= Solution
For a finite measurable partition $\xi$,
$$
H_\mu(\xi)=-\sum_{A\in\xi}\mu(A)\log\mu(A).
$$
Its <conditional entropy of finite measurable partitions> relative to $\eta$ is
$$
H_\mu(\xi\mid\eta)
=\sum_{B\in\eta}\mu(B)
\left[-\sum_{A\in\xi}\mu(A\mid B)\log\mu(A\mid B)\right].
$$
The concavity of $-t\log t$, equivalently <conditioning reduces entropy>, gives
$$
\boxed{H_\mu(\xi\mid\eta)\leq H_\mu(\xi).}
$$
The atoms of $T^{-1}\xi$ are $T^{-1}A$ with the same measures as the atoms $A$ of $\xi$. Hence $\boxed{H_\mu(T^{-1}\xi)=H_\mu(\xi)}$.
Set
$$
a_N=H_\mu\left(\bigvee_{n=0}^{N-1}T^{-n}\xi\right).
$$
The chain rule and invariance give $a_{N+M}\leq a_N+a_M$, so $(a_N)$ is a <subadditive sequence>. Therefore
$$
h_\mu(T,\xi)=\lim_{N\to\infty}\frac{a_N}{N}
=\inf_{N\geq1}\frac{a_N}{N}.
$$
The <Kolmogorov-Sinai entropy> is $h_\mu(T)=\sup_\xi h_\mu(T,\xi)$ over finite partitions.
Taking $F=\{0,\ldots,N-1\}$ immediately shows that the infimum over arbitrary finite $F$ is at most $h_\mu(T,\xi)$. For the reverse inequality, apply <Shearer's inequality> to translates of a fixed finite $F$ inside a long interval. Every interior coordinate is covered $|F|$ times, while only $O(\max F)$ boundary coordinates are lost. Subadditivity bounds the boundary contribution; division by the interval length and passage to the limit give
$$
h_\mu(T,\xi)\leq\frac1{|F|}H_\mu\left(\bigvee_{n\in F}T^{-n}\xi\right).
$$
Taking the infimum proves
$$
\boxed{h_\mu(T,\xi)=\inf_{\varnothing\ne F\subseteq\mathbb Z_{\geq0}\text{ finite}}
\frac1{|F|}H_\mu\left(\bigvee_{n\in F}T^{-n}\xi\right).}
$$
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