title: Spherical measure
text: # Spherical measure
  
  In mathematics — specifically, in geometric measure theory — spherical measure $σ^n$ is the "natural" Borel measure on the n-sphere $S^n.$ Spherical measure is often normalized so that it is a probability measure on the sphere, i.e. so that $σ^n(S^n)$ = 1.
  
  ## Definition of spherical measure
  
  There are several ways to define spherical measure. One way is to use the usual "round" or "arclength" metric $ρ_n$ on $S^n;$ that is, for points x and y in $S^n,$ $ρ_n(x,$ y) is defined to be the (Euclidean) angle that they subtend at the centre of the sphere (the origin of $R^{n+1}).$ Now construct n-dimensional Hausdorff measure $H^n$ on the metric space $(S^n,$ $ρ_n)$ and define
  
      $\sigma ^{n}={\frac {1}{H^{n}(\mathbf {S} ^{n})}}H^{n}.$
  
  One could also have given $S^n$ the metric that it inherits as a subspace of the Euclidean space $R^{n+1};$ the same spherical measure results from this choice of metric.
  
  Another method uses Lebesgue measure $λ^{n+1}$ on the ambient Euclidean space $R^{n+1}:$ for any measurable subset A of $S^n,$ define $σ^n(A)$ to be the (n + 1)-dimensional volume of the "wedge" in the ball $B^{n+1}$ that it subtends at the origin. That is,
  
      $\sigma ^{n}(A):={\frac {1}{\alpha (n+1)}}\lambda ^{n+1}(\{tx\mid x\in A,t\in [0,1]\}),$
  
  where
  
      $\alpha (m):=\lambda ^{m}(\mathbf {B} _{1}^{m}(0)).$
  
  The fact that all these methods define the same measure on $S^n$ follows from an elegant result of Christensen: all these measures are obviously uniformly distributed on $S^n,$ and any two uniformly distributed Borel regular measures on a separable metric space must be constant (positive) multiples of one another. Since all our candidate $σ^n's$ have been normalized to be probability measures, they are all the same measure.
  
  ## Relationship with other measures
  
  The relationship of spherical measure to Hausdorff measure on the sphere and Lebesgue measure on the ambient space has already been discussed.
  
  Spherical measure has a nice relationship to Haar measure on the orthogonal group. Let O(n) denote the orthogonal group acting on $R^n$ and let $θ^n$ denote its normalized Haar measure (so that $θ^n(O(n))$ = 1). The orthogonal group also acts on the sphere $S^{n−1}.$ Then, for any x ∈ $S^{n−1}$ and any A ⊆ $S^{n−1},$
  
      $\theta ^{n}(\{g\in \mathrm {O} (n)\mid g(x)\in A\})=\sigma ^{n-1}(A).$
  
  In the case that $S^n$ is a topological group (that is, when n is 0, 1 or 3), spherical measure $σ^n$ coincides with (normalized) Haar measure on $S^n.$
  
  ## Isoperimetric inequality
  
  There is an isoperimetric inequality for the sphere with its usual metric and spherical measure (see Ledoux & Talagrand, chapter 1):
  
  If A ⊆ $S^{n−1}$ is any Borel set and B⊆ $S^{n−1}$ is a $ρ_n-ball$ with the same $σ^n-measure$ as A, then, for any r > 0,
  
      $\sigma ^{n}(A_{r})\geq \sigma ^{n}(B_{r}),$
  
  where $A_r$ denotes the "inflation" of A by r, i.e.
  
      $A_{r}:=\{x\in \mathbf {S} ^{n}\mid \rho _{n}(x,A)\leq r\}.$
  
  In particular, if $σ^n(A)$ ≥ 1/2 and n ≥ 2, then
  
      $\sigma ^{n}(A_{r})\geq 1-{\sqrt {\frac {\pi }{8}}}\,\exp \left(-{\frac {(n-1)r^{2}}{2}}\right).$
  