I am studying the Lebesgue measure. I read about the Vitali set in $\Bbb{R}$. Can we use Vitali sets for $\Bbb{R}$ to construct a Vitali set in $\Bbb{R} ^k$?
I know that if we let $V$ be a Vitali set in $\Bbb{R}$ then
$$\mathbb R^n = \bigcup_{q \in \mathbb Q} (V+q) \times \mathbb R^{n-1}$$
But I am stuck here. Please help me to continue my idea.
The procedure, including proving the set is non-measurable, is only slightly different from the $1$-dimensional case. Call $x,\,y\in\Bbb R^n$ equivalent if $x-y\in\Bbb Q^n$. By the axiom of choice, a set $V$ exists containing exactly one element of each equivalence class's intersection with $[0,\,1]^n$. Fix an enumeration $q_k$ of $([-1,\,1]\cap\Bbb Q)^n$, then define $V_k:=V+q_k$. You can easily verify the $V_k$ are pairwise disjoint and $([0,\,1]\cap\Bbb Q)^n\subseteq\bigcup_kV_k\subseteq([-1,\,2]\cap\Bbb Q)^n$. At last we reach the desired contradiction: if $V$ has measure $\mu$, then $1\le\sum_{k=1}^\infty\mu\le 3^n$.