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 $h_\mu(T) \leq (n+t) h_\mu(f)$.

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