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.



Hausdorff measure on the sphere is well defined?
This [MSE question](https://math.stackexchange.com/questions/107099/volume-form-and-hausdorff-measure) 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 $vol\_g$. 


