Does any finite, atomless Borel measure on $\mathbb{R}^n$ assign measure zero to the boundary of an open ball?

Question

Let $\mu$ a finite, atomless (non-atomic) Borel measure on $\mathbb{R}^n$. Is it necessarily true that $\mu\big(\partial B_{\epsilon}(x)\big)=0$ for any open ball $B_{\epsilon}(x)$?

This is true for the Lebesgue measure but most standard arguments use translation-invariance. I also know this holds for Borel product measures with atomless marginals: is it true in general for any finite, atomless Borel measure?

Answer 1

Pick a ball $B$ and define a measure $\mu$ on its boundary $B'$. (for example the "area" of the part of the boundary of the ball).

We can now define a measure on $\mathbb R^n$ by letting $\nu(A)=\mu (A\cap B')$