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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.

Siming Tu
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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
  • It is possible, yes. For example if you chose the rational to be $0$ then all the rationals in $[-1,1]$ will appear in the union, otherwise they will not. This, however, has absolutely nothing to do with the axiom of choice. – Asaf Karagila Apr 19 '12 at 16:19
  • @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. – Siming Tu Apr 19 '12 at 16:30

2 Answers2

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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
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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 08:17
  • 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
  • @leo: I am not sure if that would depend on the representatives. My guess is that it does not matter. See, if we took the sum over all the rationals we would get $\mathbb R$. However taking only boundedly many rationals we will likely to end up with an interval + "Vitali dust" on the sides, and there might be an argument of scaling which will prove that $U$ is not measurable. – Asaf Karagila Apr 19 '12 at 17:37
  • @AsafKaragila, or is there a particular construction of $\mathcal N$ so that $U$ is measurable? – leo Apr 19 '12 at 17:39
  • let us continue this discussion in chat – leo Apr 19 '12 at 17:44
  • @leo: I have no idea about a particular choice of representatives which would imply this set is measurable. I'm somewhat fresh out of ideas. I will think about it and let you know. – Asaf Karagila Apr 19 '12 at 17:46
  • @leo: Sorry, but the chat site is blocked here so I cannot meet you there and continue the discussion... – Asaf Karagila Apr 19 '12 at 18:13
  • @AsafKaragila this was my comments: or is there a particular construction of $\mathcal N$ so that $U$ is measurable?

    What do you mean by "if we took the sum over all the rationals"

    I'm about to delete the answer. I don't see how to improve it and don't have much time.

    If you succeed in finding a particular choice of the representatives implying $U$ is measurable, it is better that you answer that question. In any case, no doubt that many valuable thoughts will appear :-)

     – leo Apr 19 '12 at 18:16
  • @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. – Asaf Karagila Apr 19 '12 at 18:39
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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 $-2\leq r_k &lt-1$ 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.

Siming Tu
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