Goldman (4) conjectured that if Z is a linear set having the property that for every (Lebesgue) measurable real function f the set f−1[Z] is a measurable set, then Z must be a Borel set. I pointed out (2) that any analytic non-Borel set provides a counterexample, and Eggleston(3) showed that a set can have the property but be neither analytic nor even an analytic complement, for example, any Luzin set. As Eggleston mentions, in the construction of Luzin sets the continuum hypothesis is assumed (compare Sierpiński(6), Chapter II), and the question arises whether it can be dispensed with in his theorem. We shall show that a non-analytic set having Goldman's property can be constructed with the help of the axiom of choice alone, without the continuum hypothesis; the problem for analytic complements remains open. We shall also generalize one of Eggleston's intermediate results.
Two Classical Surprises Concerning the Axiom of Choice and the Continuum Hypothesis
