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Does there exist an uncountable dense subgroup, $\Gamma$, of the additive group $\mathbb{R}$, such that every selector of the partition of $\mathbb{R}$ canonically associated with the equivalence relation $x \in \mathbb{R}$ & $y \in \mathbb{R}$ & $x − y \in \Gamma$ is nonmeasurable?

This question is related to a previous question of mine which has not been answered yet.

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2 Answers 2

I edited my former answer in light of a recent e-mail exchange with Sławomir Solecki.

There is a substantial literature on Vitali-like selectors. A major paper in the area is the following, available here.

Cichoń, J.; Kharazishvili, A.; Węglorz, B. On sets of Vitali's type. Proc. Amer. Math. Soc. 118 (1993), no. 4, 1243–1250.

There are many interesting results in the above paper, including the fact, observed by Joel Hamkins in his answer, that $MA+\lnot CH$ implies the existence of uncountable subgroups $G$ of $\Bbb{R}$ with no measurable $G$-selectors; where a $G$-selector is a choice function that selects one element from each $G$-coset.

Moreover, Theorem 6 of the above paper shows that under $MA+\lnot CH$ there is even a subgroup $G$ of $\Bbb{R}$ of power $2^{\aleph_0}$ that has no measurable $G$-selector.

This is to be contrasted with Theorem 2 of the paper, which asserts every analytic subgroup $G$ of reals has a measurable $G$-selector.

According to Solecki (private communication) it is apparently unknown if $ZFC$ can prove that there is an uncountable subgroup of the reals all of whose selectors are Lebesgue non-measurable.

However, if one allows extensions of measures, then there is a fundamental distinction to be made (note: the theorem below was first stated and proved with $CH$; but $CH$ was subsequently removed in the paper cited below).

Theorem (Solecki). The following are equivalent for every subgroup $G$ of reals.

(a) $G$ is countable and dense.

(b) Every $G$-selector is non-measurable with respect to any translation invariant extension of the Lebesgue measure.

Solecki's result above is a special case of Cor. 2.2. of the following paper of his:

Translation invariant ideals, Israel J. Math. 135 (2003), 93--110

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Great ! –  Joel David Hamkins Jul 8 '11 at 21:10
    
But I'm confused now about whether these results settle the OP's question even under ZFC+CH, since Solecki is insisting on non-measurability with respect to all extension measures. Couldn't there still be an uncountable group like the OP wants, but not violating Solecki's theorem because the group has a measurable selector with respect to some extension measure? –  Joel David Hamkins Jul 8 '11 at 21:34
    
Good point Joel, I agree. –  Ali Enayat Jul 8 '11 at 23:04
    
One more point: In the paper by Cichoń et al., in the paragraph following Theorem 2, it is asserted that as soon as the subgroup $G$ contains a perfect subset, it has a measurable $G$-selection. –  Ali Enayat Jul 8 '11 at 23:29
    
Thanks for the references Ali, they're very helpful. –  George Lazou Jul 9 '11 at 16:11

Let me give half an answer by pointing out that it is relatively consistent with ZFC that there is such a group. Indeed, it is relatively consistent with ZFC that there are numerous such groups, and indeed, that the continuum is very large and every dense subgroup of size less than the continuum (which would include many uncountable subgroups) has the property you mention. This is a consequence of Martin's Axiom plus $\neg$CH.

The reason is that the classical Vitali argument generalizes to uncountable subgroups, provided that they have size less than the additivity number $\text{add}(\mathcal{N})$ of the null ideal $\mathcal{N}$, and actually, it suffices to be less than the covering number $\text{cov}(\mathcal{N})$. The additivity number is the largest cardinal such that the union of fewer than $\text{add}(\mathcal{N})$ many measure zero sets still has measure zero (see this MO question for further information). We all know that the union of countably many measure zero sets has measure zero, and so $\aleph_1\leq\text{add}(\mathcal{N})\leq 2^{\aleph_0}$. But it is also known to be relatively consistent with $\text{ZFC}+\neg\text{CH}$ that one may take the union of any $\aleph_1$ many (or more) measure zero sets and still have a measure zero set. In other words, it is relatively consistent that $\text{add}(\mathcal{N})$ is much larger than $\aleph_1$. Indeed, for any ordinal $\alpha$, one can arrange that $\aleph_\alpha\leq\text{add}(\mathcal{N})=2^{\aleph_0}$. Indeed, $\text{add}(\mathcal{N})=2^{\aleph_0}$ is a consequence of Martin's Axiom MA, which is consistent with very large values of the continuum.

The point now is that the classical Vitali argument shows that if $\Gamma$ is any subgroup of $\mathbb{R}$ of size less than the additivity number $\text{add}(\mathcal{N})$, and $V$ is a selector with respect to translation by $\Gamma$, selecting one element from each equivalence class, then $V$ will be non-measurable. To see this, observe that since $\mathbb{R}$ is the union of $|\Gamma|$ many translates of $V$, it follows that $V$ cannot have measure zero, since the union of fewer than $\text{add}(\mathcal{N})$ many measure zero sets still has measure zero. And $V$ cannot have positive measure, since then it will have positive measure on an finite interval, and one can proceed just as in the Vitali case, finding infinitely many disjoint positive measure sets in a bounded interval, a contradiction.

The argument can be improved to the case of $\Gamma$ of size less than the covering number of the null ideal $\text{cov}(\mathcal{N})$, the smallest number of measure zero sets that cover $\mathbb{R}$, since we had covered $\mathbb{R}$ with the $\Gamma$-translates of $V$. This is an improvement, since it is consistent that the covering number is strictly larger than the additivity number.

In summary, what the argument shows is that it is consistent with ZFC that the continuum is very large, but every dense subgroup of $\mathbb{R}$ of size less than the continuum, and this includes many uncountable subgroups since the continuum is large, has all their selectors being non-measurable. This situation is a consequence of $\text{MA}+\neg\text{CH}$.

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It seems premature to accept this answer, since it is conceivable that someone will provide the desired group in ZFC, and not just a consistency statement as I have. (Although it is also conceivable that someone will show it is consistent that there are no such groups...) –  Joel David Hamkins Jul 8 '11 at 14:15
    
Incidently, it seems to suffice in the argument to use the covering number, rather than the additivity number, since the union of $|\Gamma|$ many translates of $V$ is covering $\mathbb{R}$. This is a slightly better result, since the covering number can be strictly larger than the additivity number. –  Joel David Hamkins Jul 8 '11 at 14:21
    
Thanks Joel. Yes I thought so too, but I was trying to be polite as I am guessing that the full answer may be an open question... –  George Lazou Jul 8 '11 at 14:26
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Meanwhile, George, I've noticed from your profile that you've never made any votes, despite having asking eight questions. Did you know that you can vote up any answer or question on this site? This is different from accepting an answer to a question you've asked---it's explained in the FAQ---and the voting of knowledgable people such as yourself is an important part of how the site maintains its quality. To vote, just click on the little triangle above the number. I would recommend that you vote up any answers that you find useful, and not just answers to your own questions. –  Joel David Hamkins Jul 8 '11 at 14:39
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Joel's second comment (the one about the covering number) implies that one can get this consistency result with an "easier" forcing than the one that yields Martin's axiom. Just add a lot of random reals. A notational issue: It would be better not to use $\mathfrak a$ for the additivity of measure, because it has become the standard symbol for the smallest cardinality of a maximal almost disjoint family of sets of natural numbers. –  Andreas Blass Jul 8 '11 at 15:25

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