Let $V\subset [0,1]$ be the Vitali set. Then for any closed subset $F$ of $V$ we have that $m(F)=0<m^{*}(V)\leq 1$ (where $m$ is the Lebesgue measure and $m^*$ is the Lebesgue outer measure on $\mathbb{R}$).
Let us define $A:=\cup_{n=1}^\infty((2n+1)+V)$, so that we have $m^{*}(A)=\sum_{n=1}^\infty m^{*}(((2n+1)+V))=\sum_{n=1}^\infty m^{*}(V)=\infty$.
Now, if $F$ is any closed subset of $A$, is it true that $m(F)=0$? If the answer is always "yes", how can we proceed to get this result?
Thanks for any hint or comment.
Suppose $F=\bar F\subset A.$ For $n\in \Bbb N$ let $F_n=F\cap [2n+1,2n+2].$
Now $F_n$ is a closed subset of $A\cap [2n+1,2n+2]=V+2n+1$
so $F_n-(2n+1)$ is a closed subset of $(V+2n+1)-(2n+1)=V$
so $0=m(F_n-(2n+1))=m((F_n-(2n+1))+2n+1)=m(F_n).$
And $F=\cup_{n\in \Bbb N}F_n$ so $m(F)\leq \sum_{n\in \Bbb N}m(F_n)=0.$