In some article about ergodic theory, it said (without proof) that if $\mu$ is a $f$-invariant measure, then $h_\mu(f^k)=k*h_\mu(f)$. I'm looking to prove this.
$h_\mu(f^k,A)=\lim_n (1/n)*H_\mu(\bigvee_{i=0}^{n-1} f^{-ik}A)$, so in some manner we should have $H_\mu(\bigvee_{i=0}^{n-1} f^{-ik}A)=k*H_\mu(\bigvee_{i=0}^{n-1} f^{-i}A)$, but not necessarily exactly that since we only need this to hold on the $sup_A$ over this quantities. It simply looks wrong to me, so im looking for some more opinions.
Call $g=f^k$ and for a finite entropy partition $\mathcal{P}$ of the space, denote by $\mathcal{P}_f^n=\displaystyle\bigvee_{i=0}^{n-1} f^{-i}(\mathcal{P})$ its n-th iteration under $f$ (and similar notation for $g$). Note that $$ \mathcal{P}_f^{km}=\bigvee_{i=0}^{km-1} f^{-i}(\mathcal{P})=\bigvee_{i=0}^{m-1} f^{-ik}\left( \bigvee_{j=0}^{k-1} f^{-j}(\mathcal{P}) \right) $$ Then $$ h_\mu(g,\mathcal{P}_f^k)=\lim_n\dfrac{1}{n}H_\mu\left(\bigvee_{i=0}^{n-1} g^{-i}(\mathcal{P_f^k})\right)=\lim_n \dfrac{1}{n}H_\mu\left(\bigvee_{i=0}^{n-1}f^{-ik} \left( \bigvee_{j=0}^{k-1} f^{-j}(\mathcal{P}) \ \right) \right)=\lim_n\dfrac{1}{n}H_\mu(\mathcal{P}_f^{kn})=k\lim_n\dfrac{1}{nk}H_\mu(\mathcal{P}_f^{kn})=kh_\mu(f,\mathcal{P}). $$ Now, since $\mathcal{P}^k$ is finer than $\mathcal{P}$, we have that $h_\mu(g,\mathcal{P})\leq kh_\mu(f,\mathcal{P})=h_\mu(g,\mathcal{P^k})\leq \sup_{\mathcal{P}}h_\mu(g,\mathcal{P})=h_\mu(g) $. Since the last inequality is true for every partition $\mathcal{P}$, taking supremum over $\mathcal{P}$ yields $h_\mu(g)=kh_\mu(f)$.