Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
- AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
Lebesgue outer measure agrees with Jordan outer content on every bounded subset of
Statement
Assume the Axiom of Countable Choice. Every bounded subset of has Lebesgue outer measure equal to its Jordan outer content.
Facts & Assumptions
Given: The Axiom of Countable Choice.
Lebesgue outer measure is at most Jordan outer content, and a bounded Jordan measurable set is Lebesgue measurable with Lebesgue measure equal to its Jordan content (Lebesgue outer measure is at most Jordan outer content, and a bounded Jordan measurable set is Lebesgue measurable with Lebesgue measure equal to its Jordan content).
Every at most countable subset of is Lebesgue null; in particular (Every at most countable subset of is Lebesgue null; in particular ).
The Jordan outer content is defined through finite rectangle covers (Jordan inner and outer content and Jordan measurable bounded sets in ).
The same lower bound on total length holds for a finite cover of an interval by bounded intervals of any of the four bounded forms (If finitely many intervals cover a closed bounded interval , the sum of their lengths is at least ).
Refutation
What is true is [L1]: one inequality holds for every bounded set, and equality holds when the set is Jordan measurable.
The witness has Lebesgue outer measure by [L2]; and if finitely many bounded intervals cover , their union is closed and contains the dense subset of , so it contains all of , and [F2] forces total length at least , while the single interval realises , so [F1] gives Jordan outer content .
So the false statement drops the Jordan measurability hypothesis from the true theorem [L1], and the witness of step 1.2 refutes it.
Depends on
- Lebesgue outer measure is at most Jordan outer content, and a bounded Jordan measurable set is Lebesgue measurable with Lebesgue measure equal to its Jordan content
- Every at most countable subset of $\mathbb{R}^n$ is Lebesgue null; in particular $\lambda_1(\mathbb{Q})=0$
- If finitely many intervals cover a closed bounded interval $[a,b]$, the sum of their lengths is at least $b - a$
- Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable
- Jordan inner and outer content and Jordan measurable bounded sets in $\mathbb{R}^m$
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
55 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
- T. Tao, An Introduction to Measure Theory (GSM 126), Exercises 1.2.1 and 1.2.8 (standard reference, not scraped)