We just finished our unit on covering lemma's in my analysis class and my professor proved both the Vitali and Besicovitch covering lemma's (for finite and infinite coverings) using balls. He mentioned that it was possible to use cubes in place of balls and the proof of how to do this with Vitali seemed pretty easy. I was wondering if anyone had access to a proof of the Besicovitch covering lemma that uses cubes instead of balls and tackles the infinite case. In other words,
Given a set $E\subset\mathbb{R}^n$ and a possibly infinite family of cubes $\{Q_{x_i}\}_{{x_i}\in E}$ such that $E\subset \bigcup_{{x_i}\in E}Q_{x_i}$ with $sup_{xi}\rho(Q_{xi})<+\infty$ where $Q=(x,\rho)$ is the cube with center x and side length $\rho$, show that there exists a sub-family of cubes $\{Q_{x_i}\}_{{x_i}\in I}$ such that $E\subset\bigcup_{x_i\in I}Q_{x_i}$ and each point in E lies in at most a finite number of cubes in the sub-family.
I have seen a couple of proofs but they all require that E be bounded. Are there any out there for a general E?
Best Answer
DiBenedatto has a proof of this theorem in his book real analysis. I tried to turn it in but our professor informed me that he did not want the centers of the cubes to be the points in E, he wanted E to be inside the middle half of the cubes. Here is the link to DiBenedetto
http://books.google.com/books?id=5ddbKSkaL8EC&pg=PA118&lpg=PA118&dq=besicovitch+covering+cubes&source=bl&ots=_M5urDKGwu&sig=96Egj3jMorCGp_5gA1q0-vt86Lg&hl=en&sa=X&ei=3buTUoL2AqvLsATo-YHoDw&ved=0CFkQ6AEwCA#v=onepage&q=besicovitch%20covering%20cubes&f=false