Working over ZF but without the Axiom of Choice (AC), assume that the Hahn–Banach Theorem holds for $\ell^\infty$. Does it follow that there exists a set of real numbers that is not Lebesgue measurable?
The motivation for this question is the old question of what would be involved in replacing ZFC by ZF+LM plus some weakened version of AC (here LM denotes the axiom "all sets of reals are Lebesgue measurable"). Now, analysts are accustomed to freely invoking theorems such as Hahn–Banach and Krein–Milman, and are going to dislike having to check some picky logical detail every time they do so. In light of Solovay's model, one naïve hope would be to define some notion of "tame" function space, and prove, using DC (the axiom of dependent choice), that all the standard theorems hold for tame function spaces, and that any function space that arises "in practice" is tame.
Unfortunately, $\ell^\infty$ is not tame. We can see this even without a precise definition of "tame" because by applying the Hahn–Banach theorem to $\ell^\infty$, one can show that $(\ell^\infty)^* \ne \ell^1$. On the other hand, it is known that ZF+DC+BP (Baire property) implies that $(\ell^\infty)^* = \ell^1$.
These inconvenient facts limit our options. One option would be to regard $(\ell^\infty)^* = \ell^1$ as a feature and not a bug; students coming to the subject for the first time might find it to be a pleasant fact. But it might be a hard sell for analysts who have grown up using Hahn–Banach freely and who "know" that $(\ell^\infty)^* \ne \ell^1$.
Another option would be to ignore AC for the moment and just examine the extent to which Hahn–Banach is compatible with LM. That brings me to the stated question.
