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Prove that there exists a bounded set $A\subset\mathbb{R}$ such that $m^*(F)\le m^*(A)-1$ for every closed set $F\subset A$.

Here $m^*$ is the outer mesure.

I used the bounded and non measurable Vitali set $V$, and defined $A=tV$, where $t>0$, but I could only prove that $$m^*(A\setminus F)\ge1$$ However, since $A$ is non measurable either, I have that $m^*(A)\le m^*(F)+m^*(A\setminus F)$, so I don't know how to finish the proof.

AlephZero
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    For example, can you get $m^*(A) = 1$ but every closed subset has measure zero? – GEdgar Dec 14 '19 at 18:36
  • I thought of that, but can I construct such example using the Vitali set? since I don't know other non-measurable sets. – AlephZero Dec 14 '19 at 18:47
  • I'm guessing something like this: Let $B$ be the biggest borel set inside $V$, then $V\setminus B$ is non-measurable and every borel set inseide $V\setminus B$ has measure zero. But, I don't know how can I justify that – AlephZero Dec 14 '19 at 19:06
  • I just checked, that is one of the Vitali's set properties. – AlephZero Dec 14 '19 at 20:51

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