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There are previous proofs showing that multiples of an irrational number $r$ form a dense subset of $[0,1]$, but is the set "uniform" in $[0,1]$. Now, since $R\equiv r\mathbb{Z}(\mod 1)$ is a countably infinite subset, it's not quite easy to define exactly what it means to be "uniform". However, let me try the following formulation (and if there's a better one, please let me know). Consider the set $[N]=\{-N,-N+1,...,N-1,N \}$ and let $R_N = r[N] (\mod 1)$ so that $R_N \nearrow R$. Since $R_N$ is finite, let us order $R_N$ as $x_1 < x_2 <\cdots <x_{2N+1}$. We can then define a cumulative distribution function $$ F_N(x) = \frac{1}{2N+1}\int_0^x \sum_i 1_{x < x_i} dx $$ In this case, does $F_N(x) \to x$ in any sense?

Andrew Yuan
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The answer is yes. The property you are asking is precisely the equidistributivity. Weyl's criterion tells that this property can be defined in one of the following equivalent ways:

Definition. A sequence $(a_n)$ taking values in $[0, 1]$ is said to be equidistributed if it satisfies any of the following equivalent conditions:

  1. We have $$\lim_{n\to\infty}\frac{\{a_1,\ldots,a_n\}\cap[a,b]}{n} = b-a $$ for any $0 \leq a < b \leq 1$.

  2. We have $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} f(a_k) = \int_{0}^{1} f(x) \, \mathrm{d}x $$ for any Riemann integrable $f$ on $[a, b]$.

  3. We have $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} e^{2\pi i \ell a_k} = 0 $$ for any $\ell \in \mathbb{Z}\setminus\{0\}$.

  4. We have $$ \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \delta_{a_k} = \mathrm{Uniform}([0, 1]) $$ in distribution, where $\delta_a$ is the unit point mass (a.k.a. Dirac delta) concentrated at $a$ and $\mathrm{Uniform}([0, 1])$ is the uniform distribution over $[0, 1]$.

Now the condition 3 immediately shows that a sequence of the form $(\alpha k \text{ mod } 1)_{k\in\mathbb{N}}$ for an irrational $\alpha$ forms an equidistributed sequence, which is also the content of the equidistribution theorem.

Sangchul Lee
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