Given the set $\mathbb R\setminus \mathbb Q.$
The interior set is the collection of all the interior points, where the interior point of a set $S$ from $\mathbb R,$ is a point $x \in S,$ such that there exists an $ \varepsilon >0 $ to make an open set U that looks like $(x - \varepsilon, x + \varepsilon)$ such that $x \in U$ and $U \subset S$.
My explanation for Interior set being a null set (Please review)
So, for any arbitrary irrational point in the given set, If I form an open interval around that point of size $|x|<\varepsilon$, but that interval has no other point than the irrational point itself, so therefore a neighbourhood does not exist for the irrational point.
Why would the Exterior set be a null set?
(Exterior Set - Collection of all the exterior points of set S)
(Exterior Point - A number $a \in\mathbb R$ is said to be an exterior point of a set $S$ from $\mathbb R$ if there exist a neighbourhood of a which is contained in $S^c$)