Generating measurable partition

For an invertible measure-preserving system, a generating measurable partition $\xi$ has $\sigma ({\bigvee }_{j\in \mathbb{Z}}{T}^{-j}\xi )=\mathcal{B}$ modulo null sets. Thus knowing the entire two-sided itinerary determines all measurable observations. A finite or countable finite-entropy generating partition computes the system entropy through the Kolmogorov-Sinai generator theorem.

A one-sided generator is a measurable partition $\xi$ with $\sigma ({\bigvee }_{j\ge 0}{T}^{-j}\xi )=\mathcal{B}$ modulo null sets. Equivalently, every measurable set can be approximated in measure by unions of atoms of finite forward-name blocks. A one-sided coordinate partition generates a one-sided Bernoulli shift; for a nontrivial base distribution, it does not generate the two-sided version, whose negative coordinates are independent of the nonnegative ones.

If an invertible probability measure-preserving system has a finite one-sided generator $\xi$, then its future sigma-algebra ${\mathcal{F}}_1=T^{-1}\mathcal{B}$ equals $\mathcal{B}$ modulo null sets. Hence $H_{\mu }(\xi \mid {\mathcal{F}}_1)=0$, and the infinite-future formula for partition entropy rate and Kolmogorov-Sinai generator theorem give $h_{\mu }(T)=0$. Invertibility matters: a fair binary one-sided Bernoulli shift has entropy $\log 2$ and a finite one-sided generator.

For an invertible probability measure-preserving system with a finite or countable finite-entropy generating measurable partition $\xi$, $h_{\mu }(T)=h_{\mu }(T,\xi )$. For a noninvertible system the same conclusion holds for a finite-entropy one-sided generator. The finite-entropy hypothesis is part of this form of the theorem.