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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
    Commented Mar 4, 2020 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
    Commented Mar 4, 2020 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
    Commented Mar 4, 2020 at 18:19
  • $\begingroup$ @fedja and NateElredge I have been naïve, thanks for the correction and sorry for my erroneous comment.. $\endgroup$
    – Benoît Kloeckner
    Commented Mar 5, 2020 at 20:02

1 Answer 1

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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
    Commented Mar 7, 2020 at 17:33
  • $\begingroup$ @Dirk Werner : Preiss and Tiser claim to show this for general Banach spaces in the title and the abstract, but the Theorem itself is formulated and proved for separable Banach spaces only. Do you know whether the general statement holds? Thanks in advance! $\endgroup$
    – IljaKlebanov
    Commented Feb 22, 2023 at 8:51
  • $\begingroup$ Wouldn't the support of a (regular finite) measure be separable? $\endgroup$
    – Dirk Werner
    Commented Feb 22, 2023 at 14:22
  • $\begingroup$ Thanks for your reply! I am not familiar with this result, could you please point me to a reference? So your argument would be: Consider the support of the (regular finite) measure, which is separable. Then take the closure of its linear span, which is a (possibly smaller) separable Banach space and work therein. Is this correct? $\endgroup$
    – IljaKlebanov
    Commented Feb 22, 2023 at 23:59
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    $\begingroup$ @iljusch: Let $E$ be the support. If it is not separable then it contains an uncountable set $S$ with no limit points. So for each $x \in S$ we can find an open ball $B(x, r_x)$ containing no other points of $S$. Then the balls $B(x, r_x/2)$ are pairwise disjoint. But since $x$ is in the support, each of these uncountably many disjoint balls must have positive measure. This is impossible for a finite or $\sigma$-finite measure. $\endgroup$
    – Nate Eldredge
    Commented Feb 23, 2023 at 3:34

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