In this answer, there's the following line:
If $m_*(\mathcal N^c)<1$, then $\forall\ \varepsilon>0$, $\exists\ U$ is measurable s.t. $\mathcal N^c \subset U \subset I$ and $m_*(U)<1-\varepsilon$.
(Here $m_*$ denotes outer measure.)
I don't see immediately why such measurable set $U$ always exists. Can someone help to give out one such construction of $U$ with arbitrarily $\epsilon$?
It should be $\exists\epsilon\gt 0$ instead of $\forall\epsilon\gt 0$.
This is because if such set $U$ exists for arbitrary $\epsilon\gt 0$ then:
$$ (1-\epsilon)=(1-\epsilon) + 0\gt m_*(U)+m_*(U^c)\ge m_*(U\cup U^c)=1, \forall\epsilon, $$
which is false.
Also $\exists\epsilon\gt 0$ is reasonable since from the assumption that $m_*(\mathcal N^c)\lt 1$, we can show that there exists such measurable set $U$. From the definition of $m_*$ we have:
$$ m_*(\mathcal N^c) =\inf\left\{ \Sigma_{k=1}^{\infty}\mathscr{l}(I_n): I_n\text{ are open intervals and }\mathcal N^c\subseteq\bigcup_{k=1}^{\infty}\ I_n\right\}\lt 1. $$
So it must be the case that there exists some $\{I_n\}$ such that $\mathcal N^c\subseteq\bigcup_{k=1}^{\infty}\ I_n$ and $\Sigma_{k=1}^{\infty}\mathscr{l}(I_n)\lt 1$, otherwise $m_*(\mathcal N^c)\ge1$ which is a contradiction to the assumption. Choose $U=\bigcup_{k=1}^{\infty}\ I_n\cap[0, 1]$ and $0\lt\epsilon\lt 1-\Sigma_{k=1}^{\infty}\mathscr{l}(I_n)$ satisfies the desired condition.