If $Atimes B$ is an element of a product $sigma$-algebra $mathcal Atimesmathcal B$ then do we have...












2












$begingroup$


If $mathcal A$ is a $sigma$-algebra on set $X$ and $mathcal B$ is a $sigma$-algebra on set $Y$ then it is well known that $mathcal Atimesmathcal B$ is the notation for a $sigma$-algebra on $Xtimes Y$ that is generated by sets $Atimes B$ with $Ainmathcal A$ and $Binmathcal B$.



So evidently we have: $$Ainmathcal Atext{ and }Binmathcal Bimplies Atimes Binmathcal Atimesmathcal Btag1$$



Now my question:




Is the converse of $(1)$ also true?




I have always believed it is without bothering, but when I tried to find a proof for this "obvious" fact I regretfully failed.






probability-theory measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    thank you for asking the question :)
    $endgroup$
    – NewMath
    Dec 4 '18 at 14:04
















2












$begingroup$


If $mathcal A$ is a $sigma$-algebra on set $X$ and $mathcal B$ is a $sigma$-algebra on set $Y$ then it is well known that $mathcal Atimesmathcal B$ is the notation for a $sigma$-algebra on $Xtimes Y$ that is generated by sets $Atimes B$ with $Ainmathcal A$ and $Binmathcal B$.



So evidently we have: $$Ainmathcal Atext{ and }Binmathcal Bimplies Atimes Binmathcal Atimesmathcal Btag1$$



Now my question:




Is the converse of $(1)$ also true?




I have always believed it is without bothering, but when I tried to find a proof for this "obvious" fact I regretfully failed.






probability-theory measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    thank you for asking the question :)
    $endgroup$
    – NewMath
    Dec 4 '18 at 14:04














2












2








2





$begingroup$


If $mathcal A$ is a $sigma$-algebra on set $X$ and $mathcal B$ is a $sigma$-algebra on set $Y$ then it is well known that $mathcal Atimesmathcal B$ is the notation for a $sigma$-algebra on $Xtimes Y$ that is generated by sets $Atimes B$ with $Ainmathcal A$ and $Binmathcal B$.



So evidently we have: $$Ainmathcal Atext{ and }Binmathcal Bimplies Atimes Binmathcal Atimesmathcal Btag1$$



Now my question:




Is the converse of $(1)$ also true?




I have always believed it is without bothering, but when I tried to find a proof for this "obvious" fact I regretfully failed.






probability-theory measure-theory







share|cite|improve this question









$endgroup$




If $mathcal A$ is a $sigma$-algebra on set $X$ and $mathcal B$ is a $sigma$-algebra on set $Y$ then it is well known that $mathcal Atimesmathcal B$ is the notation for a $sigma$-algebra on $Xtimes Y$ that is generated by sets $Atimes B$ with $Ainmathcal A$ and $Binmathcal B$.



So evidently we have: $$Ainmathcal Atext{ and }Binmathcal Bimplies Atimes Binmathcal Atimesmathcal Btag1$$



Now my question:




Is the converse of $(1)$ also true?




I have always believed it is without bothering, but when I tried to find a proof for this "obvious" fact I regretfully failed.






probability-theory measure-theory




probability-theory 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 10:12









VeraVera

2,467617




2,467617












  • $begingroup$
    thank you for asking the question :)
    $endgroup$
    – NewMath
    Dec 4 '18 at 14:04


















  • $begingroup$
    thank you for asking the question :)
    $endgroup$
    – NewMath
    Dec 4 '18 at 14:04
















$begingroup$
thank you for asking the question :)
$endgroup$
– NewMath
Dec 4 '18 at 14:04




$begingroup$
thank you for asking the question :)
$endgroup$
– NewMath
Dec 4 '18 at 14:04










2 Answers
2






active

oldest

votes


















2












$begingroup$

Fortunately I found a proof for it myself after all.




Let $p_{1}:Xtimes Yto X$ and $p_{2}:Xtimes Yto Y$ be prescribed
by $left(x,yright)mapsto x$ and $left(x,yright)mapsto y$ respectively.



