Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
How 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 density theorem

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)).

Let E⊆Rn be Lebesgue measurable. Then

  1. Θ(E,x)=1 for almost every x∈E;
  2. Θ(E,x)=0 for almost every x∈Ec.

Facts & Assumptions

Given: The Axiom of Countable Choice and a Lebesgue measurable set E⊆Rn.

[L1]

The density of E at x is Θ(E,x)=lim⁡r→0+λ(E∩B(x,r))λ(B(x,r)) when the limit exists. (Density of a measurable set at a point)

[L2]

Almost every point of a locally integrable function is a Lebesgue point. (Almost every point is a Lebesgue point of a locally integrable function)

[L3]

Every Euclidean ball has positive finite measure. (Euclidean balls have positive finite Lebesgue measure)

Proof

technique · direct
1.1

Because 0≤1E≤1 and balls have finite measure by [L3], the [L2, L3, given, algebra] indicator 1E is locally integrable. Apply [L2] to 1E. There is a null set N such that every x∉N is a Lebesgue point of 1E.

L2L3givenalgebra
2.1

Let x∈E∖N. Then 1E(x)=1, and for every r>0, [L1, step 1.1, algebra] 1λ(B(x,r))∫B(x,r)∣1E(y)−1∣ dλ(y)=1−λ(E∩B(x,r))λ(B(x,r)). Since the left-hand side tends to 0, [L1] gives Θ(E,x)=1.

L1step 1.1algebra
2.2

Let x∈Ec∖N. Then 1E(x)=0, and for every r>0, [L1, step 1.1, algebra] 1λ(B(x,r))∫B(x,r)∣1E(y)−0∣ dλ(y)=λ(E∩B(x,r))λ(B(x,r)). Again the left-hand side tends to 0, so [L1] gives Θ(E,x)=0.

L1step 1.1algebra
3.1

Steps 2.1 and 2.2 hold outside the null set N, so they prove the two [step 1.1, step 2.1, step 2.2] density conclusions.

step 1.1step 2.1step 2.2∎

Depends on

Used by

Dependency tree · two levels

21 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