Lemma 1.2.6 (Lebesgue outer measure supercedes elementary measure)

Obviously, Lebesgue outer measure is always less or equal to the elementary measure (since the elementary measure is contained in the set of box measures over which we take the infimum):

Thus, it is left to demonstrate that $m(E)\le m^{\text{∗}}(E)$. We give ourself an $𝜖>0$ of a room and demonstrate that $m(E)\le m^{\text{∗}}(E)+𝜖$. By definition of $m^{\text{∗}}(E)$, we can find a countable cover $E\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|B_n|\le m^{\text{∗}}(E)+𝜖$. Unfortunately, we cannot compare elementary measure $m(E)$ and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|B_n|$ since there is no result on countable unions for elementary measures. A very powerful theorem in mathematics, the Heine-Borel theorem, is capable of reducing countably infinite open covers into finite subcovers. However, we must satisfy two conditions: $E$ must be closed, and the cover has to be open. Therefore, we break the proof down into cases.

(i)

First case: $E$ is closed. The only problem left is that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n$ need not to be open cover; this is, however, easily dealt with by taking a “slightly” bigger open boxes and spending another $𝜖$, i.e., choosing $B_n^{\prime }$ such that $|B_n^{\prime }|\le |B_n|+𝜖∕2^n$. Then ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n^{\prime }$ cover $E$, and we can apply Heine-Borel theorem to get a finite cover $E\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{B}_n^{\prime }$. Now we can apply the monotonicity of elementary measure and conclude that $$m(E)\le \underset{n=1}{\overset{N}{\sum }}{B}_n^{\prime }\le \underset{n=1}{\overset{\infty }{\sum }}{B}_n^{\prime }\le \underset{n=1}{\overset{\infty }{\sum }}(|B_n|+𝜖∕2^n)=\underset{n=1}{\overset{\infty }{\sum }}|B_n|+𝜖\le m^{\text{∗}}(E)+2𝜖$$

as desired.

(ii)

Second case: $E$ is not closed. Partition $E$ into finite union of disjoint boxes $Q_1,\dots ,Q_k$. They are not closed, but to use the previous part we need them to be closed. Once again, we give ourselves an epsilon of the room, and approximate $Q_1,\dots ,Q_k$ by smaller closed sub boxes $Q_1^{\prime },\dots ,Q_k^{\prime }$ such that $|Q_i^{\prime }|\ge |Q_i|+𝜖∕k$. We then have by monotonicity of the Lebesgue outer measure: $$\begin{array}{llll}\hfill m^{\text{∗}}(E)& \ge m^{\text{∗}}(Q_1^{\prime }\cup \dots \cup Q_k^{\prime })& \hfill & \end{array}$$

This set is closed; we can apply the previous part to obtain

$$\begin{array}{llll}\hfill & =m(Q_1^{\prime }\cup \dots \cup Q_k^{\prime })& \hfill & \\ \hfill & =m(Q_1^{\prime })+\dots +m(Q_k^{\prime })& \hfill & \\ \hfill & \ge m(Q_1)+\dots +m(Q_k)-𝜖& \hfill & \\ \hfill & =m(E)-𝜖& \hfill & \end{array}$$

as desired.