Characteristic for product space $left(Xtimes Y,mathcal{A}timesmathcal{B}right)$
is that for any measurable space $left(Z,mathcal{C}right)$ and every function $f:Zto Xtimes Y$ we have: $$f:Zto Xtimes Ytext{ is measurable if and only if }p_{1}circ ftext{ and
}p_{2}circ ftext{ are both measurable}tag1$$



The case is trivial if $B=varnothing$ so let it be that $Bneqvarnothing$.



Now for some $yin B$ let $u_{y}:Xto Xtimes Y$ be prescribed
by $xmapstoleft(x,yright)$.



Then based on $left(1right)$ we find easily that $u_{y}$ is measurable.



Then consequently $A=u_{y}^{-1}left(Atimes Bright)inmathcal{A}$.



Similarly it can be proved that $Binmathcal{B}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we set $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53










  • $begingroup$
    @NewMath No. We start with sets $Asubseteq X$ and $Bsubseteq Y$ and arrive at the conclusion that $Ainmathcal Atext{ and }Binmathcal Biff Atimes Binmathcal Atimesmathcal B$. So we are only talking about subsets of $Xtimes Y$ that can be written as $Atimes B$. The product $sigma$-algebra $mathcal Atimesmathcal B$ also contains (lots of) sets that cannot be written like that.
    $endgroup$
    – Vera
    Dec 9 '18 at 14:50





















1












$begingroup$

Both implications are true and they are proved in the beginning of Fubini's Theorem. For example you look at the discussion of 'sections' of measurable sets in Rudin's RCA.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the information. I do not have the disposal of Rudin, and actually I would like to say a proof of it on this site. Could you provide a link to the proof of Fubini's theorem maybe?
    $endgroup$
    – Vera
    Dec 4 '18 at 10:34










  • $begingroup$
    @Vera You can find a proof on Wikipedia.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:37










  • $begingroup$
    Sorry, but the info on wikipedia about Fubini's theorem is not enough (for me) to provide in a proof. My question is still one that has not been answered yet.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:01










  • $begingroup$
    I have answered my own question. Would you be so kind to check it, please? Thank you.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:46










  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we define $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? Acctually, ${Atimes Bmid Ain mathcal A, Bin mathcal B}$ is a $sigma -$algebra, no ? So $sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}={Atimes Bmid Ain mathcal A, Bin mathcal B}$ is straightforward... why we precise $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? are there cases where it's wrong ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Fortunately I found a proof for it myself after all.




Let $p_{1}:Xtimes Yto X$ and $p_{2}:Xtimes Yto Y$ be prescribed
by $left(x,yright)mapsto x$ and $left(x,yright)mapsto y$ respectively.



Characteristic for product space $left(Xtimes Y,mathcal{A}timesmathcal{B}right)$
is that for any measurable space $left(Z,mathcal{C}right)$ and every function $f:Zto Xtimes Y$ we have: $$f:Zto Xtimes Ytext{ is measurable if and only if }p_{1}circ ftext{ and
}p_{2}circ ftext{ are both measurable}tag1$$



The case is trivial if $B=varnothing$ so let it be that $Bneqvarnothing$.



Now for some $yin B$ let $u_{y}:Xto Xtimes Y$ be prescribed
by $xmapstoleft(x,yright)$.



Then based on $left(1right)$ we find easily that $u_{y}$ is measurable.



Then consequently $A=u_{y}^{-1}left(Atimes Bright)inmathcal{A}$.



Similarly it can be proved that $Binmathcal{B}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we set $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53










  • $begingroup$
    @NewMath No. We start with sets $Asubseteq X$ and $Bsubseteq Y$ and arrive at the conclusion that $Ainmathcal Atext{ and }Binmathcal Biff Atimes Binmathcal Atimesmathcal B$. So we are only talking about subsets of $Xtimes Y$ that can be written as $Atimes B$. The product $sigma$-algebra $mathcal Atimesmathcal B$ also contains (lots of) sets that cannot be written like that.
    $endgroup$
    – Vera
    Dec 9 '18 at 14:50


















2












$begingroup$

Fortunately I found a proof for it myself after all.




Let $p_{1}:Xtimes Yto X$ and $p_{2}:Xtimes Yto Y$ be prescribed
by $left(x,yright)mapsto x$ and $left(x,yright)mapsto y$ respectively.



