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--- math105-s22:notes:lecture_12 [2022/02/23 23:02]
pzhou created
+++ math105-s22:notes:lecture_12 [2022/02/25 00:35] (current)
pzhou
@@ Line 4: / Line 4: @@
   * Redo Fubini's Theorem (Tao 8.5.1)
+  * Vitali Covering Lemma.
 ===== Upper and Lower Lebesgue integral =====
@@ Line 31: / Line 32: @@
  $\int F_+(x) dx = \int F_- (x) dx$
 hence  $F_+(x) = F_-(x)$  for almost all  $x$ . Thus, for a.e.  $x$ , we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus  $\int f(x,y) dy$  exists for a.e. x.
+==== A Lemma ====
+Suppose  $A$  is measurable, and  $B \In A$  any subset, with  $B^c = A \RM B$ . Then
+ $m(A) = m^*(B) + m_*(B^c)$
+Proof:
+ $m^*(B) = \inf \{ m(C) \mid A \supset C \supset B, C \text{measurable} \} = \inf \{ m(A) - m(C^c) \mid A \supset C \supset B, C \text{measurable} \}$
+ $= m(A) - \sup \{ m(C^c) \mid A \supset C \supset B, C \text{measurable} \} = m(A) - \sup \{ m(C^c) \mid C^c \subset B^c, C^c \text{measurable} \} = m(A) - m_*(B^c)$
+===== Pugh 6.8: Vitali Covering =====
+ $\gdef\vcal{\mathcal{V}}$
+A Vitali covering  $\vcal$  of a set  $A \In \R^n$  is such that, for any  $p \in A, r > 0$ , there is a covering set  $V \in \vcal$ , such that  $\{p\} \subsetneq V \In B_r(p)$ , where  $B_r(p)$  is the open ball of radius  $r$  around  $p$ .
+**Vitali Covering Lemma:** Let  $\vcal$  be a Vitali covering of a measurable bounded subset  $A$  by closed balls, then for any $\epsilon>0$, there is a countable disjoint subcollection $\vcal' = \{V_1, V_2, \cdots \} $, such that$ A \RM \cup_k V_k $is a null set, and$ \sum_k m(V_k) \leq m(A) + \epsilon$.
+Proof: The construction is easy, like a 'greedy algorithm'. First, using the given  $\epsilon$ , we find an open subset  $W \supset A$ , with  $m(W) \leq m(A) + \epsilon$ . Let  $\vcal_1 = \{V \in \vcal: V \In W\}$ , and  $d_1 = \sup \{diam V: V \in \vcal_1\}$ . We pick  $V_1 \in \vcal_1$  where the diameter is sufficiently large, say  $diam V_1 > d_1 /2$ . Then, we delete  $V_1$  from  $W$ , let  $W_2 = W \RM V_1$ , and consider  $\vcal_2 = \{ V \in \vcal_1, V \In W_2\}$ , and define  $d_2 = \sup \{diam V: V \in \vcal_2\}$ , and pick  $V_2$  among  $\vcal_2$  so that  $diam V_2 > d_2 /2$ . Repeat this process, we get a collection of disjoint closed balls  $\{V_i\}$ . Suffice to show that  $A \RM \cup V_i$  is a null set.
+The crucial claim is the following, for any positive integer  $N$ , we have
+ $\cup_{k=N}^\infty 5 V_k \supset A \RM (\cup_{i=1}^{N-1} V_i)$
+Suppose not, and there is a point  $a \in A \RM (\cup_{i=1}^{N-1} V_i)$ , but not in  $\cup_{k=N}^\infty 5 V_k$ , then we can find a closed ball  $B \in \vcal_N$ , such that  $a \in B$ . Since  $a \notin 5V_N$ , we have  $B \not\subset 5V_N$ . This implies  $B \cap V_N = \emptyset$ . Draw a picture. This implies  $B \in \vcal_{N+1}$ . Then, repeat the above story  $N$  replaced by  $N+1$  and same  $a,B$ , we can keep going and show that  $B \in \vcal_k$  for all  $k \geq N$ . That cannot be true, since  $d_k \to 0$ , but  $B$  has fixed radius.

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