Can it be shown, on the assumption that $ZF$ is consistent, that there is a model of $ZF$ in which the reals cannot be well-ordered but there does exist a set of reals which is not Lebesgue measurable?
$\begingroup$
$\endgroup$
2
-
3$\begingroup$ Yes. ${}{{}}{}$ $\endgroup$– Andrés E. CaicedoCommented Dec 24, 2017 at 14:01
-
6$\begingroup$ You probably want ZF+DC in order to have a decent theory of Lebesgue measure. $\endgroup$– Joel David HamkinsCommented Dec 24, 2017 at 14:03
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Shelah proved that assuming $\sf ZF+DC$, if every set of reals is Lebesgue measurable, then $\omega_1$ is inaccessible to reals.
This alone should hint you that it is easy to arrange models where $2^{\aleph_0}$ cannot be well-ordered, $\sf DC$ holds, and there are non-measurable sets.
More specifically, any free ultrafilter on $\omega$ generates a non-measurable set of reals. So working in any model where $\omega$ has free ultrafilters is enough.
(As Joel remarked, assuming $\sf DC$ is sort of essential, since otherwise the usual notion of measure could break apart, e.g. the reals can be a countable union of null sets, or even countable sets.)
answered Dec 24, 2017 at 14:36