Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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A bounded open Jordan set has an increasing exhaustion by compact finite unions of grid rectangles with vanishing content remainder

Statement

Let n1n\ge1 and let VRnV\subseteq\mathbb R^n be bounded, open, and Jordan measurable. There are compact Jordan sets K1K2V,K_1\subseteq K_2\subseteq\cdots\subseteq V, each a finite union of closed grid rectangles, such that every compact CVC\subseteq V lies in some KjK_j and cont(VKj)0.\operatorname{cont}(V\setminus K_j)\longrightarrow0.

Facts & Assumptions

Given: Bounded open Jordan set VV.

[L1]

A bounded set is Jordan measurable exactly when its boundary has content zero (A bounded set in Rm\mathbb{R}^m is Jordan measurable iff its boundary is null, equivalently of content zero).

[L2]

A finite cube cover can be replaced by sufficiently fine grid cells with controlled total volume (A finite rectangle cover admits grid control with arbitrarily small volume excess).

[L3]

Jordan content is finitely additive on interior-disjoint Jordan pieces (Jordan content is finitely additive when the overlap has content zero).

Proof

technique · exhaustion
1.1

Enclose VV in a rectangle and choose nested dyadic grids whose meshes tend to zero. Let KjK_j be the union of every closed cell of the jjth grid that is contained in VV. Only finitely many cells occur. Every child of a retained cell is retained, so KjKj+1K_j\subseteq K_{j+1}; each KjK_j is compact, Jordan, and contained in VV.

given
2.1

If compact CVC\subseteq V, the distance from CC to the closed complement of VV is positive. Once the mesh diameter is smaller than that distance, every grid cell meeting CC is contained in VV, so CKjC\subseteq K_j.

givenstep 1.1
3.1

Every unretained cell meeting VV also meets a mesh-sized neighborhood of V\partial V. By [L1], that boundary has content zero; [L2] therefore makes the total volume of all such cells arbitrarily small for fine enough grids. Finite additivity [L3] bounds cont(VKj)\operatorname{cont}(V\setminus K_j) by that volume, proving the limit.

L1L2L3

Depends on

Used by

Dependency tree · next 3 levels

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