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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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The Cantor slab C×[0,1]C\times[0,1] has content zero in R2\mathbb{R}^2

Example

For the ordinary Cantor set CC, the slab C×[0,1]C\times[0,1] has content zero in R2\mathbb R^2, and hence Jordan content 00.

Verification

technique · constructive
1.1

Given ε>0\varepsilon>0, cover CC by finitely many positive-width intervals IrI_r with rr\sum_r\ell_r and maxrr\max_r\ell_r sufficiently small. Degenerate members may be enlarged within the budget.

L1chooseconstruct
1.2

Above IrI_r, stack squares of side r\ell_r. By [L2], at most 1/r+21/\ell_r+2 squares suffice, with total area at most r+2r2\ell_r+2\ell_r^2.

L2given
2.1

Summing gives at most rr+2(maxrr)rr<ε\sum_r\ell_r+2(\max_r\ell_r)\sum_r\ell_r<\varepsilon. Thus the slab has cube-content zero.

step 1.1step 1.2given
3.1

By Jordan inner and outer content and Jordan measurable bounded sets in Rm\mathbb{R}^m, cube-content zero makes the Jordan outer content 00. The nonnegative inner content is at most the outer content, so both are 00; the slab is Jordan measurable with content 00, unlike the fat-Cantor slab The Smith–Volterra–Cantor slab S×[0,1]S\times[0,1] is compact and not Jordan measurable.

step 2.1givendischarge-construct

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