# What can be the measure of a Vitali set?

Suppose the continuum $$\mathfrak{c}$$ is real-valued measurable, i.e., there exists a countably additive probabilistic measure on $$\mathfrak{c}$$ that measures all subsets. Then by the construction on p.131 of Jech's Set Theory, there exists a measure on $$2^{\omega}$$ that extends the usual product measure. We can then translate it to a measure on $$[0,1]$$, or (more or less) equivalently on $$\mathbb{R}$$.

Such a measure is certainly not translation-invariant, but what can we say about the measure of a Vitali set $$A$$? Since $$\mathbb{R}$$ is covered by the translates $$A+q,\ q\in\mathbb{Q}$$, certainly $$A+q$$ is positive for some $$q$$. Can the measure be anything? Similarly, what's the measure of an ultrafilter on $$\omega$$, etc.

• I would suspect it can be anything between 0 and 1. Dec 7, 2023 at 19:12
• It will depend on the Vitali set. If the new measure extends the Lebesgue measure, then it will be bounded by the outer measure of $A$, and this can be made as small as desired. There are Vitali sets for $\mathbb{R}$ contained in $[0,\epsilon]$ for any $\epsilon>0$. math.stackexchange.com/a/14623/413 Dec 7, 2023 at 20:18

Let $$\mu$$ be a total extension of Lebesgue measure. For $$\lambda>0,$$ we define another total extension of Lebesgue measure by $$\mu_{\lambda}(X)=\frac{1}{\lambda}(\lambda X).$$ We will show there is $$\lambda$$ such that for all $$x \in [0,1],$$ there is a Vitali set $$V_x$$ with $$\mu_{\lambda}(V_x)=x.$$

In 21:00 - 27:00 of my talk "Paradoxes of perfectly small sets", I construct $$X\subset \mathbb{R}$$ and fields $$F' \subsetneq F$$ such that $$\mu(\mathbb{R} \setminus X)=0,$$ and for any $$x, y \in X,$$ we have the equivalence $$x - y \in F' \Leftrightarrow x-y \in F.$$ Fix $$\lambda \in F \setminus F'.$$ Then $$X':=\frac{1}{\lambda} X$$ has at most one point per Vitali equivalence class, and $$\mu_{\lambda}(X' \cap [0,1))=1.$$ Extend $$X' \cap [0,1)$$ to a Vitali set $$V_1.$$ Finally, we define

$$V_x = ([0, x) \cap V_1) \cup \left \{y+\frac{1}{2} \bmod 1: y \in [x, 1) \cap V_1 \right \}.$$

Edit: The poster has asked whether the rescaling is necessary. It is. Let $$q_n$$ enumerate $$\mathbb{Q} \cap [0, 1).$$ For $$x \in (0, 1),$$ we define a total extension $$\nu$$ of Lebesgue measure for which the set of measures of Vitali sets is $$(0, x]$$, by

$$\nu(A) = \sum_{n=0}^{\infty} x(1- x)^n \mu_{\lambda} (V_1 \cap (A + q_n \bmod 1)).$$

Edit 2: Here's the complete classification of $$V(\nu):=\{\nu(Y): Y \text{ a Vitali set}\}$$ for total extensions of Lebesgue measure (assuming one exists). For each $$\nu,$$ there is $$x \in (0, 1]$$ such that $$V(\nu) = [0, x],$$ or there is $$x \in (0, 1)$$ such that $$V(\mu) = (0, x].$$

My answer above gives constructions for $$V(\nu) = [0, 1]$$ and $$V(\nu) = (0, x].$$ Finally, $$V(\nu)=[0, x]$$ can be achieved by replacing $$q_n$$ with $$q_{n+1}$$ in the above construction.

Now we check that these are all the possibilities. First, if $$V_1$$ is a Vitali set such that $$\nu(V_1)=1,$$ then our construction of $$V_0$$ above gives a Vitali set of measure 0. Next, suppose $$V_x$$ is a Vitali set such that $$\nu(V_x) = x,$$ and let $$\epsilon \in (0, x).$$ Let $$n$$ be such that $$\nu(V_x + q_n) < \epsilon$$ (addition implicitly taken mod 1). Define a Vitali set $$U_z=([0, z) \cap V_x) \cup \left \{y+q_n : y \in [z, 1) \cap V_x \right \}.$$ The function $$z \mapsto \nu(U_z)$$ is Lipschitz of constant 1, so by the intermediate value theorem, there is $$z \in (0, 1)$$ such that $$\nu(U_z) = \epsilon.$$

Finally, it remains to verify that there is a Vitali set $$V_{\text{max}}$$ with $$\nu(V_{\text{max}} ) = s:=\sup V(\nu).$$ Recursively choose $$V_n \in \mathcal{V}_n:=\{\text{Vitali set } V: \nu(V) > (1 - 3^{-n})s\}$$ such that $$\nu(V_{n+1} \triangle V_n) < (1 + 5^{-n})\inf_{U \in \mathcal{V}_{n+1}} \nu(U \triangle V_n).$$

Let $$W = \liminf V_n.$$ It is clear that $$W$$ has at most one point per Vitali equivalence class and can thus be extended to a Vitali set $$V_{\text{max}}.$$ That $$\nu(V_{\text{max}})\ge \nu(W) \ge \nu(V_n)$$ for all $$n$$ can be verified by checking $$\nu(\{x \in V_i: \{x+q_m, x+ q_n\} \subset \limsup V_n \})=0$$ for each $$m and $$\nu(\{x \in V_i: |\{m: x+q_m \in \bigcup_{n<\omega} V_n\}|=\aleph_0\})=0.$$

• Is the rescaling $\mu_\lambda$ necessary? Dec 9, 2023 at 18:31
• Yes. I've added a note on that at the end. Dec 10, 2023 at 4:23