The usual Vitali construction of a non-Lebesgue measurable set generalizes to a proof that there are no (non-trivial) translation invariant measures on $\mathcal P\mathbb R$.

On the other hand, there are many proofs of the existence of non-Lebesgue measurable sets just relying on the ultrafilter theorem instead of AC, but I can't see they can be generalized this way. They use that Lebesgue measure is determined by the measure of the intervals, or that the Haar measure on the Cantor cube is determined by the measure of the clopen sets.

Does the ultrafilter theorem imply that there are no translation invariant measures on $\mathcal P\mathbb R$? Of course, I ask for non-trivial ones, i.e., finite sets must have measure zero.

The usual Vitali construction of a non-Lebesgue measurable set generalizes to a proof that there are no (non-trivial) translation invariant measures on $\mathcal P\mathbb R$.

On the other hand, there are many proofs of the existence of non-Lebesgue measurable sets just relying on the ultrafilter theorem instead of AC, but I can't see they can be generalized this way. They use that Lebesgue measure is determined by the measure of the intervals, or that the Haar measure on the Cantor cube is determined by the measure of the clopen sets.

Does the ultrafilter theorem imply that there are no translation invariant measures on $\mathcal P\mathbb R$? Of course, I ask for non-trivial ones, i.e., $\sigma$-finite and with finite sets having measure zero.