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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.2011-02-14
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    This is probably close enough to http://math.stackexchange.com/questions/14591/vitali-type-set-with-given-outer-measure 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 $r$.2011-02-14
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    Thank you Arturo and Jonas. It was really simple.2011-02-14
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    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.2011-02-14

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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.