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We know that any set with zero outer measure is measurable, and the vitali set on $[0,1]$ is non-measurable. So it must have a non-zero outer measure.

$\textbf{Lemma: }$ For any $q \in \mathbb{Q}$ such that $0 < q < 1$, any vitali set of $[0,q]$ has the same outer measure as a vitali set of $[0,1]$.

$\textbf{Proof: }$ Let $q \in \mathbb{Q}$ be any rational number such that $0 < q < 1$. Let $V \subset [0, q]$ and $V' \subset [0,1]$ both be vitali sets.

Assume there exists an element $v' \in V'$, such that for all $v \in V$

$$v \not\sim v' \Rightarrow v - v' \not\in Q$$

But there must exists $n \in \mathbb{N}$ such that $(v' - nq) \in [0, q]$. By the definition of a vitali set we choose one element from each equivilence class. So there must exist some element $v \in V$ such that $v \sim (v' - nq)$ implying that $v \sim v'$, giving us a contradiction.

So for every element $v' \in V'$, there exists some $v \in V$ such that $v' \sim v$. So, $V \subset [0,q]$ is also a vitali set of $[0,1]$, that is every equivalence class is covered by both. So, we could even imagine that $V = V'$, thus $m^*(V) = m^*(V')$.

$\textbf{Question: }$ Suppose that $m^*(V') = \epsilon > 0$. Let $q \in \mathbb{Q}$ such that $0 < q < \epsilon$. Let $V \subset [0,q]$ be a vitali set. This gives us

\begin{align*} V \subset [0,q] & \Rightarrow m^*(V) \leq m^*([0,q]) = q < \epsilon = m^*(V') = m^*(V)\\ & \Rightarrow m^*(V) < m^*(V) \end{align*}

This is a contradiction and so $m^*(V') = 0$, implying that it is measurable.

Where did I go wrong?

Roland Killian
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  • Your lemma is not correct. The problem is the "in fact, we could even imagine" part. Being able to imagine something does not make it true. – PhoemueX Oct 01 '21 at 04:42
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    @PhoemueX Actually the OP's lemma is correct (although that sentence of course is a complete non sequitur); every Vitali subset of $[0,q]$ is a Vitali subset of $[0,1]$ as well if $0<q<1$, and a fortiori has the same outer measure as some Vitali subset of $[0,1]$ (namely itself). The issue is that the OP needs to go the other way: take an arbitrary Vitali subset of $[0,1]$ and connect it with some Vitali subset of $[0,q]$. They can only do this with the further implicit assumption that all Vitali subsets of $[0,1]$ have the same outer measure (which they don't). – Noah Schweber Oct 01 '21 at 05:00
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    @Noah: OK, yes, i interpreted the claim "the same measure as a Vitali set of $[0,1]$" as meaning "the same measure as any Vitali set of $[0,1]$", since this is what the OP (tries to) prove in the proof. – PhoemueX Oct 01 '21 at 05:40

1 Answers1

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The following (basically your lemma) is true:

Suppose $0<q<1$. Then every Vitali subset of $[0,q]$ is also a Vitali subset of $[0,1]$. Consequently, every Vitali subset of $[0,q]$ has the same outer measure as some Vitali subset of $[0,1]$ (namely itself).

This is correct. However, this does not mean that we can conflate an arbitrary Vitali subset of $[0,q]$ and an arbitrary Vitali subset of $[0,1]$ (which is what you're trying to do first when you write "So, we could even imagine that $V=V'$," and later when you write "$m^*(V)=m^*(V')$" in your chain of claimed equalities).

I think you're implicitly assuming that any two Vitali subsets of $[0,1]$ have the same outer measure, so that in some sense "Vitali subset of $[0,1]$" is (as far as measure theory goes) a complete description of a set. But this is false. There are wildly different Vitali subsets of $[0,1]$; as an example, one can show that there is a Vitali subset of $[0,1]$ of outer measure $1$.

Noah Schweber
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