# Partition into sets of positive outer measure

Let $\mu^{\star}$ denote Lebesgue outer measure. Suppose $X \subseteq [0, 1]$ and $\mu^{\star}(X) > 0$. Can we divide $X$ into uncountably many sets $\{X_i : i \in I\}$ such that for every $i \in I$, $\mu^{\star}(X_i) > 0$?

• For $X = [0,1]$ it seems that this was proved by Sierpinski and Lusin in 1917, but I can't get hold of the paper. See math.stackexchange.com/questions/8085/… – Nate Eldredge Dec 6 '16 at 0:11
• Are people downvoting / voting to close because they think this is trivial? It doesn't seem so to me, though it may well be known. – Nate Eldredge Dec 6 '16 at 0:13
• If the additivity and the cofinality characteristics of the measure-zero ideal are equal, then there's an easy affirmative answer. But the equality of those two characteristics is quite a strong hypothesis, since they are at the bottom and top (respectively) of Cichon's diagram. – Andreas Blass Dec 6 '16 at 2:16
• @AndreasBlass, I guess this article of Shelah shows that you do need some hypothesis: arxiv.org/pdf/math/9705213v2.pdf – Ramiro de la Vega Dec 6 '16 at 12:16
• @RamirodelaVega Thanks for the Shelah reference. You should probably post it as an answer, so that the question doesn't languish in unanswered limbo. – Andreas Blass Dec 7 '16 at 3:44

Suppose that the cofinality and the uniformity number of the null ideal are equal to $\kappa$. Fix a cofinal family of null sets $\{N_\alpha: \alpha \in \kappa \}$ and a bijection $\varphi=\langle \varphi_1,\varphi_2 \rangle$ from $\kappa$ onto $\kappa \times \kappa$. We define inductively a sequence $\langle x_\alpha : \alpha \in \kappa \rangle$ of distinct elements of $X$ by choosing $x_\alpha$ from the set $X \setminus \left(N_{\varphi_2}(\alpha) \cup \{x_\beta : \beta \in \alpha\} \right)$, which is non-empty since we are taking away from $X$ a null set. Now just let $X_i=\{x_\alpha : \varphi_1(\alpha)=i\}$ for $i \in \kappa$. The $X_i$ are clearly disjoint and since $x_\alpha \in X_i \setminus N_j$ whenever $\varphi(\alpha)=\langle i,j \rangle$, they are non-null sets.
On the other hand, Shelah proved in this article that if it is consistent that there is a measurable cardinal then it is also consistent that there is a non-null set $X$ which cannot be partitioned into uncountably many non-null sets.