## IMT Exercise 1.1.4

Let ${d_1,d_2\ge1}$, and let ${E_1\subset\mathbb{R}^{d_1}}$, ${E_2\subset\mathbb{R}^{d_2}}$ be elementary sets. Show that ${E_1\times E_2\subset\mathbb{R}^{d_1+d_2}}$ is elementary , and ${m^{d_1+d_2}(E_1\times E_2)=m^{d_1}(E_1)\times m^{d_2}(E_2)}$.
Let partitions of ${E_1}$ and ${E_2}$ into finite sets of disjoint boxes be ${B^1_1\cup\ldots\cup B^1_n}$, and ${B^2_1\cup\ldots\cup B^2_m}$. Then

$\displaystyle E_1\times E_2=\bigcup_{\substack{i=1\ldots n\\ j=1\ldots m}} B^1_i\times B^2_j$

Consider ${B^1_i=I^1_{i,1}\times\ldots\times I^1_{i,d_1}}$ and ${B^2_j= I^2_{j,1}\times\ldots\times I^2_{j,d_2}}$, then
$\displaystyle B^1_i\times B^2_j=I^1_{i,1}\times\ldots\times I^1_{i,d_1}\times I^2_{j,1}\times\ldots\times I^2_{j,d_2}$

It is clear that ${B^1_i\times B^2_j\subset\mathbb{R}^{d_1+d_2}}$ and as ${d_1}$ and ${d_2}$ are both finite so therefore is ${d_1+d_2}$. Furthermore
$\displaystyle m^{d_1+d_2}(B^1_i\times B^2_j)=|I^1_{i,1}|\times\ldots\times |I^1_{i,d_1}|\times |I^2_{j,1}|\times\ldots\times |I^2_{j,d_2}|=m^{d_1}(B^1_i)\times m^{d_2}(B^2_j).$

As ${E_1}$ and ${E_2}$ are partitioned into disjoint boxes the new boxes in the Cartesian product ${E_1\times E_2}$ are also disjoint. This gives
$\displaystyle m^{d_1+d_2}(E_1\times E_2)=\sum_{\substack{i=1\ldots n\\ j=1\ldots m}} m^{d_1}(B^1_i)\times m^{d_2}(B^2_j)=m^{d_1}(E_1)\times m^{d_2}(E_2)$

${\Box}$