Example 11.1.6 (Vitali-ish).label Let $I=[0,1]$ and define an equivalence relation $\sim$ on $I$ by $x\sim y\iff x-y\in\Q$ for all $x,y\in I$. Set $E_{x}=[x]/\cups{}{y\neq x}[y]$ then $I=\cups{}{x\in I}E_{x}$. Let $N\suf [0,1]$ be a set containing precisely one element for each equivalence class which exists by AC. Set $R=Q\cap [0,1]$ and for each $r\in R$ we define $N_{r}=\curl{x+r:x\in N\cap [0,1-r)}\cup \curl{x+r-1:x\in N\cap [1-r,1]}$. By translation invariance and disjoint additivity, $\mu(N_{r})=\mu(N)$ for each $r\in R$. Let $x\in[0,1]$, if $y\in N$ with $y\in [x]$ then $x\in N_{r}$ where $r= \begin{cases}r=x-y&x\geq y\\ r=x-y+1&x<y\end{cases}$. Suppose $x\in N_{r}\cap N_{s}$ for $r\neq s\in R$ then $x-r,x-s$ are disjoint elements of $N$ (one can take $x-r+1$ instead of $x-r$ to land in $I$). This shows each $x\in [0,1]$ belongs to a unique $N_{r}$. Since $N=(N\cap [0,1-r))\sqcup (N\cap [1-r,1])$ we have $\mu(N)=\mu(N_{r})$. Furthermore $[0,1]=\dcups{}{r\in R}N_{r}$ hence $1=\mu([0,1])=\sums{}{r\in R}\mu(N_{r})=\sums{}{r\in R}\mu(N)$