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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?

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    This is false if for example $S$ is an open ball.2012-11-23
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    (I assume that "$\subset$" means "is a proper subset of" and "$D - S$" means "$D \setminus S$", though both of these usages are nonstandard)2012-11-23
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    @Chris: Yes, I realized that just after I posted. (It’s actually my preferred usage, but I don’t see it often enough to expect it!) I disagree about *non-standard*: the first is perfectly standard, and the second is merely old-fashioned.2012-11-23
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    It's old-fashioned in general, but in this context it's downright wrong, since $A-B$ means $\{a-b|a \in A, b \in B\}$.2012-11-23
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    You are right Chris, I just corrected that.2012-11-23
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    What about everyone's favourite open set: $\varnothing$?2012-11-23
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    I 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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