LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Increasing the exponent past finite measure gives zero

Statement

For $0 \leq s < t < \infty$ and finite $δ > 0$ ,

$H_{δ}^{t} (A) \leq δ^{t - s} H_{δ}^{s} (A) .$

Consequently $H^{s} (A) < \infty$ implies $H^{t} (A) = 0$ .

Facts & Assumptions

Given: The objects, conventions, and hypotheses in the statement above.

[F1]

Hausdorff values are small-scale suprema of diameter-power covering costs. Unnormalised Hausdorff measure

[F2]

For positive bases the usual real-power multiplication laws hold. The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents

Proof

1.1

For $0 < r \leq δ$ , $r^{t} = r^{s} r^{t - s} \leq r^{s} δ^{t - s}$ . At $r = 0$ the left side is zero; the right is zero for $s > 0$ and $δ^{t}$ for $s = 0$ . Thus the inequality holds for every admissible covering member. Summation and infimisation prove the fixed-scale bound; if no cover exists its right side is infinity.

F1F2

2.1

If $M = H^{s} (A) < \infty$ , then $H_{δ}^{t} (A) \leq δ^{t - s} M$ . Given any fixed $η > 0$ , use all $0 < δ \leq η$ and monotonicity to bound $H_{η}^{t} (A)$ by a quantity tending to zero. Every scale value is therefore zero and so is the supremum. This includes $M = 0$ and the empty set.

F1step 1.1∎

Depends on

Used by

Dependency tree · two levels

8 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

Bishop–Peres Lemma 1.2.5; Fremlin 264Xe (standard reference, not scraped)