For all $x,y \in (0,1]$ define the equivalence relation: $x \sim y : \Leftrightarrow x-y \in \mathbb{Q}$. Let $A \subset (0,1]$ be a set with: $\forall x \in (0,1] \, \exists_1 a \in A: x \sim a$.
Show that $A$ ist not Lebesgue measurable.
For $r \in (0,1] \cap \mathbb{Q}$ define $(r+A) \operatorname{mod} 1 := [(r+A) \cup ((r-1)+A)] \cap (0,1]$. Assuming that $A$ is Lebesgue measurable, I have to prove that:
(1): $\forall r \in (0,1] \cap \mathbb{Q}: \lambda((r+A) \operatorname{mod} 1) = \lambda(A)$
(2): $(0,1] = \sum_{r\in (0,1] \cap \mathbb{Q}} (r+A) \operatorname{mod} 1$ (sum as union of disjoint sets)
This would imply $\lambda((0,1]) = \sum_{r\in (0,1] \cap \mathbb{Q}} \lambda((r+A) \operatorname{mod} 1) = \sum_{r \in (0,1] \cap \mathbb{Q}} \lambda(A)$.
If $\lambda(A) = 0$, then $\lambda((0,1]) = 0$ and if $\lambda(A) > 0$, then $\lambda((0,1] = \infty$. Both values create a contradiction and therefore $A$ is not measurable.
My problem is to prove step (1). We defined the Lebesgue measure on intervalls and it is clear to be translation invariant under usual translations. How can I determine the Lebesgue measure of the constructed set $A$ with the translation $(r+A) \operatorname{mod} 1$ and compare to the one of $A$ though?
HINT: Let $B=(r+A)\cap(0,1]$, and let $C=(r+A)\setminus B$; then $(r+A)\bmod 1$ is the disjoint union of $B$ and $C-1$.