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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.

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    $\begingroup$ Related: mathoverflow.net/a/57108/1946. $\endgroup$ – Joel David Hamkins Oct 14 '13 at 2:36
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    $\begingroup$ Could you say a little more about what kind of measures you are considering? After all, I could define $\mu(X)$ to be $\infty$, if $X$ is uncountable and otherwise $0$. This measures every set, with no mass at points and it is translation-invariant. So I guess you want $\sigma$-finite measures. But if the unit interval gets finite measure, then this determines the measure of any interval, since we can split the interval into two halves, or thirds, etc. which must get equal measure, and so on. So if an interval has finite measure, then one is back to the arguments to which you refer. $\endgroup$ – Joel David Hamkins Oct 14 '13 at 2:41
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    $\begingroup$ Yes, I am thinking of $\sigma$-finite measures. A translation invariant $\sigma$-finite measure defined on $\mathcal P\mathbb R$ for which $[0,1]$ has measure $1$ would extend Lebesgue measure. Assuming the consistency of a measurable cardinal, it is consistent with AC the existence of an extension of Lebesgue measure to $\mathcal P\mathbb R$. AC implies that such extension cannot be translation invariant (by the Vitali argument). I am asking whether the ultrafilter theorem suffices. $\endgroup$ – Carlos Oct 14 '13 at 11:35
  • $\begingroup$ I have added the $\sigma$-finite hypothesis to the question. $\endgroup$ – Carlos Oct 14 '13 at 11:41
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    $\begingroup$ I am only speculating, but I meant looking into models of the form $L(R)[U]$, where $U$ is $P(\omega)/Fin$-generic over $L(R)$. See, for example, Andres Caicedo's posting: mathoverflow.net/questions/69615/… $\endgroup$ – Ashutosh Oct 14 '13 at 22:05
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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 $[0,1]$.

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    $\begingroup$ I have red the question more attentively and discovered that I did not answer the question which was posed. The problem was to find a model of ZFC with an ultrafilter (which implies the existence of a translation-invariant measure on each finitely generated abeliang group) but withouts translation-invariant measure on the real line (the construction of such measure requires the existnce of the ultrafilter + the compactness of the circle to the power of continuum). $\endgroup$ – Taras Banakh Jul 6 '17 at 14:18

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