# [SOLVED]Besicovitch Covering Lemma

#### ArcanaNoir

##### New member
Let {Q} be a collection of cubes covering a set E in R^n. Prove that there is a countable sub collection {Q}' of these cubes which covers E and $$\cup {\frac{1}{2}Q} \subseteq \cup {Q}'$$, and the number of cubes in the subcollection containing any given point of E is less than something depending only on the dimension.
Oh and the sup of the side lengths of the cubes is bounded, but E is not bounded.

$$\frac{1}{2}Q$$ means the cube inside Q with side length 1/2 of the side length of Q.
In other words, the countable subcovering needs to cover the middle sections of the original cover.

I really like this proof Real Analysis - Emmanuele DiBenedetto - Google Books

But in this proof the set E is the centers of the cubes, so we only end up covering centers, not middle halves, and also E is bounded in that proof. My professor said I could modify this proof but I don't see how. *sad face* He also said the first half of my most recent attempt at this proof was badly written and the second half was worse.

The idea for dealing with the unbounded set E is to take "layers" of cubes which are about the same size. So something like sup{S(Q)}=R where S(Q) is the side length of Q and then $$(1-\epsilon)^kR< S(Q)\le R$$ for the first layer, then take a maximal set of points in E which are some specific distance apart (this is a piece I messed up on, no matter what distance I separate my points by there is always some flaw in my choice). Then take the cubes containing those points and discard some cubes (which ones?). Rinse and repeat.

I tried doing some optimal arrangement of separated points around a specific point for the bound, but I was told I did that wrong too (that was the "worse" part apparently.)

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