# More general Vitali sets

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?

• I have no idea what you are saying. What is $\Omega$? What is a line set? – mathworker21 Jul 18 '18 at 11:34
• $\Omega$ is in this case $\mathbb{R}^n$. Imagine that you have a Vitali set for $\mathbb{R}$ in $\mathbb{R}^2$ this can be located on the x-axis. If you take lines which are orthogonal to the x-axis and intersect with the x-axis in a point of the chosen Vitali set you get what I called a line-set. – Rico1990 Jul 18 '18 at 11:41
• Have you tried picking representatives from $\Bbb{R^n/Q^n}$? – Asaf Karagila Jul 18 '18 at 12:07
• Thank you for your answer. – Rico1990 Jul 19 '18 at 10:14

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.

• Which equivalence relation did you use in your proof? We used groups $I_x \lbrace z \in (0,1) | z - x \in \mathbb{Q} \rbrace$ and chose for each a representative, what gave us the Vitali set. Then we continued like you did. Can you explain why we get the measure between 1 and 3 in the 2-dimensional case? I suppose that this is derived by the product measure, right? – Rico1990 Jul 19 '18 at 10:14
• @Rico1990: Your Vitali set is the same as my $A_1$, just described using (very slightly) different words, and I use the closed unit inverval intstead the open one, which matters not at all. --- In the two-dimensional case we have $[0,1]\times[0,1]\subseteq B_2 \subseteq [-1,2]\times[0,1]$ and those two rectangles have measure $1$ and $3$. This is directly derived from the specification of the Lebesgue measure: The unit square must have measure $1$ and the measure is invariant under translations. (The long rectangle is a sum of three squares). – Henning Makholm Jul 19 '18 at 10:46
• Ok, thank you again. In case further questions appear I'll give a signal. – Rico1990 Jul 19 '18 at 17:05

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.

• Thank you for your answer. – Rico1990 Jul 19 '18 at 10:14
• Can you explain for me why $(V+q)×\Bbb{R}^{n−1}$ is not measurable.? – 129492 Mar 23 at 15:31
• @129492. Do you know what a Vitali set is and why such is not measurable? – md2perpe Mar 23 at 15:38
• Vitali set in $R$ is $[0;1]$ and to show that is not a measurable we define an equivalence relation on S... Right ? I read it in Real_Analysis__Measure_Theory__Integration__and_Hilbert_Spaces__Princeton_Lectures_in_Analysis___Volume_3_.pdf – 129492 Mar 23 at 15:45
• @129492. To construct $V$ we define an equivalence relation on $[0,1)$. To show that $V$ is not measurable, we take a countable union $\bigcup_k V_k$ of translations modulo 1 of $V$ such that the union is all of $[0,1)$. Then, since Lebesgue measure is countably additive and translation invariant, we must have $$1 = m([0,1)) = m(\bigcup_k V_k) = \sum_k m(V_k) = \sum_k m(V) = \infty \times m(V).$$ The last expresion is either $0$ (if $m(V)=0$) or $\infty$ (if $m(V)>0$). Contradiction! – md2perpe Mar 23 at 16:08