Today I’ll do probably the best known Vitali Covering Lemma. I’ll take the approach of Rudin.
Statement (finite version): If W is the union of a finite collection of balls (say to N), then there is a subcollection so that
a) the balls with are disjoint.
b) , and
c) . Hmm…I guess I should say from the looks of it.
Proof: Quite simple in this case. Just order the radii in decreasing order (finite so we can list them all), . So now just take a subsequence of that where and then go down the line until you get to the first ball that doesn’t intersect . Now choose as the next one that doesn’t intersect either of the ones before it. Continue in this process to completion. (a) is done since we’ve picked a disjoint subset of the original (this is a trivial condition, though, since we could ignore (b) and (c) and just choose a single element).
Now for (b), look at any of the skipped over , then we claim it was a subset of for some i that we picked. This is clear when we note order. If we skipped some r’, then there was an early one we didn’t skip, so and since we skipped it, intersects . If we double the radius of r, then it will go cover at least half of and if we triple it, it will cover all of . So for any skipped ball we have giving us (b).
For (c), we just use the standard property of Lebesgue measure that . Sum over the set we created and we are done.
Infinite Case: Let be any collection of balls in such that . Then there exists a disjoint subcollection () such that .
Proof: Let R be the sup of the radius of the balls (which we’ve forced to be finite). Now we define subcollections. Let be the subcollection of balls with radius in . Now take the maximal disjoint subcollection of , etc. (maximal subcollection of of disjoint from …). This collection now satisfies the requirements.
Next time I’ll do Vitali’s Covering Theorem. I’m debating whether to prove it or not. Applications of it might be more interesting.