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If we take one element from each coset of $\mathbb Q$ in $\mathbb R$, we obtain a Vitali set, which is unmeasurable. What if instead of a choice of representatives, we take the union of a subset of the cosets of $\mathbb Q$? Such a set can be written as $\mathbb Q + T$, for $T\subseteq\mathbb R$.

For which $T$ is $\mathbb Q + T$ measurable? For which $T$ is it of positive measure?

I'm not sure what to say about this other than that if $T$ is of positive measure itself, $\mathbb Q + T$ is as well - provided it's measurable in the first place. I'm worried this might be independent of ZFC or something.

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    $\begingroup$ For something to be independent of ZFC, you would need a specific statement, and not a question as vague as the one you gave. There are some trivial things you can say, for instance that $\mathbb Q+T$ has measure zero iff $T$ does. Do you have any precise questions? $\endgroup$
    – Wojowu
    Commented Oct 24, 2021 at 15:15
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    $\begingroup$ @Wojowu I don't particularly find questions of the form "which objects $T$ have a given property $P$" all that vague (sometimes they seem vague until someone comes along with a clearly very precise and satisfactory characterization), but here's a more precise question: does there exist a $T$ such that both $\mathbb Q + T$ and its complement are measurable and of positive measure? $\endgroup$
    – Jack M
    Commented Oct 24, 2021 at 15:19
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    $\begingroup$ Well it was definitely way too vague to declare it as independent of ZFC. To the specific question: the answer is no, see here $\endgroup$
    – Wojowu
    Commented Oct 24, 2021 at 15:33
  • $\begingroup$ It's obvious that if $T$ is Lebesgue-measurable then so is $T+\mathbf{Q}$, and if $T$ has measure zero so does $T+\mathbf{Q}$, and if $T$ has positive measure then so does $T+\mathbf{Q}$. The link given by Wojowu says more: if $T$ has positive measure then $T+\mathbf{Q}$ has infinite measure (and has complement of measure zero). $\endgroup$
    – YCor
    Commented Oct 24, 2021 at 17:52
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    $\begingroup$ It's not really a commutativity phenomenon. If $G$ is a locally compact second countable group (with a left or right Haar measure) and $\Gamma$ is a countable dense subgroup, the discussion is pretty much the same for $T\Gamma$, when $T$ is measurable: either $T$ has measure zero and then $T\Gamma$ has measure zero, or $T$ has positive measure and then $T\Gamma$ has complement of measure zero. $\endgroup$
    – YCor
    Commented Oct 24, 2021 at 19:13

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