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A simple polygon is Jordan measurable and its content is the sum of the contents of its triangles
Statement
A simple polygon is Jordan measurable and its content is the sum of the contents of the triangles in any triangulation.
Facts & Assumptions
Given: A simple polygonal region .
Every simple polygon admits a triangulation (Every simple polygon admits a triangulation).
Every closed triangle is Jordan measurable and has the determinant content formula (A triangle has content , equal to half base times height when the chosen side is nonzero).
If bounded Jordan measurable sets have content-zero intersection, then their union is Jordan measurable and (Jordan content is finitely additive when the overlap has content zero).
The graph of a continuous function on a closed nondegenerate rectangle has content zero (The graph of a continuous function on a closed nondegenerate rectangle in has content zero in ).
Content zero passes to subsets (Measure zero and content zero in by countable and finite cube covers).
Proof
Choose a triangulation by [L1]. Each of its finitely many closed triangular faces is Jordan measurable by [L2].
Distinct faces meet only in a common edge, a common vertex, or not at all. An edge is a graph of a continuous affine function after interchanging coordinates if necessary, so [L4] gives it content zero. A common vertex or the empty set is a subset of such a graph, so [L5] gives it content zero as well. Thus every face intersection has content zero.
Apply [L3] repeatedly to the finite face family using step 1.2. The union is Jordan measurable and its content is the sum of all face contents.
The same argument applies to any triangulation, and every resulting sum equals the intrinsic number , so the sum is independent of the triangulation.
Depends on
- Every simple polygon admits a triangulation
- A triangle has content $\tfrac12|\det[B-A\ C-A]|$, equal to half base times height when the chosen side is nonzero
- Jordan content is finitely additive when the overlap has content zero
- The graph of a continuous function on a closed nondegenerate rectangle in $\mathbb{R}^m$ has content zero in $\mathbb{R}^{m+1}$
- Measure zero and content zero in $\mathbb{R}^m$ by countable and finite cube covers
Used by
Dependency tree · two levels
32 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
- Geometry: Combinatorics & Algorithms 2020, Chapter 4 (standard reference, not scraped)