TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null

Statement

Let $n \geq 1$ and assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Then:

For every $u \in R^{n}$ with $u \neq 0$ and every real $c$ , the affine hyperplane $H_{u, c} : = {x \in R^{n} : ⟨ u, x ⟩ = c}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ) is Lebesgue measurable with $λ_{n} (H_{u, c}) = 0$ , and so is every subset of it.
Every proper linear subspace $W ⊊ R^{n}$ (Linear subspace of a vector space) is Lebesgue measurable with $λ_{n} (W) = 0$ .

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

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, a nonzero $u \in R^{n}$ , a real $c$ , and a proper linear subspace $W$ of $R^{n}$ .

[L1]

Assuming countable choice, a Lipschitz self-map of $R^{n}$ carries a set of Lebesgue outer measure zero to a Lebesgue measurable set of measure zero (A Lipschitz self-map of $R^{n}$ carries Lebesgue null sets to Lebesgue null sets).

[L2]

For $i_{0} < n$ and a real $c$ , the coordinate hyperplane ${x \in R^{n} : x_{i_{0}} = c}$ is Lebesgue measurable with measure $0$ (A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in $R^{n}$ ).

[L3]

Assuming countable choice, $λ_{n}$ is a complete measure on $L (R^{n})$ , so every subset of a measurable null set is measurable of measure $0$ (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[F1]

The Euclidean inner product of $x, y \in R^{n}$ is $⟨ x, y ⟩ : = \sum_{k < n} x_{k} y_{k}$ , and it is symmetric, bilinear and positive definite, making $R^{n}$ an inner product space (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , 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 $V$ , $W^{⊥} : = {v \in V : ⟨ v, w ⟩ = 0$ for every $w \in W}$ , and ${0}^{⊥} = V$ (The orthogonal complement $W^{⊥} = {v : ⟨ v, w ⟩ = 0 for all w \in W}$ , 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 $\dim W + \dim W^{⊥} = \dim V$ ).

[F4]

For every linear $L : R^{m} \to R^{n}$ there is $K \geq 0$ with $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for every $h$ (Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ , Linear map between vector spaces over the same field).

[F5]

$f$ is Lipschitz with constant $L \geq 0$ if $d_{Y} (f (x), f (x^{'})) \leq L d_{X} (x, x^{'})$ for all $x, x^{'}$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction), and $∥ x - y ∥_{2} = d_{2} (x, y)$ (Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page, claim 3; The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ; $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

Proof

technique · direct

1.1F1

Fix $j < n$ with $u_{j} \neq 0$ and define $Ψ : R^{n} \to R^{n}$ by $Ψ (x)_{l} : = x_{l}$ for $l \neq j$ and $Ψ (x)_{j} : = (c - \sum_{l \neq j} u_{l} x_{l}) / u_{j}$ . Then $Ψ$ carries the coordinate hyperplane $P : = {x : x_{j} = 0}$ onto $H_{u, c}$ : a point of $P$ has $⟨ u, Ψ (x) ⟩ = \sum_{l \neq j} u_{l} x_{l} + u_{j} Ψ (x)_{j} = c$ , and conversely a point $y \in H_{u, c}$ is $Ψ (x)$ for the point $x$ agreeing with $y$ off the coordinate $j$ and having $x_{j} = 0$ .

1.2F4F5

$Ψ$ is Lipschitz: the difference $Ψ (x) - Ψ (x^{'})$ equals $L (x - x^{'})$ for the linear map $L$ obtained from $Ψ$ by deleting the constant $c / u_{j}$ , so $d_{2} (Ψ (x), Ψ (x^{'})) = ∥ L (x - x^{'}) ∥_{2} \leq K d_{2} (x, x^{'})$ for a real $K \geq 0$ .

2.1step 1.1step 1.2L1L2L3

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 $H_{u, c} = Ψ [P]$ is Lebesgue measurable with $λ_{n} (H_{u, c}) = 0$ ; completeness then gives the same for every subset of it, which is claim 1.

3.1step 2.1L3F1F2F3∎

If $W$ is a proper linear subspace then $W^{⊥} \neq {0}$ : otherwise $W = W^{⊥ ⊥} = {0}^{⊥} = R^{n}$ . Choosing a nonzero $u \in W^{⊥}$ puts $W$ inside $H_{u, 0}$ , so claim 1 and completeness make $W$ Lebesgue measurable of measure $0$ ; at $n = 1$ the hyperplane $H_{u, c}$ is the singleton ${c / u_{0}}$ and the only proper subspace is ${0}$ .

Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null