Lemma 1.2.9 (Outer measure of countable unions of almost disjoint boxes)

We demonstrate that the left-hand side is both bigger and equal and smaller and equal to the right-hand side.

$\text{≤}$)

From countable subadditivity of the Lebesugue outer measure, and from the fact that it coincides with elementary measure, we deduce that $$m^{\text{∗}}(E)=m^{\text{∗}}\left(\underset{n=1}{\overset{\infty }{\bigcup }}{B}_n\right)\le \underset{n=1}{\overset{\infty }{\sum }}{m}^{\text{∗}}(E)=\underset{n=1}{\overset{\infty }{\sum }}|B_n|$$

$\text{≥}$)

For each $N\in \mathbb{N}$ we have $$E\supset \underset{n=1}{\overset{N}{\bigcup }}{B}_n$$

By subadditivity of the Lebesgue outer measure we deduce

$$\begin{array}{llll}\hfill m^{\text{∗}}(E)& \ge m^{\text{∗}}\left(\underset{n=1}{\overset{N}{\bigcup }}{B}_n\right)& \hfill & \end{array}$$

But ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{B}_n$ is an elementary set; thus,

$$\begin{array}{llll}\hfill & =m\left(\underset{n=1}{\overset{N}{\bigcup }}{B}_n\right)& \hfill & \end{array}$$

Thus, if we manage to prove the theorem assertion for the elementary sets and elementary measures, then we are done by letting $N\to \infty$. Since we “axiomatically” know that (1) the volume of a box and its closure coincide, by (2) the subadditivity of the elementary measure we already have

$$m(B_1\cup \dots \cup B_N)\stackrel{2}{\text{≤}}m(B_1)+\dots +m(B_N)\stackrel{1}{\text{=}}|B_1|+\dots +|B_N|$$

On the other hand (3) monotonicity and (4) finite additivity we have

$$m(B_1\cup \dots \cup B_N)\stackrel{3}{\text{≥}}m(B_1^{\text{∘}}\cup \dots \cup B_N^{\text{∘}})\stackrel{4}{\text{=}}m(B_1^{\text{∘}})+\dots +m(B_N^{\text{∘}})\stackrel{1}{\text{=}}|B_1|+\dots +|B_N|$$

as desired.