TITLE: Partition into sets of positive outer measure
QUESTION [10 upvotes]: 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 $i \in I$, $\mu^{\star}(X_i) > 0$?

REPLY [7 votes]: 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.