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A Lebesgue measurable subgroup of (Rn,+) of positive measure is all of Rn

Statement

Let n1 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 n1, the Axiom of Countable Choice, and a Lebesgue measurable subgroup G of (Rn,+) with λn(G)>0.

[L1]

Assuming countable choice, a Lebesgue measurable ERn with λn(E)>0 has a real r>0 with B(0,r)EE (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 HG is a subgroup when eH, 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 xF there is a natural number m1 with x<m1F (Every complete ordered field is Archimedean); and for every real ε>0 there is a natural k1 with 1/k<ε (For every ε>0 in a complete ordered field there is a natural n1 with 1/n<ε).

[F3]

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

Proof

technique · direct
1.1

Since G is a subgroup, 0G and xyG whenever x,yG, so GGG; conversely G=G0GG, and therefore GG=G.

F1
1.2

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

L1L2
2.1

Let xRn. The Archimedean property gives a natural m1 with x2/r<m, so m1x2=m1x2<r and m1xB(0,r)G by steps 1.1 and 1.2.

step 1.1step 1.2F2F4
3.1

A subgroup is closed under addition, so an induction on j shows j(m1x)G for every natural j, the case j=0 being 0G; taking j=m gives x=m(m1x)G, and as x was arbitrary, G=Rn.

step 2.1F1F3

Depends on

Used by

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Sources