Let $V$ be a Vitali set on $\mathbb{R}$ and suppose that $V$ is not bounded, ie the representatives $v$ of the cosets of $\mathbb{Q}$ are chosen in a such a way that $|v|\geq M$ for all $M\in\mathbb{R}$. If such a Vitali set exists, can it contain a set with positive measure?

I know that all measurable subsets of the standard Vitali set in $[0,1]$ have measure 0, but what if $V$ cannot be contained in a bounded set?

  • $\begingroup$ If all measurable subsets of the standard Vitali set in $[0,1]$ have measure $0$, then the same will hold for unbounded Vitali sets: $V \cap [-B, B]$, for any bound $B>0$, will be contained in a bounded Vitali set, which can then be mapped by rescaling to a Vitali set in $[0, 1]$. (We might want to assert that $B$ is rational so this rescaling retains a Vitali set.) $\endgroup$ – Dustan Levenstein Oct 12 '16 at 18:20
  • $\begingroup$ ... and I guess I should add, any measurable $S \subset V$ can be measured as the supremum of the measures of $S \cap [-B, B]$. $\endgroup$ – Dustan Levenstein Oct 12 '16 at 18:21

The answer is no.

Say that a set $X\subseteq\mathbb{R}$ is pre-Vitali if $X$ contains at most one real in each $\mathbb{Q}$-equivalence class. Then any bounded pre-Vitali set contains no measurable set of positive measure, since every bounded pre-Vitali set is contained in a bounded Vitali set.

Now suppose $V$ is an unbounded Vitali set. For each $n$, the set $V_n=V\cap [-n, n]$ is a pre-Vitali set. Suppose $M$ is measurable with positive measure; we'll show $M\not\subseteq V$.

The key point is the following: for some $n$, $M_n=M\cap [-n, n]$ also has positive measure. Why? Well, the union of countably many measure-zero sets has measure zero, so if $M_n$ had measure zero for each $n$, $M$ itself would have measure zero; but we assumed $M$ had positive measure.

OK, so some $M_n$ has positive measure. Well, each $M_n$ is measurable, so $M_n\not\subseteq V_n$. But this means $M\not\subseteq V$, so we're done.

  • $\begingroup$ What? You just said that $M_n$ has positive measure and and then said $M$ is the union of measure 0 sets $M_n$. I don't get it. $\endgroup$ – UserA Oct 12 '16 at 18:39
  • $\begingroup$ @Adel No, you've misunderstood my answer (and I've clarified the language a bit). I said "suppose $M$ has positive measure" and then concluded that $M\not\subseteq V$. That's exactly what we're trying to prove: that $V$ does not contain any measurable set of positive measure. The point is that if $M$ has positive measure, we can decompose $M$ into the $M_n$s. Since $M$ has positive measure, some $M_n$ must have measure not zero. But then $M_n\not\subseteq V_n$, so $M\not\subseteq V$. $\endgroup$ – Noah Schweber Oct 12 '16 at 18:43

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