LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 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.

Continuous compactly supported functions are recovered by small ball averages

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $g \in C_{c} (R^{n})$ and let $x \in R^{n}$ . Then $\lim_{r \to 0^{+}} A_{r} g (x) = g (x) .$

Facts & Assumptions

Given: The Axiom of Countable Choice, a function $g \in C_{c} (R^{n})$ , and a point $x \in R^{n}$ .

[L1]

The class $C_{c} (R^{n})$ consists of continuous functions on $R^{n}$ with compact support. (The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$ )

[L2]

The ball average is $A_{r} g (x) = \frac{1}{λ (B (x, r))} \int_{B (x, r)} g (y) d λ (y) .$ (The average of a locally integrable function over a Euclidean ball)

Proof

technique · direct

1.1L1given

Let $ε > 0$ . Since $g$ is continuous at $x$ by [L1], there is [L1, given] $δ > 0$ such that $∥ y - x ∥_{2} < δ ⟹ ∣ g (y) - g (x) ∣ < ε .$

2.1step 1.1L2algebra

For $0 < r < δ$ , every $y \in B (x, r)$ satisfies the hypothesis of step 1.1, [step 1.1, L2, algebra] so $∣ A_{r} g (x) - g (x) ∣ \leq \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ g (y) - g (x) ∣ d λ (y) \leq ε .$

3.1step 2.1∎

Because step 2.1 holds for every $ε > 0$ , one has [step 2.1] $A_{r} g (x) \to g (x)$ as $r \to 0^{+}$ .

Continuous compactly supported functions are recovered by small ball averages

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources