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Let $E\subset[a,b]$.

How can we show that:

$\exists F \in F_\sigma$: $F\subseteq E$ and $\mu^*(F)=\mu^*(E)$

Can we conclude that $E$ is measurable?

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1 Answer 1

I think $E$ must be measurable since there is a non-measurable set $H$ in $[a,b]$ such that for any measurable subset $A$ of $H$,we have $μ(A)=μ^*(A)=0$ but $μ^*(H)>0$.

WLOG,let $a=0,b=1$.Define an equivalence relation ~ on $[0,1]$ s.t. $a$~$b \iff a-b \in \mathbb{Q}$.Let $H$ be the representatives of each equivalence class.Then $[0,1] \subseteq \bigcup_{q\in [0,1] \cap\mathbb{Q}}(H+q)\subseteq [-1,2]$.It can shown that $H$ is not measurable and for every measurable subset $A$ of $H$,we have $μ(A)=0$.However,$μ^*(H)>0$,otherwise $H$ is measurable since $μ$ is complete measurable.

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This could not happen if the measure is complete. –  Bombyx mori Dec 10 '12 at 6:43
    
I have edited my answer to explicitly find such set. –  Ben Dec 10 '12 at 7:20

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