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 4 hours ago
Aryeh Kontorovich
2,3481425
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 4 hours ago
Aryeh Kontorovich
2,3481425
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 4 hours ago
Aryeh Kontorovich
2,3481425
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 4 hours ago
Aryeh Kontorovich
2,3481425
edited 4 hours ago
YCor
27.1k380132
asked 4 hours ago
Aryeh Kontorovich
2,3481425
edited 4 hours ago
YCor
27.1k380132
edited 4 hours ago
YCor
27.1k380132
edited 4 hours ago
YCor
27.1k380132
27.1k380132
asked 4 hours ago
Aryeh Kontorovich
2,3481425
asked 4 hours ago
Aryeh Kontorovich
2,3481425
asked 4 hours ago
Aryeh Kontorovich
2,3481425
2,3481425
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 37 mins ago
Mohammad Golshani
18.9k265148
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f319895%2ffubini-without-ch%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
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 37 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 37 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 37 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 37 mins ago
Mohammad Golshani
18.9k265148
answered 37 mins ago
Mohammad Golshani
18.9k265148
answered 37 mins ago
Mohammad Golshani
18.9k265148
answered 37 mins ago
Mohammad Golshani
18.9k265148
18.9k265148
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f319895%2ffubini-without-ch%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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