Entropy of factors of Bernoulli schemes

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

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It looks like your distance $d(f,g)$ is just $\bar d$-distance between the processes. It's known that entropy is continuous in $\bar d$ (see Rudolph's book for a fairly readable proof). – Anthony Quas Jun 26 '12 at 18:46
Thanks Anthony. I believe $d(f_n,f) \to 0$ implies that the $\bar d$-distance between $Y_n$ and $Y$ goes to 0. So, your comment gives me the answer I needed. I am not sure $d(f,g)$ is equivalent to $\bar d$, since there can be distinct finitary maps $f$ and $f'$ on $X$ that both map to $Y$, and I'm not sure $d(f,g)$ always equals $d(f',g)$. Perhaps, if I always took the $f$ and $g$ that minimized $d(f,g)$, there is equivalence with $\bar d$ on the image processes. – Stephen Shea Jun 26 '12 at 19:23
I guess I agree - $d(f,g)$ is an upper bound for $\bar d$, rather than $\bar d$ itself. – Anthony Quas Jun 26 '12 at 20:41