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For ${\mathcal{F}}_L=\sigma ({\bigvee }_{j\ge L}{T}^{-j}\xi )$ and a finite measurable partition $\xi$, $H_{\mu }({\xi }_0^{L-1}\mid {\mathcal{F}}_L)=L{h}_{\mu }(T,\xi )$. The backward conditional entropy chain rule expresses the left side as ${\sum }_{j=0}^{L-1}{H}_{\mu }(T^{-j}\xi \mid {\mathcal{F}}_{j+1})$. Each term equals $h_{\mu }(T,\xi )$ by measure preservation and the infinite-future formula for partition entropy rate. No invertibility is needed.

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.

On a probability measure-preserving system, for a finite measurable partition $\xi =\{A_1,\dots ,A_r\}$, the entropy of a finite measurable partition isUse natural logarithms, so information entropy is measured in nats; another fixed logarithm base rescales all answers. The join of measurable partitions is their common refinement, and write ${\xi }_a^b={\bigvee }_{j=a}^b{T}^{-j}\xi$, with an empty join the trivial partition. The entropy rate of a measurable partition and Kolmogorov-Sinai entropy are respectivelyThe block entropies form a subadditive sequence, so the first limit exists by the Fekete lemma. For finite partitions the conditional entropy of finite measurable partitions is $H(\eta \mid \zeta )=H(\eta \vee \zeta )-H(\zeta )$.

Put $c_0=H(\xi )$ and $c_k=H(\xi \mid {\xi }_1^k)$ for $k\ge 1$. The chain rule for information entropy, applied from the last coordinate backwards, and measure preservation giveThe second equality uses invariance of the joint partition atom probabilities under the common pullback $T^{-j}$; invertibility is unnecessary. Since conditioning reduces entropy, $c_k$ decreases to a nonnegative limit $c$. The Cesaro convergence of a sequence of this convergent sequence has the same limit. ThereforeEquivalently $h_{\mu }(T,\xi )=H_{\mu }(\xi \mid {\mathcal{F}}_1)$, where ${\mathcal{F}}_1=\sigma ({\bigvee }_{j\ge 1}{T}^{-j}\xi )$. Here conditional entropy of a countable measurable partition conditioned on a sigma-algebra is computed using conditional partition atom probabilities; the Martingale convergence theorem gives continuity under increasing conditioning sigma-algebras. This is the infinite-future formula for partition entropy rate.

The Kolmogorov-Sinai generator theorem states that if a finite or countable measurable partition $\eta$ has finite entropy of a countable measurable partition and its iterates generate the whole completed sigma-algebra modulo null sets, then $h_{\mu }(T)=h_{\mu }(T,\eta )$. For an invertible system, generating means $\sigma ({\bigvee }_{j\in \mathbb{Z}}{T}^{-j}\eta )=\mathcal{B}$ modulo null sets. For a noninvertible system a one-sided generator, using $j\ge 0$, suffices. The two-sided and one-sided versions must not be confused.

For a Bernoulli shift with discrete symbol probabilities $(p_i)$, the coordinate-zero measurable partition has independent coordinate iterates. ThusFor a finite alphabet the coordinate partition is a generator of finite entropy of a finite measurable partition; on the two-sided sequence space use all integer coordinate iterates, and on the one-sided space use the nonnegative ones. The Kolmogorov-Sinai generator theorem proves the displayed answer in both cases. The same calculation applies to countably many symbols when their Shannon entropy is finite, using the countable finite-entropy version of the theorem. If the Shannon entropy is infinite, merge all but the first $r$ symbols into one cell. These finite coordinate partitions have entropy rate $-{\sum }_{i\le r}{p}_i\log p_i-p_{>r}\log p_{>r}\to \infty$, so the system entropy is infinite. Zero-probability symbols contribute zero. In particular a fair $r$-symbol shift has entropy $\log r$.

For the final assertion, let ${\mathcal{F}}_0=\sigma ({\bigvee }_{j\ge 0}{T}^{-j}\xi )$ and complete it modulo null sets. The approximation property forces ${\mathcal{F}}_0=\mathcal{B}$ modulo null sets. Indeed for each $A\in \mathcal{B}$, choose approximating sets from finite blocks with error tending to zero. Their indicators approach $1_A$ in $L^2$, and $L^2({\mathcal{F}}_0)$ is a closed vector subspace, so $A$ is ${\mathcal{F}}_0$-measurable modulo a null set.

Invertibility now gives ${\mathcal{F}}_1=T^{-1}{\mathcal{F}}_0=T^{-1}\mathcal{B}=\mathcal{B}$ modulo null sets. In particular the present partition $\xi$ is measurable with respect to its entire future, so $H(\xi \mid {\mathcal{F}}_1)=0$. The infinite-future formula gives $h_{\mu }(T,\xi )=0$. Since the given one-sided generator is also a two-sided generator, the Kolmogorov-Sinai generator theorem finishes the proof:This is finite one-sided generator of an invertible system forces zero entropy. Invertibility is essential: a fair binary one-sided Bernoulli shift has a finite one-sided generator and Kolmogorov-Sinai entropy $\log 2$.