Exercise 1.1.16 (Jordan measure and Cartesian products)

Just as in the previous exercise, we want, somehow, to squeeze the hypothesis and the result between $𝜖>0$

Using the same technique as before, we could find elementary outer covers $\overline{E_1},\overline{E_2}$ of $E_1,E_2$ (with $\overline{E_1}\times \overline{E_2}$ conveniently being elementary outer covers of $E_1\times E_2$) and elementary inner covers $\underline{E_1},\underline{E_2}$ of $E_1,E_2$ (with $\underline{E_1}\times \underline{E_2}$ conveniently being elementary inner covers of $E_1\times E_2$) which do not differ from the original measures by some small error (which we will decide for later). By monotonicity property of Jordan measure (Exercise 1.1.6) we then obtain

where the crucial step was applying the theorem hypothesis to elementary sets (which we can do, thanks to the Exercise 1.1.4). Subtracting $m(E_1)\times m(E_2)$ from both sides, we obtain

At this point (since we started the proof bottom-up) it is obvious that we should have chosen the covers in a way such that both the left- and the right-hand side of the above inequality equal to $𝜖>0$. Fix some $𝜖>0$. Notice that the following choice of elementary covers results in the desired result:

Similarly, we can choose $\underline{E_1},{\underline{E}}_2$ such that $m\left(\underline{E_1}\right)m\left(\underline{E_2}\right)-m(E_1)m(E_2)>-𝜖$, which, after inserting into the inequality, yields the desired result.

Fix $𝜀>0.$ Let $E\subset A\subset F,E^{\prime }\subset C\subset F^{\prime }$ such that $E,E^{\prime },F,F^{\prime }\in E_0$ and $m(F\setminus E)<𝜀$, $m\left(F^{\prime }\setminus E^{\prime }\right)<𝜀$. Note that $E\times E^{\prime }\subset A\times C\subset F\times F^{\prime }$ and $E\times E^{\prime },F\times F^{\prime }$ are elementary sets. Thus, since we assumed that $A,C$ are Jordan measurable,

Also, since $m(A)=m(A\setminus E)+m(E)\le m(F\setminus E)+m(E)$ and

we have

Taking $𝜀\to 0$ we have that $J^{\text{∗}}(A\times C)=J_{\text{∗}}(A\times C)=m(A)\cdot m(C)$.