Haar's theorem for the rotation-invariant distribution on the sphere

Solution 1:

Uniqueness of Haar measure is generally easier to prove than existence of Haar measure, particularly for compact groups. For instance in this case one can argue as follows. Let $\sigma$ be the left-invariant measure on $\text{SO}(n)$ which you know and love, and let $\mu$ be the usual rotation-invariant measure on $S^{n-1}$. Then for every $x\in S^{n-1}$ the map $g\mapsto gx$ sends the measure $\sigma$ to $\mu$, so if $\nu$ is an arbitrary normalized rotation-invariant measure on $S^{n-1}$ then for an arbitrary continuous function $f:S^{n-1}\to\mathbf{R}$ we have, by Fubini's theorem,

$$\int_{S^{n-1}} f(x) \,d\nu(x) = \int_{\text{SO}(n)} \int_{S^{n-1}} f(gx) \,d\nu(x) \,d\sigma(g)\\ = \int_{S^{n-1}} \int_{\text{SO}(n)} f(gx) \,d\sigma(g)\,d\nu(x)\\ = \int_{S^{n-1}} \int_{S^{n-1}} f(y) \,d\mu(y) \,d\nu(x)\\ = \int_{S^{n-1}} f(y) \,d\mu(y),$$ so $\nu=\mu$.

It wasn't important in the above argument to know that $\sigma$ is the unique Haar measure on $\text{SO}(n)$, only that it exists, which you can prove with Lie theory. However you could prove uniqueness of $\sigma$ with a similar argument.

Solution 2:

I've found an answer for my question: Theorem 3.4 of Pertti Mattila's Geometry of sets and measures in Euclidean spaces: Fractals and rectifiability.

Solution 3:

You could use the uniqueness of the Lebesgue measure on $\mathbb{R}^n$.

If $\mu$ is your rotation-invariant measure, equip $\mathbb{R}^n\setminus\{0\} = S^{n-1} \times (0,\infty) $ with the product measure $\mu \times m$ where $m$ is the Lebesgue measure. Then this measure is invariant under the group of rigid motions and is hence the Lebesgue measure on $\mathbb{R}^n$.

Now, for $A \subset S^{n-1}$, this will imply that $\mu(A) = m(A\times (0,1))$, which gives uniqueness.

Edit: as pointed out below, the correct measure should be $\mu \times m'$, where $m'((0,r)) = r^{n-1}$.