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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.

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    @Asaf Karagila : Yes, axiom of choice can only assure that we can choose one representative in one class, but it can do nothing with which representative to be chosen,so it maybe not appropriate to use the tag [axiom-of-choice] and I delete it.2012-04-19

2 Answers 2

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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 <-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.

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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.

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    @leo: If the union was $\bigcup_{q\in\mathbb Q}\mathcal N+q$ then the result was $\mathbb R$. I suspect that the real argument why doing that over a bounded interval of rationals will be unmeasurable would have to be related to scaling rather than translations.2012-04-19