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CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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Q∩[0,1] has measure zero and not content zero, although it is bounded

Statement refuted

Refuted claim: every set of measure zero has content zero (FALSE: every set of measure zero has content zero).

The witness is E:=QR∩[0,1], the rationals of the unit interval (The rationals embed densely in the reals, Intervals of R: the nine order-convex forms, nondegeneracy, and length). It is at most countable, hence null; it is bounded; and every finite family of intervals covering it has total length at least 1, because the union of finitely many closed intervals is closed and contains the closure of E, which is all of [0,1]. The refutation is carried out in full in FALSE: every set of measure zero has content zero; this item records the witness and says what makes it work.

Facts & Assumptions

Given: The set E=QR∩[0,1].

[A1]

The refuted claim: every subset of R of measure zero has content zero.

[L3]

A finite family of closed intervals covering [0,1] has total length at least 1 (If finitely many intervals cover a closed bounded interval [a,b], the sum of their lengths is at least b−a).

[L4]

Content zero means a finite cover of total length below every positive ε; on compact sets content zero and measure zero coincide (Measure zero (a countable cover by intervals of total length below every ε) and content zero (a finite such cover), For a compact subset of R, measure zero and content zero coincide).

Counterexample

technique · direct
1.1

E has measure zero by [L1], and E⊆[0,1] is bounded.

L1
1.2

Any finite family of closed intervals covering E has total length at least 1: its union is closed by [L2] and contains E, hence contains [0,1] by [L2], and [L3] applies.

L2L3
2.1

So E does not have content zero, since a witness at ε=2−1 would give a finite cover of total length at most 2−1<1; E therefore witnesses the failure of [A1].

step 1.1step 1.2A1L4∎

Remarks

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