We know that the consistency of ZFC+"Exists an inaccessible cardinal" implies the consistency of ZF+DC+"All sets are Lebesgue measurable"; and DC proves the existence of non-Borel sets.

J. Truss proved that repeating Solovay's construction by collapsing any limit cardinal to be $\aleph_1$ we obtain a model of ZF+"All sets are Lebesgue measurable", and in that model DC holds if and only if we collapsed an inaccessible. If we collapsed a singular cardinal then the resulting model has the property that all sets are Borel.

If we assume ZF+"All sets are Lebesgue measurable"+"There exists a non-Borel set", can we conclude that there is an inner model with an inaccessible cardinal?

Some clarifications:

- When I say Borel sets, I mean elements of the $\sigma$-algebra generated by the open sets.
- When I say Lebesgue sets, I mean elements of the $\sigma$-algebra generated by completing the Borel $\sigma$-algebra with respect to the null ideal.
As the Borel measure may fail to be $\sigma$-additive, we can as the following to complement the above question:

Assume that ZF+"All sets are Lebesgue measurable"+"The Borel measure is $\sigma$-additive", can we conclude that there is an inner model with an inaccessible cardinal?

Now, taking the Borel and Lebesgue sets as defined above makes more sense.

The above leads to the next question:

If the Borel measure is

*not*$\sigma$-additive, can we represent $\mathbb R$ as a countable union of null sets?(It is tempting to say immediately yes, but remember that countable unions of countable sets need not be countable anymore.)

mightmean a set with a Borel code (i.e., a real that encodes how to build the set from open sets by iterated complementation and countable union). The two meanings can differ in the absence of choice. – Andreas Blass Nov 1 '12 at 12:00