Fubini without CH
$begingroup$
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 Jan 2 at 0:43
YCor
27.3k480132
asked Jan 2 at 0:20
Aryeh KontorovichAryeh Kontorovich
2,4881527
$endgroup$
add a comment |
$begingroup$
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 Jan 2 at 0:43
YCor
27.3k480132
asked Jan 2 at 0:20
Aryeh KontorovichAryeh Kontorovich
2,4881527
$endgroup$
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
add a comment |
$begingroup$
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 Jan 2 at 0:43
YCor
27.3k480132
asked Jan 2 at 0:20
Aryeh KontorovichAryeh Kontorovich
2,4881527
$endgroup$
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 Jan 2 at 0:43
YCor
27.3k480132
asked Jan 2 at 0:20
Aryeh KontorovichAryeh Kontorovich
2,4881527
edited Jan 2 at 0:43
YCor
27.3k480132
asked Jan 2 at 0:20
Aryeh KontorovichAryeh Kontorovich
2,4881527
edited Jan 2 at 0:43
YCor
27.3k480132
edited Jan 2 at 0:43
YCor
27.3k480132
edited Jan 2 at 0:43
YCor
27.3k480132
27.3k480132
asked Jan 2 at 0:20
Aryeh KontorovichAryeh Kontorovich
2,4881527
asked Jan 2 at 0:20
Aryeh KontorovichAryeh Kontorovich
2,4881527
asked Jan 2 at 0:20
Aryeh KontorovichAryeh Kontorovich
2,4881527
2,4881527
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
add a comment |
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
4
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
$begingroup$
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
1
$begingroup$
Does this thing have anything to do with this?: jdh.hamkins.org/…
$endgroup$
– Michael Hardy
Jan 2 at 0:40
$begingroup$
Does this thing have anything to do with this?: jdh.hamkins.org/…
$endgroup$
– Michael Hardy
Jan 2 at 0:40
4
4
$begingroup$
Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
$endgroup$
– Gerald Edgar
Jan 2 at 1:45
$begingroup$
Martins axiom (consistent with not-CH) will be enough to do Sierpinski's example.
$endgroup$
– Gerald Edgar
Jan 2 at 1:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
See Cardinal Conditions for Strong Fubini Theorems,
Joseph Shipman
Transactions of the American Mathematical Society
Vol. 321, No. 2 (Oct., 1990), pp. 465-481.
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 Jan 2 at 4:42
Mohammad GolshaniMohammad Golshani
19.1k267149
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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votes
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oldest
votes
$begingroup$
See Cardinal Conditions for Strong Fubini Theorems,
Joseph Shipman
Transactions of the American Mathematical Society
Vol. 321, No. 2 (Oct., 1990), pp. 465-481.
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 Jan 2 at 4:42
Mohammad GolshaniMohammad Golshani
19.1k267149
$endgroup$
add a comment |
$begingroup$
See Cardinal Conditions for Strong Fubini Theorems,
Joseph Shipman
Transactions of the American Mathematical Society
Vol. 321, No. 2 (Oct., 1990), pp. 465-481.
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 Jan 2 at 4:42
Mohammad GolshaniMohammad Golshani
19.1k267149
$endgroup$
add a comment |
$begingroup$
See Cardinal Conditions for Strong Fubini Theorems,
Joseph Shipman
Transactions of the American Mathematical Society
Vol. 321, No. 2 (Oct., 1990), pp. 465-481.
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 Jan 2 at 4:42
Mohammad GolshaniMohammad Golshani
19.1k267149
$endgroup$
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 Jan 2 at 4:42
Mohammad GolshaniMohammad Golshani
19.1k267149
answered Jan 2 at 4:42
Mohammad GolshaniMohammad Golshani
19.1k267149
answered Jan 2 at 4:42
Mohammad GolshaniMohammad Golshani
19.1k267149
answered Jan 2 at 4:42
Mohammad GolshaniMohammad Golshani
19.1k267149
19.1k267149
add a comment |
add a comment |
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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