Fubini without CH












9














In Real and Complex Analysis, Rudin gives an example (due to Sierpinski) of a function $f:[0,1]^2to[0,1]$ separately Lebesgue-measurable in each argument, such that
$$
int_0^1 dxint_0^1f(x,y),dy
neq
int_0^1 dyint_0^1f(x,y),dx
$$
(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 $f:[0,1]^2to[0,1]$ separately Lebesgue-measurable in each argument satisfy the conclusion of Fubini's theorem?






set-theory lo.logic measure-theory integration







share|cite|improve this question




















  • 3




    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
    6 hours ago






  • 1




    Does this thing have anything to do with this?: jdh.hamkins.org/…
    – Michael Hardy
    6 hours ago






  • 3




    Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
    – Gerald Edgar
    5 hours ago
















9














In Real and Complex Analysis, Rudin gives an example (due to Sierpinski) of a function $f:[0,1]^2to[0,1]$ separately Lebesgue-measurable in each argument, such that
$$
int_0^1 dxint_0^1f(x,y),dy
neq
int_0^1 dyint_0^1f(x,y),dx
$$
(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 $f:[0,1]^2to[0,1]$ separately Lebesgue-measurable in each argument satisfy the conclusion of Fubini's theorem?






set-theory lo.logic measure-theory integration







share|cite|improve this question




















  • 3




    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
    6 hours ago






  • 1




    Does this thing have anything to do with this?: jdh.hamkins.org/…
    – Michael Hardy
    6 hours ago






  • 3




    Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
    – Gerald Edgar
    5 hours ago














9












9








9







In Real and Complex Analysis, Rudin gives an example (due to Sierpinski) of a function $f:[0,1]^2to[0,1]$ separately Lebesgue-measurable in each argument, such that
$$
int_0^1 dxint_0^1f(x,y),dy
neq
int_0^1 dyint_0^1f(x,y),dx
$$
(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 $f:[0,1]^2to[0,1]$ separately Lebesgue-measurable in each argument satisfy the conclusion of Fubini's theorem?






set-theory lo.logic measure-theory integration







share|cite|improve this question















In Real and Complex Analysis, Rudin gives an example (due to Sierpinski) of a function $f:[0,1]^2to[0,1]$ separately Lebesgue-measurable in each argument, such that
$$
int_0^1 dxint_0^1f(x,y),dy
neq
int_0^1 dyint_0^1f(x,y),dx
$$
(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 $f:[0,1]^2to[0,1]$ separately Lebesgue-measurable in each argument satisfy the conclusion of Fubini's theorem?






set-theory lo.logic measure-theory integration




set-theory lo.logic measure-theory integration






share|cite|improve this question















share|cite|improve this question













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edited 6 hours ago









YCor

27.1k380132




27.1k380132










asked 6 hours ago









Aryeh Kontorovich

2,3681425




2,3681425








  • 3




    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
    6 hours ago






  • 1




    Does this thing have anything to do with this?: jdh.hamkins.org/…
    – Michael Hardy
    6 hours ago






  • 3




    Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
    – Gerald Edgar
    5 hours ago














  • 3




    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
    6 hours ago






  • 1




    Does this thing have anything to do with this?: jdh.hamkins.org/…
    – Michael Hardy
    6 hours ago






  • 3




    Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
    – Gerald Edgar
    5 hours ago








3




3




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
6 hours ago




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
6 hours ago




1




1




Does this thing have anything to do with this?: jdh.hamkins.org/…
– Michael Hardy
6 hours ago




Does this thing have anything to do with this?: jdh.hamkins.org/…
– Michael Hardy
6 hours ago




3




3




Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
– Gerald Edgar
5 hours ago




Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
– Gerald Edgar
5 hours ago










1 Answer
1






active

oldest

votes


















6














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






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    6














    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






    share|cite|improve this answer


























      6














      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






      share|cite|improve this answer
























        6












        6








        6






        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






        share|cite|improve this answer












        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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Mohammad Golshani

        19k265148




        19k265148






























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