math105-s22:notes:lecture_13

$\gdef\vcal{\mathcal V}$

Last time we were in the middle of proving Vitali Covering Lemma, which informally says: “given a Vitali cover of a bounded measurable set $A$ using closed balls and given any $\epsilon>0$, we can find an almost cover of $A$ by countably many disjoint closed balls, where we waste no more than $\epsilon$ in the cover, and we miss only a null set in the cover”.

The construction goes by 'greedy algorithm' (which is always pick the biggest among all possible choices), however, since we have infinitely many balls, we cannot pick the biggest, so we pick some ball whose radius is 'good enough'. More precisely, we first pick an open set $W \supset A$ with $m(W) < m(A) + \epsilon$. We set $\vcal_0 = \vcal$, $W_1=W$. Then we run the iteration. starting at $n=1$

- $\vcal_n = \{ V \in \vcal_{n-1} | V \In W_n \}$, then $\vcal_n$ is still a Vitali cover of $A \cap W_n$ (why?)
- $d_n = \sup{diam V | V \in \vcal_n\} $
- choose $V_n$ so that $diam (V_n) > d_n/2$.
- set $W_{n+1} = W_n \RM V_n$.
- increase $n$ by 1 and repeat

After we run the algorithm and obtain a collection of disjoint balls $V_n$, we need to show that these balls cover $A$ up to a null set. Last time, we proved that, for any positive integer $N$, we have $$ \cup_{k \geq N} 5 V_k \supset A \RM (V_1 \cup \cdots \cup V_{N-1}) $$ (recall the proof)

Why is this useful? It allows us to say, for any $\delta > 0$, there exists an $N$, so that $m(A \RM (V_1 \cup \cdots \cup V_{N-1}) ) < \cup_{k \geq N} 5 V_k) < 5^n \sum_{k=N}^\infty m(V_k) < \delta$. (This last inequality is always achievable by choosing $N$ large enough, since $\sum_{k=1}^\infty m(V_k) < m(W) < \infty$. )

We decompose $\R^n$ into a grid of size $1$ cubes, throw aways the boundaries (which is a measure zero set). We enumerate these cubes as $\{C_n\}$, then for any $\epsilon>0$, we find Vitali cover for $A \cap C_i$ with excess $\epsilon / 2^i$. Then we union together the solution to the sub-problemes (countable union of countable collection is still a countable collection).

General shape of the ball. Actually, we can work with non-Euclidean norm. (can the shape be more general?)

Suppose $E \in \R^n$ is measurable, for any $p \in \R^n$, we define density of $E$ at $p$ to be $$\delta(p,E) = \lim_{Q \downarrow x} m(E \cap Q) $$ remark:

- here we don't have a sequence, so what does convergence mean? (limit indexed by a poset)
- we could define lower density $\underline\delta(p,E)$ using $\liminf$; similarly upper density..

Example, if $E = (0,1) \In \R$, does density exists at the boundary of $E$?

If $\delta(p,E)=1$, we say $p$ is a density point of $E$.

Lebesgue density theorem. Almost all points of $E$ are density point.

Proof: define for any $0 \leq a < a$, let $E_a = \{p \in E \mid \underline\delta(p,E) < a \}$. Claim $m^*(E_a) =0$. Given the claim, then $E_{<1} := \cup_{0 \leq a<1} E_a = \cup_{n=1}^\infty E_{1/n}$ is null set. And any nondensity point $p$ would belong to $E_{<1}$, then we are done.

To prove the claim $m^*(E_a)=0$, we take the collection of cubes $Q$, such that $[Q: E] = m(Q \cap E) / m(Q) < a$. Such cubes form a Vitali covering of $E_a$, indeed, for any $p \in E_a$ and any $r >0$, there are some cube $Q$ contained in $B_r(p)$ (containing p), with $[Q:E] < a$. Then, using Vitali covering by such cubes, and for any $\epsilon>0$, we can get disjoint almost cover of $E_a$ by such $Q$ with margin size $\epsilon$. Then we have $$ m^*(E_a) = \sum_{i=1}^\infty m^*(E_a \cap Q_i) \leq \sum_{i=1}^\infty m^*(E \cap Q_i) \leq \sum_{i=1}^\infty a m^*(Q_i) = a \sum_i m(Q_i) \leq a(m(E_a) + \epsilon) $ Thus, $m^*(E_a) \leq (a/ (1-a)) \epsilon$, for any $\epsilon$, hence $m^*(E_a)=0$.

math105-s22/notes/lecture_13.txt · Last modified: 2022/03/01 00:10 by pzhou