TITLE: Does there exist a non-zero signed finite borel measure which is zero on all balls?
QUESTION [12 upvotes]: Let $(X,d)$ be a compact separable metric space. Let $\mu$ be a Borel, regular, finite, signed measure on $X$ such that for all $x\in X$, for all $r>0$, $\mu(B(x,r))=0$, where $B$ denotes the (either open or closed) ball w.r.t $d$. 
Is $\mu$ zero?
If $\mu$ is positive one can show that $\mu=0$ using the Borel-Lebesgue theorem, but what if $\mu$ is signed?

REPLY [11 votes]: Stealing an answer from "user940" on Math.SE, the answer is yes, such measures can exist.  In the paper

Davies, Roy O., Measures not approximable or not specifiable by means of balls, Mathematika, Lond. 18, 157-160 (1971). ZBL0229.28005.

the author constructs a compact metric space $X$ and two distinct Borel probability measures $\mu_1, \mu_2$ that agree on every closed ball.  (They must therefore also agree on open balls, because an open ball is a countable increasing union of closed balls.)  Taking $\mu = \mu_1 - \mu_2$ provides your desired signed measure.