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Does there exist a characterization of

  • subsets of a segment such that the Lebesgue measure of its $\epsilon$-neighborhood tends to $0$ as $\epsilon\searrow 0$;
  • bounded functions that are continuous on the complement of a set of this type?

I mean either standard terminology or something which could be more convenient to work with.

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    You may say it has Jordan content zero. These are the "null" sets for the Riemann integral.2011-11-12
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    So the Riemann-integrable functions are bounded functions that are continuous on the complement of a set of this type?2011-11-12
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    What I meant was: the characteristic function of such a set is Riemann integral, with integral zero. But I believe yours is true, too. At least for a bounded function that vanishes outside a bounded set.2011-11-12
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    It seems it fails for countable sets. I have updated my question.2011-11-13
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    wikipedia [says](http://en.wikipedia.org/wiki/Riemann_integral) (subsection 'Integrability'): A function on a compact interval is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure)2011-11-13
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    What you wrote may not be what you mean, since even the Lebesgue measure of the super-set $\epsilon$-interval approaches $0$ as $\epsilon$ approaches $0.$ Perhaps you mean the measure of the intersection of the set with the $\epsilon$-interval divided by $\epsilon$ approaches $0$? Look up "Lebesgue density", specifically points of $0$-density. For the function part of your question, you might be taking about the collection of approximately continuous functions (google "approximate continuity").2011-11-14
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    @Dave: What do you mean, what is wrong? Example: the $\epsilon$-neighborhood of the countable set (and thus having Lebesgue measure zero) of all rational numbers is the set of all real numbers.2011-11-14
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    I thought you were saying this, which is trivial: Let $E \subseteq {\mathbb R}$ and $I$ be an interval of length $\epsilon$. Then the measure of $E \cap I$ approaches $0$ as $\epsilon$ approaches $0$. However, in re-reading your question and the comments, it seems you're talking about Peano-Jordan content. H. Fast studied certain aspects of this in ${\mathbb R}^2$ in Fund. Math. 46 (1958), pp. 137-163 (in Russian) at http://matwbn.icm.edu.pl/tresc.php?wyd=1&tom=46 See also pp. 93-99 of http://www.ramanujanmathsociety.org/docs/mnlVol15No4Mar06.pdf2011-11-14
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    The following paper might also be of interest: Orrin Frink, *Jordan Measure and Riemann Integration*, Annals of Mathematics (2) 34 #3 (July 1933), 518-526. http://math.uga.edu/~pete/Frink33.pdf2011-11-14

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