Union of non-measurable sets Let $\mathcal{N}$ be a Vitali non-measurable set in [0,1], and $\{r_k\}_{k=1}^{\infty}$ be an enumeration of all the rationals in [-1,1]. Consider the sets $$\mathcal{N}_k=\mathcal{N}+r_k.$$ My question is that, whether the union of all the $\mathcal{N}_k$'s, $$\mathop{\cup}_{k=1}^{\infty}\mathcal{N}_k$$ is measurable.
 A: Edit: it seems that this answer is not correct as it is. See the comments below. 
I suppose that you refer to the Vitali set of $[0,1]$ constructed by choosing one element of each equivalence classes of the relation defined on $[0,1]$ by $$x\sim y\iff x-y\in\mathbb Q.$$
Let $U=\bigcup_{k=1}^{\infty}\mathcal{N}_k$. Taking $d=1$ in the Theorem stated here, if we can prove that the set of differences $U-U$ contains no interval then $U$ have measure $0$ or is not measurable.
Take $x,y\in U$. The sets $\mathcal{N}_k$ are disjoint, so there are two cases:


*

*$x,y\in \mathcal{N}_k$:  in this case $x=n_1+r_k$ and $y=n_2+r_k$, so $x-y=n_1-n_2$, with $n_1,n_2\in \mathcal{N}$, by the construction of $\mathcal{N}$, $x-y\in\mathbb{R}\setminus\mathbb{Q}$.

*$x\in \mathcal{N}_k$, $y\in\mathcal{N}_j$: in this case $x-y=(n_1-n_2)+(r_k-r_j)$, for some $n_1,n_2\in \mathcal{N}$. If $x-y\in\mathbb{Q}$ then, as you can see, $n_1-n_2\in\mathbb{Q}$ and again by the construction of $\mathcal{N}$ it can not be.


Therefore the set $U-U$ only contains irrational numbers and then it can not contains intervals.
If $U$ has measure $0$ then $\mathcal{N}_k$ too, in that case $\mathcal{N}=\mathcal{N}_k-r_k$ has measure $0$, in particular $\mathcal N$ is measurable and that's contradictory.
The only remaining possibility is that $U$ is not measurable.
A: Let $A=\mathop{\cup}_{k=1}^{\infty}\mathcal{N}_k$. If A is measurable, then $A_1=A\cap [-1,0]$ and $A_2=A\cap [1,2]$ are both measurable. Let $B_1=A_1+\{1\}$ and $B_2=A_2-\{1\}$, we claim that $$B_2=([0,1]\backslash B_1)\cup \mathcal{N}$$ and this is disjoint union, which implies that $B_1$ and $B_2$ can not both be measurable. So at least one of the sets $A_1$ and $A_2$ is non-measurable, which is a contradiction.
Proof of the claim：
1）First we show that $[0,1]\backslash B_1$ and $\mathcal{N}$ are disjoint.
If $x\in([0,1]\backslash B_1)\cap \mathcal{N}$, then From the fact that $x\in \mathcal{N}$ we have that $x-1\in A_1$ then $x=(x-1)+1\in B_1$, which is a contradiction.
2) Second we show that $B_2\subset([0,1]\backslash B_1)\cup \mathcal{N}.$
If $x\in B_2$, then $x+1\in A_2$, so there exists a rational number $0\leq r \leq 1$ such that $x+1-r\in \mathcal{N}$. If $r=1$, then $x\in \mathcal{N}$. If $r\neq 1$, then $$x-1=(x+1-r)-(2-r)\notin A_1$$ since $2-r>1$, and so $x\in [0,1]\backslash B_2.$
3)Finally we prove that $([0,1]\backslash B_1)\cup \mathcal{N} \subset B_2.$
If $x\in \mathcal{N}$ then $x+1\in A_2$ and so $x\in B_2$. If $x\in [0,1]\backslash B_1$ then there exists a rational number $-2\leq r_k &lt-1$ such that $x-1-r\in \mathcal{N}.$ Note that $x+1=(x-1-r)+(2+r)$ and $x+1 \in [1,2]$ so $x+1\in A_2$ which means that $x\in B_2$.
So the proof of the claim is completed and we have that $A$ is non-measurable.
