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It is well known that the Vitali set is unmeasurable. When given a measure space $(\mathbb{R},\mathcal{M},\mu)$, can we say that the Vitali set as a subset of $\mathbb{R}$ is never measurable regardless of the choice of the measure $\mu$?

I know the Vitali set is not Lebesgue measurable, but is the Vitali set unmeasurable in every measure space?

Asaf Karagila
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    Certainly not: the function that assigns value $0$ to every subset of $\Bbb R$ is a perfectly good measure, for example, and in this measure every set is measurable. In general, measurability is much more about the measure, not the specific set. – Greg Martin May 08 '21 at 19:15
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    @Greg: Or the one that assigns $0$ or $1$ based on whether or not $\pi$ is an element of the set. – Asaf Karagila May 08 '21 at 19:17
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    Counting measure is another example for which every set is measurable. –  May 08 '21 at 19:33
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    For the record, there is no such thing as the Vitali set. We usually refer to a Vitali set when we want to indicate a set $S\subseteq \Bbb R$ such that 1) for all $\alpha\in \Bbb R$ there is some $q\in\Bbb Q$ such that $\alpha-q\in S$ and 2) for all $x,y\in S$, either $x=y$ or $x-y\notin \Bbb Q$. –  May 08 '21 at 19:48
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    Another comment. If you want a set that is non-measurable for many measures, try instead a Bernstein set: https://math.stackexchange.com/q/169714/442 – GEdgar May 09 '21 at 14:20

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The proof that a Vitali set is not Lebesgue measurable relies on the fact that Lebesgue measure is invariant under translation. Unless the measure space $(\mathbb R, \mathcal M, \mu)$ has the property:

if $A \in \mathcal M$ and $\mu(A) > 0$, then $A+x \in \mathcal M$ and $\mu(A+x) > 0$ for all $x$,

then we cannot expect that argument to work.

GEdgar
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