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Reflection invariance and vanishing first moment of the sphere measure

Statement

Assume the Axiom of Countable Choice. Let $n \geq 1$ and let $σ$ be the polar surface measure on the unit sphere $S^{n - 1} \subseteq R^{n}$ (The polar surface set function on the unit sphere). Then the reflection $R ω = - ω$ preserves $σ$ , and the first moment vanishes: $\int_{S^{n - 1}} ω d σ (ω) = 0 \in R^{n}, so \int_{S^{n - 1}} a \cdot ω d σ (ω) = 0 (a \in R^{n}) .$ The integrals are finite because $σ$ is a finite Borel measure.

Facts & Assumptions

Given: the Axiom of Countable Choice, an integer $n \geq 1$ and the polar surface measure $σ$ on $S^{n - 1}$ .

[F1]

$σ (E) = n λ_{n} {r ω : ω \in E, 0 < r \leq 1}$ for Borel $E \subseteq S^{n - 1}$ (The polar surface set function on the unit sphere).

[F2]

For an invertible linear $T : R^{n} \to R^{n}$ with matrix $A$ and every Lebesgue measurable $E$ , $λ_{n} (T [E]) = ∣ \det A ∣ λ_{n} (E)$ ; in particular $λ_{n} (- E) = λ_{n} (E)$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F3]

Under Countable Choice, $σ$ is a finite Borel measure on $S^{n - 1}$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

Proof

1.1F1F2algebra

Reflection invariance. For Borel $E \subseteq S^{n - 1}$ the set $- E$ is Borel and ${r θ : θ \in - E, 0 < r \leq 1} = - {r ω : ω \in E, 0 < r \leq 1}$ , so [F1] and [F2] give $σ (- E) = n λ_{n} (- {r ω : ω \in E, 0 < r \leq 1}) = n λ_{n} {r ω : ω \in E, 0 < r \leq 1} = σ (E)$ .

1.2F3algebra

Vanishing of the moment. Each coordinate function $g_{i} (ω) = ω_{i}$ is Borel and bounded by $1$ on $S^{n - 1}$ , so $\int g_{i} d σ$ is a finite real number by [F3]. Because $σ$ is invariant under the bijection $ω \mapsto - ω$ , the substitution formula for a measure-preserving bijection — valid for indicators by definition, for simple functions by linearity, and for bounded Borel functions by the supremum definition of the integral — gives $\int_{S^{n - 1}} g_{i} d σ = \int_{S^{n - 1}} g_{i} (- ω) d σ (ω) = - \int_{S^{n - 1}} g_{i} d σ$ ; hence $\int g_{i} d σ = 0$ for every $i$ , that is $\int_{S^{n - 1}} ω d σ (ω) = 0$ .

2.1algebra∎

For $a \in R^{n}$ , linearity of the integral in the integrand gives $\int_{S^{n - 1}} a \cdot ω d σ (ω) = a \cdot \int_{S^{n - 1}} ω d σ (ω) = 0$ .

Depends on

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Sources

John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2025 archived author manuscript, AMS Graduate Studies in Mathematics) (standard reference, not scraped)