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

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Hausdorff measure is metric and measures every Borel set

Statement

Assume the Axiom of Countable Choice. On every metric space, $H^{s}$ is a metric outer measure for each finite $s \geq 0$ . In particular, if nonempty $A, B$ have $d (A, B) > 0$ ,

$H^{s} (A \cup B) = H^{s} (A) + H^{s} (B) .$

The equality also holds if either set is empty. Every Borel set is Carathéodory measurable, and the restriction to the full Carathéodory sigma-algebra is a complete measure.

Facts & Assumptions

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

[F1]

$H^{s}$ is an outer measure under Countable Choice. Hausdorff measure is an outer measure

[F2]

A metric outer measure is additive on nonempty positively separated sets. Metric outer measures

[F3]

Every Borel subset of a metric space is Carathéodory measurable for every metric outer measure. Every Borel set is Carathéodory measurable for a metric outer measure

[F4]

The Carathéodory domain of an outer measure is a sigma-algebra and its restriction is complete. Carathéodory's theorem: measurable sets form a sigma-algebra carrying a complete measure

Proof

1.1

Let $h = d (A, B) > 0$ . A covering set of diameter at most $δ < h$ cannot meet both $A$ and $B$ . Partition any cover of their union by which set it meets, discarding sets meeting neither. Its cost is at least $H_{δ}^{s} (A) + H_{δ}^{s} (B)$ ; if no cover exists the inequality still holds.

given

2.1

Take small-scale suprema in that inequality. The supremum of the sums of the two nondecreasing scale values is the sum of their suprema: approximate both finite lower bounds at one common scale; this also proves the assertion when one supremum is infinite. Subadditivity gives the opposite inequality. Empty sets use the outer-measure zero axiom. Thus the metric condition holds, also at $s = 0$ .

F1F2step 1.1

3.1

The metric criterion gives Borel measurability, and the Carathéodory theorem gives completeness on the full measurable domain.

F3F4step 2.1∎

Depends on

Used by

Zero-dimensional Hausdorff measure is counting measure Proposition
Euclidean Hausdorff measure is proportional to Lebesgue measure Theorem
Hausdorff measure is Borel regular Theorem

Dependency tree · two levels

19 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

Fremlin, Measure Theory, 264C,E (standard reference, not scraped)
Bishop–Peres, Theorem 1.2.4 (standard reference, not scraped)