Can a Vitali set be Lebesgue measurable? (ZF)

Question

Here is the definition of Lebesgue measure.

The standard proof that Vitali sets are not Lebesgue measurable uses countable additivity of Lebesgue measure, which is not a theorem of ZF. (In particular, it is consistent that the real line is a countable union of countable sets, and thus a countable union of measure zero sets.) Since ZF does prove that Lebesgue measure is super-additive, that proof can be easily adapted to show in ZF that if a Vitali set is measurable, then its measure is zero. By the Caratheodory construction, this is equivalent to having outer measure zero.


Does ZF prove that all Vitali sets have positive outer measure?

If no, does ZF prove "if there exists a Vitali set, then there exists a Vitali set with positive outer measure"?

Answer 1

The answer to your second question is yes. Let $G=(\mathbb{Q}[\sqrt{2}], +).$ This is a countable abelian group, so $\mathbb{R}/G$ is a hyperfinite Borel equivalence relation. In particular, it embeds into $\mathbb{R}/\mathbb{Q}.$ (See section 7 here for a summary of these results: Kechris - The theory of countable Borel equivalence relations, preliminary version May 8, 2019). Thus we can use a transversal for the latter to construct a transversal for the former. Let $X$ be a transversal for $\mathbb{R}/G.$

Let $V=\{(x + q\sqrt{2}) \text{ mod 1}: x \in X, q \in \mathbb{Q}\} \subset [0, 1).$ This is a Vitali set closed under translation by $\mathbb{Q}\sqrt{2}.$ We will show its outer measure is at least $\frac{1}{4}$ (in fact, the argument can be extended to show its outer measure is 1). Suppose towards contradiction that $U$ is an open cover of $V$ with measure less than $\frac{1}{4}.$ For each $n \ge 2,$ we can canonically find an open cover of $V \cap [0, \frac{\sqrt{2}}{n}]$ of measure less than $\frac{\sqrt{2}}{2n}$ by considering the least $m$ such that $\lambda(U \cap [\frac{m\sqrt{2}}{n},\frac{(m+1)\sqrt{2}}{n}])<\frac{\sqrt{2}}{2n}.$ With this we can recursively construct open covers $U_n$ of $V$ of measure less than $2^{-n-2}.$ Letting $\{q_n\}$ enumerate the rationals, we see $\{q_n+U_n\}$ covers $\mathbb{R},$ so $\lambda^*(\mathbb{R}) < 1,$ contradiction.

Also, here's an observation relevant to your first question. If $\mathbb{R}=\bigcup_{n<\omega} X_n$ is a countable union of countable sets, then there is a null subset of $[0, 1]$ which meets every mod $\mathbb{Q}$ class, namely $W=\bigcup_{n<\omega} \{x \in [0, 2^{-n}]: \exists q \in \mathbb{Q} (x+q \in X_n)\}.$ This is null since it countable outside of any $[0, \epsilon].$ Furthermore, if there is a Vitali set, that could be used to separate out a Vitali subset of $W,$ so a model satisfying these two properties would be a counterexample to the first question.

Answer 2

The answer to your first question is no. We will show there is a null Vitali set in Cohen's first model $M.$

Recall that in $M,$ there is an infinite (but Dedekind finite) set $A$ of mutually Cohen reals, with the property that for any $r \in \mathcal{P}^M(\omega),$ there is $a \in A^{<\omega}$ such that $r \in L[a].$ Let $X_n$ be the set of $r \in \mathcal{P}^M(\omega)$ for which there are $c_1, \ldots, c_n \in A$ such that $r \in L[c_1, \ldots, c_n].$ We will show $X_n$ is null.

Fix $a_1, \ldots, a_{n+1} \in A.$ Identify $a_i$ as an element of $\omega^{\omega}$ and let $\{I_q^m\}_{m=1}^{\infty}$ enumerate the rational open intervals of length $q.$ Let $U_j^{a_i}=\bigcup_{k<\omega} I_{1/(j2^k)}^{a_i(k)},$ an open set of measure at most $\frac{2}{j}.$

Now fix $x \in X_n$ and $c_1, \ldots, c_n \in A$ such that $x \in L[c_1, \ldots, c_n].$ There is an $a_i \not \in \{c_1, \ldots, c_n\}.$ Since $a_i$ is Cohen over $L[c_1, \ldots, c_n],$ we have that $x \in U_j^{a_i}$ for all $j.$ Thus, $X_n \subset \bigcup_{i=1}^{n+1} \bigcap_{j=1}^{\infty} U_j^{a_i},$ so it is null.

Let $W=\bigcup_{n<\omega} \{x \in [0, 2^{-n}]: \exists i \in \mathbb{Z} \cap [-2^n, 2^n] (x+\frac{i}{2^n} \in X_n)\} \subset [0, 1].$ This is a null set since outside any $[0, \epsilon]$ it is a finite union of null sets. Furthermore, it is easy to check that $W$ meets every mod $\mathbb{Q}$ class.

It is an old theorem of Feferman that there is a Vitali set in $M,$ with which we can canonically enumerate each mod $\mathbb{Q}$ class. Thus there is a Vitali set $V \subset W,$ which is clearly null.