Characteristic for product space $left(Xtimes Y,mathcal{A}timesmathcal{B}right)$
is that for any measurable space $left(Z,mathcal{C}right)$ and every function $f:Zto Xtimes Y$ we have: $$f:Zto Xtimes Ytext{ is measurable if and only if }p_{1}circ ftext{ and
}p_{2}circ ftext{ are both measurable}tag1$$



The case is trivial if $B=varnothing$ so let it be that $Bneqvarnothing$.



Now for some $yin B$ let $u_{y}:Xto Xtimes Y$ be prescribed
by $xmapstoleft(x,yright)$.



Then based on $left(1right)$ we find easily that $u_{y}$ is measurable.



Then consequently $A=u_{y}^{-1}left(Atimes Bright)inmathcal{A}$.



Similarly it can be proved that $Binmathcal{B}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we set $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53










  • $begingroup$
    @NewMath No. We start with sets $Asubseteq X$ and $Bsubseteq Y$ and arrive at the conclusion that $Ainmathcal Atext{ and }Binmathcal Biff Atimes Binmathcal Atimesmathcal B$. So we are only talking about subsets of $Xtimes Y$ that can be written as $Atimes B$. The product $sigma$-algebra $mathcal Atimesmathcal B$ also contains (lots of) sets that cannot be written like that.
    $endgroup$
    – Vera
    Dec 9 '18 at 14:50
















2












2








2





$begingroup$

Fortunately I found a proof for it myself after all.




Let $p_{1}:Xtimes Yto X$ and $p_{2}:Xtimes Yto Y$ be prescribed
by $left(x,yright)mapsto x$ and $left(x,yright)mapsto y$ respectively.



Characteristic for product space $left(Xtimes Y,mathcal{A}timesmathcal{B}right)$
is that for any measurable space $left(Z,mathcal{C}right)$ and every function $f:Zto Xtimes Y$ we have: $$f:Zto Xtimes Ytext{ is measurable if and only if }p_{1}circ ftext{ and
}p_{2}circ ftext{ are both measurable}tag1$$



The case is trivial if $B=varnothing$ so let it be that $Bneqvarnothing$.



Now for some $yin B$ let $u_{y}:Xto Xtimes Y$ be prescribed
by $xmapstoleft(x,yright)$.



Then based on $left(1right)$ we find easily that $u_{y}$ is measurable.



Then consequently $A=u_{y}^{-1}left(Atimes Bright)inmathcal{A}$.



Similarly it can be proved that $Binmathcal{B}$.






share|cite|improve this answer









$endgroup$



Fortunately I found a proof for it myself after all.




Let $p_{1}:Xtimes Yto X$ and $p_{2}:Xtimes Yto Y$ be prescribed
by $left(x,yright)mapsto x$ and $left(x,yright)mapsto y$ respectively.



Characteristic for product space $left(Xtimes Y,mathcal{A}timesmathcal{B}right)$
is that for any measurable space $left(Z,mathcal{C}right)$ and every function $f:Zto Xtimes Y$ we have: $$f:Zto Xtimes Ytext{ is measurable if and only if }p_{1}circ ftext{ and
}p_{2}circ ftext{ are both measurable}tag1$$



The case is trivial if $B=varnothing$ so let it be that $Bneqvarnothing$.



Now for some $yin B$ let $u_{y}:Xto Xtimes Y$ be prescribed
by $xmapstoleft(x,yright)$.



Then based on $left(1right)$ we find easily that $u_{y}$ is measurable.



Then consequently $A=u_{y}^{-1}left(Atimes Bright)inmathcal{A}$.



Similarly it can be proved that $Binmathcal{B}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 11:41









VeraVera

2,467617




2,467617












  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we set $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53










  • $begingroup$
    @NewMath No. We start with sets $Asubseteq X$ and $Bsubseteq Y$ and arrive at the conclusion that $Ainmathcal Atext{ and }Binmathcal Biff Atimes Binmathcal Atimesmathcal B$. So we are only talking about subsets of $Xtimes Y$ that can be written as $Atimes B$. The product $sigma$-algebra $mathcal Atimesmathcal B$ also contains (lots of) sets that cannot be written like that.
    $endgroup$
    – Vera
    Dec 9 '18 at 14:50




















  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we set $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53










