Exercise 14, § 1

Let $\varphi : \mathbf{R} \to \mathbf{T}$ be the canonical homomorphism; let $\theta$ be an element of infinite order in $\mathbf{T}$ and let $\theta_0$ be an (irrational) real number such that $\varphi(\theta_0) = \theta$. For each integer $n > 0$, let $S_n$ be the set of elements $k\theta$ of $\mathbf{T}$ for which $1 \leq k \leq n$, and for each interval $I$ of $\mathbf{R}$ let $N(I, n)$ be the number of elements of the set $I \cap \overline{\varphi^{-1}(S_n)}$.

a) If we take $I = [0, \theta_0[$, show that $N(I, n) = [n\theta_0]$ [note that $N(I, n)$ is then the number of pairs of integers $(x, y)$ such that $1 \leq y \leq n$ and $x \leq y\theta_0 < x + \theta_0$].

b) More generally, if we take $I = [0, m\theta_0[, m$ being an integer, we have $m.[(n - m)\theta_0] \leq N(I, n) \leq m.[n\theta_0]$ (same method).

c) Deduce that, if $I$ is any interval, the number $N(I, n)/n$ tends to a limit equal to the length of $I$, as $n \to + \infty$. (Prove the result first when $I$ is of the form $[m\theta_0 + a, m'\theta_0 + a']$, where $m, m', a$ and $a'$ are integers, by using b); then pass to the general case by approximating to the endpoints of $I$ by numbers of the form $m\theta_0 + a$.) ["Equipartition of the sequence $(k\theta) \mod 1$"].