ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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.

Summable exceptional measures in the kolmogorov induction

Example

If measurable subsets $E_{j}$ of the normalized torus satisfy $m (E_{j}) \leq 2^{- j}$ for every $j \geq 1$ , then $m ({lim sup}_{j} E_{j}) = 0$ . Thus almost every point belongs to only finitely many exceptional sets.

Facts & Assumptions

[F1]

A sequence of measurable sets with summable measures has null limsup, without independence The first Borel-Cantelli lemma for measures.

Verification

Given: The measurable sets and geometric measure bounds in the example.

1.1given

Finite geometric cancellation gives $\sum_{j = 1}^{J} 2^{- j} = 1 - 2^{- J}$ , hence $\sum_{j \geq 1} m (E_{j}) \leq 1 < \infty$ . More precisely, the tail satisfies $\sum_{j \geq J} m (E_{j}) \leq 2^{1 - J}$ .

2.1F1step 1.1∎

F1 applies to these measurable sets and the finite sum proved in step 1.1; it gives $m (⋂_{J} ⋃_{j \geq J} E_{j}) = 0$ . Equivalently, outside this null intersection, some J has the point outside every $E_{j}$ for $j \geq J$ , which is the asserted eventual good-set property. The proof of F1 bounds the limsup measure by each tail, here explicitly $2^{1 - J} \to 0$ . No independence or choice assumption on the supplied sequence is needed.

Depends on

The first Borel-Cantelli lemma for measures

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Grafakos, Classical Fourier Analysis, third edition (standard reference, not scraped)