CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26‡ rests on unproved material (inherited)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Assuming countable choice and a well-ordering of the real line, a Bernstein set is dense, has inner measure $0$ , and is not Lebesgue measurable

Statement refuted

Refuted claim: every dense subset of $R$ of inner measure $0$ is Lebesgue measurable.

Assume the Axiom of Countable Choice and that the real line can be well ordered. Then a Bernstein set refutes the claim: it is dense in $R$ , has inner measure $0$ , and is not Lebesgue measurable.

Facts & Assumptions

Given: The Axiom of Countable Choice and a well-ordering of the real line.

[L1]

Assuming the real line can be well ordered, a Bernstein set exists (Assuming the real line can be well ordered, a Bernstein set exists).

[L2]

Assuming countable choice, a Bernstein set has inner measure $0$ , and in every nondegenerate bounded interval its intersection has full outer measure (A Bernstein set has inner measure $0$ , and in every nondegenerate interval its intersection has full outer measure).

[L3]

Assuming countable choice, a Bernstein set is not Lebesgue measurable (Assuming the Axiom of Countable Choice, a Bernstein set is not Lebesgue measurable).

Counterexample

technique · direct

1.1

By [L1] choose a Bernstein set $B$ . If some nonempty open interval were disjoint from $B$ , it would contain a nondegenerate closed subinterval, hence a nonempty perfect subset of $R$ , contradicting the Bernstein property. So $B$ meets every nonempty open interval and is dense in $R$ .

L1algebra

2.1

The given Axiom of Countable Choice supplies the hypothesis of [L2] and [L3]. Hence $λ_{*} (B) = 0$ by [L2], and $B$ is not Lebesgue measurable by [L3].

step 1.1L2L3

3.1

Therefore $B$ is a dense subset of $R$ of inner measure $0$ that is not Lebesgue measurable, so it refutes the claim.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

32 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

Bernstein set (Wikipedia) (standard reference, not scraped)

Assuming countable choice and a well-ordering of the real line, a Bernstein set is dense, has inner measure 00, and is not Lebesgue measurable