# 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.