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 ${0}$ .

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

John K. Hunter, Measure Theory (UC Davis lecture notes), Chapter 2 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory (GSM 126), Section 1.2 (standard reference, not scraped)

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