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--- math105-s22:notes:lecture_6 [2022/02/03 07:43]
pzhou
+++ math105-s22:notes:lecture_6 [2026/02/21 14:41] (current)
@@ 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 $m(E \times F) = m(E) m(F)$.
@@ Line 33: / Line 33: @@
 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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