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So the question the I am working on is: given $S\subseteq [0,1]$ and that $\lambda^*(S)+\lambda^*(I\setminus S) =1$, show $S$ is $\lambda$-measurable. Where $\lambda^*$ denotes the Lesbegue outer measure, and $\lambda$ its associated measure.

I have tried this problem supposing we have some $A$ such that $\lambda^*(A\cap S)+\lambda^*(A\setminus S)> \lambda^*(A)$ and trying to reach a contradiction somehow about the original hypothesis from there, but get stuck, not seeing a direction in which to proceed.

Any help or direction would be greatly appreciated.

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Hint: Given any set $S\subset[0,1]$, there exists a measurable set $E\subset[0,1]$($E$ can be chosen as a $G_\delta$ set), such that $S\subset E$ and $\lambda^*(S)=\lambda(E)$. Then with the condition $\lambda^*(S)+\lambda^*([0,1]\setminus S)=1$ you may try to show that $\lambda^*(E\setminus S)=0$.

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    @user48459:You are welcome.2012-11-08