# 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.

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There are papers by Katok and by Brin and Katok that give similar definitions of entropy to the Bowen ball definition. –  Anthony Quas Sep 7 '13 at 16:14
I think it's best to understand the formula you've written (at-least intuitively) by looking at a Bernoulli system, there $T^{n}$ just means to shift $n$ places to the left instead of $1$, hence you need to factor it in the amount of information you get from each atom in your partition (and consequently, the refinement of your generating partition would look differently then the usual one). –  Asaf Sep 7 '13 at 17:29
Another way to interpret metric entropy - you expect (say from Bernoulli systems) that near orbits should diverge in exponential rate. If you calculate the rate to capture every orbit, you get the topological entropy (this is the definition with the separation condition of topological entropy), but if you focus only on generic orbits (=orbits of generic points), you will calculate the metric entropy. From this POV, the formula is clear, you speed up the time by a factor of $n$, hence the orbits will separate quicker than in the regular time flow. –  Asaf Sep 8 '13 at 17:24
@Asaf Thanks for your opinion, for the case of T with respect to Bernoulli measure (assume Bernoulli measure can is T invariant) I think it is quite clear. However, for the Markov systems, it is not clear in intuition. –  yaoxiao Sep 9 '13 at 8:58

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 $h(T^n,m)=nh(T,m)$.
Thanks for your interesting explanation. I think I can brief summary about what did you say. We can consider this equality in the ergodic measure. (because of decomposition of invariant measure.) First let us consider a special case: symbolic space. By Shannon-Mcmillan-Breiman theorem. $$\frac{\mu[\omega_1\dots\omega_k]}{k}\rightarrow h(\mu,\sigma) \qquad \mu a.e. \omega$$ $$\frac{\mu[\omega_1\dots\omega_{nk}]}{k}\rightarrow h(\mu,\sigma^n)\qquad \mu a.e. \omega$$ It is ovious $$h(\mu,\sigma^n)=nh(\mu,\sigma)$$ –  yaoxiao Sep 9 '13 at 15:55