Exercise 1.2.6 (Inner non-regularity of the Lebesgue outer measure)

We use the usual "entangled" set $ℝ∕\mathbb{Q}$. We have already shown that the Lebesgue outer measure of this set is 1. We now demonstrate that

Pick an arbitrary open set $U\subseteq E$. By definition, for any point $x\in U\subseteq E$ we can find an open interval $(a,b)$ small enough such that $(a,b)\subseteq U$. But this is impossible, since any interval around $x$ would necessarily contain some rational number. Thus, any open set contained in $E$ is empty, and thus the $\sup <!--\; FUNCTION\; APPLICATION\; -->$ must be equal to zero.

Take $E=[0,1]\setminus \mathbb{Q}$. If $U\subset E$ is an open set, then $U=\varnothing$, because $E$ does not contain any open intervals. Thus, $\underset{U\subset E,U}{\sup <!--\; FUNCTION\; APPLICATION\; -->}$ open $m^{\text{∗}}(U)=0$. However, if $E\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{I}_n$ where ${\left(I_n\right)}_n$ are disjoint intervals, then it must be that $[0,1]\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{I}_n$. So ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n}m\left(I_n\right)\ge m([0,1])=1$. Thus, $m^{\text{∗}}(E)=1$.