can open sets be covered with another open set not much bigger? Is that correct that, for any open set $S \subset \mathbb{R}^n$, there exists an open set $D$ such that $S \subset D$ and $D \setminus S$ has measure zero?
I think it is correct and I guess I have seen the proof somewhere before, but I cannot find it in any of my books, if it is wrong, please give me a counter-example.
Also is the same correct for a closed set such that it's interior is not of measure zero?

Assume that $S\subset D$ with $S,D$ open and $\mu(D\setminus S)=0$.
If $D$ is not contained in $\overline S$, then the nonempty open set $D\setminus \overline S$ has positive measure. Therefore,  $D\subseteq \overline S$. Therefore any open set $S$ with the property that $$\tag1\partial S\subseteq \partial(\mathbb R^n\setminus \overline S)$$ is a counterexample to your conjecture: Any open ball araound a point in $\partial S$ then intersects $\mathbb R^n\setminus \overline S$ in a nonempty open set of positive measure.
For example, an open ball or virtually any open set with a smooth boundary has property (1).
