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 4 hours ago
YCor
27.1k380132
asked 5 hours ago
Aryeh Kontorovich
2,3581425
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 4 hours ago
YCor
27.1k380132
asked 5 hours ago
Aryeh Kontorovich
2,3581425
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
4 hours ago
1
Does this thing have anything to do with this?: jdh.hamkins.org/…
– Michael Hardy
4 hours ago
3
Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
– Gerald Edgar
3 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 4 hours ago
YCor
27.1k380132
asked 5 hours ago
Aryeh Kontorovich
2,3581425
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 4 hours ago
YCor
27.1k380132
asked 5 hours ago
Aryeh Kontorovich
2,3581425
edited 4 hours ago
YCor
27.1k380132
asked 5 hours ago
Aryeh Kontorovich
2,3581425
edited 4 hours ago
YCor
27.1k380132
edited 4 hours ago
YCor
27.1k380132
edited 4 hours ago
YCor
27.1k380132
27.1k380132
asked 5 hours ago
Aryeh Kontorovich
2,3581425
asked 5 hours ago
Aryeh Kontorovich
2,3581425
asked 5 hours ago
Aryeh Kontorovich
2,3581425
2,3581425
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
4 hours ago
1
Does this thing have anything to do with this?: jdh.hamkins.org/…
– Michael Hardy
4 hours ago
3
Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
– Gerald Edgar
3 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
4 hours ago
1
Does this thing have anything to do with this?: jdh.hamkins.org/…
– Michael Hardy
4 hours ago
3
Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
– Gerald Edgar
3 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
4 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
4 hours ago
1
1
Does this thing have anything to do with this?: jdh.hamkins.org/…
– Michael Hardy
4 hours ago
Does this thing have anything to do with this?: jdh.hamkins.org/…
– Michael Hardy
4 hours ago
3
3
Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
– Gerald Edgar
3 hours ago
Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
– Gerald Edgar
3 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 44 mins ago
Mohammad Golshani
18.9k265148
add a comment |
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1 Answer
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active
oldest
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 44 mins ago
Mohammad Golshani
18.9k265148
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 44 mins ago
Mohammad Golshani
18.9k265148
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 44 mins ago
Mohammad Golshani
18.9k265148
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 44 mins ago
Mohammad Golshani
18.9k265148
answered 44 mins ago
Mohammad Golshani
18.9k265148
answered 44 mins ago
Mohammad Golshani
18.9k265148
answered 44 mins ago
Mohammad Golshani
18.9k265148
18.9k265148
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
4 hours ago
1
Does this thing have anything to do with this?: jdh.hamkins.org/…
– Michael Hardy
4 hours ago
3
Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
– Gerald Edgar
3 hours ago