DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-29

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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The polar surface set function on the unit sphere

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix $n \geq 1$ and write $S^{n - 1} : = {ω \in R^{n} : ∣ ω ∣ = 1} .$ Let $Φ : R^{n} ∖ {0} ⟶ (0, \infty) \times S^{n - 1}, Φ (x) = (∣ x ∣, x / ∣ x ∣) .$ This map is continuous. If $E \subseteq S^{n - 1}$ is Borel, then $(0, 1] \times E$ is Borel and ${r ω : ω \in E, 0 < r \leq 1} = Φ^{- 1} ((0, 1] \times E)$ is Borel in $R^{n} ∖ {0}$ by A continuous map has Borel preimages of Borel sets, hence Borel in $R^{n}$ by The Borel sigma-algebra of a subspace is the trace of the ambient Borel sigma-algebra. It is therefore Lebesgue measurable by Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable. Define $σ (E) : = n λ_{n} ({r ω : ω \in E, 0 < r \leq 1}) .$

The next theorem proves that this is a Borel measure on $S^{n - 1}$ and that it is exactly the surface measure needed for polar coordinates.

Depends on