Is the Haar measure on the orthogonal group uniform on preimages of balls in orbits? Let $G$ be the orthogonal group on $\mathbb{R}^n$. The norm on $G$ is the operator norm. $G$ is a compact (Lie) group, so it has a unique Haar measure $\mu$ of total measure 1 which is both left and right invariant. 
Fix a unit vector $x_0 \in \mathbb{R}^n$. The orbit of $x_0$ is $Gx_0 = S^{n-1} \subseteq \mathbb{R}^n$. The map $\alpha: g \mapsto g x_0$ is a surjection from $G$ to $S^{n-1} \subseteq \mathbb{R}^n$.   
Cover $S^{n-1}$ by balls $B_i$ in $\mathbb{R}^n$ of radius $r$ whose centers lie on $S^{n-1}$. 
I think the preimages $\alpha^{-1}(B_i)$ of the balls should all have the same measure (say up to a constant factor).
How can I prove (or disprove) this?
EDIT: 
https://en.wikipedia.org/wiki/Spherical_measure#Relationship_with_other_measures
According to that wikipedia article, the image measure of the Haar measure $\mu$ on $G$ through $\alpha$ is exactly the uniform measure $\sigma$ on the sphere:
$$
\sigma(A)=\mu(\{ g: \alpha(g)=gx_0 \in A\}) = \mu(\alpha^{-1}(A))
$$
In particular, 
$$
\sigma(B_i \cap S^{n-1}) = \sigma(B_i) = \mu(\alpha^{-1}(B_i \cap S^{n-1}))
$$
Since $\sigma$ is uniform and since the balls have the same radius with centers on the sphere, $\sigma(B_i \cap S^{n-1})=\sigma(B_j \cap S^{n-1})$ for all $i,j$.
Thus my question can be reduced to:
Why is the image measure $\mu(\alpha^{-1}( \; \cdot \; ))$ exactly the uniform measure $\sigma$ on the sphere?  
 A: Consider two $B_1$ and $B_2$ balls of radius $r$ with centers on $S^{n-1}$. So the centers are $g_1 x_0$ and $g_2 x_0$ for some $g_1,g_2 \in G$. 
Note 
\begin{align}
B_1 \cap S^{n-1} &= \{ gx_0 : g \in G, \; |gx_0 - g_1x_0| < r \} \\
B_2 \cap S^{n-1} &= \{ gx_0 : g \in G, \; |gx_0 - g_2x_0| < r \}
\end{align}
and so 
\begin{align}
\alpha^{-1}(B_1 \cap S^{n-1}) &= \{ g \in G : |gx_0 - g_1x_0| < r \} \\
\alpha^{-1}(B_2 \cap S^{n-1}) &= \{ g \in G : |gx_0 - g_2x_0| < r \}
\end{align}
Claim: $\alpha^{-1}(B_1 \cap S^{n-1})$ and $\alpha^{-1}(B_2 \cap S^{n-1})$ are translates. 
Since $\mu$ is translation invariant, the claim implies
$$
\mu(\alpha^{-1}(B_1 \cap S^{n-1})) = \mu(\alpha^{-1}(B_2 \cap S^{n-1})).
$$
This proves $\mu(\alpha^{-1}(\; \cdot \;))$ must be the uniform measure on the sphere, since it assigns the same measure to ''caps'' of equal radius.
Proof of Claim: Since $g_2 = (g_2 g_1^{-1})g_1$, we have 
\begin{align*}
g_2 g_1^{-1} \alpha^{-1}(B_1 \cap S^{n-1}) 
&= \{ (g_2 g_1^{-1}) g \in G : |gx_0 - g_1x_0| < r \} \\
&= \{ (g_2 g_1^{-1}) g \in G : |(g_2 g_1^{-1})gx_0 - (g_2 g_1^{-1})g_1x_0| < r \} \\
&= \{ h \in G : |hx_0 - (g_2 g_1^{-1})g_1x_0| < r \} \\
&= \{ h \in G : |hx_0 - g_2x_0| < r \} \\
&= \alpha^{-1}(B_2 \cap S^{n-1})
\end{align*}
