# Evolution of a density under the doubling angle map

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 $$\frac12$$?

• Up to now, I am able to prove it if the density w.r.t. Lebesgue measure is Lipschitz on all the support of $\mu$. – Michal Kupsa Dec 4 at 9:22
• Approximate your density by $C^\infty$ function with $L^1$-error less than $\varepsilon$. Since you said you can prove it for Lipschitz (and, a fortiori, smooth) densities, you can do a standard procedure and prove that limsup of difference of measure and $1/2$ is at most $\varepsilon$. Since $\varepsilon$ is arbitrary the claim follows. – Aleksei Kulikov Dec 4 at 13:36

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 $$\int \mathbf 1_{[0,\frac 12]}\,d\lambda \int f\,d\lambda=\frac 12$$.
• The motivation for the question comes from the course on cryptography, where I needed to show how to generate a random bit from an arbitrary continuous random variable (not necessary concentrated on $[0,1]$). Following Antony's anwser, I can give a more general statement about generating the sequence of random bits (not only one bit) of "high quality" using the double angle map iteration on the measure that is absolutely continuous w.r.t. some $T$-invariant mixing measure on $[0,1]$. – Michal Kupsa Dec 5 at 9:37