More general Vitali sets

Question

Let $\mu^n$ be the n-dimensional Lebesgue measure. I want to show that the transformation $\mu^n: \mathcal{P}(\Omega)\rightarrow [0,\infty]$ doesn't exist. In other word I want to use Vitali sets to demonstrate that there are sets which aren't measurable.

I spoke with a friend of mine who said that we might use the Vitali sets for $\mathbb{R}$ we got in a proof and just attach to them a line in so that we get a similar set in $\mathbb{R}^2$.

My first question would be: How can I show that this new "line"-set isn't measurable?

When we go further then we might deduce for a given dimension n that we attach to the points of a Viatli set lines and have sets which aren't measurable. Is this correct?

Answer 1

For concreteness, let's first sketch a Vitali-based proof that not all sets of reals are measurable:

Let $A_1$ be a subset of $[0,1]$ that contains exactly one representative for each equivalence class in $\mathbb R/\mathbb Q$. The set $$ B_1 = \bigcup_{q\in\mathbb Q\cap[-1,1]} (A_1+q) $$ then satisfies $$ [0,1] \subseteq B_1 \subseteq [-1,2] $$ so if it is measurable its measure must be between $1$ and $3$. But it is a disjoint union of countably many translated copies of $A_1$. This means that $A_1$ cannot have measure $0$ (because then $B_1$ would have measure $0$ too), nor can it have measure $>0$ (because then $B_1$ would have infinite measure). So $A_1$ is not measurable.

In two dimensions you can simply set $$ A_2 = A_1 \times [0,1]$$ $$ B_2 = \bigcup_{q\in\mathbb Q\cap[-1,1]} (A_2+\langle q,0\rangle) = B_1 \times [0,1] $$ and then repeat the same argument: $B_2$ should have measure between $1$ and $3$, but that cannot be a countably infinite sum of identical terms.

The generalization to higher dimensions should now be clear.

Answer 2

Let $V$ be a Vitali set and consider $V \times \mathbb R^{n-1}$. Then $$\mathbb R^n = \bigcup_{q \in \mathbb Q} (V+q) \times \mathbb R^{n-1}$$ and it's obvious that each $(V+q) \times \mathbb R^{n-1}$ is not measurable.