Adding the first two to ZF + DC (dependent choice) is consistent (assuming that ZFC + Con(ZFC) is consistent, as just about everybody believes)
This isn’t quite right. The consistency of ZF + DC + “every subset of R is Lebesgue measurable” is equivalent to the consistency of an inaccessible cardinal, which is a much stronger assumption then the consistency of ZFC + Con(ZFC).
This isn’t quite right. The consistency of ZF + DC + “every subset of R is Lebesgue measurable” is equivalent to the consistency of an inaccessible cardinal, which is a much stronger assumption then the consistency of ZFC + Con(ZFC).
Sorry, my mistake. Still, set theorists usually believe this.