# Construction of an non-measurable set

The statement of my problem is as follows: An equivalence relation on $\mathbb{R}^N$ is defined by $x \sim y$ if $x-y \in \mathbb{Q}^N$, so that the equivalence classes are defined by $[x] = x + \mathbb{Q}^N$. By the Axiom of Choice we can construct the following set C: Let B denote the unit ball in $\mathbb{R}^N.$ For each equivalence class $[x]$ with $[x] \cap B \neq \emptyset$, choose exactly one $y \in [x] \cap B$ to be in C. Let $I = 2B \cap \mathbb{Q}^N.$

i) Show the family of sets $(q+C)_{q \in I}$ is pairwise disjoint.

ii) Show $B \subseteq \displaystyle\bigcup_{q\in I} (q+C) \subseteq 3B$

iii) Show that $m^*(C) > 0$ and deduce that $C$ is not measurable. (Here $m^*$ is the $N$-dimensional Lebesgue outer measure).

I am unsure how to approach i) at all.

For ii) I have no idea how to show $B\subseteq \displaystyle\bigcup_{q\in I} (q+C).$ For the second part of ii) since $I \subseteq 2B$ and $C \subseteq B$ I can see why it should hold, but don't know to justify it rigorously.

Here is my attempt for iii):

Suppose $m^*(C) = 0.$ By ii) $B\subseteq \displaystyle\bigcup_{q\in I} (q+C)$ so $m^*(B) \leq m^*\left(\displaystyle\bigcup_{q\in I} (q+C)\right)$. Since $I \subseteq \mathbb{Q}^N$ is countable, by countable sub-additivity of outer measures $\displaystyle m^*\left(\bigcup_{q\in I} (q+C)\right) = \sum_{q\in I} m^*(q+C)$. By translation invariance of the Lebesgue outer measure $m^*(q+C) = m^*(C)=0$ so $\displaystyle m^*\left(\bigcup_{q\in I} (q+C)\right) = 0$ implying $m^*(B) =0$, which is clearly false. Thus $m^*(C) > 0.$

Now suppose C is measurable, so $m^*(C)= m(C)$. By translation invariance, $q+C$ is measurable with $m(q+C) = m(C)$.

From ii), we deduce $m^*\left(\displaystyle\bigcup_{q\in I} (q+C)\right) \leq m^*(3B)$. By part i), and Countable additivity of disjoint sets, $\displaystyle m^*\left(\bigcup_{q\in I} (q+C)\right)= \sum_{q\in I} m(C) = \infty$ since $m(C) > 0.$ But it is easy to show $m^*(3B) < \infty$, which is again a contradiction.

I am relatively sure by working for iii) is correct. If someone could check it, and also give me help on how to proceed for parts i) and ii), it would be very helpful. Thank you.

-
One usually says non-measurable, not immeasurable. – Mariano Suárez-Alvarez Oct 9 '11 at 17:44

A hint for i):

By contradiction. Assume they are not pairwise disjoint. Then $\exists q_1 , q_2 \in Q$ such that $x \in (q_1 + C) \cap (q_2 + C)$. Then $q_1 + c = x = q_2 + c^\prime$. So $q_1 - q_2 = c^\prime - c$.

Edit $C$ is a set of representatives of equivalence classes. Two elements are equivalent if they differ by a rational number. So no two elements in $C$ differ by a rational because otherwise they would be in the same equivalence class and therefore be the same element in $C$.

To finish the argument from above: the thing on the left hand side is a rational (because $\mathbb{Q}$ is a group) and the thing on the right hand side is not a rational because if it was, you would have $c = c^\prime$ and therefore $q_1 = q_2$. But we assumed that $q_1 + C$ and $q_2 + C$ were two different equivalence classes, i.e. we assumed $q_1 \neq q_2$.

Now you have a contradiction because "something rational" $=$ "something irrational", so your assumption that they share an element must have been wrong.

Edit 2 Regarding ii):

To argue for the second $\subset$, note that $C \subset B = [-1,1]$. The smallest $q \in I = 2B \cap \mathbb{Q}$ is $-2$, the largest is $2$. So it's save to assume that the smallest element $q + c$ for $c \in C$ and $q \in B \cap \mathbb{Q}$ is $-2 - 1 = -3$ and the largest is $2 + 1 = 3$. Therefore $C \subset [-3, 3]$. The same argument for $N$ dimensions.

Here I'm not sure about $2 B \cap \mathbb{Q}^N$. I think it should be $B \cap \mathbb{Q}^N$.

Edit 3 To argue for the first inclusion pick an $x \in B$ and let $c$ be the representative of $[x]$. Then $\exists q \in \mathbb{Q} \cap [-1,1]$ such that $x-c = q$ and therefore $x = c + q$ for some $q$.

Regarding iii): Looks good to me.

By the way: This is called the Vitali set.

-
I see. Since $q_1 - q_2 = c' - c$, $c' \sim c$ and since we picked only 1 representative from each equivalence class to be in $C$, we must have $c' = c$ so then this forces $q_1 = q_2$ as required. Does that sound ok? Thank you. – Mani Oct 9 '11 at 16:45
Yes! That's right! I added more details to my answer. – Rudy the Reindeer Oct 9 '11 at 17:24
I've just woken up (yup, only 3 hours sleep tonight =[) and seen your expanded answer. I'll try to work through part ii) with your hints in a few hours. Could you tell me why $C \subset [0,1]$? I can only see $C \subset [-1,1]$. I think the question has $I = 2B \cap \mathbb{Q}^N$ instead of $B \cap \mathbb{Q}^N$ to make the second inclusion possible to prove with weaker estimates? Now that I see it, I feel like I should have been able to prove the first inclusion at least! Thank you so much for your help, I wish I could up vote your answer more times. – Mani Oct 9 '11 at 20:58
My pleasure! Actually, $C$ is only in $[0,1]$ if you take $B$ to be $[0,1]$ or in your case $[0,1]^N$. That's how the Vitali set is usually constructed. I'll correct my answer to use $B = [-1,1]^N$ as in your exercise. – Rudy the Reindeer Oct 9 '11 at 21:41