Skip to main content
added 2 characters in body
Source Link
Asaf Karagila
  • 39.8k
  • 8
  • 135
  • 283

Shelah proved that assuming $\sf ZF+DC$, if every set of reals is Lebesgue measurable, then $\omega_1$ is inaccessible in $L$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.)

Shelah proved that assuming $\sf ZF+DC$, if every set of reals is Lebesgue measurable, then $\omega_1$ is inaccessible in $L$.

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.)

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.)

Source Link
Asaf Karagila
  • 39.8k
  • 8
  • 135
  • 283

Shelah proved that assuming $\sf ZF+DC$, if every set of reals is Lebesgue measurable, then $\omega_1$ is inaccessible in $L$.

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.)