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For any integer $n\geq 2$ there is one and only one (up to rescaling) rotation-invariant, finitely-additive measure on the Lebesgue $\sigma$-algebra of $S^n$.

The proof of this statement I'm aware of treats two cases separately: $n\geq 4$ and $n=2, 3$ (the latter in a bizarre turns of events ends up relying on the Ramanujan conjecture). I wonder if there is an argument that is uniform in $n$.

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First, you don't need the full Ramanujan Conjecture: property $(\tau)$ is enough, so in that sense you can get a uniform proof (it's just that for $n\geq 4$ we have also property (T), which is easier to prove and stronger). Second, you can in fact induce property $(\tau)$ up the ladder (see Burger--Sarnak). Thirdly, his book [1] Sarnak makes this explicit, showing how to convert the spectral solution for the problem for $O(n)$ to a solution for $O(n+1)$.

[1] Sarnak, Peter, Some applications of modular forms, Cambridge Tracts in Mathematics, 99. Cambridge: Cambridge University Press. x, 111 p. \textsterling 17.50; $ 29.95 (1990). ZBL0721.11015.

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