's property (N) does not imply absolute continuity

Statement refuted

Luzin's property (N) implies absolute continuity.

Facts & Assumptions

Given: Countable choice, F(0)=0, and F(x)=xsin⁡(1/x) for 0<x≤1.

Counterexample

technique · direct
1.1

On each [1/(n+1),1/n], F is C1, hence maps null sets to null sets. Together with the singleton {0} this countable cover proves that F has (N) in the sense of Luzin's property (N) on a compact interval.

given
2.1

F′(x)=sin⁡(1/x)−cos⁡(1/x)/x for x>0; alternating subintervals again give infinite variation. Thus F is not BV.

step 1.1algebra
3.1

The reverse implication in Banach--Zarecki characterisation of absolute continuity requires BV, so this (N) function is not AC.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources