Lebesgue outer measure

Question

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)?

Answer 1

If $\mu^\star|P(X)$ would be a measure, then we could define a $\sigma$-additive measure $\lambda:P([0,1])\to[0,1]$ by the formula $\lambda(A)=\mu^\star(A\cap X)$ for $A\subset [0,1]$. This would imply that the continuum is real-valued measurable, which is not the case under some set-theoretic assumptions (like CH).

The answer is also negative if $non(\mathcal L)$ equals continuum (i.e., each subset of cardinality $<\mathfrak c$ in $[0,1]$ has Lebesgue measure zero). In this case it is possible to mimic the classical construction of a Bernstein set and construct two disjoint subsets $Y,Z$ of $X$ such that $\mu^*(Y)=\mu^*(Z)=\mu^*(X)$. To construct such sets $Y,Z$, find a $G_\delta$-subset $G$ of $[0,1]$ such that $X\subset G$ and $\mu(G)=\mu^*(X)$. Let $\mathcal K$ be the family of all compact subsets of positive Lebesgue measure in $G$. It is clear that $\mathcal K$ has cardinality $\mathfrak c$ and hence can be enumerated as $\{K_\alpha\}_{\alpha<\mathfrak c}$. It can be shown that for any compact set $K\in\mathcal K$ we get $\mu^*(K\cap X)=\mu(K)>0$ and hence $|K\cap X|\ge non(\mathcal L)=\mathfrak c$. This allows us to choose for every ordinal $\alpha<\mathfrak c$ two distinct points $y_\alpha,z_\alpha$ in the set $K_\alpha\cap X\setminus\{y_\beta,z_\beta\}_{\beta<\alpha}$. It is clear that the sets $Y=\{y_\alpha\}_{\alpha<\mathfrak c}$ and $Z=\{z_\alpha\}_{\alpha<\mathfrak c}$ are disjoint. We claim that $\mu^*(Y)=\mu^*(X)=\mu^*(Z)$.

Assuming that $\mu^*(Y)<\mu^*(X)$, we could find a Borel subset $B\subset G$ such that $Y\subset B\subset G$ and $\mu(B)=\mu^*(Y)<\mu^*(X)=\mu(G)$. By the regularity of the Lebesgue measure, the Borel set $G\setminus A$ (of positive measure) contains a compact subset $K$ of positive measure. Then $K=K_\alpha$ for some ordinal $\alpha<\mathfrak c$ and $K_\alpha\cap Y=\emptyset$, which contradicts $y_\alpha\in K_\alpha\cap Y$. This contradiction shows that $\mu^*(Y)=\mu^*(X)$. By analogy we can prove that $\mu^*(Z)=\mu^*(X)$. Then $\mu^*|\mathcal P(X)$ is not additive and hence not a measure.

Answer 2

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].