14

Pretty much the question in the title: can there be a well-ordering the reals $\langle r_\alpha\mid \alpha<\kappa \rangle$, all of whose initial segments $I_\beta := \{r_\alpha\mid \alpha<\beta<\kappa\}$ are Lebesgue-measurable?

Observation: if the Continuum Hypothesis holds, then any well-ordering of the reals in ordertype $\omega_1$ will have this property. So it's consistent that such a well-ordering exists.

ikrto
  • 985

1 Answers1

4

Partial answer: if we additionally require that the order-type of the well-ordering is a cardinal, then this is independent of ZFC. (Edit: see comment below by Farmer S on how to turn this into a full answer)

For example, add $\aleph_2$ random reals to $L$. In the extension, $\mathbb{R}^L$ is a set of size $\aleph_1$, with positive outer measure. Any well-ordering of the reals in type $\omega_2$ will have a proper initial segment containing all of $\mathbb{R}^L$. This means that the initial segment (if measurable) needs to have positive measure and hence cardinality $2^{\aleph_0}$. So no well-ordering can have all its initial segments being measurable in this model.

Another way to see that there can be no such well-ordering in this model is by noticing that the set of random reals added must be a Sierpinski set (every measurable subset is countable). Again, since any well-ordering in type $\omega_2$ must have an initial segment containing $\aleph_1$ many random reals, any such initial segment will have positive measure (if measurable) and hence cardinality $2^{\aleph_0}$.

Jason Zesheng Chen
  • 1,601
  • 4
    This can be converted into a full answer, by partitioning $\kappa$ into minimal sub-intervals $[\alpha,\beta)$, such that each set ${r_\gamma}{\alpha\leq\gamma<\beta}$ has positive measure (excluding maybe a tail interval which might produce a measure zero set). (I.e. choose the intervals so that ${r\gamma}_{\alpha\leq\gamma<\beta'}$ has measure zero when $\beta'<\beta$.) There can only be countably many such intervals, which implies that $\aleph_2$-many random reals get into one of the pieces of the partition, which produces a contradiction like you got above. – Farmer S Dec 10 '23 at 02:12
  • That's a great observation, thanks! – Jason Zesheng Chen Dec 10 '23 at 02:15