Posted by: hilbertthm90 | September 1, 2008

Covering Lemma 2

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 B(x_i, r_i) (say to N), then there is a subcollection S\subset \{1,\ldots , N\} so that

a) the balls B(x_i, r_i) with i\in S are disjoint.

b) W\subset \bigcup_{i\in S} B(x_i, 3r_i), and

c) m(W)\leq 3^k\sum_{i\in S} m(B(x_i, r_i)). Hmm…I guess I should say W\subset\mathbb{R}^k 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), r_1\geq r_2\geq \cdots \geq r_N. So now just take a subsequence of that \{r_{i_k}\} where i_1=1 and then go down the line until you get to the first ball that doesn’t intersect B_{i_1}. Now choose B_{i_3} 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 B_j, then we claim it was a subset of B(x_i, r_i) 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 r'\leq r and since we skipped it, B(x', r') intersects B_r. If we double the radius of r, then it will go cover at least half of B_r' and if we triple it, it will cover all of B_r'. So for any skipped ball we have B(x', r')\subset B(x, 3r) giving us (b).

For (c), we just use the standard property of Lebesgue measure that m(B(x, 3r))=3^km(B(x,r)). Sum over the set we created and we are done.

Infinite Case: Let \{B_j : j\in J\} be any collection of balls in \mathbb{R}^k such that \sup_j diam(B_j)<\infty. Then there exists a disjoint subcollection (J'\subset J) such that \bigcup_{j\in J}B(x_j, r_j) \subset \bigcup_{j\in J'}B(x_j, 5r_j).

Proof: Let R be the sup of the radius of the balls (which we’ve forced to be finite). Now we define subcollections. Let Z_i be the subcollection of balls with radius in \left(\frac{R}{2^{i+1}}, 2^iR \right]. Now take the maximal disjoint subcollection Z_0' of Z_0, etc. (maximal subcollection of Z_i' of Z_i disjoint from Z_{i-1}'…). 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.

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