Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every affine hyperplane of Rn, and hence every proper linear subspace, is Lebesgue null

Statement

Let n1 and assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Then:

  1. For every uRn with u0 and every real c, the affine hyperplane Hu,c  :=  {xRn:u,x=c} (The Euclidean inner product x,y=k<nxkyk on Rn) is Lebesgue measurable with λn(Hu,c)=0, and so is every subset of it.
  2. Every proper linear subspace WRn (Linear subspace of a vector space) is Lebesgue measurable with λn(W)=0.

At n=1 a hyperplane is the singleton {c/u0} and the only proper linear subspace is {0}.

Facts & Assumptions

Given: A natural number n1, the Axiom of Countable Choice, a nonzero uRn, a real c, and a proper linear subspace W of Rn.

[L1]

Assuming countable choice, a Lipschitz self-map of Rn carries a set of Lebesgue outer measure zero to a Lebesgue measurable set of measure zero (A Lipschitz self-map of Rn carries Lebesgue null sets to Lebesgue null sets).

[L2]

For i0<n and a real c, the coordinate hyperplane {xRn:xi0=c} is Lebesgue measurable with measure 0 (A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in Rn).

[L3]

Assuming countable choice, λn is a complete measure on L(Rn), so every subset of a measurable null set is measurable of measure 0 (Assuming countable choice, L(Rn) is a sigma-algebra containing every elementary set and λn is a complete measure extending elementary volume).

[F1]

The Euclidean inner product of x,yRn is x,y:=k<nxkyk, and it is symmetric, bilinear and positive definite, making Rn an inner product space (The Euclidean inner product x,y=k<nxkyk on Rn, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[F2]

For a linear subspace W of an inner product space V, W:={vV:v,w=0 for every wW}, and {0}=V (The orthogonal complement W={v:v,w=0 for all wW}, Linear subspace of a vector space).

[F3]

For every subspace W of a finite-dimensional inner product space V, W=W (In finite dimension, W=W and dimW+dimW=dimV).

[F4]

For every linear L:RmRn there is K0 with Lh2Kh2 for every h (Every Euclidean linear map has a unique matrix and satisfies Lh2Kh2 for some K0, Linear map between vector spaces over the same field).

Proof

technique · direct
1.1

Fix j<n with uj0 and define Ψ:RnRn by Ψ(x)l:=xl for lj and Ψ(x)j:=(cljulxl)/uj. Then Ψ carries the coordinate hyperplane P:={x:xj=0} onto Hu,c: a point of P has u,Ψ(x)=ljulxl+ujΨ(x)j=c, and conversely a point yHu,c is Ψ(x) for the point x agreeing with y off the coordinate j and having xj=0.

F1
1.2

Ψ is Lipschitz: the difference Ψ(x)Ψ(x) equals L(xx) for the linear map L obtained from Ψ by deleting the constant c/uj, so d2(Ψ(x),Ψ(x))=L(xx)2Kd2(x,x) for a real K0.

F4F5
2.1

The coordinate hyperplane P is Lebesgue measurable of measure 0, hence of outer measure 0, so steps 1.1 and 1.2 with the Lipschitz lemma give that Hu,c=Ψ[P] is Lebesgue measurable with λn(Hu,c)=0; completeness then gives the same for every subset of it, which is claim 1.

step 1.1step 1.2L1L2L3
3.1

If W is a proper linear subspace then W{0}: otherwise W=W={0}=Rn. Choosing a nonzero uW puts W inside Hu,0, so claim 1 and completeness make W Lebesgue measurable of measure 0; at n=1 the hyperplane Hu,c is the singleton {c/u0} and the only proper subspace is {0}.

step 2.1L3F1F2F3

Depends on

Used by

Dependency tree · two levels

96 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