Let $R: [0,1]\to[0,1]$, with $ R(x)=x+\alpha \mod 1 $ and $ \alpha = \frac{p}{q} $ a rational rotation map on the unit interval. I want to proof the following assertion
the ergodic measures for $R$ are of the form $ \mu =\sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q} $ for some $x_1,\dots,x_n\in[0,1]$ (and where $\delta_{\{x_i\}}$ denotes the dirac measure).
My understanding is, that to be $R$- ergodic, the measure $\mu$ has to be atomic and concentrated on a finite number of atoms, which - I'm guessing - are going to be the the measures that are concentrated on the finite orbits of points. My proof so far:
Let $ A $ be a $ R $-invariant set, i.e. $ R^{-1}A=A $. I can assume, that $ A $ is not empty (the empty set is a trivial case). So there exists a point $ x\in A $. Since $ \alpha $ is rational, it's clear, that the orbit of $ x $ is a finite set $ orb(x)=\{x_1,\dots,x_q\} $. Since $ R^{-1}A=A $ and clearly the points of the orbit map onto each other, it follows that $ \{x_1,\dots,x_n\}\subset A $. Since $ R $ is ergodic, $ \mu(A)\in\{0,1\} $. First, we assume, that $ \mu(A) = 1 $. It's easy to see, that the $ R $-Orbits of points in $ [0,1] $ are $ R $-invariant sets, this means, that $ \mu(\{x_1,\dots,x_q\})\in \{0,1\} $.
If $ \mu(\{x_1,\dots,x_q\}) = 1 $, then $ \mu = \sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q} $.
If $ \mu(\{x_1,\dots,x_q\})= 0 $, then $ \mu(A\setminus\{x_1,\dots,x_q\}) =1 $. This means there is another $ \hat{x}\in A\setminus\{x_1,\dots,x_q\} $. And since $ A\setminus\{x_1,\dots,x_q\} $ is still $ R $-invariant, the whole finite orbit of $ \hat{x} $ is a subset of $ A\setminus\{x_1,\dots,x_q\} $: $ orb(\hat{x})=\{\hat{x}_1,\dots,\hat{x}_q\}\subset A\setminus\{x_1,\dots,x_q\}. $
Now we can proceed the same way as before:
$ \{\hat{x}_1,\dots,\hat{x}_q\} $ is a $ R $-invariant set, this means $ \mu(\{\hat{x}_1,\dots,\hat{x}_q\})\in\{0,1\} $. If $ \mu(\{\hat{x}_1,\dots,\hat{x}_q\})=1 $, then $ \mu= \sum\limits_{i=1}^q\frac{\delta_{\{\hat{x_i}\}}}{q} $. If $ \mu(\{\hat{x}_1,\dots,\hat{x}_q\})=0 $, we look again at $ A\setminus(\{x_1,\dots,x_n\}\cup\{\hat{x}_1,\dots,\hat{x}_q\}) $ and repeat the same procedure on this set.
So my questions are
- is the proof correct so far? In particular: is it valid, that $ \mu(\{x_1,\dots,x_q\}) = 1 $ implies $ \mu = \sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q} $ ?
- how can I handle the case $ \mu(A) = 0 $? (Then $ \mu(\{x_1,\dots,x_q\})= 0 $ as well, so I guess $ \mu $ has to be concentrated on some other orbit, but I don't know how to formalize that part).
$\mu(\{x_1,\dots,x_q\}) = 1$ implies $\mu = \sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q}$ is correct. But I do not understant how this close the case $\mu (A)=1$ since you can repeat infinitely times your procedure without reaching a atom of the measure.
If $A$ is invariant and $\mu(A)=0$ then $A^c$ the complement, is also invariant and $\mu(A^c)=1$. That might help.
I think a working strategy is to do a dichotomy. Take a invariant ergodic measure $\mu$. If $A\subset[0,1/q[$ we will call $\hat{A}=\cup_{i=0}^{q-1}R^i(A)$. $\hat{A}$ is always invariant for $R$.
Let's called $A^0=[0,1/q[$. We have $\mu(A^0)=1/q$ (because of invariance and $\hat{A^0}$ is $[0,1]$ .
Now you can cut $A^0$ in two parts $A^1_1=[0,1/2q[$ and $A^1_2=[1/2q,1/q[$, $\hat{A^1_1}$ and $\hat{A^1_2}$ are a partition of $[0,1]$ and are disjoint, by ergodicity there is $i_1 \in \{0,1 \}$ such that $\mu(\hat{A^1_{i_1}})=1$.
Doing this again and again you will find a sequence of set $A^k \subset [0,1/q[$ with the following properties :
Taking the intersection of all this set you find just a point $x$, and then $\mu=\sum\limits_{i=1}^q\frac{\delta_{\{x_i\}}}{q}$.