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Gauss–Kuzmin–Wirsing operator
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=== 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.
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