Vitali Covering Theorem for Arbitrary Subsets of Doubling Metric Spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:31:00Z http://mathoverflow.net/feeds/question/110713 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110713/vitali-covering-theorem-for-arbitrary-subsets-of-doubling-metric-spaces Vitali Covering Theorem for Arbitrary Subsets of Doubling Metric Spaces David 2012-10-25T23:29:06Z 2012-10-26T07:05:39Z <p>Hello,</p> <p>I've been wondering today around the following exercise in J.Heinonen's "Lectures on Analysis on Metric Spaces" (see below for terminology): <em>prove that the statement of Vitali Covering Theorem for bounded subsets $A\subset X$ implies the statement for all subsets $A\subset X$</em>.</p> <p>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).</p> <p>Let me quickly recall the actual statement of the theorem: <em>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 $A$.</em></p> <p>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.</p> <p>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.</p> <p>So before proceeding with trying the problem. Do you know whether this is true or have a hint for proving it?</p> <p>Best</p> http://mathoverflow.net/questions/110713/vitali-covering-theorem-for-arbitrary-subsets-of-doubling-metric-spaces/110729#110729 Answer by Tapio Rajala for Vitali Covering Theorem for Arbitrary Subsets of Doubling Metric Spaces Tapio Rajala 2012-10-26T07:05:39Z 2012-10-26T07:05:39Z <p>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 $$A_R \setminus \bigcup_{i=1}^NB_i \subset \bigcup_{5B_i \cap A_R \ne \emptyset, i > N}5B_i.$$</p>