Say that $A\subset \mathbb{R}^n$ is measurable and of positive, finite measure.
I'm trying to see if the following is true.
If $A$ is invariant under all orthogonal reflections across $(n-1)$ dimensional subspaces in $\mathbb{R}^n$, then $A$ is a ball (centered at the origin).
I can see that for $n=1$ this is false. But I've been told its true for $n>1$. Can somebody please shed a bit of light on this one for me? There must be some kind of a elegant symmetry argument that I don't see.
If needed we can also assume that $A$ is open.
By the Cartan–Dieudonné theorem, reflections around $(n-1)$-dimensional hyperplanes generate the entire orthogonal group $O(n)$ of $\mathbb R^n$. So, asking that a subset $A \subset \mathbb R^n$ is closed under reflections is the same as asking that it is closed under the natural operation by $O(n)$. Thus, $A$ is a union of $O(n)$-orbits. Now $O(n)$ contains the special orthogonal group $SO(n)$ as subgroup. The orbit of a point under $SO(n)$ is simple an $(n-1)$-sphere, and an $(n-1)$-sphere is also closed under reflections. Hence the $O(n)$-orbit of a point is an $(n-1)$-sphere. It follows that $A$ is a union of $(n-1)$-spheres.
Taking polar coordinates, we have more or less $\mathbb R^n \cong S^{(n-1)} \times \mathbb R_{\ge 0}$. (Technically, we have to remove some measure zero-sets to get a diffeomorphism.) Under this map $A$ takes the form $S^{(n-1)} \times A_0$ with $A_0 \subset \mathbb R_{\ge 0}$. From this it should follow that $A$ has positive, finite measure if and only if $A_0$ has.
Altogether, $A$ is a union of $(n-1)$-spheres with the radii forming a measureable subset of finite, positive measure of $\mathbb R_{\ge 0}$.