CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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The Cantor function fails Luzin's property $(N)$

Statement refuted

Every continuous function of bounded variation has Luzin's property $(N)$ .

Facts & Assumptions

Given: Countable choice and the standard Cantor--Lebesgue function $C : [0, 1] \to [0, 1]$ .

Counterexample

technique · direct

1.1

$C$ is continuous, nondecreasing, and maps the ternary Cantor set $K$ onto $[0, 1]$ .

given

2.1

The set $K$ has Lebesgue measure zero, but $λ (C (K)) = 1$ .

step 1.1

3.1

Thus $C$ fails $(N)$ as defined in Luzin's property $(N)$ on a compact interval; Banach--Zarecki confirms it cannot be AC.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Donald L. Cohn, Measure Theory, 2nd ed., §6.3, Exercise 5 (standard reference, not scraped)

The Cantor function fails Luzin's property (N)(N)