Measurability of Interior and closure of a non-measurable set.

Question

I was thinking about this question -

Let $E$ be a non-measurable subset of $[0,1]$, then what can be said about the measurability of $int(E)$ and $\overline{E}$ ? .

I know the fact that any subset of a measurable set need not be measurable (example - Vitali's construction of non-measurable set.)

Also as per the definition of $int(E)$ is the union of open sets contained in $E$ and I think as the open sets are measurable and union of measurable sets being measurable so $nt(E)$ is measurable.

Also similarly for $\overline{E}$ since closure of $E$ is the intersection of all closed sets containing $E$, and closed sets are measurable,also intersection of measurable sets being measurable implies $\overline{E}$ is also measurable.

I doubt my method as I have nowhere used the non-measurability of $E$.
But how can we proceed if the set is non-measurable?

Answer 1

As you say open and closed sets are measurable, so the interior and the closure of any set are measurable.

Answer 2

Since the phrasing asks "what can be said" about the Lesbegue measurability of $int(E)$ and $\overline{E}$ for a non-measurable set $E \subset [0,1]$, I would want to add the observation that the measures of $int(E)$ and $\overline{E}$ must differ.

This is easily seen because $int(E)\subseteq E\subseteq \overline{E}$. Therefore were it the case that $m(int(E)) = m(\overline{E})$, we would have $m(\overline{E}\setminus int(E))=0$. Then it would follow from the completeness of Lebesgue measure that the set $E$ would be measurable (with measure equal to both that of $int(E)$ and $\overline{E}$).

But since $E$ is assumed to be non-measurable, it must be that: $$m(int(E)) \neq m(\overline{E}) $$