ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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.

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The Smith-Volterra-Cantor set has Lebesgue measure exactly $1 / 2$

Example

Assume the Axiom of Countable Choice and let $S$ be the Smith-Volterra-Cantor set. Then

$λ_{1} (S) = \frac{1}{2} .$

This is the exact value behind the published statement that $S$ is not null.

Facts & Assumptions

Given: The Axiom of Countable Choice and the stage lengths $(λ_{n})_{n \in N}$ and stage sets $(S_{n})_{n \in N}$ of The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals.

[F1]

$λ_{0} = 1, λ_{n + 1} = (λ_{n} - 4^{- n - 1}) \cdot 2^{- 1} (n \in N) .$ (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals)

[L1]

Assuming countable choice, a box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F2]

Let $(E_{n})_{n \in N}$ be a decreasing sequence of measurable sets for a measure $μ$ . If $μ (E_{n_{0}}) < + \infty$ for some $n_{0}$ , then $μ (⋂_{n} E_{n}) = \inf_{n} μ (E_{n})$ (Continuity from above when one set has finite measure).

[F3]

$S$ is closed and bounded, hence compact (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero).

[F4]

If $∣ r ∣ < 1$ then the series $\sum r^{k}$ converges (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

Verification

technique · direct

1.1

At stage $n$ , the set $S_{n}$ is a disjoint union of $2^{n}$ closed intervals of common length $λ_{n}$ , so $λ_{1} (S_{n}) = 2^{n} λ_{n}$ .

F1F3L1algebra

2.1

Put $u_{n} : = 2^{n} λ_{n}$ . Then $u_{0} = 1$ and $u_{n + 1} = u_{n} - 2^{- n - 2}$ by [F1], so an induction together with [F4] gives $u_{n} = \frac{1}{2} + 2^{- n - 1}$ for every $n \in N$ .

step 1.1F1F4algebra

3.1

The sets $S_{n}$ decrease to $S$ , and $λ_{1} (S_{0}) = 1 < + \infty$ , so [F2] yields $λ_{1} (S) = \inf_{n} λ_{1} (S_{n}) = \lim_{n} (\frac{1}{2} + 2^{- n - 1}) = \frac{1}{2}$ .

step 2.1F2F3algebra∎

Depends on

Used by

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Dependency tree · two levels

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The Smith-Volterra-Cantor set has Lebesgue measure exactly 1/21/2