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--- math105-s22:notes:lecture_6 [2022/02/02 23:42]
pzhou
+++ math105-s22:notes:lecture_6 [2022/02/03 16:28] (current)
pzhou [Lecture 6]
@@ Line 1: / Line 1: @@
 ====== Lecture 6 ======
+[[https://berkeley.zoom.us/rec/share/E9YcGb4kx_xuNLrBAVWTWhkZt0Y70DbhrtUMeBjTk6gZYhA8393qWN6W5UO9Yl0_.bMFQSrS0aqiGTY6X| video]]
 ===== Theorem 21 =====
 If  $E \In \R^n, F \In \R^k$  are measurable, then  $E \times F$  is measurable, with  $m(E) \times m(F) = m(E \times F)$ .
-Let's first treat some special case. If  $m(E)=0$ , and  $m(F) = \infty$ , what is  $m(E \times F)$ ? You have seen a special case as  $m(\{0\} \times \R)=0$  in  $\R^2$ . The general proof is similar, for each  $\epsilon$ , and each  $n \in \N$ , we can find a countable collection of boxes that covers  $E \times B(0, n)$  with total volume less than $\epsilon/2^n $. Then, we let$ n=1,2,\cdots$, and put together these collection of boxes into a bigger collection (still countable), that gives a cover of  $E \times \R^k$  with total area less than  $\epsilon$ .
+Let's first treat some special case. If  $m(E)=0$ , and  $m(F) = \infty$ , what is  $m(E \times F)$ ? You have seen a special case as  $m ( \{ 0 \} \times \R)=0$  in  $\R^2$ . The general proof is similar, for each  $\epsilon$ , and each  $n \in \N$ , we can find a countable collection of boxes that covers  $E \times B(0, n)$  with total volume less than $\epsilon/2^n $. Then, we let$ n=1,2,\cdots$, and put together these collection of boxes into a bigger collection (still countable), that gives a cover of  $E \times \R^k$  with total area less than  $\epsilon$ .
 Next, let's prove some nice cases, that  $m(E \times F) = m(E) m(F)$ .
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 By inner regularity, we may replace  $E$  by a closed set  $K$ . Since E is bounded, hence  $K$  is compact. Now, we try to cover  $K$  by open boxes of total area less than  $\epsilon$ . Let  $K_1= \pi (K)$  the projection to the first factor, than  $K_1$  is compact.
   * For each  $x \in I$ , we cover  $K_x$  by an open set  $V(x)$  of $m(V(x))<\epsilon $. We can find$ U(x) \supset x $, that$ U(x) \times V(x) \supset \pi_1^{-1}(U(x))  $. This is possible since$ K$ is compact.
-  * We know  $K \subset \cup_x U(x) \times V(x)$ , but that's uncountably many set. We can pass to a finite subcover, indexed by  $x_1, \cdots, x_N$ . Let $U_i = U(x_i) \RM (\cup_{j<i} U(x_j)) $,$ V_i = V(x_i) $, then we still have$ U_i \times V_i \supset \pi^{-1} U_i $. Thus,$ U_i $are disjoint, and we have$ m(\cup U_i) \leq 1 $and$ m(K) \leq \sum_i m(U_i)\times m(V_i) \leq \epsilon$.
+  * We know  $K \subset \cup_x U(x) \times V(x)$ , but that's uncountably many set. We can pass to a finite subcover, indexed by  $x_1, \cdots, x_N$ . Let $U_i = U(x_i) \RM (\cup_{j<i} U(x_j)) $,$ V_i = V(x_i) $, then we still have$ U_i \times V_i \supset \pi^{-1} U_i $. Thus,$ U_i $are disjoint, and we have$ m (\cup U_i) \leq 1 $and$ m (K) \leq \sum_i m(U_i)\times m(V_i) \leq \epsilon$.
+===== Discussion =====
+  - Can you prove that  $\{y=x\} \In \R^2$  has measure  $0$ ?
+  - In both of the two proofs above, we assumed  $E$  was bounded, how to deal with the general case?
+  - Prove that every closed subset (e.g. your favorite Cantor set is a closed set) in  $\R$  is a  $G_\delta$ -set. Is it true that every open set is a  $F_\sigma$ -set?

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