1.33

Recall two things. First, that $m^*(E) = 0 \hspace{0.25cm} \Rightarrow E \in \mathcal{M}$. Second, recall from problem 32, part a, that any measurable subset to a non-measurable set must have measure zero. Thus, since $\mathcal{N}^c \cup \mathcal{N} = [0,1]$, it suffices to show that $m(\mathcal{N}^c) = 1$.

Assume to the contrary that for some $\epsilon \in (0,1)$, $\hspace{0.25cm} \mathcal{N}^c \subset U \subset [0,1]$, measurable, such that $m(U) = 1 - \epsilon$. It directly follows that $([0,1] \cap U^c) \subset \mathcal{N}$. This is a contradiction, because $m \big(([0,1] \cap U^c) \big) = \epsilon > 0$.

Therefore, $m^*(\mathcal{N}) + m^*(\mathcal{N}^c) > m(\mathcal{N} \cup \mathcal{N}^c)$.