Exercise 1.2.5 (Lebesgue and Jordan measures agree on countable unions of almost disjoint boxes)

In the case that $E$ is unbounded, both quantities are infinite. Thus, suppose that $E$ is bounded.

$\text{≤}$)

We have to demonstrate that (in general, not only for this exercise) we have $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{m(A):\begin{array}{c}A\text{ is elementary}\hfill \\ A\subseteq E\hfill \end{array}\right\}\le m^{\text{∗}}(E)$$

Pick an arbitrary elementary inner cover $A\subset E$ with the elementary measure $m(A)=m^{\text{∗}}(A)$ by Lemma 1.2.6. By monotonicity of the Lebesgue outer measure, we have $m^{\text{∗}}(A)\le m^{\text{∗}}(E)$. Since our choice of $A$ was arbitrary, taking the supremums yields the result.

$\text{≥}$)

We give ourselves an $𝜖>0$ of a room to show that $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{m(A):\begin{array}{c}A\text{ is elementary}\hfill \\ A\subseteq E\hfill \end{array}\right\}\ge \underset{n=1}{\overset{\infty }{\sum }}|B_n|-𝜖$$

Since ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|B_n|-𝜖$ is a series of positive values, we for our $𝜖>0$ we can find a $N\in \mathbb{N}$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|B_n|-{\sum }_{n=1}^N|B_n|\le 𝜖$. But the finite sum describes the elementary measure of some inner cover of $E$: we have $|B_1|+\dots +|B_N|=|B_1^{\text{∘}}|+\dots +|B_N^{\text{∘}}|=m({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{B}_n)$. Thus set $A:={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{B}_n$ is a disjoint union of boxes inside $E$; thus $A$ is a elementary inner cover of $E$. In other words, we have

$$m(A)+𝜖\ge m^{\text{∗}}(E)$$

Taking supremums for $A$, and letting $𝜖\to 0$ we get the desired result.