Well-ordering of R is enought to create unmeasurable set.
Does a well-ordering on the constructable version of R provably do this? I fear I can’t tell if you’ve addressed my question yet.
Constructible version of R (if it is inside R and not the whole R) is just like Q: dense, small part of the whole, and usually zero-measure. So, this construction will yield something of measure zero if you define measure on the whole R.
Does a well-ordering on the constructable version of R provably do this? I fear I can’t tell if you’ve addressed my question yet.
Constructible version of R (if it is inside R and not the whole R) is just like Q: dense, small part of the whole, and usually zero-measure. So, this construction will yield something of measure zero if you define measure on the whole R.