Exercise 1.2.8 (Lebesgue measure supercedes Jordan measure)

By the Jordan measurability criteria, we can find disjoint boxes $B_1,\dots ,B_N$ with ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^N{B}_i\supset E$ and disjoint boxes $C_1,\dots ,C_n$ with ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{C}_i\subset E$ such that

where we disregard the difference between the elementary and the Lebesgue outer measure by Lemma 1.2.6. In other words, since ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{C}_i\subseteq {\bigcup }_{i=1}^N{B}_i$, and since both are Lebesgue measurable, we have

Thus, we have found an elementary set which covers $E$ at a total cost which does not exceed $𝜖$ in measure. It is easy to make ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^N{B}_i$ open by extending it a little bit, but equivalent criteria for measurability Exercise 1.2.7. Part (vi) assures us that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^N{B}_i$ being Lebesgue-measurable (as an elementary set) is already enough.