Singular stationary measures are not always fractal (Q1314316)

Let \(G\) be some transformation semigroup of a space \(M\) and \(\mu\) some probability measure of \(G\) supported on two points \(T_ 1\) and \(T_ 2\) of \(G\), say \(\mu (T_ 1)=p\) and \(\mu(T_ 2)=1-p\), then a probability measure \(\nu\) on \(M\) is a stationary measure for \(\mu\) iff \[ \int f(x)dx=p \int f(T_ 1(x))d \nu(x)+(1-p) \int f(T_ 2(x))d \nu (x). \] For \(M=[0,1]\), \(G\) the semigroup of transformations of \([0,1]\), \(T_ 1,T_ 2\) two Lipschitz transformations satisfying some extra condition amongst them \(T_ 1([0,1]) \cap T_ 2 ([0,1])\neq\emptyset\) the author proves that 1. \(\nu\) is unique, 2. \(\nu\) has no atoms, 3. \(\nu\) is singular with respect to Lebesgue measure on \([0,1]\), 4. \(\nu\) has \(M\) as its support. The analogous result is obtained for \(M\) as the real projective one- space, \(G=SL_ 2(\mathbb{R})\), the group of \(2 \times 2\) real matrices of determinant 1 and \[ T_ 1=\left( \begin{matrix} a & b \\ 0 & a^{-1} \end{matrix} \right),\;T_ 2=\left( \begin{matrix} c & 0 \\ d & c^{-1} \end{matrix} \right) \] as special transformations for \(a,b,c,d \in \mathbb{R}\).