We have the transformation $T: [0,1) \rightarrow [0,1)$ given by $Tx = \beta x \text{ mod } 1$ with $\beta = \frac{1+ \sqrt{5}}{2}$. Calculate the entropy $h_{\mu}(T)$ of $T$ wrt the invariant measure $\mu$ given by \begin{equation} \mu(A)= \int_A g(x)d\lambda(x), \end{equation} with \begin{align} g(x) &= \frac{5+3\sqrt{5}}{10}, \text{ } 0\leq x < \frac{\sqrt{5}-1}{2} \\ &=\frac{5+\sqrt{5}}{10} \text{ , } \frac{\sqrt{5}-1}{2} \leq x <1 \end{align}
I have to use Shannon-McMillan-Breiman theorem and the fact that if we replace $\mu$ by $\lambda$, i.e. \begin{equation} \lim_{n \to \infty}- \frac{\log\lambda(P_n(x))}{n}=h_{\mu}(T) \text{a.e. with respect to $\lambda$} \end{equation}
Can someone help me with this problem?
Using beta-expansions, one checks that the transformation $T$ above is equivalent to the Markov shift on the set of all $\{0,1\}$-valued sequences without consecutive 1. The transition matrix is given by $p(0,0)=\beta^{-1}$, $p(0,1)=\beta^{-2}$, $p(1,0)=1$, $p(1,1)=0$, whereas the stationary measure is given by $\pi(0) = \beta^2/(\beta^2+1)$, $\pi(1) = 1/(\beta^2+1)$. Therefore, the entropy is $$-\sum_{i,j}\pi(i) p(i,j)\log p(i,j) = \frac{\beta^2}{\beta^2+1}(\beta^{-1}\log\beta+2\beta^{-2}\log\beta) + \frac{1}{\beta^2+1} \times 0 = \log\beta,$$ since $\beta^{-1}+2\beta^{-2} = 1 + \beta^{-2}$.