ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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.

Egorov for x^k on the unit interval

Example

On $[0, 1]$ with Lebesgue measure, the functions $f_{k} (x) : = x^{k}$ converge pointwise to the function $f (x) = {\begin{cases} 0, & 0 \leq x < 1, \\ 1, & x = 1. \end{cases}$

For every $ε \in (0, 1)$ , the exceptional set $(1 - ε, 1]$ has measure $ε$ , and on the closed core $[0, 1 - ε]$ the convergence is uniform with estimate $∣ x^{k} - f (x) ∣ \leq (1 - ε)^{k}$ .

Facts & Assumptions

Given: Lebesgue measure on $[0, 1]$ , the sequence $f_{k} (x) : = x^{k}$ , and $ε \in (0, 1)$ .

[L1]

Egorov's theorem says that on a finite measure space, almost-everywhere convergence implies almost-uniform convergence. (Egorov's theorem)

Verification

technique · direct

1.1

If $0 \leq x < 1$ , then $x^{k} \to 0$ , while $1^{k} = 1$ for every $k$ . Thus $f_{k} \to f$ pointwise on $[0, 1]$ .

given

2.1

Put $E : = (1 - ε, 1]$ . Then $λ (E) = ε$ , and for $x \in [0, 1 - ε]$ one has $f (x) = 0$ and $∣ f_{k} (x) - f (x) ∣ = x^{k} \leq (1 - ε)^{k}$ . Since $(1 - ε)^{k} \to 0$ , the convergence is uniform on $[0, 1 - ε]$ .

step 1.1algebra

3.1

This is the almost-uniform conclusion predicted abstractly by [L1], and here the exceptional set and the uniform estimate are explicit.

step 2.1L1∎

Depends on

Egorov's theorem

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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.