Why do rotationally ordered configurations have well defined distributon function? Let $u=(u_{j})_{j \in \mathbb{Z}}$ where $u_{j}\in \mathbb{R}$ for all $j \in \mathbb{Z}$ be a rotationally ordered configuration i.e. $S_{n,m}u>u$ or $S_{n,m}u<u$ or $S_{n,m}u=u$ where 
$u<v$ iff $u_{j}<v_{j}$ for all $j \in \mathbb{Z}$ and where $(S_{n.m}u)_{j}=u_{j+n}+m$ for $n,m \in \mathbb{Z}$. 
Let $F_{u}(x)=\lim_{N,M \to \infty}\frac{\#\{j \in \mathbb{Z}: -N \leq j \leq M, \{u_{j}\} \in [0,x \rangle\}}{N+M}$ where $\{u_{j}\}$ is fractional part od $u_{j}$ and $x \in [0,1]$. My question is why do rotationally ordered configurations have well defined $F_{u}$? 
 A: I was hoping for a more elegant answer than the one I will present. I will demostrate the case when mean spacing $\rho$ is irrational. Every rotationally ordered configuration $u_{j}$ has mean spacing. That is there exists a number $\rho$ so that $u_{j}/j \to \rho$ when $|j|\to \infty$. Suppose that $\rho$ is irrational then $u_{j}=\rho j +\alpha_{j}$. For rotationally ordered configuration $(u_{j})$ we have an estimate 
$$|u_{j+m}-u_{j}-\rho m|\leq 1$$
So we get that there exists natural number $C>0$ so that $|\alpha_{j}|\leq C$ for all $j \in \mathbb{Z}$. We have two sequences $\mathbb{N}\ni j \mapsto \alpha_{j}$, $\mathbb{N}\ni j \mapsto (\alpha_{-j})$. We will prove that the mentioned sequences are convergent. The claim will be proved only for the first sequence the other follows similarly. 
Suppose that first sequence is not convergent then because $[-C,C]$ is a compat space the set of all accumulation points $A=\{a_{i}: i \in I\}$ of that sequence has more than one element. Let us denote by $X'$ set of all accumulation points of some set $X$. Then we know that $A' \subseteq A$ so there exists without loss of generality $a_{1} \in A$ so that $\varepsilon=\inf\{|a_{i}-a_{1}|: i\in I\setminus\{1\}\}>0$. Let us define $$M=\{ m \in \mathbb{N}: \exists k \in \mathbb{Z} \hbox{ so that }|\rho m-k|<\varepsilon/3\}$$
$$A_{i}=\{j \in \mathbb{N}: |\alpha_{j}-a_{i}|<\varepsilon/3\}$$
As $u_{j}$ is rotationally ordered we have that $|\rho m -k|>|\alpha_{j+m}-\alpha_{j}|$. Now if $m \in M$, $j+m \in A_{i}$ and $j \in A_{1}$ then we have a contradiction. Therefore we have $(A_{1}+M)\cap A_{i}=\emptyset$. Let's define $B$ to be the smallest set such that $\mathbb{N}=B \cup \bigcup_{i\in I}A_{i}$ then $B$ is finite. As we have
$$(A_{1}+M)\cap (B \cup\bigcup_{i \in I}A_{i})=A_{1}+M$$ there exists $k \in \mathbb{N}$ such that $(A_{1}^{k}+M) \subseteq A_{1}^{k}$ where $A_{1}^{k}=A_{1}\cap \{k,k+1,k+2,\dots\}$. But then $(A_{1}^{k}+M+M) \subseteq A_{1}^{k}+M \subseteq A_{1}^{k}$. 
As there exist $a,b \in \mathbb{N}$ such that $a,ab+1 \in M$ then for $H=M \cup (M+M) \cup (M+M+M) \cup \dots$ we have that there exists $l \in \mathbb{N}$ such that $H \cap \{l,l+1,l+2,\dots\}=\{l,l+1,l+2,\dots\}$ and that $A_{1}^{k}+H \subseteq A_{1}^{k}$. We get that $A$ has only one element naimly $a_{1}$.   
By Weil's theorem sequence $\{x_{j}\}$ is equidistributed iff 
$$\frac{1}{N}\sum_{j=1}^{N}e^{2\pi i k x_{j}}\to 0$$
for every $k \in \mathbb{Z}\setminus \{0\}$ as $N \to \infty$. 
As the sequences $\mathbb{N}\ni j \mapsto \rho j$ and $\mathbb{N}\ni j \mapsto \rho(-j)$
are equidistributed and the sequneces $(\alpha_{j})_{j\in\mathbb{N}}$ and $(\alpha_{-j})_{j \in \mathbb{N}}$ are convergent we have that the sequences $(\rho j+\alpha_{j})_{j \in \mathbb{N}}$ and $(\rho (-j)-\alpha_{-j})_{j \in \mathbb{N}}$ are equidistributed by Weil's theorem but that means that 
$$\frac{\#\{j \in [-N,M]\cap \mathbb{Z}: \{u_{j}\} \in [0,x\rangle\}}{N+M} \to x$$ 
as $N,M \to \infty$. So not only the limit exists but we know it's value. 
