Intuition of Kolmogorov-Sinai entropy For a measurable entropy of measurable transformation $T$ from $(X,\mathcal{B},m)$ to itself.
For each finite measurable partition $\mathcal{A}=\{A_i\}_{i=1}^{m}$ of $X$, we can define
$h(\mathcal{A},T,m)$ as $\lim_{n\rightarrow \infty}\frac{1}{n}H(\bigvee_{i=0}^{n-1}T^{-i}\mathcal{A})$.
and measurable h(T,m) is to take sup of all possible finite measurable partition.
It is subtle as least to me to understand $h(T^{n},m)=n h(T,m) (n>0)$. Of course I can prove it almost by definition. However I can not feel the essence of measurable entropy.
For topological entropy, we can understood $h(T^n)=nh(T)$ very well using Bowen balls related to numbers of equivalence of (cut-off) orbits.
I wondered whether there exist similar explanation for measurable entropy. Any reference and commnents will be greatly appreciated.
 A: A good way to understand measurable entropy is via the Shannon-McMillan-Breiman Theorem. Roughly speaking it says that there is a constant $c$ so that most atoms $A$ in $\bigvee_{i=0}^{n-1} T^{-i}\mathcal{A}$ have measure $m(A)\approx e^{-cn}$, and the value of $c$ is the measure entropy $h(\mathcal{A},T,m)$. More precisely, given $\epsilon>0$, for all sufficiently large $n$ there is a collection of atoms in $\bigvee_{i=0}^{n-1} T^{-i}\mathcal{A}$ whose union has measure a least $1-\epsilon$, and such that each atom in this collection has measure between $e^{-(c-\epsilon)n}$ and $e^{-(c+\epsilon)n}$. Thus measure entropy tells the exponential decay rate of the measure of atoms in $\bigvee_{i=0}^{n-1} T^{-i}\mathcal{A}$. From this it should be clear that $h(T^n,m)=nh(T,m)$.
This interpretation is the starting point for proofs of fundamental theorems such as Ornstein's result that Bernoulli shifts with the same entropy are measurably isomorphic. With good control of the exponential sizes of atoms, a partial isomorphism can be built using Rohlin towers, and then successively refined with combinatorial input from the Marriage Lemma to converge to an isomorphism. A lucid account of this proof, starting from the basics, is in Paul Shields' book "The Theory of Bernoulli Shifts", now available for free and easy to find via web search.
