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?
can open sets be covered with another open set not much bigger?
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real-analysis
general-topology
measure-theory
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1I believe what you may be thinking of are certain regularity conditions on Lebesgue measure. They are: if $E$ is a Lebesgue measurable set, then for every \epsilon > 0: 1) There is an open set $U$ containing $E$ with |U \setminus E| < \epsilon 2) There is a closed set $F$ inside of $E$ with |E\setminus F| < \epsilon 3) If |E| < \infty, then there is a finite union of closed cubes so that the symmetric difference with $E$ has measure less than/equal to $\epsilon$. – 2012-11-23
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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).