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I am looking for a succinct way to describe a subset of the natural numbers which has ``measure zero" in the following sense: Let X_1 \subset X_2 \subset .... be any strictly nesting sequence of finite subsets of natural numbers, I want to describe a subset S \subset \N which has the property that the limit of the usual counting measures applied to S restricted to X_i is zero.

I don't know that much about measure theory and am having trouble finding any measures on countable sets such as the natural numbers. There should be a natural measure on them such that the sets I describe above have zero measure with respect to it. Is there a name for such a natural measure on the natural numbers?

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    $\begingroup$ The word "measure" in this context is not the appropriate one. You are probably looking for subsets of the naturals having zero density, even if your definition is not the correct definition of density. In fact, you should divide the counting measure by the size of $X_i$ before taking the $\liminf$ (or the $\limsup$). See en.wikipedia.org/wiki/Natural_density for more details. $\endgroup$ Commented Jun 12, 2011 at 12:33
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    $\begingroup$ I hate to say it, but I feel obliged to: this site is for mathematics approximately at a research level. Please read the faq for appropriate places to ask this type of question. $\endgroup$ Commented Jun 12, 2011 at 13:47
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    $\begingroup$ While I agree that the question is not really suited to MO, I would at least offer some encouragement: this kind of question is both natural and good when one is trying to get to grips with notions of `size' for infinite subsets of the natural numbers. Hopefully the wikipedia link in Francesco Polizzi's comment may be of use. $\endgroup$
    – Yemon Choi
    Commented Jun 13, 2011 at 2:52

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A measure, here, isn't going to work. You can modify your idea to get near the idea of density, as Francesco pointed out. But, since you're new with measure theory, remember that a measure is countably additive, and (assuming you do want to divide by the size of the $X_i$ in your explanation, as was suggested) you're going to get the measure of any single point to be zero. Then by countable additivity, you're going to get the measure of any set to be zero.

This is why you're having trouble finding measures on countable sets; if you want small sets to be measure zero, you're going to get the zero measure, which is not what you want.

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  • $\begingroup$ Yes this seems to be precisely why finding such a measure will fail. If one wishes to weight all natural numbers equally, then only the trivial `zero measure' can assign a finite measure to an infinite set. $\endgroup$
    – Spencer
    Commented Jun 12, 2011 at 23:52

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