  • $begingroup$
    @NewMath No. We start with sets $Asubseteq X$ and $Bsubseteq Y$ and arrive at the conclusion that $Ainmathcal Atext{ and }Binmathcal Biff Atimes Binmathcal Atimesmathcal B$. So we are only talking about subsets of $Xtimes Y$ that can be written as $Atimes B$. The product $sigma$-algebra $mathcal Atimesmathcal B$ also contains (lots of) sets that cannot be written like that.
    $endgroup$
    – Vera
    Dec 9 '18 at 14:50


















$begingroup$
So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we set $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ?
$endgroup$
– NewMath
Dec 4 '18 at 18:53




$begingroup$
So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we set $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ?
$endgroup$
– NewMath
Dec 4 '18 at 18:53












$begingroup$
@NewMath No. We start with sets $Asubseteq X$ and $Bsubseteq Y$ and arrive at the conclusion that $Ainmathcal Atext{ and }Binmathcal Biff Atimes Binmathcal Atimesmathcal B$. So we are only talking about subsets of $Xtimes Y$ that can be written as $Atimes B$. The product $sigma$-algebra $mathcal Atimesmathcal B$ also contains (lots of) sets that cannot be written like that.
$endgroup$
– Vera
Dec 9 '18 at 14:50






$begingroup$
@NewMath No. We start with sets $Asubseteq X$ and $Bsubseteq Y$ and arrive at the conclusion that $Ainmathcal Atext{ and }Binmathcal Biff Atimes Binmathcal Atimesmathcal B$. So we are only talking about subsets of $Xtimes Y$ that can be written as $Atimes B$. The product $sigma$-algebra $mathcal Atimesmathcal B$ also contains (lots of) sets that cannot be written like that.
$endgroup$
– Vera
Dec 9 '18 at 14:50













1












$begingroup$

Both implications are true and they are proved in the beginning of Fubini's Theorem. For example you look at the discussion of 'sections' of measurable sets in Rudin's RCA.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the information. I do not have the disposal of Rudin, and actually I would like to say a proof of it on this site. Could you provide a link to the proof of Fubini's theorem maybe?
    $endgroup$
    – Vera
    Dec 4 '18 at 10:34










  • $begingroup$
    @Vera You can find a proof on Wikipedia.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:37










  • $begingroup$
    Sorry, but the info on wikipedia about Fubini's theorem is not enough (for me) to provide in a proof. My question is still one that has not been answered yet.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:01










  • $begingroup$
    I have answered my own question. Would you be so kind to check it, please? Thank you.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:46










  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we define $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? Acctually, ${Atimes Bmid Ain mathcal A, Bin mathcal B}$ is a $sigma -$algebra, no ? So $sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}={Atimes Bmid Ain mathcal A, Bin mathcal B}$ is straightforward... why we precise $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? are there cases where it's wrong ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53


















1












$begingroup$

Both implications are true and they are proved in the beginning of Fubini's Theorem. For example you look at the discussion of 'sections' of measurable sets in Rudin's RCA.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the information. I do not have the disposal of Rudin, and actually I would like to say a proof of it on this site. Could you provide a link to the proof of Fubini's theorem maybe?
    $endgroup$
    – Vera
    Dec 4 '18 at 10:34










  • $begingroup$
    @Vera You can find a proof on Wikipedia.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:37










  • $begingroup$
    Sorry, but the info on wikipedia about Fubini's theorem is not enough (for me) to provide in a proof. My question is still one that has not been answered yet.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:01










  • $begingroup$
    I have answered my own question. Would you be so kind to check it, please? Thank you.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:46










  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we define $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? Acctually, ${Atimes Bmid Ain mathcal A, Bin mathcal B}$ is a $sigma -$algebra, no ? So $sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}={Atimes Bmid Ain mathcal A, Bin mathcal B}$ is straightforward... why we precise $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? are there cases where it's wrong ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53
















1












1








1





$begingroup$

Both implications are true and they are proved in the beginning of Fubini's Theorem. For example you look at the discussion of 'sections' of measurable sets in Rudin's RCA.






share|cite|improve this answer









$endgroup$



Both implications are true and they are proved in the beginning of Fubini's Theorem. For example you look at the discussion of 'sections' of measurable sets in Rudin's RCA.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 10:20









