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math105-s22:notes:lecture_3 [2022/01/22 10:27]
pzhou created 		math105-s22:notes:lecture_3 [2022/01/25 14:17] (current)
pzhou
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 ====== Lecture 3 ====== ====== Lecture 3 ======
 +
 +{{ :math105-s22:notes:note_jan_25_2022_math105_3.pdf |note}} [[https://berkeley.zoom.us/rec/share/54Qr7VHBSiXzfMp6LH0TkV6BpqMBqrIja6onUg-DV6a4W4nt_4XK5FstD1u6xh4b.NZK87fNfKNqq9iaT | video]]
  
 Today we continue going over Tao's sequence of Lemma 7.4.2 - 7.4.11 Today we continue going over Tao's sequence of Lemma 7.4.2 - 7.4.11
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 Let $E = (0,+\infty)betheopenhalfspacein be the open half space in \R.Foranysubset. For any subset A \In \R$, we need to prove that  Let $E = (0,+\infty)betheopenhalfspacein be the open half space in \R.Foranysubset. For any subset A \In \R$, we need to prove that 
 m∗(A)=m∗(A∩Ec)+m∗(A∩E). m^*(A) = m^*(A \cap E^c) + m^*(A \cap E).  m∗(A)=m∗(A∩Ec)+m∗(A∩E). m^*(A) = m^*(A \cap E^c) + m^*(A \cap E).
-Let A+=A∩EA_+ = A \cap E and A−=A∩EcA_- = A \cap E^c, note that if 0∈A0 \in A, then 0∈A−0 \in A_-. First, we show that m∗(A)≤m∗(A−)+m∗(A+) m^*(A) \leq m^*(A_-) + m^*(A_+). This is because any open cover of A−A_- and open cover of A+A_+, union together form an open cover of AA. Next, we need to show that, for any $\epsilon>0$, we have +Let A+=A∩EA_+ = A \cap E and A−=A∩EcA_- = A \cap E^c, note that if 0∈A0 \in A, then 0∈A−0 \in A_-. First, we note that m∗(A)≤m∗(A−)+m∗(A+) m^*(A) \leq m^*(A_-) + m^*(A_+) by finite sub-additivity of outer-measure.  (This is because any open cover of A−A_- and open cover of A+A_+, union together form an open cover of AA.) Next, we need to show that, for any $\epsilon>0$, we have 
 m∗(A)+2ϵ≥m∗(A−)+m∗(A+). m^*(A) + 2\epsilon \geq m^*(A_-) + m^*(A_+). m∗(A)+2ϵ≥m∗(A−)+m∗(A+). m^*(A) + 2\epsilon \geq m^*(A_-) + m^*(A_+).
 The plan is the following. Given an open covering of AA by intervals {Bj}j=1∞\{B_j\}_{j=1}^\infty, such that m∗(A)+ϵ>∑j∣Bj∣m^*(A) + \epsilon > \sum_j |B_j|. We define The plan is the following. Given an open covering of AA by intervals {Bj}j=1∞\{B_j\}_{j=1}^\infty, such that m∗(A)+ϵ>∑j∣Bj∣m^*(A) + \epsilon > \sum_j |B_j|. We define
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 m∗(A)+2ϵ≥∑j∣Bj∣+ϵ≥∑j∣Bj+∣+∑j∣Bj−∣≥m∗(A+)+m∗(A−). m^*(A) + 2\epsilon \geq \sum_j |B_j| + \epsilon \geq \sum_j |B_j^+| + \sum_j |B_j^-| \geq m^*(A_+) + m^*(A_-).  m∗(A)+2ϵ≥∑j∣Bj∣+ϵ≥∑j∣Bj+∣+∑j∣Bj−∣≥m∗(A+)+m∗(A−). m^*(A) + 2\epsilon \geq \sum_j |B_j| + \epsilon \geq \sum_j |B_j^+| + \sum_j |B_j^-| \geq m^*(A_+) + m^*(A_-).
    
-That finishes the proof for n=1n=1 case. How to generalize to higher dimension? In the above discussion, we used the trick of 1=∑j1/2j1 = \sum_j 1/2^j, which is same trick as we prove outer-measure has countable sub-additivity. We can prove the general $n$ case in two steps, first treat the case of $A$ being an open box, then for general subset $A \In \R^n$, and given an open box cover $\{B_j\}$ of AA with $\sum_j |B_j| < m^*(A) +\epsilon$, we have +That finishes the proof for n=1n=1 case. How to generalize to higher dimension? In the above discussion, we used the trick of 1=∑j1/2j1 = \sum_j 1/2^j, which is same trick as we prove outer-measure has countable sub-additivity. See the hint in Tao's Exercise for a two-step prove, that utilizes many results that we have proven.  
-$$ m^*(A) + \epsilon > \sum_j |B_j| = \sum_j m^*(B_j)  = \sum_j m^*(B_j \cap E) + m^*(B_j \RM E) \geq m^*(\cup_j(B_j \cap E)) + m^*(\cup(B_j \RM E)) \gep m^*(A \cap E) + m^*(A \RM E) + 
 +==== Lemma 7.4.4 ==== 
 +We only prove some part here.  
 +  * (c) If $E_1, E_2$ are measurable, then $E_1 \cup E_2$ and E1∩E2E_1 \cap E_2 are measurable.  
 +  * (e) every open box, and every closed box, is measurable  
 +  * (f) if a set has outer-measure zero, then it is measurable.  
 + 
 +Proof:  
 +  * (c) Try the case first with $A = \{ x^2 + y^2 < 1 \} \In \R^2and and E_1 = \{x>0\}$ and $E_2 = \{y>0\}$, then we can partition $$ A = A_{++} \sqcup A_{+-} \sqcup A_{-+} \sqcup A_{--}  (cut the pie to 4 pieces), then we have finite additivity of outer-measure in this case. Such conclusion holds in general, assuming E1,E2E_1, E_2 are measurable.  
 +  * (e) express open box as intersections of translated half-spaces.  
 +  * (f) Assume EE has outer-measure 0, then $0 \leq m^*(A \cap E) \leq m^*(E) = 0$, we can prove $m^*(A) \geq m^*(A \RM E) + m^*(A \cap E) = m^*(A \RM E)$ again by monotonicity. 
 + 
 +==== Lemma 7.4.5 ==== 
 +finite additivity: if $\{E_j\}_{j=1}^Nisafinitecollectionof∗∗disjoint∗∗measurablesets,thenforanysubset is a finite collection of **disjoint** measurable sets, then for any subset A$, we have  
 + m^*(A \cap (\cup_j E_j)) = \sum_j m^*(A\cap E_j)  
 + 
 +Proof: again, we try the case of N=2N=2 first, to get intuition. Let $B = A \cap (E_1 \cup E_2),and, and B_1 = A \cap E_1 = B \cap E_1,and, and B_2 = A \cap E_2 = B\cap E_2$, so, we are trying to prove that 
 + m^*(B) = m^*(B_1) + m^*(B_2)  
 +since E1E_1 and E2E_2 are disjoint, we notice that $B_2 = B \RM E_1$, thus the above holds by measurability of $E_1appliedtothetestset applied to the test set B$ 
 + 
 + 
 +===== Discussion Time ===== 
 +Let's fill in the details of the above sketches.  
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math105-s22/notes/lecture_3.1642876074.txt.gz · Last modified: 2022/01/22 10:27 by pzhou

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