TITLE: Lebesgue outer measure
QUESTION [8 upvotes]: Denote the Lebesgue outer measure by $\mu^{\star}$. Is there a subset $X \subseteq [0, 1]$ such that $\mu^{\star}(X) > 0$ and $\mu^{\star} \upharpoonright \mathcal{P}(X)$ is a measure (countably additive set function)?

REPLY [3 votes]: This question has a negative answer (given by Gregorz Plebanek), which follows from the following theorem of Gitik and Shelah.
Theorem (Gitik-Shelah, 1989): If a set $X$ admits an atomless probability $\sigma$-additive measure $\mu:\mathcal P(X)\to[0,1]$ defined on the algebra of all subsets of $X$, then the Banach space $L_1(\mu)$ has density $>|X|$.
On the other hand, if for some $X\subset\mathbb R$ with $\nu=\mu^*(X)$ the restriction $\mu^*|\mathcal P(X)$ is a measure, then $L_1(\nu)$ is separable (since the $\sigma$-algebra of Borel subsets of $X$ is countably generated), which contradicts Gitik-Shelah Theorem.
A combinatorial proof of Gitik-Shelah Theorem was given in 
[A. Kamburelis, A new proof of the Gitik-Shelah theorem. 
Israel J. Math. 72:3 (1990) 373–380].