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math105-s22:notes:lecture_12

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math105-s22:notes:lecture_12 [2022/02/23 23:58]
pzhou [Pugh 6.8: Vitali Covering]
math105-s22:notes:lecture_12 [2022/02/25 00:35] (current)
pzhou
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 F+(x)dx=F(x)dx \int F_+(x) dx = \int F_- (x) dx  F+(x)dx=F(x)dx \int F_+(x) dx = \int F_- (x) dx
 hence F+(x)=F(x)F_+(x) = F_-(x) for almost all xx. Thus, for a.e. xx, we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus f(x,y)dy\int f(x,y) dy exists for a.e. x.  hence F+(x)=F(x)F_+(x) = F_-(x) for almost all xx. Thus, for a.e. xx, we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus f(x,y)dy\int f(x,y) dy exists for a.e. x. 
 +
 +==== A Lemma ====
 +Suppose AA is measurable, and BAB \In A any subset, with Bc=A\BB^c = A \RM B. Then
 +m(A)=m(B)+m(Bc) m(A) = m^*(B) + m_*(B^c)
 +Proof: 
 +m(B)=inf{m(C)ACB,Cmeasurable}=inf{m(A)m(Cc)ACB,Cmeasurable} 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(Cc)ACB,Cmeasurable}=m(A)sup{m(Cc)CcBc,Ccmeasurable}=m(A)m(Bc) = 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 =====
math105-s22/notes/lecture_12.1645689500.txt.gz · Last modified: 2022/02/23 23:58 by pzhou