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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 $\epsilon$.

Next, let's prove some nice cases, that $m(E \times F) = m(E) m(F)$.

• $E, F$ are open boxes.
• $E, F$ are open sets (each open set is a disjoint union of open boxes and a measure zero set. See Pugh's dyadic proof).

Now, how to prove that $ E \times F$ is measurable? We could use Caratheodory criterion, or, we could use Lebesgue criterion, by constructing outer and inner approximations. Again, we may assume $E$ is bounded and $F$ is bounded (otherwise, they can be written as disjoint union of bounded measurable pieces, and we can deal with them pieces by pieces, and do countable union in the end). We may assume $E \In B_1, F \In B_2$, for $B_i$ some open boxes.

Let $H_E$ be a $G_\delta$-set and $K_E$ be $F_\delta$, such that $H_E \supset E \supset K_E$, and $m(H_E \RM K_E) = 0$. Define $H_F, K_F$ for $F$ similarly. Then, by downward monotone continuity of measure, we have $$ m(H_E \times H_F) = \lim m(H_{E,n} \times H_{F,n}) = \lim m(H_{E,n}) \times m(H_{F,n}) = m(H_E) \times m(H_F) = m(E) \times m(F) $$ where $H_{E,n}$ are open sets, with $H_{E,n} \supset H_{E,n+1}$, and $H_E = \cap_n H_{E,n}$ for all $n$.

Also, we have $$ H_E \times H_F \RM (K_E \times K_F) \subset (H_E \times K_E) \times B_2 \cup B_1 \times (H_F \times K_F) $$ where the last term is a null set, hence $m(K_E \times K_F) = m(H_E \times H_F) = m(E) \times m(F)$.

If $E \In \R^n \times \R^k$ is measurable, then $E$ is a zero set if and only if almost( = up to a zero set) every slice $E_x$, ($x \in \R^n$) is measure zero.

Pf: as usual, we may assume $n=k=1$ and $E$ is contained in the unit square. Suppose $E$ is measurable, and $m(E_x)=0$ for almost all $x \in I$, then we want to show $m(E)=0$. Let $Z \In I$ be the set of $x$ where $m(E_x) \neq 0$. Then, $m(Z)=0$. Since $Z \times I$ is measureable and has measure 0, we may replace $E$ by $E \RM (Z \times I)$, and assume for all $x \in I$, $m(E_x)=0$.

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 $m (K) \leq \sum_i m(U_i)\times m(V_i) \leq \epsilon$.

1. Can you prove that $\{y=x\} \In \R^2$ has measure $0$?
2. In both of the two proofs above, we assumed $E$ was bounded, how to deal with the general case?
3. 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?