Vitali sets of full outer measure

Question

Is there a "standard" construction to get a Vitali set in $\mathbb R$ of full outer Lebesgue measure?

Answer 1

See e.g. https://groups.google.com/forum/?fromgroups=#!topic/sci.math/ofkao7iugNg

Answer 2

Take $V$ a Vitali set invariant under multiplication by rational numbers $V = \mathbb{Q}^{\times} V$. Then the set of fractional parts of elements in $V$ is a Vitali set contained in $[0,1)$ of exterior measure $1$.

You can find such a $V$ in a naive way: first choose a system of representatives $S$ under the action of the group $x \mapsto a x + b$, $a \in \mathbb{Q}^{\times}$, $b\in \mathbb{Q}$; take $V = \bigcup_{q\in \mathbb{Q}^{\times}} \ q\cdot S$. Or you can take $V$ a complement of $\mathbb{Q}$ inside $\mathbb{R}$, that is $\mathbb{R} = V\oplus \mathbb{Q}$.

Proof of the statement:

Since $\bigcup_{t\in \mathbb{Q}}(V + t) = \mathbb{R}$, we have $\mu^{*}(V)>0$. Therfore, for every $\epsilon > 0$ there exists an interval $[\frac{m}{n}, \frac{m+1}{n}]$ such that $\mu^*(V\cap I) > (1-\epsilon) \mu (I)$. Since $V$ is invariant under multiplication by $n$, we have the above inequality for the interval $I=[m,m+1]$. It should be easy now.