Ultrafilter theorem and translation invariant measures 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.
 A: Theorem. If there exists a free ultrafilter $\mathcal U$ on $\omega$, then there exists a non-measurable subset of the real line.
Proof. First observe that the free ultrafilter $\mathcal U$ is a non-measurable subset of the Cantor cube $\{0,1\}^\omega$ with respect to the standard product measure on $\{0,1\}^\omega$ (here we identify subsets of $\omega$ with their characteristic functions, which are elements of the Cantor cube $\{0,1\}^\omega$). 
Next, consider the standard Cantor ladder map $$f:\{0,1\}^\omega\to[0,1]\subset\mathbb R,\;\;f:(x_n)_{n\in\omega}\mapsto\sum_{n\in\omega}\frac{x_n}{2^{n+1}}$$
and observe that $f$ is measure-preserving in the sense that for any measurable subset $B\subset [0,1]$ the set $f^{-1}(B)$ is measurable in $\{0,1\}^\omega$ and has product measure equal to the Lebesgue measure of $B$.
Since the symmetric difference $\mathcal U\triangle f^{-1}(f(\mathcal U))$ is at most countable, the set $f^{-1}(f(\mathcal U))$ is not measurable in $\{0,1\}^\omega$ and hence its image $f(\mathcal U)$ is not measurable in $[0,1]$.
