19
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Jan 2 at 0:32
  • 1
    $\begingroup$ Does this thing have anything to do with this?: jdh.hamkins.org/… $\endgroup$ – Michael Hardy Jan 2 at 0:40
  • 4
    $\begingroup$ Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example. $\endgroup$ – Gerald Edgar Jan 2 at 1:45
15
$\begingroup$

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

enter image description here

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.