Suppose that we are given a collection of dyadic, mutually disjoint, open cubes in $\mathbf{R}^n$ in which the the union of all the cubes has finite measure. Is it necessary that the boundary of the cubes has zero measure? I think this is may not necessarily the case as the boundary of all the cubes, even if they do overlap, may possibly create a fractal. Or is this true? I'm not quite sure. Note that these cubes come from the Calderón–Zygmund decomposition and so each cube necessarily touches the boundary of another.
The reason I ask is that I recently worked through the proof of the boundedness of Calderón–Zygmund operators in which the Calderón–Zygmund decomposition is applied. Part of the proof relies on showing that you can move the sum of the integrals of the "bad" functions to the inside of the integral by applying the Lebesgue dominated convergence theorem. The bound has to be uniform in the integration variable, so in order to obtain this bound we need to take $x$ outside of the closure of the union of all the cubes. However, the union of the boundaries of the cubes may not have zero measure? What if we assume the cubes are closed? Does that change anything.
Any help is appreciated. Thanks in advance!
The cubes in the Calderón–Zygmund decomposition are dyadic cubes. There are countably many dyadic cubes. The boundary of each cube has measure zero. Countable union of sets of measure zero has measure zero.