The fullest version of the paradox (that is known to me) says that if you have two sets $A, B \subseteq \Bbb R^3$, both of which have non-empty interior, then you can divide $A$ up into some finite number $n$ of disjoint subsets, then move those subsets around isometrically so that they remain disjoint and their union is now $B$.

More formally, if $A, B$ have non-empty interior, there exist sets $A_k, k = 1, ..., n$ such that $A = \bigcup_{k=1}^n A_k$ and $A_j\cap A_k = \emptyset$ when $j \ne k$. And there exist isometries $f_k, k = 1, ..., n$ of $\Bbb R^3$ such that $B = \bigcup_{k=1}^n f_k(A_k)$ and $f_j(A_j) \cap f_k(A_k) = \emptyset$ when $j \ne k$.

Because of the paradox, we know that if $V : \scr P(\Bbb R^3) \to \Bbb R$ is a set function, then one of these 3 conditions must hold:

- there are disjoint sets $A, B \subseteq \Bbb R^3$ such that $V(A\cup B) \ne V(A) + V(B)$, or
- there are isometries $f$ of $\Bbb R^3$ and sets $A \subseteq \Bbb R^3$ such that $V(A) \ne V(f(A))$, or
- $V$ is constant on all sets with interior.

**Edit** Adding explanation why at least one of the three conditions must hold:

The first two conditions are just "$V$ is not additive" and "$V$ is not preserved under isometries". So if neither of those hold, then $V$ is additive and is preserved by isometries. In this case, let $A$ and $B$ be two sets with interior. Then per the BTP result I gave, we can write $A = \bigcup_{k=1}^n A_k$ and $B = \bigcup_{k=1}^n f_k(A_k)$ for disjoint $A_k$ and $f_k(A_k)$. Hence
$$V(A) = V\left(\bigcup_{k=1}^n A_k\right) = \sum_{k=1}^n V(A_k) = \sum_{k=1}^n V(f_k(A_k)) = V\left(\bigcup_{k=1}^n f_k(A_k)\right) = V(B)$$
Thus if $V$ is additive and preserved under isometries, it must have the same value for all sets with interior (and in fact, that value must be $0$, since any set with interior is the disjoint union of two other sets with interior).

The problem is that all three are violations of properties that any concept of "volume" should have:

- Volume is additive: if two sets are disjoint, the volume of their union should be the sum of their volumes.
- Volume is unchanged by isometries. Volume is supposed to depend only on size and shape, not location or orientation. So moving a set around rigidly should not change its volume.
- Volume is obviously not constant on sets with interior. The only way it could be both additive and constant is if it was always $0$. But the volume of the unit cube is $1$.

The conclusion is therefore that it is impossible to define a concept of volume that works for all subsets of $\Bbb R^3$.