Kavi Rama MurthyKavi Rama Murthy

54.8k32055




54.8k32055












  • $begingroup$
    Thank you for the information. I do not have the disposal of Rudin, and actually I would like to say a proof of it on this site. Could you provide a link to the proof of Fubini's theorem maybe?
    $endgroup$
    – Vera
    Dec 4 '18 at 10:34










  • $begingroup$
    @Vera You can find a proof on Wikipedia.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:37










  • $begingroup$
    Sorry, but the info on wikipedia about Fubini's theorem is not enough (for me) to provide in a proof. My question is still one that has not been answered yet.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:01










  • $begingroup$
    I have answered my own question. Would you be so kind to check it, please? Thank you.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:46










  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we define $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? Acctually, ${Atimes Bmid Ain mathcal A, Bin mathcal B}$ is a $sigma -$algebra, no ? So $sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}={Atimes Bmid Ain mathcal A, Bin mathcal B}$ is straightforward... why we precise $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? are there cases where it's wrong ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53




















  • $begingroup$
    Thank you for the information. I do not have the disposal of Rudin, and actually I would like to say a proof of it on this site. Could you provide a link to the proof of Fubini's theorem maybe?
    $endgroup$
    – Vera
    Dec 4 '18 at 10:34










  • $begingroup$
    @Vera You can find a proof on Wikipedia.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 10:37










  • $begingroup$
    Sorry, but the info on wikipedia about Fubini's theorem is not enough (for me) to provide in a proof. My question is still one that has not been answered yet.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:01










  • $begingroup$
    I have answered my own question. Would you be so kind to check it, please? Thank you.
    $endgroup$
    – Vera
    Dec 4 '18 at 11:46










  • $begingroup$
    So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we define $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? Acctually, ${Atimes Bmid Ain mathcal A, Bin mathcal B}$ is a $sigma -$algebra, no ? So $sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}={Atimes Bmid Ain mathcal A, Bin mathcal B}$ is straightforward... why we precise $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? are there cases where it's wrong ?
    $endgroup$
    – NewMath
    Dec 4 '18 at 18:53


















$begingroup$
Thank you for the information. I do not have the disposal of Rudin, and actually I would like to say a proof of it on this site. Could you provide a link to the proof of Fubini's theorem maybe?
$endgroup$
– Vera
Dec 4 '18 at 10:34




$begingroup$
Thank you for the information. I do not have the disposal of Rudin, and actually I would like to say a proof of it on this site. Could you provide a link to the proof of Fubini's theorem maybe?
$endgroup$
– Vera
Dec 4 '18 at 10:34












$begingroup$
@Vera You can find a proof on Wikipedia.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 10:37




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@Vera You can find a proof on Wikipedia.
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– Kavi Rama Murthy
Dec 4 '18 at 10:37












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Sorry, but the info on wikipedia about Fubini's theorem is not enough (for me) to provide in a proof. My question is still one that has not been answered yet.
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– Vera
Dec 4 '18 at 11:01




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Sorry, but the info on wikipedia about Fubini's theorem is not enough (for me) to provide in a proof. My question is still one that has not been answered yet.
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– Vera
Dec 4 '18 at 11:01












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I have answered my own question. Would you be so kind to check it, please? Thank you.
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– Vera
Dec 4 '18 at 11:46




$begingroup$
I have answered my own question. Would you be so kind to check it, please? Thank you.
$endgroup$
– Vera
Dec 4 '18 at 11:46












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So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we define $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? Acctually, ${Atimes Bmid Ain mathcal A, Bin mathcal B}$ is a $sigma -$algebra, no ? So $sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}={Atimes Bmid Ain mathcal A, Bin mathcal B}$ is straightforward... why we precise $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? are there cases where it's wrong ?
$endgroup$
– NewMath
Dec 4 '18 at 18:53






$begingroup$
So at the end, $mathcal Atimes mathcal B={Atimes Bmid Ain mathcal A, Bin mathcal B}$... so why do we define $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? Acctually, ${Atimes Bmid Ain mathcal A, Bin mathcal B}$ is a $sigma -$algebra, no ? So $sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}={Atimes Bmid Ain mathcal A, Bin mathcal B}$ is straightforward... why we precise $mathcal Atimes mathcal B=sigma {Atimes Bmid Ain mathcal A, Bin mathcal B}$ ? are there cases where it's wrong ?
$endgroup$
– NewMath
Dec 4 '18 at 18:53




















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