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--- math105-s22:notes:lecture_4 [2022/01/27 00:37]
pzhou
+++ math105-s22:notes:lecture_4 [2022/01/27 14:29] (current)
pzhou
@@ Line 1: / Line 1: @@
 ====== Lecture 4 ======
+{{ :math105-s22:notes:note_jan_27_2022_math105_4.pdf |note}}, [[https://berkeley.zoom.us/rec/share/FpG9GNaABIipr523cZa_36y6w6O9We_zWhb-llmVIusvAyLlFmVqtiBUpjp9Q2hZ.-dvCEBgnfAdGXV7e | video ]]
 ==== Cor 7.4.7 ====
@@ Line 10: / Line 11: @@
 ==== Lemma 7.4.8: Countable addivitivty ====
 Let  $\{E_j\}_{j=1}^\infty$  be a countable collection of **disjoint** subsets. Then, their union  $E$  is measurable, and we have
- m^*(E) = \sum_{j=1}^\infty m^*(E_j}
+ m^*(E) = \sum_{j=1}^\infty m^*(E_j)
 Proof: Tao's proof is really clever, let's first try to go through the proof, then discuss how we can come up with it ourselves.
 First, we start by showing  $E$  is measurable from the definition: we want to show for any subset  $A$ , we have
- $m^*(A) = m^*(A \cap E) + m^(A \RM E)$
+ m^*(A) = m^*(A \cap E) + m^*(A \RM E)
 Suffice to show  $\geq$  direction. Let  $F_N = \sum_{j=1}^N E_j$ . We have two expressions
   *  $m^*(A \cap E) \leq \sum_{j=1}^\infty m(A \cap E_j) = \sup_{N > 1} \sum_{j=1}^N m^*(A \cap E_j) = \sup_{N > 1} m^*(A \cap F_N)$
@@ Line 22: / Line 24: @@
  $m^*(A \RM E) + m^*(A \cap E) \leq \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM E)) = \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM F_N)) = \sup_{N > 1}  m^*(A) = m^*(A)$
+OK, that shows  $E$  is measurable. To finish off, we need to show countable addivity
+ $m^*(E) = \sum_{j=1}^\infty m^*(E_j)$
+Since  $E = \cup E_j$ , we have  $\leq$  from countable sub-addivity. Then, by monotonicity, we have
+ $m^*(E) \geq m^*(F_N) = \sum_{j=1}^N m^*(E_j)$
+since this is true for all  $N$ , we can sup over  $N$ , and get
+ $m^*(E) \geq \sup_N \sum_{j=1}^N m^*(E_j) = \sum_{j=1}^\infty m^*(E_j)$
+----
+One slogan is to approximate  $E$  by  $F_N$ . We want to prove
+  $m^*(A) \geq  m^*(A \cap E) + m^*(A \RM E)$
+If we have
+  * (1)  $m^*(A \cap E) \leq m^*(A \cap F_N)$  and
+  * (2)  $m^*(A \RM E) \leq m^*(A \RM F_N)$ ,
+then we can write
+ $m^*(A \cap E) + m^*(A \RM E) \leq m^*(A \cap F_N) + m^*(A \RM F_N) = m^*(A)$
+but unfortunately, (1) is wrong. One way to remedy this, is to show that (assuming  $m^*(A)< \infty$ ),  for any  $\epsilon > 0$ , there exists an  $N$ , such that  $m^*(A \cap E) \leq m^*(A \cap F_N) + \epsilon$  holds. (see if you can make this approach work). Another more elegant approach is done as above, using countable subadditivity to get  $\leq$ , then introduce a  $\sup$  to get to finite  $N$ .
+Try to forget this proof, and come up with your own. It might be fun.
+==== Lemma 7.4.9 ====
+The  $\sigma$ -algebra property.
+Given a countable collection of measurable set  $\Omega_j$ , one need to prove that  $\cup \Omega_j$  and  $\cap \Omega_j$  are measurable.
+We only need to prove the case of  $\Omega = \cup_j \Omega_j$ , since the  $\cap$  operation can be obtained by taking complement and  $\cup$ . The hint is to define
+ $\Omega_N = \cup_{j=1}^N \Omega_j$
+and  $E_N = \Omega_N \RM \Omega_{N-1}$ , then  $\{E_j\}$  are measurable, mutually disjoint, and  $\cup_j E_j = \cup_j \Omega_j = \Omega$ .
+==== Lemma 7.4.10 ====
+Every open set can be written as a finite or countable union of open boxes.
+I will leave this as discussion problem.
+  * A subset  $U$  is open, if for every point  $x \in U$ , there exists an open ball  $B(x,r) \In U$ .
+  * claim: a subset  $U$  is open, iff  $\forall x\in U$ , there exists an open box  $B$ , such that  $x \in B \In U$ .
+  * claim: a subset  $U$  is open, iff  $\forall x\in U$ , there exists an open box  $B$  with rational boundary coordinates, such that  $x \in B \In U$ .
+  * There are countably many open boxes with rational boundary coordinates.
+==== Lemma 7.4.11 ====
+All open sets are measurable.
+Since open boxes are measurable, and countable union of measurable sets are measurable.
+==== Discussion Problem ====
+An alternative definition for measurable set is the following:
+=== Def 2 ===
+A subset  $E$  is measurable, if for any $\epsilon>0$, there exists an open set  $U\supset E$ , such that  $m^*(U \RM E) < \epsilon$ .
+Can you show that this definition is equivalent to the Caratheodory criterion (the one we had been using)?

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