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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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I think you choose one representative from every coset of the rationals, so $\mathcal N + \mathcal Q$ wants to contain the whole interval. –  mike Apr 18 '12 at 11:25
    
There are many many Vitali sets. Saying "the" implies some sort of uniqueness. –  Asaf Karagila Apr 18 '12 at 13:33
    
@Asaf Karagila : Yes. So I change the word "the" by "a" to denote by $\mathcal{N}$ one of the vitali sets. –  Siming Tu Apr 18 '12 at 13:47
    
Why is [axiom-of-choice] needed here? You're not asking about the need for choice or what happens without the axiom of choice. You simply use a result which require the axiom of choice to begin with. –  Asaf Karagila Apr 19 '12 at 16:08
    
@Asaf Karagila : I wonder if the choice of representative will affect the answer. –  Siming Tu Apr 19 '12 at 16:17

2 Answers 2

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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In the case "$x\in \mathcal{N}_k$, $y\in\mathcal{N}_j$", it is possible that $n_1=n_2$. So it is possible that $x-y\in\mathbb{Q}$. –  Siming Tu Apr 19 '12 at 8:17
    
Given $x\in[0,1]$ we have that there is a unique $x'\in\mathcal N$ such that $x'-x\in\mathbb Q$. Since $0\in U$ we only need to show that $x\in U$. Indeed $|x'-x|\le1$ therefore there is some $r_k\in[-1,1]$ which is rational such that $x+r_k=x'$ so $x\in\mathcal N_k$ and therefore $x\in U$. We have shown that $[0,1]\subseteq U-U$. –  Asaf Karagila Apr 19 '12 at 16:18
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Even simpler is to see that $[0,1]\subseteq U$ and $0\in U$ therefore $[0,1]\subseteq U\subseteq U-U$ (which is what I wrote above, actually). –  Asaf Karagila Apr 19 '12 at 16:24
    
@SimingTu yep you're right. Let me see if there is a possibility to improve this. –  leo Apr 19 '12 at 17:20
    
@AsafKaragila I see, don't know why I don't see that before. Is there some assumptions that allow us to employs this arguments to show that depending on the $\mathcal N$, $U$ is not measurable? –  leo Apr 19 '12 at 17:33

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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