Let $X$ be a Bernoulli scheme. A factor $\psi: X \to Y$ is finitary if for almost every $x \in X$ there exist integers $m \leq n$ such that the zero coordinates of $\psi(x)$ and $\psi(x')$ agree for almost all $x' \in X$ with $x[m,n]=x'[m,n]$.

Let $f:X \to Y$ and $g: X\to Y'$ be finitary factor maps. Define a map $\phi_{f,g}: X \to Z$ where $(\phi_{f,g}(x))_i=0$ if $(f(x))_i=(g(x))_i$, and $(\phi_{f,g}(x))_i=1$ otherwise. Since $f$ and $g$ are finitary factor maps, $\phi$ is a finitary factor map. Let $Z$ have measure $\nu$. Now define $d(f,g)=\nu(1)$.

Again, let $X$ be a Bernoulli scheme and $f: X \to Y$ a finitary factor map. Suppose there exists a sequence of finitary factors $f_n: X \to Y_n$ where $lim_{n \to \infty} d(f_n,f)=0$. Here $Y_n$ refers to the $n$th process in the sequence and not the $n$th coordinate of $Y$. Is it always the case that $lim_{n \to \infty} h(Y_n)=h(Y)$? Here $h$ denotes the entropy.

Thanks

upper boundfor $\bar d$, rather than $\bar d$ itself. – Anthony Quas Jun 26 '12 at 20:41