> [!definition] > > Let $(\Omega, \cf, \bp, T)$ be a [[Measure System|measure system]], then the system is **ergodic** if every $T$-invariant set is [[Null Set|null]] or co-null. > [!theorem] > > Let $(\Omega, \cf, \bp, T)$ be a measure system, then $T$ is ergodic if and only if for all $E \in \cf$ with $\bp\bracs{E} > 0$, its saturation with respect to the orbit equivalence relation is co-null. > [!theorem] > > Let $(Y, \cb(Y))$ be a standard Borel space, then $T$ is ergodic if and only every $T$-invariant function is constant a.s. > > *Proof*. Suppose that $T$ is ergodic, and let $f: \Omega \to \bracs{0, 1}^\nat$, then $f^{-1}(E)$ is null or co-null for every Borel set $E \in \cb(Y)$. Use the 0-1 law argument. > [!definition] > > An ergodic system is **mixing** if for any $E, F \in \cf$, > $ > \limv{n}\bp\bracs{E \cap T^nF} = \bp\bracs{E} \cdot \bp\bracs{F} > $ > [!theorem] > > Let $\Omega = \real/\integer$ equipped with the [[Lebesgue Measure|Lebesgue measure]], and $\alpha \in \Omega \setminus \rational$. Let $T_\alpha: \Omega \to \Omega$ be the translation map by $\alpha$, then $(\Omega, T_\alpha)$ is an ergodic measure system. > > *Proof*. Let $E \subset \Omega$ such that $m(E) > 0$ and suppose that $E$ is $T_\alpha$-invariant. Let $\eps > 0$, $I \subset \Omega$ be intervals such that $\frac{m(I \cap E)}{m(I)} > 1 - \eps$, then $\frac{m(T_\alpha^n I \cap T_\alpha^n E)}{m(T_\alpha^n I)} > 1 - \eps$ for all $n \in \nat$. In addition, use truncation to assume that $m(I) \le \eps$. Using irrational rotations, cover $\Omega$ with disjoint copies of $I$, with measure adding up to $1 - \eps$, then $m(E) \ge (1 - \eps)^2$. > [!definition] > > Let $\Sigma$ be a finite set, $\pi$ be a probability measure on $\Sigma$, and $S$ be the left shift operator > $ > S: \Sigma^\integer \to \Sigma^\integer \quad S(x)(n) = x(n+1) > $ > then the system $(\Sigma^\integer, \pi^\integer, S)$ is an ergodic system, known as the **Bernoulli shift**. Two Bernoulli shifts are isomorphic if and only if they have the same [[Entropy|entropy]]. > [!theorem] > > Let $\Sigma = \bracs{0, 1}$, $\pi$ be the uniform measure on $\Sigma$, then there exists $T: [0, 1]^2 \to [0, 1]^2$ such that $(\Sigma^\integer, \pi^\integer, S)$ is isomorphic to $([0, 1]^2, m, T)$. > > *Proof*. Define a mapping > $ > f: \Sigma^\integer \to [0, 1]^2 \quad x \mapsto \braks{\sum_{n = 0}^\infty 2^{-n}x_n, \sum_{n = 1}^\infty 2^{-n+1}x_{-n}} > $ > then $f \circ S \circ f^{-1}$ is a measure preserving map on the cylinder sets. By [[Dynkin's Uniqueness Theorem]], $f \circ S \circ f^{-1}$ is measure preserving. > [!theorem] > > Let $\Sigma = \bracs{0, 1}$, $\pi$ be the uniform measure, and $S$ be the Bernoulli shift, then the system $(\Sigma^\nat, \pi^\nat, S)$ is mixing. > > *Proof*. Let $A, B \subset \Sigma$ be measurable. Assume that $A = \bigcap_{j = 1}^n \pi_j^{-1}(a_j)$, $B = \bigcap_{j = 1}^m \pi_j^{-1}(b_j)$, and identify them as $\seqf{a_j}$, $\{b_j\}_1^m$. Let $k \ge m, n$, then $\bp\bracs{A \cap T^kB} = \bp\bracs{A} \cdot \bp\bracs{B}$.