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 measure is a Radon measure on R^n
Statement
Let and assume the Axiom of Countable Choice. Lebesgue measure on is a Radon measure, and it is regular in the stronger compact-inner-regular-on-all-Borel-sets convention.
Facts & Assumptions
Given: , the Axiom of Countable Choice, and Lebesgue measure on .
Under these hypotheses, Lebesgue measure is finite on bounded sets, outer regular on arbitrary sets, and compact-inner-regular on measurable sets. (Lebesgue measure is sigma-finite, and every metrically bounded subset of has finite outer measure, Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of is the infimum of the measures of the open sets containing it, Assuming countable choice, the Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets)
Proof
Compact subsets of are bounded, so they have finite measure by [L1]. Outer regularity on Borel sets and compact inner regularity on open sets are also direct instances of [L1]. These are precisely the Radon clauses.
The last part of [L1] applies to every Borel set, so the stronger regularity assertion also holds.
Depends on
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- Radon measure on an LCH space
- $\mathbb{R}^n$ is locally compact and $\sigma$-compact
- Lebesgue measure is sigma-finite, and every metrically bounded subset of $\mathbb{R}^n$ has finite outer measure
- Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of $\mathbb{R}^n$ is the infimum of the measures of the open sets containing it
- Assuming countable choice, the Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets
Used by
- A locally integrable density functional is represented by g dlambda Example
- The Riemann integral functional is represented by Lebesgue measure on an interval Example
- C_c(X) is dense in Lⁱnfinity(mu) for every Radon measure False statement
- Every positive linear functional on C_c(X) is uniformly sup-norm bounded False statement
Dependency tree · two levels
51 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
- Donald L. Cohn, Measure Theory, 2nd ed., Chapter 7 (standard reference, not scraped)