0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.