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  "body": "\n## Counterexample\n\nLet $S^2 = \\set{x \\in \\R^3 : \\norm{x} = 1}$ be the unit sphere.\nAssume the [[axiom of choice|0HTGB13FE2]].\nThere are a [[countable set|01TQB88H4C]] $D \\subseteq S^2$ and pairwise disjoint sets $A$, $B$, $C$ with $S^2 \\setminus D = A \\cup B \\cup C$.\nThey can be chosen so that the four sets $A$, $B$, $C$ and $B \\cup C$ are congruent to each other by [[rotations|0HFREYRSZP]] of $\\R^3$.\nHence there is no finitely additive, rotation-invariant function $\\mu$ on all subsets of $S^2$ with values in $[0, \\infty)$ and $\\mu(S^2) = 1$.\n\n## Remarks\n\nThe proof uses two rotations $\\varphi$ and $\\psi$ in the [[special orthogonal group|0CGYNDCWRR]] $SO(3)$ with $\\varphi^2 = \\psi^3 = \\id$.\nThey can be chosen to generate a free product of a group of order $2$ and a group of order $3$.\nThe set $D$ collects the fixed points of the non-identity elements of this group, and the group acts freely on the rest of the sphere.\nChoosing one point in each orbit transfers a paradoxical splitting of the group to the sphere.\nFor the consequence, finite additivity and invariance give $\\mu(A) = \\mu(B) = \\mu(C) = \\mu(B) + \\mu(C)$.\nSo all three are $0$, and $\\mu(D) = 1$.\nBut for a countable $D$ there is a rotation $\\rho$ such that the sets $D, \\rho(D), \\rho^2(D), \\ldots$ are pairwise disjoint, which forces $\\mu(D) \\le 1/n$ for every $n$.\nHausdorff used this to show that the measure problem has a negative answer in dimension $3$ and higher; the Banach–Tarski paradox later strengthened the result.\n\n## Explanations\n\n### L1 human\n\nApart from a countable set of points, the sphere can be cut into three pieces $A$, $B$, $C$ with a strange property.\nRotating $A$ gives $B$, rotating it again gives $C$, and some rotation also turns $A$ into $B$ and $C$ together.\nSo $A$ looks like one third of the sphere and also like one half of it.\nThis shows that no notion of area can be defined for all subsets of the sphere, if it adds up correctly and does not change under rotation.\nThe pieces cannot be drawn; their existence uses the axiom of choice.\n",
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