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--- math105-s22:notes:lecture_12 [2022/02/24 07:02]
pzhou created
+++ math105-s22:notes:lecture_12 [2026/02/21 14:41] (current)
@@ 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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