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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)?

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    $\begingroup$ @FedorPetrov: Question seems clear to me? I'm pretty sure the answer is no. $\endgroup$ – Nate Eldredge Dec 5 '16 at 20:52
  • $\begingroup$ $\mathcal P(X)$ = power set of $X$. $\endgroup$ – Gerald Edgar Dec 5 '16 at 20:58
  • $\begingroup$ $\upharpoonright$ = restricted to $\endgroup$ – Gerald Edgar Dec 5 '16 at 21:09
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    $\begingroup$ I don't think this answers the question, but surely is related: every set of positive Lebesgue measure contains a non-(Lebesgue-)measurable subset. $\endgroup$ – Wojowu Dec 5 '16 at 21:20
  • $\begingroup$ I think the answer is no, and I suspect that the fact that this measure should come from the Lebesgue outer measure si not useful to prove it (no translations available in X). I think one has to use fully Ulam's results on measurable cardinals: en.wikipedia.org/wiki/… $\endgroup$ – Pietro Majer Dec 5 '16 at 21:37
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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.

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    $\begingroup$ But isn't it also consistent that the continuum is real-valued measurable? So this doesn't resolve the question in ZFC. $\endgroup$ – Nate Eldredge Dec 6 '16 at 0:16
  • $\begingroup$ I added a proof under $non(L)=c$. But ZFC-question is still not resolved. It would be interesting to understand what happens for a set $X$ of cardinality $|X|=non(L)<\mathfrak c$ with $\mu^*(X)>0$. $\endgroup$ – Taras Banakh Dec 6 '16 at 22:48
  • $\begingroup$ A natural attempt to construct a (consistent) example of a set $X$ with $\mu^*|\mathcal P(X)$ being a measure is to find a non-measurable subset $X$ of the real line such that all subsets of $X$ are Borel in $X$. It is not clear if such a set can exist at all: by a result of Fletcher (1978), mentioned in the Miller;s survey "Special subsets of the real line" a subset $X$ of the real line has is Lebesgue null if all subsets of $X$ are relative $F_\sigma$-sets in $X$. $\endgroup$ – Taras Banakh Dec 8 '16 at 8:12
  • $\begingroup$ The problem of @Lebesgue is equivalent to the following one: Is there an atomless Borel probability measure $\mu$ on a metrizable separable space $X$ such that each subset $A$ of $X$ is $\mu$-measurable (in the sense that $B_1\subset A\subset B_2$ for some Borel subsets $B_1,B_2$ of $X$ such that $\mu(B_1)=\mu(B_2)$). I have a strong feeling that such a problem could not be open. Maybe Fremlin in his fundamental "Measure Theory" writes something essential on this topic? $\endgroup$ – Taras Banakh Dec 8 '16 at 8:17
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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].

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  • $\begingroup$ The fact that any set of reals can be divided into two sets of same outer measure was first proved in N. Lusin: Sur la d´ecomposition des ensembles, C. R. Acad. Sci. Paris 198 (1934), 1671-1674. $\endgroup$ – Ashutosh Dec 26 '16 at 20:38

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