# Fubini without CH

In Real and Complex Analysis, Rudin gives an example (due to Sierpinski) of a function $$f:[0,1]^2\to[0,1]$$ separately Lebesgue-measurable in each argument, such that $$\int_0^1 dx\int_0^1f(x,y)\,dy \neq \int_0^1 dy\int_0^1f(x,y)\,dx$$ (all integrals are w.r.t. the Lebesgue measure on $$[0,1]$$). The construction of $$f$$ requires the Continuum Hypothesis, and my question is: What happens if we negate CH? Does it then follow that all functions $$f:[0,1]^2\to[0,1]$$ separately Lebesgue-measurable in each argument satisfy the conclusion of Fubini's theorem?

• This paper by Friedman appears to show that a slightly weaker statement is consistent with ZFC: if both iterated integrals make sense then they are equal. – Nate Eldredge Jan 2 at 0:32
• Does this thing have anything to do with this?: jdh.hamkins.org/… – Michael Hardy Jan 2 at 0:40
• Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example. – Gerald Edgar Jan 2 at 1:45

See Cardinal Conditions for Strong Fubini Theorems, Joseph Shipman Transactions of the American Mathematical Society Vol. 321, No. 2 (Oct., 1990), pp. 465-481.

In general: Let $$(X,A,μ)$$ and $$(Y,B,ν)$$ be $$σ$$-finite measure spaces. The strong Fubini axiom ($$SFA^∗$$) asserts that whenever the iterated integrals for some $$f:X×Y→[0,∞)$$ are defined then they must be equal. It is known that for $$X=Y=R$$ and $$μ=ν=$$ Lebesgue measure, $$CH$$ implies not-$$SFA^∗$$ and the above paper shows that non(Lebesgue null)$$<$$Cov(Lebesgue null) implies $$SFA^∗$$.

You may also look at Strong Fubini axioms from measure extension axioms for extensions