Let $\mu$ be a probability measure on $I=[0,1]$, absolutely continuous with respect to Lebesgue measure. Denote by $T$ the "doubling angle map" on $I$, where $T(x)=2x \text{ mod }1$. Is it true, in general, that
 $\mu\left(T^{-n}\left[0,\frac12\right)\right)$ converges to $\frac12$?



Evolution of a density under the doubling angle map
Write $d\mu(x)=f(x)\,d\lambda(x)$ where $\lambda$ is Lebesgue measure. Then you're asking about $\int \mathbf 1\_{[0,\frac 12]}\circ T^n\,d\mu = \int \mathbf 1\_{[0,\frac 12]}\circ T^n f(x)\,d\lambda(x)$. Since $T$ is mixing with respect to the invariant measure $\lambda$, this converges to $\int \mathbf 1\_{[0,\frac 12]}\,d\lambda \int f\,d\lambda=\frac 12$. 


