I'm studying a paper from Michael Keane on Interval Exchange Transformations.
Before I ask my question I explain a little about the subject:
Let $X=[0,1)$ and $n \geq 2$ an integer. For each probability vector $\alpha=(\alpha_1, \alpha_2, \cdots, \alpha_n) $ with $\alpha_i >0$ for $1 \leq i \leq n$ we set : \begin{align} &\beta_0 =0\\ &\beta_i=\sum\limits_{j=1}^i\alpha_j\\&X_i=[\beta_{i-1},\beta_i) \end{align} Let $\tau$ be a permutation of the symbols $\{1,2,\cdots,n\}$. Then \begin{align} \alpha^{\tau}=(\alpha_{\tau^{-1}(1)},\alpha_{\tau^{-1}(2)},\cdots, \alpha_{\tau^{-1}(n)}) \end{align} is a probability vector with positive components, and we can form the corresponding $\beta_i^{\tau}$ and $X_i^{\tau}$, $1\leq i \leq n$. Now define $T:X \to X$ by setting \begin{align} Tx=x-\beta_{i-1}+\beta_{\tau(i)-1}^\tau \end{align} For each $ x \in X_i$ and each $ 1 \leq i \leq n$. $T$ maps each interval $X_i$ isometrically onto the corresponding interval $X_{\tau(i)}^\tau$.
We call $T$ an $(\alpha,\tau)$-Interval Exchange Transformation.
$T$ possesses the following properties:
1- $T$ is invertible, and its inverse is the $(\alpha^\tau, \tau^{-1})$ interval exchange transformation.
2- $T$ is continuous except at the points of the set $D=\{\beta_1,\beta_2,\cdots,\beta_{n-1}\}$ and $T$ is continuous from the right at these points.
3- $\lim\limits_{x\to \beta_i}Tx= \beta_{\tau(i)}^\tau$ for $1 \leq i \leq n$
4- $T\beta_i=\beta_{\tau(i+1)-1}^\tau$ for $0 \leq i \leq n-1$.
My question is to prove 3 and 4.
My try:
For 3: \begin{align} \lim\limits_{x \to \beta_i}Tx=T\beta_i &= \beta_i - \beta_{i-1}+\beta_{\tau(i)-1}^\tau\\ &= \alpha_i+ \beta_{\tau(i)-1}^\tau\\ &=\alpha_{\tau^{-1}(\tau(i))}+(\alpha_{\tau^{-1}(1)} + \cdots +\alpha_{\tau^{-1}(\tau(i)-1)})\\ &=\alpha_{\tau^{-1}(1)} + \cdots +\alpha_{\tau^{-1}(\tau(i)-1)}+\alpha_{\tau^{-1}(\tau(i))}\\ &=\beta_{\tau(i)}^\tau \end{align} Now how could I compute number 4? Where should I do a different thing in my calculation?
