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Show that for any $r\in(0,1]$, there is a Vitali set $\mathbb{V}\subseteq[0,1]$ such that $\mu^*(\mathbb{V})=r$

So my idea was to construct a Vitali set around r or something like that in order to have the outer measure of the Vitali set is less or equal than r and also trying to get the other way

Jojo98
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  • Do the usual Vitali set construction, but in $[0 ,r]$. i.e, declare that $a \sim b$ iff their difference is rational, and show that $[a] \cap [0, r] \not = \emptyset$, for all $a \in [0, 1]$ ($[a]$ being the equivalence class of $a$). Then you get a set $V \subseteq [0, r]$ that is non measurable for the usual reasoning. Moreover you can show that $\mu^*[V] = 0$ (since otherwise its translates would cover $[0, r]$ and $[0, r]$ would have infinite measure). Thus, $[0, r] \setminus V$ is also non measurable and must have measure $r$. – Vercingetorix Dec 03 '20 at 04:45
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    Does this answer your question? Vitali-type set with given outer measure – HallaSurvivor Dec 03 '20 at 05:25

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