<s>[INST] Prove $0, \frac{1}{2}, 0, \frac{1}{3}, \frac{2}{3}, 0, \frac{1}{4}, \frac{2}{4}, ...$ equidistributed in $[0, 1)$ Prove
$0, \frac{1}{2}, 0, \frac{1}{3}, \frac{2}{3}, 0, \frac{1}{4}, \frac{2}{4}, ...$
equidistributed in $[0, 1)$.

A sequence of numbers $\xi_1, \xi_2, \xi_3, ...$ in $[0, 1)$ is said to be equidistributed if for every
  interval $(a, b) \subset [0, 1)$
$$\lim\limits_{N\to\infty} \frac {\bigl|\{1\le n\le N: \xi_n \in (a, b)\}\bigr|} {N}
= b-a$$

It's from Chapter 4 in the book Fourier Analysis: An Introduction
 [/INST] Let $a < b$ be given. We have for $0 \le p < q$ that $\frac pq \in (a,b)$ iff $qa < p < qb$ iff $\lfloor qa\rfloor < p < \lceil qb \rceil$. 
For $N \in \mathbb N$, choose the maximal $q$ with $\frac 12q(q+1) < N$, then all fractions with denominator $\le q$ have apperead under $(\xi_n)_{n\le N}$. We have 
\begin{align*}
  \left|\{1 \le n \le N: \xi_n \in(a,b)\}\right| &\ge \sum_{r=2}^q \bigl(\lceil rb \rceil - \lfloor ra\rfloor - 1\bigr)\\
  &\ge \sum_{r=2}^q \bigl(r(b-a) - 3\bigr)\\
  &= \left(\frac 12q(q+1) - 1\right)(b-a) - 3(q-1)
\end{align*}
But now 
\[ \frac 12 q(q+1) < N \le \frac 12 (q+1)(q+2) \]
hence
\[ 1 \le \frac N{\frac 12q(q+1)} \le \frac{(q+1)(q+2)}{q(q+1)} \to 1, q \to \infty \]
So, for $N \to \infty$
\begin{align*}
 \frac 1N \left|\{1 \le n \le N: \xi_n \in(a,b)\}\right|
   &\ge \frac 1N\left(\frac 12q(q+1) - 1\right)(b-a) - \frac 3N(q-1)\\
   &\to b-a.
\end{align*}
</s>