# Can a Banach measure be consistent with Friedman's Fubini-type theorem for non-measurable functions?

I read on Wikipedia that Harvey Friedman proved that the following is consistent with ZFC+¬CH:

For all functions $f:[0,1]^2 \mapsto \mathbb{R}^+$ such that both $\int_0^1 \left ( \int_0^1 f(x,y) dy \right ) dx$ and $\int_0^1 \left ( \int_0^1 f(x,y) dx \right ) dy$ are well-defined and exist, these integrals are equal.

Is it consistent with (or even provable from) ZFC + "the Fubini theorem for non-measurable functions" that there is a Banach measure $\mu$ on $\mathbb{R}^2$ with the following property:

For all subsets $S$ of $[0,1]^2$ such that both iterated integrals for $\chi_S$ are well-defined and exist, $$\mu(S)=\int_0^1 \left ( \int_0^1 \chi_S(x,y) dy \right ) dx =\int_0^1 \left ( \int_0^1 \chi_S(x,y) dx \right ) dy.$$

If there is such a Banach measure, then it would seem to satisfy most of the properties a total measure on $\mathbb{R}^2$ "should" have (other than $\sigma$-additivity of course), and could provide a Platonistic argument for ¬CH (in the vein of the axiom of symmetry, but more convincingly in my opinion).