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{\bf Question}

Let $A\subseteq{\bf R^n}$ and $B\subseteq{\bf R^m}$.  Let $m^*_n$
denote Lebesgue outer measure in ${\bf R^n}$.  Show that
$m^*_{n+m}(A\times B)=m^*_n(A)\cdot m^*_n(B)$ where $A\times B$ is
the cartesian product.



\vspace{0.25in}

{\bf Answer}

Let $\bigcup R_i \supseteq A$ and $\bigcup S_j \supseteq B$.

Suppose $m^*(A)<\infty$ and $m^*(B)<\infty$

Then $\ds \bigcup_{ij}R_i\times S_j\supseteq A\times B$

Choose $\{R_i\}$ so that $\sum|R_i|\leq m^*(A)+\epsilon$

Choose $\{S_j\}$ so that $\sum|S_j|\leq m^*(B)+\epsilon$

Then $\sum|R_i\times S_j|=\sum|R_i||S_j|=\sum|R_i|\sum|S_j|$

$\leq m^*(A)m^*(B)+\epsilon_1$

Therefore $m^*(A\times B)\leq m^*(A)m^*(B)+\epsilon$

Now cover $(A\times B)$ by $R_i\subseteq {\bf R^{m+n}}$

Let $S_i$ be the projection of $R_i$ onto ${\bf R^n}$

Let $T_i$ be the projection of $R_i$ onto ${\bf R^m}$

Then $\bigcup S_i\supseteq A$, and $\bigcup T_i\supseteq B$

Choose $R_i$ so that $\sum|R_i|\leq m^*(A\times B)+\epsilon$

$m^*(A)m^*(B)\leq \sum|S_i|\sum|T_j|=\sum|R_i|\leq m^*(A\times
B)+\epsilon$

Hence the result.  Deal with infinite measure cases using
$\sigma$-finiteness arguments.


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