Product measure : difference between $mathcal Mtimes mathcal M$ and ${Atimes Bmid A,Bin mathcal M}$












1












$begingroup$


Consider the Lebesgue measure space $mathscr M^1=(mathbb R,mathcal M,m)$. So, the Product space is given by $$mathscr M^2=(mathbb R^2, mathcal Mtimes mathcal M, m_2).$$



Even if $mathscr M^1$ is complete, the space $mathscr M^2$ is not complete. For instance, if $mathcal N$ is the Vitali set, then ${0}times mathcal N$ is not in $mathcal Mtimes mathcal M$. But I'm not really sure how to show that. My question are the following :



1) I know that $$mathcal Mtimes mathcal M=sigma {Atimes Bmid A,Bin mathcal M}.$$
But do we have that $${Atimes Bmid A,Bin mathcal M}=sigma {Atimes Bmid A,Bin mathcal M} ?$$
I guess it's not true (I can't find a counter example, do you have one ?)



2) So if $mathcal Mtimes mathcal Mneq {Atimes Bmid A,Bin mathcal M}$, an element $Ctimes D$ can be in $mathcal Mtimes mathcal M$ but neither $C$ nor $D$ is in $mathcal M$. So how can I prove that ${0}times mathcal Nnotin mathcal Mtimes mathcal M$ ?










share|cite|improve this question









$endgroup$












  • $begingroup$
    An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:51
















1












$begingroup$


Consider the Lebesgue measure space $mathscr M^1=(mathbb R,mathcal M,m)$. So, the Product space is given by $$mathscr M^2=(mathbb R^2, mathcal Mtimes mathcal M, m_2).$$



Even if $mathscr M^1$ is complete, the space $mathscr M^2$ is not complete. For instance, if $mathcal N$ is the Vitali set, then ${0}times mathcal N$ is not in $mathcal Mtimes mathcal M$. But I'm not really sure how to show that. My question are the following :



1) I know that $$mathcal Mtimes mathcal M=sigma {Atimes Bmid A,Bin mathcal M}.$$
But do we have that $${Atimes Bmid A,Bin mathcal M}=sigma {Atimes Bmid A,Bin mathcal M} ?$$
I guess it's not true (I can't find a counter example, do you have one ?)



2) So if $mathcal Mtimes mathcal Mneq {Atimes Bmid A,Bin mathcal M}$, an element $Ctimes D$ can be in $mathcal Mtimes mathcal M$ but neither $C$ nor $D$ is in $mathcal M$. So how can I prove that ${0}times mathcal Nnotin mathcal Mtimes mathcal M$ ?










share|cite|improve this question









$endgroup$












  • $begingroup$
    An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:51














1












1








1





$begingroup$


Consider the Lebesgue measure space $mathscr M^1=(mathbb R,mathcal M,m)$. So, the Product space is given by $$mathscr M^2=(mathbb R^2, mathcal Mtimes mathcal M, m_2).$$



Even if $mathscr M^1$ is complete, the space $mathscr M^2$ is not complete. For instance, if $mathcal N$ is the Vitali set, then ${0}times mathcal N$ is not in $mathcal Mtimes mathcal M$. But I'm not really sure how to show that. My question are the following :



1) I know that $$mathcal Mtimes mathcal M=sigma {Atimes Bmid A,Bin mathcal M}.$$
But do we have that $${Atimes Bmid A,Bin mathcal M}=sigma {Atimes Bmid A,Bin mathcal M} ?$$
I guess it's not true (I can't find a counter example, do you have one ?)



2) So if $mathcal Mtimes mathcal Mneq {Atimes Bmid A,Bin mathcal M}$, an element $Ctimes D$ can be in $mathcal Mtimes mathcal M$ but neither $C$ nor $D$ is in $mathcal M$. So how can I prove that ${0}times mathcal Nnotin mathcal Mtimes mathcal M$ ?










share|cite|improve this question









$endgroup$




Consider the Lebesgue measure space $mathscr M^1=(mathbb R,mathcal M,m)$. So, the Product space is given by $$mathscr M^2=(mathbb R^2, mathcal Mtimes mathcal M, m_2).$$



Even if $mathscr M^1$ is complete, the space $mathscr M^2$ is not complete. For instance, if $mathcal N$ is the Vitali set, then ${0}times mathcal N$ is not in $mathcal Mtimes mathcal M$. But I'm not really sure how to show that. My question are the following :



1) I know that $$mathcal Mtimes mathcal M=sigma {Atimes Bmid A,Bin mathcal M}.$$
But do we have that $${Atimes Bmid A,Bin mathcal M}=sigma {Atimes Bmid A,Bin mathcal M} ?$$
I guess it's not true (I can't find a counter example, do you have one ?)



