How does the Vitali set violate the definition of measurable sets? 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 $A$?
 A: $V$ 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 
$$|A \cap V|_e+|A \cap V^C|_e=|V|_e+1 > 1 = |A|_e$$
A: 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 $1$.
