Entropy of $f^{m(x)+n}$ of full shift

Question

Let $(X,\mu,f)$ be a two-sided full shift system. Assume that there is $t \in \mathbb{N}$ such that for every $n \in \mathbb{N}$ and $x \in X$, we can define $T(x)=f^{n+m(x)}(x)$, where $m(x) \leq t; $ $m(x) \in \mathbb{N}$ and that depends on $x$. I also assume that $T$ is measure preserving.

It is well-known that $h_{\mu}(f^n)=nh_{\mu}(f)$, where $h_{\mu}$ is the measure-theoretic entropy.

Can we use the above fact to say $h_{\mu}(T) \leq (n+t)h_{\mu}(f)$?

Edit: As it was mentioned in comments, $T$ is not necessarily measure preserving. I add the assumption that $T$ is measure preserving.

Answer 1

If I understand right, your function $m$ (for a fixed $n$) takes on only finitely many values, which are all measurable sets. You can then define the partition $\{m^{-1}(i)\}$ and the associated finite $\sigma$-algebra $\mathcal{B}$.

If we define $\mathcal{A}$ to be any finite sub-$\sigma$-algebra of your original $\sigma$-algebra of measurable sets, then it's well-known (Corollary 4.10 of Walters) that $k^{-1} H_\mu\left(\bigvee_{i=0}^{k-1} T^{-i} \mathcal{A} \right)$ converges (in fact decreases) to $h_\mu(T, \mathcal{A})$.

In your case, I think by definition $\bigvee_{i = 0}^k T^{-i} \mathcal{A}$ is contained in $\bigvee_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B})$. Therefore, $k^{-1} H_\mu\left(\bigvee_{i=0}^{k-1} T^{-i} \mathcal{A} \right) \leq k^{-1} H_\mu\left(\bigvee_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B}) \right) = (n+t) (k(n+t))^{-1} H_\mu\left(\bigvee_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B}) \right).$

But by the above, the first quantity approaches $h_\mu(T, \mathcal{A})$ and the final quantity approaches $(n+t) h_\mu(f, \mathcal{A} \vee \mathcal{B})$. Therefore, $h_\mu(T, \mathcal{A}) \leq (n+t) h_\mu(f, \mathcal{A} \vee \mathcal{B})$, and then taking the supremum over $\mathcal{A}$ yields $h_\mu(T) \leq (n+t) h_\mu(f)$.