A question about extending Lebesgue measure to all subsets of the real numbers. Is it known to be consistent with just ZF set theory-without the Axiom of Choice-that there exists an extension of Lebesgue measure to all subsets of real numbers, which is countably additive and isometrically invariant?.....I am confused about this because some articles in the literature, which seem to imply that my question has a "yes" answer, also seem to add a Large Cardinal Axiom to the axioms of ZF. But my question specifically rules out any axioms beyond those of ZF.
 A: It is known that the consistency of "All sets of reals are Lebesgue measurable" over $ZF+DC$ has large cardinal strength; however, note the role of $DC$, Dependent Choice, in the base theory. With many related results, dropping the requirement of $DC$ also drops the consistency strength of the related regularity property. For example, I believe $ZF$+"All sets have the perfect set property" is equiconsistent with $ZFC$ (Truss, although I can't find the citation right now), whereas $ZF+DC$+"All sets have the perfect set property" has the same consistency strength as an inaccessible (Shelah).
The problem is that - unlike e.g. the perfect set property - it's not $100\%$ clear how to define Lebesgue measurability if we drop the axiom of dependent choice, since now the countable union of countable sets need not be countable! So my instinct is that you'll get wildly varying answers, depending on what definition of "Lebesgue measurable" you use, reflecting the fact that these definitions are not equivalent over mere $ZF$.
