Week 9 Worksheet Solutions

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1. Since $$\mu (B\mathrm{△}{T}^{-1}B)=\int |1_B-T{1}_B|\mathsf{d}\mu$$ we see that $1_B$ is fixed by $T$ in ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$. By ergodicity it is constant and therefore $\mu (B)\in \{0,1\}$.
2. If $0\le \varphi \le 1_{X\setminus B}\cdot |f|$ is a simple functions then we must have $$\int \varphi \mathsf{d}\mu =0$$ and therefore $$\int 1_{X\setminus B}\cdot |f|\mathsf{d}\mu =0$$ by the definition of integration. Then $$|\int f\mathsf{d}\mu -\int 1_B\cdot f\mathsf{d}\mu |\le \int 1_{X\setminus B}\cdot |f|\mathsf{d}\mu =0$$ as needed.
3. 1. Define $T(x)=x+\alpha mod1$ on $[0,1)$. It is measure-preserving and ergodic with respect to Lebesgue measure on $[0,1)$. The pointwise ergodic theorem provides a set $\mathrm{\Omega }\subset [0,1)$ with $\mu (\mathrm{\Omega })=1$ such that $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(x+n\alpha mod1)=\int f\mathsf{d}\lambda =0$$ for all $x\in \mathrm{\Omega }$.
   2. Fix $x\in [0,1]$. Since $x+n\alpha$ can only be rational for one value of $n$ we have $$f(x+n\alpha mod1)=0$$ 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 $$\begin{array}{rl}& |\{1\le n\le N:x(n)+x(n+1)\ge 2x(n+2)\}|\\ =& \sum _{n=0}^{N-1}{1}_{[11]\cup T^{-2}[0]}(T^n(x))\end{array}$$ and the pointwise ergodic theorem using the fair coin measure gives $$\begin{array}{rl}& \underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{|\{1\le n\le N:x(n)+x(n+1)\ge 2x(n+2)\}|}{N}}\\ =& \int 1_{[11]\cup T^{-2}[0]}\mathsf{d}\mu \\ =& \mu ([11])+\mu (T^{-2}[0])-\mu ([110])\\ =& \frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{5}{8}\end{array}$$ 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 $k\in \mathbb{N}$ a set ${\mathrm{\Omega }}_k\subset X$ with $\mu ({\mathrm{\Omega }}_k)=1$ and $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{f}_k(T^{n}x)=\int f_k\mathsf{d}\mu$$ for all $x\in {\mathrm{\Omega }}_k$. Put $$\mathrm{\Omega }=\bigcap _{k\in \mathbb{N}}{\mathrm{\Omega }}_k$$ and fix $x\in \mathrm{\Omega }$. Fix $ϵ>0$. Fix also $g:X\to \mathbb{R}$ continuous. There is $k\in \mathbb{N}$ with $$f_k(y)-ϵ\le g(y)\le f_k(y)+ϵ$$ for all $y\in X$ and therefore $$\int f_k\mathsf{d}\mu -ϵ=\underset{N\to \mathrm{\infty }}{lim}\sum _{n=0}^{N-1}{f}_k(T^{n}x)-ϵ\le \underset{N\to \mathrm{\infty }}{lim inf}\sum _{n=0}^{N-1}g(T^{n}x)$$ and $$\underset{N\to \mathrm{\infty }}{lim sup}\sum _{n=0}^{N-1}g(T^{n}x)\le \underset{N\to \mathrm{\infty }}{lim}\sum _{n=0}^{N-1}{f}_k(T^{n}x)+ϵ=\int f_k\mathsf{d}\mu +ϵ$$ because $x\in {\mathrm{\Omega }}_k$. Since $\Vert g-f_k{\Vert }_{\mathsf{u}}\le ϵ$ we also have $$|\int f_k-g\mathsf{d}\mu |\le ϵ$$ giving $$\begin{array}{rl}\int g\mathsf{d}\mu -2ϵ& \le \underset{N\to \mathrm{\infty }}{lim inf}\sum _{n=0}^{N-1}g(T^{n}x)\\ & \le \underset{N\to \mathrm{\infty }}{lim sup}\sum _{n=0}^{N-1}g(T^{n}x)\le \int g\mathsf{d}\mu +2ϵ\end{array}$$ and therefore $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}g(T^n(x))=\int g\mathsf{d}\mu$$ for all $x\in \mathrm{\Omega }$ since $ϵ>0$ was arbitrary. As a countable intersection of sets with full measure, the set $\mathrm{\Omega }$ itself has full measure.