Fubini without CH
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
edited 6 hours ago
YCor
27.1k380132
asked 6 hours ago
Aryeh Kontorovich
2,3681425
add a comment |
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
edited 6 hours ago
YCor
27.1k380132
asked 6 hours ago
Aryeh Kontorovich
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
add a comment |
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
edited 6 hours ago
YCor
27.1k380132
asked 6 hours ago
Aryeh Kontorovich
2,3681425
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
edited 6 hours ago
YCor
27.1k380132
asked 6 hours ago
Aryeh Kontorovich
2,3681425
edited 6 hours ago
YCor
27.1k380132
asked 6 hours ago
Aryeh Kontorovich
2,3681425
edited 6 hours ago
YCor
27.1k380132
edited 6 hours ago
YCor
27.1k380132
edited 6 hours ago
YCor
27.1k380132
27.1k380132
asked 6 hours ago
Aryeh Kontorovich
2,3681425
asked 6 hours ago
Aryeh Kontorovich
2,3681425
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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
answered 2 hours ago
Mohammad Golshani
19k265148
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
answered 2 hours ago
Mohammad Golshani
19k265148
add a comment |
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
answered 2 hours ago
Mohammad Golshani
19k265148
add a comment |
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
answered 2 hours ago
Mohammad Golshani
19k265148
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
answered 2 hours ago
Mohammad Golshani
19k265148
answered 2 hours ago
Mohammad Golshani
19k265148
answered 2 hours ago
Mohammad Golshani
19k265148
answered 2 hours ago
Mohammad Golshani
19k265148
19k265148
add a comment |
add a comment |
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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