Week 9 Worksheet Solutions

Home | Assessment | Notes | Worksheets | Tutorials

1. Since \[ \mu(B \triangle T^{-1} B) = \int | 1_B - T 1_B | \intd \mu \] we see that $1_B$ is fixed by $T$ in $\mathsf{L}^{\!\mathsf{2}}(X,\B,\mu)$. By ergodicity it is constant and therefore $\mu(B) \in \{0,1\}$.
2. If $0 \le \phi \le 1_{X \setminus B} \cdot |f|$ is a simple functions then we must have \[ \int \phi \intd \mu = 0 \] and therefore \[ \int 1_{X \setminus B} \cdot |f| \intd \mu = 0 \] by the definition of integration. Then \[ \left| \int f \intd \mu - \int 1_B \cdot f \intd \mu \right| \le \int 1_{X \setminus B} \cdot |f| \intd \mu = 0 \] as needed.
3. 1. Define $T(x) = x + \alpha \bmod 1$ on $[0,1)$. It is measure-preserving and ergodic with respect to Lebesgue measure on $[0,1)$. The pointwise ergodic theorem provides a set $\Omega \subset [0,1)$ with $\mu(\Omega) = 1$ such that \[ \lim_{N \to \infty} \dfrac{1}{N} \sum_{n=0}^{N-1} f(x + n \alpha \bmod 1) = \int f \intd \lambda = 0 \] for all $x \in \Omega$.
   2. Fix $x \in [0,1]$. Since $x + n \alpha$ can only be rational for one value of $n$ we have \[ f(x + n \alpha \bmod 1) = 0 \] for all but possibly one value of $n$. The average of these values is therefore zero.
   3. No: we cannot apply our uniform distribution result to $f$ as it is not continuous. (In fact $f$ is not even Riemann integratble.)
4. We have $x(n) + x(n+1) \ge 2x(n+2)$ if and only if either $x(n+2) = 0$ or $x(n) = 1 = x(n+1)$. Thus \[ \begin{align*} & | \{ 1 \le n \le N : x(n) + x(n+1) \ge 2x(n+2) \} | \\ = {}& \sum_{n=0}^{N-1} 1_{[11] \cup T^{-2} [0]}(T^n(x)) \end{align*} \] and the pointwise ergodic theorem using the fair coin measure gives \[ \begin{align*} & \lim_{N \to \infty} \dfrac{| \{ 1 \le n \le N : x(n) + x(n+1) \ge 2x(n+2) \} |}{N} \\ = {}& \int 1_{[11] \cup T^{-2} [0]} \intd \mu \\ = {}& \mu([11]) + \mu(T^{-2}[0]) - \mu([110]) \\ = {}& \tfrac{1}{4} + \tfrac{1}{2} - \tfrac{1}{8} = \tfrac{5}{8} \end{align*} \] which is reasonable as five of the eight strings of length 3 over $\{0,1\}$ have the desired property.
5. By the Stone-Weierstrass theorem there is a sequence $f_1,f_2,\dots$ of continuous functions on $X$ that is dense uniformly among all continuous functions on $X$. By the pointwise ergodic theorem there is for each $k \in \N$ a set $\Omega_k \subset X$ with $\mu(\Omega_k) = 1$ and \[ \lim_{N \to \infty} \dfrac{1}{N} \sum_{n=0}^{N-1} f_k(T^n x) = \int f_k \intd \mu \] for all $x \in \Omega_k$. Put \[ \Omega = \bigcap_{k \in \N} \Omega_k \] and fix $x \in \Omega$. Fix $\epsilon \g 0$. Fix also $g : X \to \R$ continuous. There is $k \in \N$ with \[ f_k(y) - \epsilon \le g(y) \le f_k(y) + \epsilon \] for all $y \in X$ and therefore \[ \int f_k \intd \mu - \epsilon = \lim_{N \to \infty} \sum_{n=0}^{N-1} f_k(T^n x) - \epsilon \le \liminf_{N \to \infty} \sum_{n=0}^{N-1} g(T^n x) \] and \[ \limsup_{N \to \infty} \sum_{n=0}^{N-1} g(T^n x) \le \lim_{N \to \infty} \sum_{n=0}^{N-1} f_k(T^n x) + \epsilon = \int f_k \intd \mu + \epsilon \] because $x \in \Omega_k$. Since $\| g - f_k \|_\mathsf{u} \le \epsilon$ we also have \[ \left| \int f_k - g \intd \mu \right| \le \epsilon \] giving \[ \begin{align*} \int g \intd \mu - 2 \epsilon &{} \le \liminf_{N \to \infty} \sum_{n=0}^{N-1} g(T^n x) \\ &{} \le \limsup_{N \to \infty} \sum_{n=0}^{N-1} g(T^n x) \le \int g \intd \mu + 2 \epsilon \end{align*} \] and therefore \[ \lim_{N \to \infty} \dfrac{1}{N} \sum_{n=0}^{N-1} g(T^n (x)) = \int g \intd \mu \] for all $x \in \Omega$ since $\epsilon \g 0$ was arbitrary. As a countable intersection of sets with full measure, the set $\Omega$ itself has full measure.