# Vitali Covering Theorem for Arbitrary Subsets of Doubling Metric Spaces

Hello,

I've been wondering today around the following exercise in J.Heinonen's "Lectures on Analysis on Metric Spaces" (see below for terminology): prove that the statement of Vitali Covering Theorem for bounded subsets $A\subset X$ implies the statement for all subsets $A\subset X$.

Here $X$ is a metric space equipped with a Borel regular outer measure $\mu$ which is doubling (balls have finite measure and doubling the radius of a ball increases its measure at most by a constant factor).

Let me quickly recall the actual statement of the theorem: in a doubling metric space $(X,\mu)$ with $\mu$ Borel regular, any family $\mathcal{F}$ of closed balls centerd at a set $A\subset X$ with the property that about every $a\in A$ there are elements in $\mathcal{F}$ of arbitrarily small radii admits a countable disjointed subfamily $\mathcal{G}$ covering almost all of $A$.

The actual proof of Heinonen works for sets of finite measure. The conclusion for unbounded sets is trivial for Lebesgue measure in $\mathbb{R}^n$, and whenever boundaries of balls have measure zero (so we can a.e partition $A$ by subsets admiting the conclusion of Vitali by mutually disjoint families). If boundaries of balls are not negligible (at least for some family of concentric balls with radii going to infinity) I can't find the argument bringing me there for arbitrary subsets.

I have seen a couple of references: Wikipedia (ok, not the best one maybe) works with Lebesgue measure, and Evans "Measure Theory and Fine Properties of Functions" does it for Radon Measures on $\mathbb{R}^n$, but only on sets of finite measure.

So before proceeding with trying the problem. Do you know whether this is true or have a hint for proving it?

Best

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if it can help, in doubling length spaces the boundaries of balls are always negligible –  Simo_the_Wolf Oct 26 '12 at 0:23
Surprisingly I got a (conditional) answer in my way back home: provided the Borelians on $X$ are measurable you can actually prove the existence of a collection of balls as the above (concentric, going to infinity, with negligible boundary) using the (other) properties of $\mu$ (its finiteness in balls). Measurability of the borelians should be a consequence of regularity (but I need to take some book and see whether this happens in this abstract setting). –  David Oct 26 '12 at 0:25
@Simo_the_Wolf thanks. Think things will go that way as you see. –  David Oct 26 '12 at 0:26
Edition made: $\mu$ is (only) an outer measure. –  David Oct 26 '12 at 0:27
@Simo_the_Wolf: In the claim there is no assumption that the space $X$ would be a length space. In particular, it may very well happen that the collection $\mathcal{F}$ contains only balls with boundaries having positive measure. –  Tapio Rajala Oct 26 '12 at 7:11

## 1 Answer

The case where $A$ is unbounded does not really need much modifications. First you use the basic covering theorem to get a disjointed collection of balls $B_i$ with $A \subset \bigcup_i 5B_i$. Then, for example, fix a point $x_0 \in X$ and take $R>0$ and consider $A_R = A \cap B(x_0,R)$. Now $$\sum_{5B_i \cap A_R \ne \emptyset, i \ge N}\mu(5B_i) \rightarrow 0, \quad \text{as }N \rightarrow \infty$$ and $$A_R \setminus \bigcup_{i=1}^NB_i \subset \bigcup_{5B_i \cap A_R \ne \emptyset, i > N}5B_i.$$

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