[constantcrying]

Completely wrong. From the Wikipedia article: "Given a solid ball in three-dimensional space, there exists a decomposition of the ball into a finite number of disjoint subsets, which can then be put back together in a different way to yield two identical copies of the original ball."

There is absolutely no issue with uncountability here. The issue is with the particular shape of the parts, where V is not reasonably definable.

[emblaegh]

You first split the sphere into an uncountable number of subsets, then group these into a finite number of subsets, whose measure sum to twice the measure of the original set.

(Un)countability is at the core of most of the counterintuitive results of measure theory, exactly because of the the third property of measure.

[adroniser]

To be clear, the construction given here violates the finite additivity property of measure. It's got nothing to do with the countable/uncountable additivity property.

[emblaegh]

Yeah but this is only possible if the sets in question are uncountably infinite.

[adroniser]

Yes but you are not taking an uncountable union. You are taking a finite union.

[afiori]

IIRC correctly first you split the surface into an uncountable partition then you use the axiom of choice to "color" each point of each partition of a color and then define your finite number of subsets as all the points of each given color.