Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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A Lebesgue measurable subset of R with empty interior has measure zero

Statement

Assume the Axiom of Countable Choice. Every Lebesgue measurable subset of R with empty interior has measure zero.

Facts & Assumptions

Given: The Axiom of Countable Choice and the Smith-Volterra-Cantor set S.

[F1]

A is nowhere dense when the interior of its closure is empty (Nowhere dense, meager (first category), residual, and second category subsets of R).

[L2]

A subset of R has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-interval covers (A subset of R has Lebesgue outer measure zero if and only if it has measure zero in the sense of countable closed-interval covers).

Refutation

technique · direct
1.1

The Smith-Volterra-Cantor set is nowhere dense by [L1], so [F1] gives that it has empty interior.

L1F1
1.2

The same source item [L1] says that S is not null, so [L2] gives λ1(S)0; since Lebesgue measure is nonnegative, this means λ1(S)>0.

L1L2algebra
2.1

So a measurable set can have empty interior and still have positive Lebesgue measure; the Smith-Volterra-Cantor set refutes the statement.

step 1.1step 1.2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources