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Suppose I have been given a sequence $\{s_n\}_{n=1}^{\infty}$ of real numbers in $[c,d]$. Then we say that this sequence in $[c,d]$ is equidistributed if $$\lim_{N \rightarrow \infty} \frac{\# \{1 \leq n \leq N : s_n \in [a,b]\}}{N} = \frac{b-a}{d-c}$$ for every subinterval $[a,b] \subset [c,d]$. Here $\#A$ denotes the cardinality of $A$.

Now suppose this sequence is a sequence of rational numbers in some interval. Since the rationals constitute a set of measure zero, so I was wondering how would they fit in this property of equidistribution? I know from the 2nd answer of Uniformly distributed rationals that there does exist equidistributed sequence of rationals. So, suppose for a given rational numbers sequence, what is the largest possible interval for which that sequence in that interval is equidistributed?

My motivation for the above question was the following example - $$\{\langle nx\rangle\}_{n=1}^{\infty}$$ where $\langle x\rangle$ denotes the fractional part of $x$. Let $nx \in \mathbb{Q}$. Then, in that case there are only finitely many distinct terms in the sequence. So, is this sequence equidistributed in some interval? If yes then what can be said about the largest possible interval of equidistribution?

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Dark_Knight
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  • I am not sure I understand your question. For some generic rational sequence $s$, do you want a method to find a maximal interval in which $s$ is equidistributed? If that's the case, I can't even begin to think on it. – Fimpellizzeri Sep 15 '16 at 06:28
  • Pretty much. I am not sure but intuitively I think that "rationals sets are of measure zero" has something to do with this equidistribution property. – Dark_Knight Sep 15 '16 at 13:21

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Sequences of rational numbers would or would not fit this definition just like any other sequences. Measure 0 has nothing to do with it (in particular, because the definition never mentions any measure; also, because any sequence, rational or not, constitutes a set of measure zero).

Rational sequences mentioned in the referenced question are equidistributed in $(0,1)$ (and not in any larger interval), because they were constructed precisely for that purpose. Multiply any of them by 100, and the result would be equidistributed in $(0,100)$. Need a larger interval? Multiply by a larger number.

As for $\{\langle nx\rangle\}$ with rational $x$, of course it is not equidistributed in any interval. The same is true for any other sequence with finitely many distinct terms. (Take any subinterval between adjacent terms; there are no terms in it, so the said limit is 0, but the length is not 0, so no luck.)

Ivan Neretin
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