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In my textbook the Vitali set is shown as a classic example of non-measurable sets. The proof is done by showing that you can derive an impossible measure of this set if it is measurable. I also searched the internet and there are other proofs more or less the same. One proof can be found here "Is outer measure a measure?".

However, no material I read directly shows how the Vitali set (denoted as $V$) violates the definition of measurable sets, that is for any set $A$ the equation $|A|_e=|A\bigcap V|_e+|A\bigcap V^C|_e$ holds, where $|A|_e$ denotes outer measure. Since the Vitali set is non-measurable, there must exist some set $A$ that does not satisfy the equation. Can anyone help find an example of such an $A$?

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    $\begingroup$ Your $A$ is bound to depend on $V$, and I suspect constructing $A$ requires the axiom of choice. $\endgroup$ – user21820 Jul 19 '15 at 5:57
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$V$ must have positive outer measure: if $V$ had outer measure $0$, then it would follow from the countable subadditivity of outer measure that $[0,1]$ also had outer measure $0$, which is nonsense.

On the other hand, $[0,1]-V$ has outer measure $1$. It's easiest to see this by showing that $V$ has inner measure $0$. Inner measure is superadditive, so if $V$ had positive inner measure, then we could take the union of some of its mod-$1$ rational translates in $[0,1]$ to have arbitrarily large inner measure, which is also nonsense.

So if $A=[0,1]$ then $$|A \cap V|_e+|A \cap V^C|_e=|V|_e+1 > 1 = |A|_e$$

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  • $\begingroup$ Great solution! Follow up question. Why is the other argument used far more often than this one? It seems like this gets to the point much more directly and is less complicated. $\endgroup$ – user223391 Jul 19 '15 at 6:54
  • $\begingroup$ Thank you for your example! Are you assuming all representatives of equivalent classes are chosen from $[0,1]$? Otherwise it seems $|A\bigcap V|_e$ could be an empty set if all representatives are chosen from an interval disjoint with $[0,1]$, say $[2,3]$. $\endgroup$ – Tony Jul 19 '15 at 14:30
  • $\begingroup$ @Tony: I am assuming that, yes. In my experience most definitions of the Vitali set explicitly restrict it to lying in some bounded interval, to make this kind of proof easier. If your definition of $V$ involves taking equivalence class representatives from all of $\Bbb{R}$, you'd have to do something like argue that $V \cap [n, n+1]$ must have positive outer measure for some $n$, and then use that $[n, n+1]$ as $A$... $\endgroup$ – Micah Jul 19 '15 at 16:04
  • $\begingroup$ @avid19 The other argument goes more into the direction of the basic problem of measure. It shows that the unit interval can be partitioned into a countable number of sets that are basically cyclic translations of the Vitali sets and hence, there can be no countable additive and translation invariant measure for the Vitali set and the unit interval. $\endgroup$ – Dirk Jul 20 '15 at 5:14
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A comment on the example of @Micah:

If for a set $V$ there exists a measurable set $A\supset V$ of finite measure such that $\mu^{\star}(A) = \mu^{\star}(V) + \mu^{\star}(A\backslash V)$ then $V$ itself is measurable ( hence, we do not need to check for all possible $A$, measurable or not).

Now, for a set $W$ of $\mathbb{R}$ we have $\mu^{\star}(W)= \inf \mu(U)$ for $U$ a disjoint union of open intervals containing $W$. The set $V$ does not contain any intervals of length $>0$ as any two points in $V$ have an irrational difference. Therefore, any open $U$ containing $[0,1]\backslash V$ will have as complement in $[0,1]$ finitely many points, hence will be of measure at least $1$.

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