I have a homework problem I am having trouble on. Could someone please help me?
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.