A Covering Lemma for Arbitrary Measures In the book "Harmonic Measure" by Garnett and Marshall, we have the following result:
Lemma I.2.3 Let $\mu$ be a positive Borel measure on $\partial{\mathbb{D}}$ and let $\{I_{j}\}$ be a finite sequence of open intervals in $\partial{\mathbb{D}}$. Then $\{I_{j}\}$ contains a pairwise disjoint subfamily $\{J_{k}\}$ such that
$\sum\limits_{k}\mu(J_{k})\geq{\frac{1}{3}\mu\big(\cup_{j}I_{j}\big)}$
By repeating the argument in Garnett and Marshall we get the same result for positive measures $\mu$ on $\mathbb{R}$, only with a better constant- $1/2$ instead of $1/3$. Does an analogous result hold in $\mathbb{R}^{n}$ for $n>1$? Namely, does there exist an absolute constant $c=c(n)>0$ such that if $\mu$ is a positive Borel measure on $\mathbb{R}^{n}$ and $\{Q_{i}\}$ is a finite sequence of balls or cubes, then there exists a pairwise disjoint subfamily of balls or cubes $\{J_{k}\}$ such that
$\sum\limits_{k}\mu(J_{k})\geq{c\cdot{}\mu\big(\cup_{i}Q_{i}\big)}$
If this fails in higher dimensions, does there exist a characterization of the measures for which such a condition holds?
 A: It doesn't answer the question, but this is too long for a comment. I just want to point out that this is not possible with general rectangles even for Lebesgue measure on $\mathbb{R}^2$. Since in one dimension we find that balls, cubes and rectangles are all the same this seems to indicate something more than the one-dimensional argument would be needed to prove the statement.
To see that it is not possible for general rectangles, consider a set of $n$ points evenly spaced points on the unit circle and draw all edges between distinct points, so we obtain a copy of the complete graph on $n$ vertexes. Now, let $\epsilon > 0$ and for each edge consider the smallest rectangle that contains its $\epsilon$-neighborhood. As $\epsilon \downarrow 0$ this union of these rectangles has area $\epsilon n^2 - O(\epsilon^2)$, so for sufficiently small $\epsilon$ the area of the union is at least $\frac{1}{2}\epsilon n^2$. But it is clear that any pairwise disjoint collection of these rectangles can have at most one representative for each vertex, hence its union cannot have area larger than $\epsilon n$.
A: The answer is no for $n \ge 2$. Consider open cubes $Q(c,1)=\{c_i <x_i <c_i+1\}$ of side $1$. Starting from $Q(0,1)$ and moving $c$ along the diagonal joining $0$ with $(1,\dotsc,1)$ one constructs $N$ cubes $Q(c_i,1)$ such that two of them always intersect and no cube is contained in the union of the others. Take $$z_i \in Q(c_i,1) \setminus \bigcup_{j \neq i}Q(c_j,1)$$ and $\mu$ equal to the sum of Dirac masses at $z_i$. Then $\mu(\bigcup Q(c_i,1))=N$ but any disjoint subcollection of the cubes consist of only $1$ cube and has measure $1$.
