Covering/packing lemma for a Polish space with a Borel probability measure

Question

Let $(X,d)$ be a complete separable metric space (Polish space) with a Borel probability measure $\mu$. Given $\varepsilon_1, \varepsilon_2 > 0$ can one find a finite set of disjoint open balls $B_1,\ldots, B_n$ of diameter $< \varepsilon_1$ such that $\mu(B_1) + \ldots + \mu(B_n) > 1-\varepsilon_2$?

I am embarrassed I am asking this. I hope it is not as trivial as I fear!

Answer 1

After a lot of searching on the web, the general answer is no.

There exists a compact metric space $\Omega$ [hence Polish] and a probability measure $P$ on $\mathcal{B}(\Omega)$ such that for any countable set $\mathcal{C}$ of disjoint closed balls of radius not exceeding unity, $\sum_{B\in\mathcal{C}} P(B) < 1/2$.

(Notice, this also applies to disjoint open balls (replacing $< 1/2$ with $\leq 1/2$) by slightly shrinking the open balls to be closed.) This is Example 4.49 in Counterexamples in Probability and Real Analyisis. (It is great to know such a book exists!) They cite Davies, "Measures are not approximable or specifiable by means of balls" 1971. In that paper Davies shows an even more interesting theorem.

There exists a compact metric space $\Omega^*$, and two distinct probability Borel measures $\mu_1$ and $\mu_2$ on $\Omega^*$, such that $\mu_1(S)=\mu_2(S)$ for every closed ball $S$ in $\Omega^*$ of radius less than unity.

Hence on $\Omega^*$, the measures of the balls alone don't determine the structure of the measure. (Instead one needs to know the measures of the finite unions of balls as well.)

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Positive results:

• For the space $[0,1]^n$ with the Lebesgue measure, such a covering theorem does exist via the Vitali covering theorem.

• The same is true for any Borel probability measure $\mu$ on $\mathbb{R}^n$ via the Besicovitch covering theorem.

• Davies also mentions that the result holds when the space is finite dimensional in some sense.