ex. (not exactly vitali but similarly non-Lebesgue) suppose $Θ$ is an irrational (i.e. $\Theta/{2\pi}$ is not rational) angle like $\pi \sqrt 2$.

define a graph $\mathcal G$ on the unit circle $S^1$ by $x,y$ are adjacent if one is the other rotated by $\Theta$

then the connected components of $\mathcal G$ are bi-infinite line graphs with countably many vertices

since these components have countably many, but whole graph has uncountably many vertices, then there must be uncountably many components

intuitively, there should be a 2-coloring of each component and thus the whole graph

claim: $\chi(G)=2$

proof. from each component of $\mathcal G$, choose a point to be red. alternate from there.

claim. let $A=\{x:c(x)=R\}$, $B=\{x:c(x)=B\} =V(\mathcal G)\backslash A$. $A$ and $B$ are non-Lebesgue measurable.

proof. otherwise, say $A$ and $B$ are measurable. notice that rotating $A$ by $\Theta$ gives $B$. this is a measure-preserving operation. also

$$ m(A)+m(B)=m(S^1)=1 $$

so $m(A)=m(B)=\frac 12$.

rotating $A$ by $2\Theta$, we get $A$

rotation by $2\Theta$ is “ergodic”, so the measure of $A$ must be either $0$ or $1$.

contradiction.