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Define counting function $A(E; N; \omega)$ as the number of terms $x_n, 1\leq n\leq N$, for which $\{x_n\}\in E$.

Then the sequence $\omega=(x_n), n=1,2,...,$ of real numbers is said to be uniformly distributed modulo 1 (abbreviated u.d. mod 1) if for every pair $a, b$ of real numbers with $0<a<b<1$ we have $$\lim\limits_{N\rightarrow\infty}\frac{A([a,b);N;\omega)}{N}=b-a.$$

Assume we have a real number sequence $(a_n):n\in\mathbb N$, which is u.d. mod 1. It is obvious that for smaller intervals the proportion of fractional parts in those intervals will be smaller. Hence, intuition suggests that some kind of measure $P$ could be introduced to get

$$P\left(\{a_n\}>\frac 1n\right)=1, \quad\text{when }\quad n\rightarrow\infty.$$

I am looking for a reference where this idea is explored. Disproving my intuition for different kind of measures would be welcome too.

I do not want to reinvent the wheel and I am looking for a most common and conventional approach in building this bridge between number and probability theory.

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  • $\begingroup$ This seems like it should be trivial from the definition of "u.d. mod 1". No? (I assume you mean something equivalent to "equidistributed sequence"). $\endgroup$
    – user44191
    Commented Aug 23, 2021 at 13:30
  • $\begingroup$ My intuition says yes, but I don't want to reinvent the wheel, if someone already did it. $\endgroup$
    – TheGrandDuke
    Commented Aug 23, 2021 at 13:38
  • $\begingroup$ I don't think this really appropriate for mathoverflow, but the answer is yes: for each $\frac{1}{2} > \varepsilon > 0$ we have $\mathbb{P} \left( \{ a_n \} > \varepsilon \right) \to 1 - 2 \varepsilon$. Taking $\varepsilon$ arbitrarily small we get the desired result. $\endgroup$
    – Random
    Commented Aug 23, 2021 at 13:47
  • $\begingroup$ Can you give me a reference to this fact? $\endgroup$
    – TheGrandDuke
    Commented Aug 23, 2021 at 13:49
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    $\begingroup$ In case this is the point of confusion: what do you mean by "probability that $a_n > \epsilon$" here? A sequence doesn't inherently come with a probability! Generally, one would associate the Lebesgue measure, since it's the limit of the empirical measures $\frac{1}{n} \sum_{i=0}^{n-1} \delta_{a_n}$ when $(a_n)$ is uniformly distributed. Then, indeed, the answer is trivial, since $P(x > \epsilon) = 1-\epsilon$. $\endgroup$
    – Ronnie Pavlov
    Commented Aug 23, 2021 at 16:39

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