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The boundary of a convex set is Lebesgue null

Statement

Assume countable choice. For every convex set $C \subseteq R^{d}$ , its topological boundary $\partial C = \overline{C} ∖ C^{\circ}$ has $d$ -dimensional Lebesgue measure zero. If $C$ has empty interior, then $C$ itself is contained in a proper affine subspace and is Lebesgue null.

Facts & Assumptions

Given: Countable choice, a finite integer $d \geq 1$ , and a convex set $C \subseteq R^{d}$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable choice permits the countable approximations used by the Lebesgue measure suppliers.

[F1]

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

[F2]

The geometric simplex spanned by affinely independent vertices, Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ , and The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : Affine independence of $c_{0}, x_{1}, \dots, x_{d}$ means the differences $x_{i} - c_{0}$ are linearly independent. Their closed simplex contains the image of ${t_{i} > 0 : \sum_{i = 1}^{d} t_{i} < 1}$ under $t \mapsto c_{0} + \sum_{i} t_{i} (x_{i} - c_{0})$ . The linear part and its inverse are continuous, so this image is a nonempty open set; balls use the Euclidean norm.

[F3]

A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not: An invertible linear map $T$ scales Lebesgue measure by $∣ \det T ∣$ .

[F4]

Measure of a set difference when the smaller set has finite measure and Measures are monotone: For nested measurable sets of finite measure, the measure of the difference is the difference of their measures, and measure is monotone.

[F5]

Finite and countable subadditivity of measures: A countable union of null measurable sets is null.

[F6]

A finite-dimensional normed subspace is closed and Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation: Finite-dimensional linear subspaces are closed, and translation preserves Lebesgue measurability and measure.

[F7]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line and Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure: A closed bounded subset of $R^{d}$ is compact and has finite Lebesgue measure.

Proof

technique · affine-hull reduction and homothetic contraction

1.1F1F2F6givenchoosealgebra

If $C = \emptyset$ , the claim is immediate. Otherwise choose $c_{0} \in C$ and let $W$ be the linear span of ${x - c_{0} : x \in C}$ . If $W = R^{d}$ , finite-dimensional linear algebra selects $d$ linearly independent differences $x_{i} - c_{0}$ from this spanning family. By definition these $d + 1$ points are affinely independent; their simplex lies in $C$ by convexity and contains a nonempty open set by [F2], contradicting empty interior. Thus $W$ is proper and $C \subseteq c_{0} + W$ . By [F1], $W$ is null; [F6] makes $c_{0} + W$ closed and null. Since $\overline{C} \subseteq c_{0} + W$ , both $C$ and its boundary are null. This proves the empty-interior case.

1.2A1givenalgebra

Suppose $C$ has nonempty interior. Choose $x_{0} \in C^{\circ}$ and $ρ > 0$ with $B (x_{0}, ρ) \subseteq C$ . If $y \in \overline{C}$ and $0 < t < 1$ , [A1] selects $y_{j} \in C$ with $y_{j} \to y$ . Convexity gives $B (x_{0} + t (y_{j} - x_{0}), (1 - t) ρ) \subseteq C .$ The centers converge to $x_{0} + t (y - x_{0})$ , so that limit point has a ball of radius $(1 - t) ρ / 2$ contained in $C$ . Therefore $x_{0} + t (\overline{C} - x_{0}) \subseteq C^{\circ}$ .

2.1F3F4F6F7step 1.2algebra

For each integer $r$ large enough that $K_{r} = \overline{C} \cap \overline{B} (0, r)$ contains $x_{0}$ , [F7] makes $K_{r}$ compact with finite measure. The contraction $x_{0} + t (K_{r} - x_{0})$ lies in $C^{\circ}$ by step 1.2 and in $K_{r}$ by convexity. Write this contraction as $t K_{r} + (1 - t) x_{0}$ . Applying [F3] to $t I_{d}$ and then translation invariance [F6] gives its measure as $t^{d} λ_{d} (K_{r})$ . Hence $\partial C \cap K_{r} \subseteq K_{r} ∖ (x_{0} + t (K_{r} - x_{0})),$ and [F4] gives $λ_{d} (\partial C \cap K_{r}) \leq (1 - t^{d}) λ_{d} (K_{r}) .$ Let $t ↑ 1$ to obtain $λ_{d} (\partial C \cap K_{r}) = 0$ .

3.1A1F5step 2.1∎

The bounded sets $K_{r}$ cover $\overline{C}$ as $r$ ranges over sufficiently large integers. Therefore $\partial C$ is a countable union of the null sets in step 2.1 and is null by [F5].

Depends on

Used by

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Sources

Wilfrid Gangbo and Robert J. McCann, The Geometry of Optimal Transportation, proof of Proposition 3.4(iv), printed p. 20 (convex-boundary nullity); local proof supplied here (standard reference, not scraped)