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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?

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    $\begingroup$ The set of balls generates the Borel sigma-algebra, so that $\mu$ is characterized by its values on balls and must be zero (if you are more comfortable with positive measures, think about the decomposition of $\mu$ into its positive and negative part, and observe that they have the same values on all balls and thus coincide). $\endgroup$ – Benoît Kloeckner Mar 4 at 17:17
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    $\begingroup$ @BenoîtKloeckner Erm... How do you show that the family of zero measure sets is a $\sigma$-algebra? Usually you can do it if you start with a semi-ring using the monotone class lemma, but balls do not form one. Am I missing anything? $\endgroup$ – fedja Mar 4 at 17:44
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    $\begingroup$ @BenoîtKloeckner: The usual argument for this needs a collection which not only generates the Borel $\sigma$-algebra but is also closed under finite intersections (a $\pi$-system), which the balls do not satisfy. $\endgroup$ – Nate Eldredge Mar 4 at 18:19
  • $\begingroup$ @fedja and NateElredge I have been naïve, thanks for the correction and sorry for my erroneous comment.. $\endgroup$ – Benoît Kloeckner Mar 5 at 20:02
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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.

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  • $\begingroup$ On the positive side, the answer to whether $\mu$ is zero is yes for Banach spaces; as proved by D. Preiss and J. Tiser, Measures in Banach spaces are determined by their values on balls. Mathematika 38, No. 2, 391-397 (1991). Zbl 0755.28006 $\endgroup$ – Dirk Werner Mar 7 at 17:33

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