$f(x)=\|x\|_2$, $f_{\#}\mu=f_{\#}\nu\implies\mu=\nu$

Question

I want to show that if $\mu$ and $\nu$ are probability measures on $\mathbb{R}^d$ that are invariant under rotations and $f(x)=\|x\|_2=\left(\sum\limits_{i=1}^d x_i^2\right)$ sucht that the pushforwards are the same,i.e. $f_{\#}\mu=f_{\#}\nu$ then $\mu=\nu$. I understand that if the pushforwards are the same than the measure on the "annuli" of $\mathbb{R}^d$ will be the same. Also I understand that since the measures are invariant under rotations that if we cut the annulus (A) into $n$ small pieces (with one of these pieces called $P$) then $n\cdot\nu(P)=\nu(A)=\mu(A)=n\cdot\mu(P)$ so we get that $\mu(P)=\nu(P)$. So intuitively I can imagine that $\mu$ and $\nu$ will be the same. But I don't see how to make a rigorous argument.

Answer 1

If you are familiar with Lebesgue differentiation theorem for finite Borel measures on $\mathbb{R}^n$, the claim is a quick corollary of this result.

If you are not, one way to prove the claim is as follows: let $\mathcal{A}$ be the family of Borel sets such that $\mu(A)=\nu(A)$. You proved that this holds for annuli and "slices" of annuli. It's easy to prove that $\mathcal{A}$ is a Dynkin system. Thus, from Dynkin's theorem it follows that the $\sigma-$algebra generated by slices of annuli (let's call it $\mathcal{B}$) is contained in $\mathcal{A}$. It's not hard to see that $\mathcal{B}$ contains the open balls, and thus it's the Borel $\sigma-$algebra, and so the two measures are equal.

Edit:

To see why $\mathcal{B}$ contains the open balls, one can use the following: let $B$ be a open ball. First, suppose that $0\not\in\overline{B}$. Now, let $P:\Omega\to [0,2\pi)\times[0,\pi]^{n-1}\times \mathbb{R}$ be the polar coordinates map (where $\Omega$ is a suitable region containing $B$ such that $P_{|\Omega}$ is a diffeomorphism). Its easy to see that the slices of annuli get mapped to open $n-$cells and viceversa, while $B$ gets mapped to an open set $P(B)$. Since open $n-$cells generate the topology, we can write $P(B)=\cup_\mathbb N C_i$ (with $C_i$ the cells). Taking the inverse (which is well defined because $P$ is a diffeomorphism) we get that $B=\cup_\mathbb{N} P^{-1}(C_i)$ and since $P^{-1}(C_i)$ are slices of annuli, we are done.

Now, let $B$ be general. If it contains $0$, we can cover a neighbourhood of it by a small ball centered in $0$ (which is in $\mathcal{B}$ as we now from the start), let us call it $B(0)$. For every $x\in \mathbb{Q}^n-\{0\}:x\in B$, let $r>0$ be such that $B_{r/2}(x)\subset B_r(x)\subseteq B$. It is easy to see that if $0\not \in B=\cup_\mathbb N B(r_i/2,x_i)$, otherwise $B=B(0)\cup_\mathbb N B(r_i/2,x_i)$ and we are done