Week 2 Tutorial Solutions

Measure

1. 1. Since the emptyset has cardinality zero we have $\mu (\varnothing )=0$. 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 $${\displaystyle \frac{|B_{r(1)}|+\cdots +|B_{r(w)}|}{|G|}}={\displaystyle \frac{|B_{r(1)}\cup \cdots \cup B_{r(w)}|}{|G|}}$$ which gives $$\mu \left(\bigcup _{n\in \mathbb{N}}{B}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (B_n)$$ as desired.
   2. If $H<G$ then $$\mu (H)={\displaystyle \frac{|H|}{|G|}}={\displaystyle \frac{1}{[G:H]}}$$ 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 $\nu (E)=\mathrm{\infty }$ for all non-empty $E\subset G$ or $\nu$ is a scaling of $\mu$. There are no other invariant meausres on $G$.
2. 1. $\nu (\mathbb{N})=1$ and we have $${\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2N}}\le {\displaystyle \frac{|\{2\mathbb{N}\cap \{1,\dots ,N\}|}{N}}\le {\displaystyle \frac{1}{2}}$$ for all $N\in \mathbb{N}$ so $\nu (2\mathbb{N})=1/2$. Since $${\displaystyle \frac{|\{\{n^2:n\in \mathbb{N}\}\cap \{1,\dots ,N\}|}{N}}\le {\displaystyle \frac{2\sqrt{N}}{N}}$$ for all $N\in \mathbb{N}$ we have $\nu (\{n^2:n\in \mathbb{N}\})=0$.
   2. $\nu$ is not countable additive, because $\nu (\{n\})=0$ for all $n\in \mathbb{N}$ but $\nu (\mathbb{N})=1$.
   3. Since $\mathbb{N}$ is a countable union of its singletons every measure $\mu$ on $\mathbb{N}$ is determined by its values on each of the sets $\{n\}$. We can write $$\mu =\sum _{n\in \mathbb{N}}\mu (\{n\}){\delta }_n$$ no matter which measure on $\mathbb{N}$ we have started with. Note this is not true more generally: Lebesgue measure $\lambda$ satisfies $\lambda (\{t\})=0$ for every $t\in \mathbb{R}$ but we expect that $\lambda ([0,1])=1$ i.e. $\lambda$ is not determined by its values on the singleton sets.