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?
2
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real-analysis
general-topology
measure-theory
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1This is false if for example $S$ is an open ball. – 2012-11-23
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0(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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2@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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0It'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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0You are right Chris, I just corrected that. – 2012-11-23
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2What about everyone's favourite open set: $\varnothing$? – 2012-11-23
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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