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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

  • 1
    @takecare All the numbers in the sequence have to go _somewhere._ If some interval has more than it should, then its complement (one or two intervals) will have fewer than it should. (Needs some work to make it precise, and to account for the endpoints of the intervals.) – 2016-09-22

2 Answers 2

2

For each integer $n>1$ let $F_n=\left\{\frac{k}n:k=0,\dots,n-1\right\}\;;$

for $0\le a, $|(a,b)\cap F_n|\ge\lceil n(b-a)\rceil-1$.

The first $2$ terms of the sequence are the members of $F_2$; the next $3$ terms are the members of $F_3$, and so on. Thus, the first $\sum_{k=2}^nk=\frac12n(n+1)-1$ terms are the members of blocks $F_2$ through $F_n$ in the obvious order. For $n\ge 2$ let $T_n=\frac12n(n+1)-1$, and suppose that $T_m\le N. Then

$\begin{align*} |\{n\le N:\xi_n\in(a,b)\}&\ge\sum_{k=2}^m|(a,b)\cap F_k|\\ &\ge\sum_{k=2}^m\Big(\lceil k(b-a)\rceil-1\Big)\\ &=\sum_{k=2}^m\lceil k(b-a)\rceil-m+1\\ &\ge\sum_{k=2}^mk(b-a)-m+1\\ &=(b-a)T_m-m+1\;, \end{align*}$

so

$\frac{|\{n\le N:\xi_n\in(a,b)\}}N\ge(b-a)\frac{T_m}N-\frac{m-1}N\;.$

Now $\frac{T_m}N>\frac{T_m}{T_{m+1}}=\frac{\frac12m(m+1)-1}{\frac12(m+1)(m+2)-1}\to 1\quad\text{ as }\quad m\to\infty\;,$ and

$\frac{m-1}N\le\frac{m-1}{T_m}=\frac{m-1}{\frac12m(m+1)-1}\to 0\quad\text{ as }\quad m\to\infty\;,$

so $\lim_{N\to\infty}\frac{|\{n\le N:\xi_n\in(a,b)\}|}N\ge b-a\;.$

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    ... its complement in $(0,1)$. From there it’s plain sailing. – 2016-09-24
1

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*}

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    How do we get the other inequality? – 2016-09-22