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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
- 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
- On a small cube, a C¹ diffeomorphism distorts Jordan content by factors arbitrarily close to its linearized absolute determinant Lemma
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 98 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)