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 ( ${A C}_{ω}$ )).

Let $E \subseteq R^{n}$ be Lebesgue measurable. Then

$Θ (E, x) = 1$ for almost every $x \in E$ ;
$Θ (E, x) = 0$ for almost every $x \in E^{c}$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a Lebesgue measurable set $E \subseteq R^{n}$ .

[L1]

The density of $E$ at $x$ is $Θ (E, x) = \lim_{r \to 0^{+}} \frac{λ (E \cap B (x, r))}{λ (B (x, r))}$ 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

1.1L2L3givenalgebra

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

2.1L1step 1.1algebra

Let $x \in E ∖ N$ . Then $1_{E} (x) = 1$ , and for every $r > 0$ , [L1, step 1.1, algebra] $\frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ 1_{E} (y) - 1 ∣ d λ (y) = 1 - \frac{λ (E \cap B (x, r))}{λ (B (x, r))} .$ Since the left-hand side tends to $0$ , [L1] gives $Θ (E, x) = 1$ .

2.2L1step 1.1algebra

Let $x \in E^{c} ∖ N$ . Then $1_{E} (x) = 0$ , and for every $r > 0$ , [L1, step 1.1, algebra] $\frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ 1_{E} (y) - 0 ∣ d λ (y) = \frac{λ (E \cap B (x, r))}{λ (B (x, r))} .$ Again the left-hand side tends to $0$ , so [L1] gives $Θ (E, x) = 0$ .

3.1step 1.1step 2.1step 2.2∎

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.

Lebesgue density theorem

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources