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












2














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






















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
















2














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






















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














2












2








2







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













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









Vera

2,467617




2,467617












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


















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
















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




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










2 Answers
2






active

oldest

votes


















2














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





















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










  • @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.
    – Vera
    Dec 9 '18 at 14:50





















1














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





















  • 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?
    – Vera
    Dec 4 '18 at 10:34










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










  • 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.
    – Vera
    Dec 4 '18 at 11:01










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










  • 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 ?
    – 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














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





















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










  • @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.
    – Vera
    Dec 9 '18 at 14:50


















2














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





















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










  • @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.
    – Vera
    Dec 9 '18 at 14:50
















2












2








2






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












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









Vera

2,467617




2,467617












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










  • @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.
    – Vera
    Dec 9 '18 at 14:50




















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










  • @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.
    – Vera
    Dec 9 '18 at 14:50


















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




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












@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.
– Vera
Dec 9 '18 at 14:50






@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.
– Vera
Dec 9 '18 at 14:50













1














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





















  • 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?
    – Vera
    Dec 4 '18 at 10:34










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










  • 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.
    – Vera
    Dec 4 '18 at 11:01










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










  • 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 ?
    – NewMath
    Dec 4 '18 at 18:53


















1














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





















  • 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?
    – Vera
    Dec 4 '18 at 10:34










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










  • 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.
    – Vera
    Dec 4 '18 at 11:01










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










  • 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 ?
    – NewMath
    Dec 4 '18 at 18:53
















1












1








1






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












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 Murthy

51.2k31855




51.2k31855












  • 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?
    – Vera
    Dec 4 '18 at 10:34










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










  • 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.
    – Vera
    Dec 4 '18 at 11:01










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










  • 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 ?
    – NewMath
    Dec 4 '18 at 18:53




















  • 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?
    – Vera
    Dec 4 '18 at 10:34










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










  • 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.
    – Vera
    Dec 4 '18 at 11:01










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










  • 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 ?
    – NewMath
    Dec 4 '18 at 18:53


















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?
– Vera
Dec 4 '18 at 10:34




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?
– Vera
Dec 4 '18 at 10:34












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




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












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.
– Vera
Dec 4 '18 at 11:01




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.
– Vera
Dec 4 '18 at 11:01












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




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












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 ?
– NewMath
Dec 4 '18 at 18:53






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 ?
– NewMath
Dec 4 '18 at 18:53




















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