Can a Vitali set be Lebesgue measurable? (ZF) - MathOverflow

Can a Vitali set be Lebesgue measurable? (ZF)

2011-08-15T04:41:16Z

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 by Jonathan Hoyle for Can a Vitali set be Lebesgue measurable? (ZF)

2013-03-03T16:26:45Z

The Vitali set can never be "made measurable", primarily due to translation invariance. As you recall, the Vitali construction yield a countable collection of sets whose union is [0,1). However, through simple translations of these same sets, they can be redistributed so that their union is now [0,2), or [0,n), or any rational length. Since the union of these sets can have differing measures, the measure of these sets must therefore remain undefined.

Answer by Jonathan Hoyle for Can a Vitali set be Lebesgue measurable? (ZF)

2013-03-04T19:38:16Z

Sigma additivity is not the crucial issue issue here. In R^3 for example, the Banach-Tarski decomposition can divide the sphere up into five pieces (four non-measurable parts, and one point), and via translations and rotations only, create two spheres the same size as the original. This instance involves only finite additivity. Sigma additivity is required for R^1 and R^2, but that is not the essential issue of non-measurability.