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It is well-known that any set $E \subseteq \mathbb{R}$ with positive outer measure contains a nonmeasurable subset $V$. I know that $0 < m^*(V) \le m^*(E)$. Nevertheless, my question is the following: given $r \in \mathbb{R}$ such that $r>0$, is there a nonmeasurable subset of $\mathbb{R}$ whose outer measure is exactly $r$?

Thank you in advance.

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Take a Vitali nonmeasurable subset of $[0,1]$, and scale it appropriately. – Arturo Magidin Feb 14 '11 at 2:33
This is probably close enough to… to close as duplicate. Note however that this problem is a bit easier, because given a nonmeasurable set $V$ with finite outer measure $s\gt 0$, the set $\frac{r}{s}V=\{\frac{r}{s}\cdot x:x\in V\}$ is a nonmeasurable set with outer measure $r$. – Jonas Meyer Feb 14 '11 at 2:35
Thank you Arturo and Jonas. It was really simple. – ragrigg Feb 14 '11 at 6:18
I'll put my and Jonas's answer as a Community Wiki answer so you can mark it as "accepted" and the question can be marked as answered, in case it doesn't get the votes to be closed. – Arturo Magidin Feb 14 '11 at 14:42
up vote 2 down vote accepted

On can take a Vitali nonmeasurable subset of $[0,1]$, which has positive and finite outer measure, and just scale it appropriately.

As Jonas points out, this is closely related to this previous question, but much easier.

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