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

Week 2 Tutorial Solutions

    1. Since the emptyset has cardinality zero we have $\mu(\emptyset) = 0$. Fix $B_1,B_2,\dots$ subsets of $G$ that are pairwise disjoint. As $G$ is finite only finitely many $B_n$ say \[ B_{r(1)},\dots,B_{r(w)} \] are non-empty and we have \[ \dfrac{|B_{r(1)}| + \cdots + |B_{r(w)}|}{|G|} = \dfrac{|B_{r(1)} \cup \cdots \cup B_{r(w)}|}{|G|} \] which gives \[ \mu \left( \, \bigcup_{n \in \N} B_n \right) = \sum_{n=1}^\infty \mu(B_n) \] as desired.
    2. If $H \l G$ then \[ \mu(H) = \dfrac{|H|}{|G|} = \dfrac{1}{[G:H]} \] where $[G:H]$ is the index of $H$.
    3. For any $E \subset G$ and any $g \in G$ the sets $g^{-1} E$ and $E$ have the same cardinality because left multiplication in a group is a bijection. Thus $\mu(g^{-1} E) = \mu(E)$ for all $g \in G$ and all $E \subset G$.
    4. Fix an invariant measure $\nu$ on $G$. Then \[ \nu(\{g\}) = \nu(\{1\}) \] for all $g \in G$ because $\nu$ is invariant. But this means that \[ \nu(E) = \nu(\{1\}) |E| \] for all $E \subset G$. Thus either $\nu(E) = 0$ for all $E \subset G$ or $\nu(E) = \infty$ for all non-empty $E \subset G$ or $\nu$ is a scaling of $\mu$. There are no other invariant meausres on $G$.
    1. $\nu(\N) = 1$ and we have \[ \dfrac{1}{2} - \dfrac{1}{2N} \le \dfrac{|\{ 2\N \cap \{1,\dots,N\}|}{N} \le \dfrac{1}{2} \] for all $N \in \N$ so $\nu(2\N) = 1/2$. Since \[ \dfrac{|\{ \{ n^2 : n \in \N \} \cap \{1,\dots,N\}|}{N} \le \dfrac{2 \sqrt{N}}{N} \] for all $N \in \N$ we have $\nu(\{ n^2 : n \in \N \}) = 0$.
    2. $\nu$ is not countable additive, because $\nu(\{n\}) = 0$ for all $n \in \N$ but $\nu(\N) = 1$.
    3. Since $\N$ is a countable union of its singletons every measure $\mu$ on $\N$ is determined by its values on each of the sets $\{n\}$. We can write \[ \mu = \sum_{n \in \N} \mu(\{n\}) \delta_n \] no matter which measure on $\N$ we have started with. Note this is not true more generally: Lebesgue measure $\lambda$ satisfies $\lambda(\{t\}) = 0$ for every $t \in \R$ but we expect that $\lambda([0,1]) = 1$ i.e. $\lambda$ is not determined by its values on the singleton sets.