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math105-s22:notes:lecture_12 [2022/02/24 07:58]
pzhou 		math105-s22:notes:lecture_12 [2026/02/21 14:41] (current)
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 $$ \int F_+(x) dx = \int F_- (x) dx $$ $$ \int F_+(x) dx = \int F_- (x) dx $$
 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.  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 ===== ===== Pugh 6.8: Vitali Covering =====
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 **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 $\sum_k m(V_k) \leq m(A) + \epsilon$.  **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. +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 The crucial claim is the following, for any positive integer $N$, we have
math105-s22/notes/lecture_12.1645689482.txt.gz · Last modified: 2026/02/21 14:43 (external edit)

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