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?


  • $\begingroup$ if it can help, in doubling length spaces the boundaries of balls are always negligible $\endgroup$ Oct 26, 2012 at 0:23
  • $\begingroup$ 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). $\endgroup$
    – David
    Oct 26, 2012 at 0:25
  • $\begingroup$ @Simo_the_Wolf thanks. Think things will go that way as you see. $\endgroup$
    – David
    Oct 26, 2012 at 0:26
  • $\begingroup$ Edition made: $\mu$ is (only) an outer measure. $\endgroup$
    – David
    Oct 26, 2012 at 0:27
  • $\begingroup$ @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. $\endgroup$ Oct 26, 2012 at 7:11

1 Answer 1


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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