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CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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A Lebesgue measurable subgroup of (Rn,+) of positive measure is all of Rn

Statement

Let n≥1 and assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Let G be a subgroup of the additive group (Rn,+) (Subgroup, Group and abelian group) that is Lebesgue measurable with λn(G)>0. Then

G=Rn.

Equivalently, in the contrapositive form the sources state: a Lebesgue measurable proper subgroup of (Rn,+) has measure zero. Nothing is asserted about subgroups that are not Lebesgue measurable.

Facts & Assumptions

Given: A natural number n≥1, the Axiom of Countable Choice, and a Lebesgue measurable subgroup G of (Rn,+) with λn(G)>0.

[L1]

Assuming countable choice, a Lebesgue measurable E⊆Rn with λn(E)>0 has a real r>0 with B(0,r)⊆E−E (If a Lebesgue measurable subset of Rn has positive measure, its difference set contains an open ball about the origin, Open ball, closed ball and sphere in a metric space).

[F1]

A subset H⊆G is a subgroup when e∈H, H is closed under the operation, and H is closed under inverses (Subgroup, Group and abelian group).

[F2]

Every complete ordered field F is Archimedean: for every x∈F there is a natural number m≥1 with x<m⋅1F (Every complete ordered field is Archimedean); and for every real ε>0 there is a natural k≥1 with 1/k<ε (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

[F3]

Let S⊆N; if 0∈S and σ(m)∈S whenever m∈S, then S=N (The principle of mathematical induction).

Proof

technique · direct
1.1F1

Since G is a subgroup, 0∈G and x−y∈G whenever x,y∈G, so G−G⊆G; conversely G=G−0⊆G−G, and therefore G−G=G.

1.2L1L2

Steinhaus applied to G supplies a real r>0 with B(0,r)⊆G−G.

2.1step 1.1step 1.2F2F4

Let x∈Rn. The Archimedean property gives a natural m≥1 with ∥x∥2/r<m, so ∥m−1x∥2=m−1∥x∥2<r and m−1x∈B(0,r)⊆G by steps 1.1 and 1.2.

3.1step 2.1F1F3∎

A subgroup is closed under addition, so an induction on j shows j (m−1x)∈G for every natural j, the case j=0 being 0∈G; taking j=m gives x=m (m−1x)∈G, and as x was arbitrary, G=Rn.

Depends on

Used by

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Sources