# Hausdorff measure on the sphere is well defined?

Given $n\in\mathbb{N}$, consider the $\ell_2$ unit sphere $\mathbb{S}^{n}\subset\mathbb{R}^{n+1}$ equipped with its "geodesic" metric $\rho_n$ defined as:

$\rho_n(x,y)=\arccos \Big(\langle x,y\rangle\Big)$,

where $\langle\cdot,\cdot\rangle$ is the usual Euclidean inner product in $\mathbb{R}^{n+1}$. I have seen many authors assert that one can define a uniform probability measure $\sigma_n$ on the Borel sets of $\mathbb{S}^n$ using the "normalized Hausdorff measure". By this, I assume that the authors mean that for any borel set $A\subset\mathbb{S}^n$, we define

$\displaystyle\sigma_n(A)=\frac{\mathscr{H}^n(A)}{\mathscr{H}^n(\mathbb{S}^n)}$,

where $\mathscr{H}^n$ is the $n$-dimensional Hausdorff measure on the metric space $(\mathbb{S}^n,\rho_n)$.

The fact that the above is well defined (i.e. $\mathscr{H}^n(\mathbb{S}^n)$ is neither 0 nor $\infty$) and uniform does not seem trivial to me, yet I have never come across a reference where this is demonstrated.

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This MSE question contains a slick proof of the fact that

If $i:M^n\hookrightarrow \mathbb{R}^{n+1}$ is an embedded submanifold, then writing $g=i^*\delta$ as the induced metric on $M$, the volume form $vol_g$ and restriction Hausdorff measure $\mu$, defined by $\mu(A) = \mathcal{H}^n(i(A))$ agree as measures.

This answers your question because it implies $vol_g(S^n) = \mathcal{H}^n(i(S^n))$. That it is "uniform" is now obvious by thinking about $vol_g$.

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Let me describe a way to construct the desired measure directly giving uniformity, which is, in my opinion, much more elegant than using Riemannian geometry or the Hausdorff measure. It also gives a simple characterisation directly proving that it is equivalent to the other constructions. If you are just interested in the question why the Hausdorff measure is finite and non-trivial, you can jump to the last paragraph.

The sphere $S^n$ is homeomorphic to the quotient space $SO(n+1)/SO(n)$, where $SO(n)$ denotes the special orthogonal group. We can regard $S^n$ as a homogeneous space on which $SO(n+1)$ acts by left-multiplication. $SO(n+1)$ is a compact group thus there exists a unique normalised Haar measure on $SO(n+1)$. By compactness of the subgroup $SO(n)$ (actually by agreement of the modular function of the Haar measure of $SO(n+1)$ and $SO(n)$) we get that there exists a unique normalised Radon-measure on $S^n$ invariant under the action of $SO(n+1)$. Since all the groups are comact, the measure is particularly easy to define: Let $p\colon SO(n+1)\to SO(n+1)/SO(n)=S^n$ be the canonical surjection. For every continuous function $f\in C(S^n)$ we can define the integral as

$$\int_{S^n} f = \int_{SO(n+1)} f\circ p$$

inducing the measure we are interested in. We did not regard $S^n$ as a metric space yet, but if we do by regarding it as the $L^2$ unit sphere embedded into $R^{n+1}$, then the action of $SO(n+1)$ can be described as rotation of the space which is isometrically and transitively on $S^n$ (and $SO(n)$ is just the isotropy group of a point in $S^n$, using this observation the isomorphy of $SO(n+1)/SO(n)$ and $S^n$ regarded as homogeneous spaces follows). We get the desired result.

For a relation to the Hausdorff measure consider the homeomorphism projecting the $n$-dimensional unit disk onto a hemisphere of the $n$-sphere. This map is Lipschitz with a Lipschitz constant $\pi$. The Hausdorff measure of the (flat) unit disk is well-known and finite, thus the Hausdorff measure of the hemisphere is finite (since Lipschitz functions can only increase Hausdorff measure by a constant factor) and we get finiteness of the Hausdorff measure of the sphere. The inverse of the homeomorphism is a contraction, thus the Hausdorff measure of the hemishere is non-zero. We get that the Hausdorff measure is a non-trivial $SO(n+1)$-invariant Radon measure, thus it is (up to a constant) the same as the measure constructed above.

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Nice point! Is there any way to use this type of argument to prove that the sphere has $n$-dimensional Hausdorff measure in $(0,\infty)$? –  Otis Chodosh Jun 20 '13 at 13:56
I have extended my answer, you really do not need Riemannian geometry. –  The User Jun 20 '13 at 15:20
Very interesting! Thank you for posting this. –  user34424 Jun 22 '13 at 13:56