Exercise 1.2.14

Let $m^{\text{∗}}(E)$ be the outer measure of $E$, i.e.,

By the properties of infimum, for each $𝜖>0$ we can find a collection of boxes $E\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in \mathbb{N}}{B}_i$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }|B_i|-m^{\text{∗}}(E)\le 𝜖$. For each $n\in \mathbb{N}$ make a countable choice ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in \mathbb{N}}{B}_i^{(n)}$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }|B_i^{(n)}|-m^{\text{∗}}(E)\le 1∕n$. Then the set

has all the properties we need:

• contains $E$.
  since for any $x\in E$ we have by assumption $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{B}_i^{(n)}$.
• is Lebesgue measurable.
  since it is a countable intersection of countable unions (Lemma 1.2.13) of Lebesgue measurable boxes (Lemma 1.2.8).
• has a Lebesgue measure of $m(E^{\prime })=m^{\text{∗}}(E)$.
  Since $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:m({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{B}_i^{(n)})-m^{\text{∗}}(E)\le 1∕n$ implies $\forall <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}:m({\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{\bigcup }_{i=1}^{\infty }{B}_i^{(n)})-m^{\text{∗}}(E)\le 1∕N⟹m(E^{\prime })=m^{\text{∗}}(E)$.

Assume $m^{\text{∗}}(E)=\infty$, then $E$ is contained in ${ℝ}^d$ with $m({ℝ}^d)=\infty$. So let’s go on with the case $m^{\text{∗}}(E)$ is finite.

Consider $\bar{E}$, the closure of $E$. $E\subseteq \bar{E}$ and since $\bar{E}$ is closed, it is Lebesgue measurable, thus we have $m(\bar{E})=m^{\text{∗}}(\bar{E})$. $m^{\text{∗}}(E)$ is defined as $\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}\mathrm{vol}<!--\; FUNCTION\; APPLICATION\; -->(B_j):{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}(B_j)\text{ covers }E,B_j\text{is a box},J\text{ at most countable}\right\}$.

For any set of boxes ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}(B_j)$ that covers $E$, we have that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}(\overline{B_j})$ covers $\bar{E}$ with $m^{\text{∗}}\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}(B_j)\right)=m^{\text{∗}}\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}(\overline{B_j})\right)$ and so $m(\bar{E})=m^{\text{∗}}(E)$.