I am trying to take some inspiration from this post, but I am getting stuck with some of the details.
I have proved that the interior $E^o$ and closure $\overline{E}$ of a set $E$ are Lebesgue measurable. As far as I understand, the basic idea is to use the fact that $E^o\subseteq E\subseteq \overline{E}$ and the fact that $\lambda(E^o)=\lambda(\overline{E})$ to argue that $E$ has a Lebesgue measure. Perhaps my understanding is too limited, but there are a couple things that confuse me:
- Why is it the case that $\lambda(E^o)=\lambda(\overline{E})$?
- Why is it the case that, if $A\subseteq B\subseteq C$ and $A, C$ are Lebesgue measurable (with equal Lebesgue measure), then $B$ is Lebesgue measurable?
For 1, take $E=\mathbb{Q}\subset\mathbb{R}$. We have $\lambda(E_0)=\lambda(\varnothing)=0\neq\infty=\lambda(\mathbb{R})=\lambda(\overline{E})$.
For 2 (assuming the axiom of choice), take $C=\mathbb{R}$, $A=\mathbb{R}\setminus [0,1]$. Furthermore let $B=A\cup V$ where $V$ is the Vitali set. Evidently $A\subset B\subset C$ and $A,C$ are measurable and $\lambda(A)=\infty=\lambda(C)$, however $B$ is not measurable since $V$ is not.
Note that this depends on the fact that we can choose a set $C$ such that $\lambda(C)=\infty$. If we add the restriction that $\lambda(A)=\lambda(C)<\infty$, then $\lambda^*(B\setminus A)\leq\lambda^*(C\setminus A)=\lambda(C\setminus A)=\lambda(C)-\lambda(A)=0$. It is easy to show that if a set satisfies $\lambda^*(D)=0$, then $D$ is measurable. $B\setminus A$ is thus measurable. Since $A\subset B$, we have $B=(B\setminus A)\cup A$ which shows that $B$ is measurable.