# Equidistribution theorem. Let $(x_{n})$ be a real sequence in $[0,1)$. We say $(x_{n})$ is **equidistributed** in $[0,1)$ if for any interval $[a,b) \subset [0,1)$, we have $$ \lim_{ N \to \infty } \frac{1}{N} \#\{n \le N: x_{n} \in [a,b)\} = b-a. $$ That is to say, the fraction of sequence that lies in the interval $[a,b)$ is $b-a$. Note, we have an equivalent definition: > For $(x_{n})$ a real sequence in $[0,1)$, $(x_{n})$ is equidistributed in $[0,1)$ if and only if for all $b \in [0,1)$, > $$ \lim_{ N \to \infty } \frac{1}{N}\#\{n \le N : x_{n} < b \}=b $$ $\blacktriangleright$ Indeed, fix a real sequence $(x_{n}) \in [0,1)$. If for any $[a,b) \subset [0,1)$ we have $$ \lim_{ N \to \infty } \frac{1}{N}\#\{n\le N : x_{n} \in [a,b)\} = b-a, $$ then for any $b \in [0,1)$ we take $a =0$ and get $$ \lim_{ N \to \infty } \frac{1}{N} \# \{n\le N : x_{n} < b\}= b-0=b. $$ Conversely, if for any $b \in [0,1)$ we have $$ \lim_{ N \to \infty } \frac{1}{N} \# \{n \le N : x_{n} < b \} = b, $$ for any $[a,b) \subset [0,1)$ we have $$ \begin{eqnarray} &&\frac{1}{N} \# \{n\le N : x_{n} \in [a,b) \} \\ &=& \frac{1}{N} \# \{n\le N : x_{n} < b \} - \frac{1}{N}\# \{n\le N : x_{n} < a \}. \end{eqnarray} $$ So taking limit as $N \to \infty$ we have $$ \lim_{ N \to \infty } \frac{1}{N} \# \{n \le N : x_{n} \in [a,b)\} = b - a. \blacksquare $$ A classic result is one of Weyl's, > **Weyl's equidistrubtion theorem.** > Let $\gamma$ be any irrational number. Then the sequence $(x_{n})$ with $x_{n} = n\gamma$ is equidistributed in $[0,1)$. Let us try to prove this. ## Some preliminary results. For any real number $x$, denote $\lfloor x \rfloor$ to be the greatest integer not exceeding $x$. And denote $\left< x \right> = x-\lfloor x \rfloor$ to be the fractional part of $x$. Note that $\left< x \right> \in [0,1)$ for all real $x$. Note if $x \in \mathbb{Z}$, then $\left< x \right> = 0$ and $\left< -x \right> = 0$. However if $x \not\in \mathbb{Z}$, then $\left< -x \right> = ?$ Let us think. If $x \not\in \mathbb{Z}$, then $x = \lfloor x \rfloor + \left< x \right>$ with $\left< x \right> \in (0,1)$. And $$-x = -\lfloor x \rfloor - \left< x \right> = -\lfloor x \rfloor -1 + (1 -\left< x \right>) $$ with $(1-\left< x \right>) \in (0,1)$. So $\left< -x \right> = 1- \left< x \right>$. Now, note if $x-y \in \mathbb{Z}$, then $x = y +m$ for some $m \in \mathbb{Z}$. So $x = \lfloor y \rfloor +\left< y \right> + m$, and so $\left< x \right> = \left< y \right>$. So $\left< x-y \right> = \left< x \right> - \left< y \right> = 0$. Now take any two real numbers $x,y$. What is $\left< x-y \right> = ?$ If $\left< x \right> \ge \left< y \right>$, there are two cases. If $x-y \in \mathbb{Z}$, then $\left< x-y \right> = \left< x \right>-\left< y \right> = 0$. If $x-y \not\in \mathbb{Z}$, then we have $x-y=\lfloor x \rfloor + \left< x \right> -\lfloor y \rfloor -\left< y \right>$. So $\left< x-y \right> =\left< x \right> - \left< y \right>$. If $\left< x \right> < \left< y \right>$, then we cannot have $x-y \in \mathbb{Z}$. So $x-y \not\in \mathbb{Z}$, and we have $\left< x-y \right> = 1+ \left< x \right> - \left< y \right>$. In summary, > For any two real numbers $x,y$, we have > $$ \left< x-y \right> = \cases{\left< x \right> -\left< y \right> & if \(\left< x \right> \geq \left< y \right> \)\\1+ \left< x \right> -\left< y \right> & if \()\left< x \right> < \left< y \right> \) } .$$ ## An equivalence to equidistribution. Let $(x_{n})$ be any sequence, we claim the following > **Lemma.** > The sequence $(\left< x_{n} \right>)$ is equidistributed in $[0,1)$ if and only if for any real $t$, we have > $$ \lim_{ N \to \infty } \frac{1}{N} \sum_{n=1}^{N} \left< x_{n} + t \right> - \left< x_{n} \right> = 0 .$$ $\blacktriangleright$ Proof. Take any real $t$, and write $b = \left< -t \right>$. Notice that $$ \left< x_{n} + t \right> =\left< x_{n} - (-t) \right> $$ which equals $\left< x_{n} \right> - b$ if $\left< x_{n} \right> \ge b$, or $1 + \left< x_{n} \right> -b$ if otherwise. Let us denote $A$ to be the set of indices $n \le N$ such that $\left< x_{n} \right> \ge b$ and let $B$ to be the set of indices $n \le N$ otherwise. So $A \cup B = \{1,\dots, N\}$. Then $$ \begin{eqnarray} && \frac{1}{N} \sum_{n=1}^N \left< x_{n} + t \right> \\ &=& \frac{1}{N} \sum_{n \in A} (\left< x_{n} \right> -b) + \frac{1}{N} \sum_{n \in B} (1+\left< x_{n} \right> - b) \\ &=& \left( \frac{1}{N} \sum_{n=1}^N \left< x_{n} \right> \right) -b + \frac{1}{N}\#\{n \le N : \left< x_{n} \right> < b \}. \end{eqnarray} $$ In other words, we get key equality for any real $t$ with $b = \left< -t \right>$, $$ \begin{eqnarray} &&\frac{1}{N} \sum_{n=1}^N \left< x_{n} + t \right> -\left< x_{n} \right> \\ &=& \frac{1}{N}\#\{n \le N : \left< x_{n} \right> < b \} -b . \end{eqnarray} $$ So, suppose for any real $t$ we have $$ \lim_{ N \to \infty } \frac{1}{N} \sum_{n=1}^N \left< x_{n} + t \right> - \left< x_{n} \right> = 0. $$ Take any $b \in [0,1)$ and set $t =- b$ so that $\left< -t \right> = \left< b \right> = b$. By the key equality we have $$ \lim_{ N \to \infty } \frac{1}{N} \# \{n \le N : \left< x_{n} \right> < b\} = b. $$ Conversely, suppose for any $b \in [0,1)$ we have $$ \lim_{ N \to \infty } \frac{1}{N} \# \{n \le N : \left< x_{n} \right> < b \} = b. $$ Then for any real $t$, set $b = \left< -t \right>$, and by the key equality we have $$ \lim_{ N \to \infty } \frac{1}{N} \sum_{n=1}^N \left< x_{n} + t \right> -\left< x_{n} \right> =0. $$ So we have as claimed. $\blacksquare$ ## Proof of Weyl's equidistrubtion theorem. ??????? /