\[ \newcommand{\C}{\mathbb{C}} \newcommand{\haar}{\mathsf{m}} \newcommand{\P}{\mathcal{P}} \newcommand{\R}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\d}{\mathsf{d}} \newcommand{\g}{>} \newcommand{\l}{<} \newcommand{\intd}{\,\mathsf{d}} \newcommand{\Re}{\mathsf{Re}} \newcommand{\area}{\mathop{\mathsf{Area}}} \newcommand{\met}{\mathop{\mathsf{d}}} \newcommand{\emptyset}{\varnothing} \]

Week 1 Tutorial Solutions

  1. $C_2 = [\tfrac{0}{9},\tfrac{1}{9}] \cup [\tfrac{2}{9},\tfrac{3}{9}] \cup [\tfrac{6}{9},\tfrac{7}{9}] \cup [\tfrac{8}{9},\tfrac{9}{9}]$ $C_2 = [\tfrac{0}{27},\tfrac{1}{27}] \cup [\tfrac{2}{27},\tfrac{3}{27}] \cup [\tfrac{6}{27},\tfrac{7}{27}] \cup [\tfrac{8}{27},\tfrac{9}{27}]$ $\qquad\cup\, [\tfrac{18}{27},\tfrac{19}{27}] \cup [\tfrac{20}{27},\tfrac{21}{27}] \cup [\tfrac{24}{27},\tfrac{25}{27}] \cup [\tfrac{26}{27},\tfrac{27}{27}]$
  2. Every closed interval $[a,b]$ is a Borel set. Each set $C_n$ is a finite union of closed intervals and therefore also a Borel set. Finally $\mathcal{C}$ is a countable intersection of the $C_n$ and thus also a Borel set.
  3. We will prove that $x$ belongs to $\mathcal{C}$ if and only if it can be written in the stated form with $x(n) \in \{0,2\}$ for all $n \in \N$. First, suppose that $x$ belongs to $\mathcal{C}$. Define $d(1)$ to be $0$ or $2$ according to which of the intervals in $C_1$ the points $x$ belongs. Thus define \[ d(1) = \begin{cases} 0 & 0 \le x \le \tfrac{1}{3} \\ 2 & \tfrac{2}{3} \le x \le 1 \end{cases} \] and put $y(2) = 3x - d(1)$. Inductively, define \[ d(n+1) = \begin{cases} 0 & 0 \le y(n+1) \le \tfrac{1}{3} \\ 2 & \tfrac{2}{3} \le y(n+1) \le 1 \end{cases} \] and $y(n+1) = 3y(n) - d(n)$ for all $n \in \N$. By induction $x$ and \[ \sum_{n=1}^N \dfrac{d(n)}{3^n} \] are in the same closed subinterval of $C_N$ for every $N \in \N$ so we must have \[ x = \sum_{n=1}^\infty \dfrac{d(n)}{3^n} \] and $x$ has the desired representation.

    Conversely, if \[ x = \sum_{n=1}^\infty \dfrac{d(n)}{3^n} \] and every $d(n)$ belongs to $\{0,2\}$ then each of the truncations \[ \sum_{n=1}^N \dfrac{d(n)}{3^n} \] belongs to $C_N$ and therefore the limit must belong to $\mathcal{C}$.
  4. Define $\phi : \{0,1\}^\N$ by \[ \phi(\alpha) = \sum_{n=1}^\infty \dfrac{2\alpha(n)}{3^n} \] for every $\alpha \in \{0,1\}^\N$. By the previous question $\phi$ is surjective. It is injective because if $\alpha,\beta$ in $\{0,1\}^\N$ are not equal then there is $N \in \N$ minimal with $\alpha(N) \ne \beta(N)$. We must then have $\phi(\alpha)$ and $\phi(\beta)$ is different sub-intervals of $C_N$ and $\phi$ is therefore injective.
  5. We have $\mathcal{C} \subset C_n$ for all $n \in \N$ and therefore $\Lambda(\mathcal{C}) \le \Lambda(C_n)$ for all $n \in \N$. As $C_n$ is a union of $2^n$ disjoint intervals of length $3^{-n}$ we can say (using Problem 3 from Worksheet 1) that $\Lambda(C_n) \le 2^n/3^n$. Thus $\Lambda(\mathcal{C} \le (2/3)^n$. Since $n \in \N$ was arbitrary we must have $\mathcal{C} = 0$.