Rajchman measures via strong mixing systems

A Rajchman measure on the unit circle $\mathbb{T}$ is a Borel probability measure $\mu$ with $\lim_{n\to\infty}\hat{\mu}(n)=0$. Where $\hat{\mu}(n)=\mu(z^n)$ for $n\in\mathbb{Z}$ are Fourier coefficients of $\mu$.

Suppose $(X,\mathcal{B},T,\mu)$ is a measure-theoretic dynamical system consisting of a measure space $(X,\mathcal{B},\mu)$ and an invertible measure preserving map $T:X\to X$. A dynamical system $(X,\mathcal{B},T,\mu)$ is called strong mixing if $$\lim_{n\to\infty} \mu(T^{-n}A\cap B)=\mu(A)\mu(B)$$ for all $A,B\in\mathcal{B}$.

Denote the space $\{f\in L^2(X,\mu)|\int_X f\,d\mu=0\}$ by $L^2_0(X,\mu)$. The dynamical system $(X,\mathcal{B},T,\mu)$ is strong mixing iff $$\lim_{n\to\infty}\int_X f(T^nx)\overline{f(x)}\,d\mu(x)=0$$ for every unit $f\in L^2_0(X,\mu)$.

For every unit $f\in L^2_0(X,\mu)$, one can define a Borel probability measure $\mu_f$ on $\mathbb{T}$ by $$\widehat{\mu_f}(n)=\int_X f(T^nx)\overline{f(x)}\,d\mu(x).$$ So a dynamical system $(X,\mathcal{B},T,\mu)$ is strong mixing iff $\mu_f$ is a Rajchman measure for every unit $f\in L^2_0(X,\mu)$.

Through the above observation, given a strong mixing dynamical system $(X,\mathcal{B},T,\mu)$, for instance, Bernoulli shift $(\{0,1\}^\mathbb{Z},\mathcal{B},S,\mu_p)$ where $\mu_p$ is the product measure of the measure on $\{0,1\}$ giving $\{0\}$ measure $p$ and $\{1\}$ measure $1-p$, each $f\in L^2_0(X,\mu)$ gives rise to a Rajchman measure on $\mathbb{T}$.

Question: For which strong mixing dynamical system $(X,\mathcal{B},T,\mu)$, one can find unit $f\in L^2_0(X,\mu)$ such that $\mu_f$ is a Rajchman measure singular to the Lebesgue measure?

• I think you're exactly asking for the strong mixing dynamical systems with singular spectral type. One comment: these must have entropy 0 as any positive entropy measure has Lebesgue spectrum. – Anthony Quas Aug 14 '15 at 9:13