2) So if $mathcal Mtimes mathcal Mneq {Atimes Bmid A,Bin mathcal M}$, an element $Ctimes D$ can be in $mathcal Mtimes mathcal M$ but neither $C$ nor $D$ is in $mathcal M$. So how can I prove that ${0}times mathcal Nnotin mathcal Mtimes mathcal M$ ?







measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 9:11









NewMathNewMath

4059




4059












  • $begingroup$
    An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:51


















  • $begingroup$
    An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:51
















$begingroup$
An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
$endgroup$
– Vera
Dec 4 '18 at 11:51




$begingroup$
An unsuccesful effort to answer your question inspired me to ask this question myself. Later I found an answer on it too.
$endgroup$
– Vera
Dec 4 '18 at 11:51










1 Answer
1






active

oldest

votes


















0












$begingroup$

If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].






share|cite|improve this answer











$endgroup$













  • $begingroup$
    For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
    $endgroup$
    – NewMath
    Dec 4 '18 at 9:58












  • $begingroup$
    In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:07












  • $begingroup$
    @NewMath I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:10










  • $begingroup$
    But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:12








  • 1




    $begingroup$
    @NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:17













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: "69"
};
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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025324%2fproduct-measure-difference-between-mathcal-m-times-mathcal-m-and-a-time%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









0












$begingroup$

If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].






share|cite|improve this answer











$endgroup$













  • $begingroup$
    For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
    $endgroup$
    – NewMath
    Dec 4 '18 at 9:58












  • $begingroup$
    In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:07












  • $begingroup$
    @NewMath I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:10










  • $begingroup$
    But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:12








  • 1




    $begingroup$
    @NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:17


















0












$begingroup$

If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].






share|cite|improve this answer











$endgroup$













  • $begingroup$
    For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
    $endgroup$
    – NewMath
    Dec 4 '18 at 9:58












  • $begingroup$
    In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:07












  • $begingroup$
    @NewMath I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:10










  • $begingroup$
    But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:12








  • 1




    $begingroup$
    @NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:17
















0












0








0





$begingroup$

If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].






share|cite|improve this answer











$endgroup$



If $A$ is any set in $mathbb R$ which is not Lebesgue measurable than $Atimes mathbb {0}$ does not belong to $mathcal M times mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ mathbb R times {0}$ which has measure $0$. This proves that the product measure is not complete.
In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $Ctimes D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $mathcal M$. [An example of a subset of $mathbb R^{2}$ which is not of the form $C times D$ is the diagonal ${(x,y)in mathbb R^{2}:x=y}$].







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 '18 at 10:09

























answered Dec 4 '18 at 9:18









Kavi Rama MurthyKavi Rama Murthy

62.7k42262




62.7k42262












  • $begingroup$
    For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
    $endgroup$
    – NewMath
    Dec 4 '18 at 9:58












  • $begingroup$
    In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:07












  • $begingroup$
    @NewMath I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:10










  • $begingroup$
    But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:12








  • 1




    $begingroup$
    @NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:17




















  • $begingroup$
    For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
    $endgroup$
    – NewMath
    Dec 4 '18 at 9:58












  • $begingroup$
    In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:07












  • $begingroup$
    @NewMath I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:10










  • $begingroup$
    But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
    $endgroup$
    – NewMath
    Dec 4 '18 at 10:12








  • 1




    $begingroup$
    @NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:17


















$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58






$begingroup$
For 1), if it was true, then I could apply Fubini-Tonelli to $boldsymbol 1_{{0}times A}(x,y)$ where $A$ is non measurable. So $iint boldsymbol 1_{{0}times A}dxdy=0$ but by tonelli, $$iint boldsymbol 1_{{0}times A}dxdy=int_{{0}}left(int_A dyright)dx$$ and the integral inside doesn't make sense... I don't get the point here.
$endgroup$
– NewMath
Dec 4 '18 at 9:58














$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07






$begingroup$
In the french wiki they say : let $(A,mathcal A, mu)$ and $(B,mathcal B, nu)$ two $sigma -$finite measure and let $(Atimes B, mathcal Atimes mathcal B, nutimes mu)$ the product measure. If $f:Atimes Bto [0,infty ]$ is $mathcal Atimes mathcal B$ measurable then... But ${0}times A$ is $mathcal Mtimes mathcal M$ measurable, so tonelli should apply on $boldsymbol 1_{{0}times A}$.
$endgroup$
– NewMath
Dec 4 '18 at 10:07














$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10




$begingroup$
@NewMath I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:10












$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12






$begingroup$
But this is all the question : It's not because $A$ or $B$ is not in $mathcal M$ that $Atimes B$ won't be in $sigma {Ctimes Dmid C,Din mathcal M}$. So how can you prove that ${0}times A$ not in $sigma {Ctimes Dmid C,Din mathcal M}$ ? Just for example $[a,b[$ is not open but is in $sigma {Usubset mathbb Rmid U open}$... So, $A$ is indeed not in $mathcal M$, but ${0}times A$ could be in $sigma {Ctimes Dmid C,Din mathcal M}$
$endgroup$
– NewMath
Dec 4 '18 at 10:12






1




1




$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:17






$begingroup$
@NewMath You have to look at the proof of Fubini's Theorem for this. In the beginning of the proof of this theorem (in Rudin's book, for example) it is shown that if a set $C$ in the product belongs to the product $mathcal M times mathcal M$ then ${x: (x,y) in C}$ is in $mathcal M$ for each $y$ and ${y: (x,y) in C}$ is in $mathcal M$ for each $x$.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:17




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • 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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025324%2fproduct-measure-difference-between-mathcal-m-times-mathcal-m-and-a-time%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

How to change which sound is reproduced for terminal bell?

Can I use Tabulator js library in my java Spring + Thymeleaf project?