In this argument, one would have to use the axiom of choice since one needs a non-Lebesgue measurable set.

Is it possible without using AC to show that any bounded set is not necessarily a countable union of Jordan measurable sets?

]]>The first/only thing that has empty interior I can come up with is the Cantor set . But it seems that it could not be a desired example: there indeed exists an open set (namely the empty set) contained in such that . Also since is closed, one has . :-(

]]>and

Is it still true if we switch “open” and “closed” in the statement above? (Do we have approximation by open sets from inside and closed sets from outside?)

]]>