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



Vitali Covering Theorem for Arbitrary Subsets of Doubling Metric Spaces
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
$$
A\_R \setminus \bigcup\_{i=1}^NB\_i \subset \bigcup\_{5B\_i \cap A\_R \ne \emptyset, i > N}5B\_i.
$$


