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Jordan content is finitely additive when the overlap has content zero
Statement
If bounded Jordan measurable sets have of content zero, then In particular Jordan content is additive on disjoint finite families.
Facts & Assumptions
Given: as stated.
Indicator integrals equal Jordan content (A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content).
Content zero means that for every positive there is a finite cube cover of total volume below (Measure zero and content zero in by countable and finite cube covers); Jordan inner and outer content are the inscribed supremum and covering infimum (Jordan inner and outer content and Jordan measurable bounded sets in ).
Proof
Pointwise, . Cube-cover content zero makes the Jordan outer content of at most every positive , hence zero; its nonnegative inner content is no larger, so it too is zero. Thus is Jordan measurable with content zero, and [L1] gives .
The finite-family formula is immediate for a family of length one.
Assume it holds for a disjoint family of length .
Integrate and apply [L2] to obtain the two-set formula.
Apply the two-set formula to the union of that family and the next set. Their intersection is empty, so this adds the next content and proves the formula at length .
Hence Jordan content is additive on every finite disjoint family.
Depends on
- A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content
- Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $\mathbb{R}^m$
- Jordan inner and outer content and Jordan measurable bounded sets in $\mathbb{R}^m$
- Measure zero and content zero in $\mathbb{R}^m$ by countable and finite cube covers
- Laws of finite sums and finite products
Used by
- The washer formula for a solid of revolution between two nonnegative profiles Corollary
- The shell and washer methods both give 8π/3 for a rotated parabolic cap Example
- A bounded open Jordan set has an increasing exhaustion by compact finite unions of grid rectangles with vanishing content remainder Lemma
- A compact subset of an open Euclidean set has a compact Jordan neighborhood inside that open set Lemma
- Change of variables for a C¹ map injective and regular only on the interior of a compact Jordan set Lemma
- On a small cube, a C¹ diffeomorphism distorts Jordan content by factors arbitrarily close to its linearized absolute determinant Lemma
- A simple polygon is Jordan measurable and its content is the sum of the contents of its triangles Theorem
- The cylindrical-shell formula for a solid of revolution about the y-axis Theorem
- The volume of a three-ball by Cavalieri's cylinder-minus-cones proof Theorem
Dependency tree · two levels
37 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
- J. Lebl, Basic Analysis, Jordan Measurable Sets (standard reference, not scraped)
- J. Lebl, Basic Analysis, Outer Measure and Null Sets (standard reference, not scraped)