Haar's theorem for the rotation-invariant distribution on the sphere 
Let $\mu$ be a Borel probability measure on $\mathcal S^{n-1}$. Suppose $\mu$ is rotation-invariant, that is, $\mu(UA) = \mu(A)$ for any Borel set $A\subset\mathcal S^{n-1}$ and orthogonal matrix $U\in\mathbb R^{n\times n}$. Then $\mu$ is the
  uniform measure.

Could you recommend a reference (preferably a book) for this simple version of Haar's theorem?
 A: 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.
A: 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.
A: 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 $m'((0,r)) = r^{n-1}$. 
