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The rational points of form a bounded null set that is not Jordan measurable
Statement refuted
Every bounded null subset of is Jordan measurable.
Facts & Assumptions
Given: .
and are countable ( is countably infinite, , Finite, countably infinite, countable, uncountable).
Rationals and irrationals are dense (Both and are dense in , and every nonempty open subset of is uncountable).
Jordan measurability is equivalent to a null boundary (A bounded set in is Jordan measurable iff its boundary is null, equivalently of content zero).
Counterexample
The set is countable by [L1]. Enumerate its points and cover the -th singleton by a square with volume below ; countable-union closure gives nullity.
By [L2], both and its complement are dense at every point of the unit square, so .
The unit square has content , by The unit box in has volume , and the integral of a constant over it is and A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content, hence is not null (For compact subsets of , measure zero and content zero coincide).
The boundary criterion [L3] makes non-Jordan-measurable, despite nullity.
Depends on
- A bounded set in $\mathbb{R}^m$ is Jordan measurable iff its boundary is null, equivalently of content zero
- Measure zero and content zero in $\mathbb{R}^m$ by countable and finite cube covers
- Subsets and countable unions of null subsets of $\mathbb{R}^m$ are null
- $\mathbb{Q}$ is countably infinite
- $\mathbb{N} \times \mathbb{N} \approx \mathbb{N}$
- Finite, countably infinite, countable, uncountable
- Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable
- Interior, closure, boundary, limit point, isolated point and dense subset of a metric space
- For compact subsets of $\mathbb{R}^m$, measure zero and content zero coincide
- A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content
- The unit box in $\mathbb{R}^m$ has volume $1$, and the integral of a constant $c$ over it is $c$
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
83 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)
- A. Treibergs, MATH 3225 final solutions (standard reference, not scraped)