Exercise 1.1.4 (Elementary measure of a product is product of elementary measures)

First we demonstrate that $E_1\times E_2$ is elementary in ${ℝ}^{d_1+d_2}$. By Lemma 1.1.2 we can write $E_1=B_1\cup \dots \cup B_k$ and $E_2=C_1\cup \dots \cup C_{k^{\prime }}$ for some disjoint boxes $B_i,C_j$. Then

Since the product of two boxes ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{d_1}[a_i,b_i]\times {\prod }_{j=1}^{d_2}[c_j,d_j]$ can be written as a box itself ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{l=1}^{d_1+d_2}[e_l,f_l]$ for $e_l,f_l=a_l,b_l$ for $l=1,\dots ,d_1$ and $e_{l+d_1},f_{l+d_1}=c_l,d_l$ for $l=1,\dots ,d_2$, we conclude that the Cartesian product of two elementary sets is again elementary. Furthermore,

by finite additivity

As usual, we reduced the general theorem assertion made about elementary sets to a simpler assertion made about boxes. In other words, we have to demonstrate that for two boxes $B,C$ we have $m^{d_1+d_2}(B\times C)=m^{d_1}(B)\times m^{d_2}(C)$. By the definition of box measure and by the previous argument

Using the result in the last equation proves the assertion.