Editing Gauss–Kuzmin–Wirsing operator (section)

== Ergodicity ==
The Gauss map is in fact much more than ergodic: it is exponentially mixing,<ref>{{Cite journal |date=2004-03-30 |title=Kuzmin, coupling, cones, and exponential mixing |url=https://www.degruyter.com/document/doi/10.1515/form.2004.021/html |language=en |volume=16 |issue=3 |pages=447–457 |doi=10.1515/form.2004.021 |issn=1435-5337 |last1=Zweimüller |first1=Roland |journal=Forum Mathematicum |url-access=subscription }}</ref><ref>{{Citation |last=Pollicott |first=Mark |title=Exponential Mixing: Lectures from Mumbai |date=2019 |url=https://doi.org/10.1007/978-981-15-0683-3_4 |work=Geometric and Ergodic Aspects of Group Actions |pages=135–167 |editor-last=Dani |editor-first=S. G. |access-date=2024-01-13 |series=Infosys Science Foundation Series |place=Singapore |publisher=Springer |language=en |doi=10.1007/978-981-15-0683-3_4 |isbn=978-981-15-0683-3 |s2cid=214272613 |editor2-last=Ghosh |editor2-first=Anish|url-access=subscription }}</ref> but the proof is not elementary.

=== Entropy ===
The Gauss map, over the Gauss measure, has entropy <math>\frac{\pi^2}{6\ln 2} </math>. This can be proved by the Rokhlin formula for entropy. Then using the [[Shannon–McMillan–Breiman theorem]], with its equipartition property, we obtain [[Lochs's theorem|Lochs' theorem]].<ref>''[https://web.archive.org/web/20240117051216/https://www.mat.univie.ac.at/~bruin/ET1_lect15.pdf The Shannon-McMillan-Breiman Theorem]''</ref>

=== Measure-theoretic preliminaries ===
A covering family <math>\mathcal C</math> is a set of measurable sets, such that any open set is a ''disjoint'' union of sets in it. Compare this with [[Base (topology)|base in topology]], which is less restrictive as it allows non-disjoint unions.

'''Knopp's lemma.''' Let <math>B \subset [0, 1)</math> be measurable, let <math>\mathcal C</math> be a covering family and suppose that <math>\exists \gamma > 0, \forall A \in \mathcal C, \mu(A \cap B) \geq \gamma \mu(A)</math>. Then <math>\mu(B) = 1</math>.

'''Proof.''' Since any open set is a disjoint union of sets in <math>\mathcal C</math>, we have <math>\mu(A \cap B) \geq \gamma \mu(A)</math> for any open set <math>A</math>, not just any set in <math>\mathcal C</math>.

Take the complement <math>B^c</math>. Since the Lebesgue measure is [[Regular measure|outer regular]], we can take an open set <math>B'</math> that is close to <math>B^c</math>, meaning the symmetric difference has arbitrarily small measure <math>\mu(B' \Delta B^c) < \epsilon</math>.

At the limit, <math>\mu(B' \cap B) \geq \gamma \mu(B')</math> becomes have <math>0 \geq \gamma \mu(B^c)</math>.

=== The Gauss map is ergodic ===
Fix a sequence <math>a_1, \dots, a_n</math> of positive integers. Let <math>\frac{q_n}{p_n} = [0;a_1, \dots, a_n]</math>. Let the interval <math>\Delta_n</math> be the open interval with end-points <math>[0;a_1, \dots, a_n], [0;a_1, \dots, a_n+1]</math>. 

'''Lemma.''' For any open interval <math>(a, b) \subset (0, 1)</math>, we have<math display="block">\mu(T^{-n}(a,b) \cap \Delta_n) = \mu((a,b)) \mu(\Delta_n) \underbrace{\left(\frac{q_n(q_n + q_{n-1})}{(q_n + q_{n-1}b)(q_n + q_{n-1}a) } \right)}_{\geq 1/2} </math>'''Proof.''' For any <math>t \in (0, 1)</math> we have <math>[0;a_1, \dots, a_n + t] = \frac{q_n + q_{n-1}t}{p_n + p_{n-1}t}</math> by [[Continued fraction#Some useful theorems|standard continued fraction theory]]. By expanding the definition, <math>T^{-n}(a,b) \cap \Delta_n</math> is an interval with end points <math>[0;a_1, \dots, a_n + a], [0;a_1, \dots, a_n+ b]</math>. Now compute directly. To show the fraction is <math>\geq 1/2</math>, use the fact that <math>q_n \geq q_{n-1}</math>.

'''Theorem.''' The Gauss map is ergodic.

'''Proof.''' Consider the set of all open intervals in the form <math>([0;a_1, \dots, a_n], [0;a_1, \dots, a_n+1])</math>. Collect them into a single family <math>\mathcal C</math>. This <math>\mathcal C</math> is a covering family, because any open interval <math>(a, b)\setminus \Q</math> where <math>a, b</math> are rational, is a disjoint union of finitely many sets in <math>\mathcal C</math>.

Suppose a set <math>B</math> is <math>T</math>-invariant and has positive measure. Pick any <math>\Delta_n \in \mathcal C</math>. Since Lebesgue measure is outer regular, there exists an open set <math>B_0</math> which differs from <math>B</math> by only <math>\mu(B_0 \Delta B) < \epsilon</math>. Since <math>B</math> is <math>T</math>-invariant, we also have <math>\mu(T^{-n}B_0 \Delta B) = \mu(B_0 \Delta B) < \epsilon</math>. Therefore, <math display="block">\mu(T^{-n}B_0 \cap \Delta_n) \in \mu(B\cap \Delta_n) \pm \epsilon</math>By the previous lemma, we have<math display="block">\mu(T^{-n}B_0 \cap \Delta_n) \geq \frac 12 \mu(B_0) \mu(\Delta_n) \in \frac 12 (\mu(B) \pm \epsilon) \mu(\Delta_n) </math>Take the <math>\epsilon \to 0</math> limit, we have <math>\mu(B \cap \Delta_n) \geq \frac 12 \mu(B) \mu(\Delta_n)</math>. By Knopp's lemma, it has full measure.