Measurability of the diagonal in the product space












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Let $(E,mathcal E)$ be a measurable space. Under which assumption on $(E,mathcal E)$ can we show that $Delta:=left{(x,x):xin Eright}inmathcal Eotimesmathcal E$? Note that this doesn't hold in general. Is it correct, for example, if $E$ is a Polish space and $mathcal E=mathcal B(E)$? A reference with a proof would be enough for me.










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  • $begingroup$
    Looking at the accepted answer to the question you linked, my guess is that $Deltainmathcal{E}otimesmathcal{E}$ iff $|E|leq2^{aleph_0}$ and ${x}inmathcal{E}$ for all $xin E$.
    $endgroup$
    – SmileyCraft
    Jan 2 at 0:29
















2












$begingroup$


Let $(E,mathcal E)$ be a measurable space. Under which assumption on $(E,mathcal E)$ can we show that $Delta:=left{(x,x):xin Eright}inmathcal Eotimesmathcal E$? Note that this doesn't hold in general. Is it correct, for example, if $E$ is a Polish space and $mathcal E=mathcal B(E)$? A reference with a proof would be enough for me.










share|cite|improve this question









$endgroup$












  • $begingroup$
    Looking at the accepted answer to the question you linked, my guess is that $Deltainmathcal{E}otimesmathcal{E}$ iff $|E|leq2^{aleph_0}$ and ${x}inmathcal{E}$ for all $xin E$.
    $endgroup$
    – SmileyCraft
    Jan 2 at 0:29














2












2








2





$begingroup$


Let $(E,mathcal E)$ be a measurable space. Under which assumption on $(E,mathcal E)$ can we show that $Delta:=left{(x,x):xin Eright}inmathcal Eotimesmathcal E$? Note that this doesn't hold in general. Is it correct, for example, if $E$ is a Polish space and $mathcal E=mathcal B(E)$? A reference with a proof would be enough for me.










share|cite|improve this question









$endgroup$




Let $(E,mathcal E)$ be a measurable space. Under which assumption on $(E,mathcal E)$ can we show that $Delta:=left{(x,x):xin Eright}inmathcal Eotimesmathcal E$? Note that this doesn't hold in general. Is it correct, for example, if $E$ is a Polish space and $mathcal E=mathcal B(E)$? A reference with a proof would be enough for me.







measure-theory reference-request






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asked Jan 2 at 0:19









0xbadf00d0xbadf00d

1,88741530




1,88741530












  • $begingroup$
    Looking at the accepted answer to the question you linked, my guess is that $Deltainmathcal{E}otimesmathcal{E}$ iff $|E|leq2^{aleph_0}$ and ${x}inmathcal{E}$ for all $xin E$.
    $endgroup$
    – SmileyCraft
    Jan 2 at 0:29


















  • $begingroup$
    Looking at the accepted answer to the question you linked, my guess is that $Deltainmathcal{E}otimesmathcal{E}$ iff $|E|leq2^{aleph_0}$ and ${x}inmathcal{E}$ for all $xin E$.
    $endgroup$
    – SmileyCraft
    Jan 2 at 0:29
















$begingroup$
Looking at the accepted answer to the question you linked, my guess is that $Deltainmathcal{E}otimesmathcal{E}$ iff $|E|leq2^{aleph_0}$ and ${x}inmathcal{E}$ for all $xin E$.
$endgroup$
– SmileyCraft
Jan 2 at 0:29




$begingroup$
Looking at the accepted answer to the question you linked, my guess is that $Deltainmathcal{E}otimesmathcal{E}$ iff $|E|leq2^{aleph_0}$ and ${x}inmathcal{E}$ for all $xin E$.
$endgroup$
– SmileyCraft
Jan 2 at 0:29










1 Answer
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True for the Borel sigma algebra of any second countable space (in partular separable metric space).






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    Do you have a reference at hand?
    $endgroup$
    – 0xbadf00d
    Jan 2 at 10:15










  • $begingroup$
    It is easy to show that the product sigma field is nothing but the Borel sigma field of the product space under the assumption of second countability. Since the Diagonal is a closed set its complement is open, hence Borel in the product, hence belongs to the product sigma filed.
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 11:40











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

True for the Borel sigma algebra of any second countable space (in partular separable metric space).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you have a reference at hand?
    $endgroup$
    – 0xbadf00d
    Jan 2 at 10:15










  • $begingroup$
    It is easy to show that the product sigma field is nothing but the Borel sigma field of the product space under the assumption of second countability. Since the Diagonal is a closed set its complement is open, hence Borel in the product, hence belongs to the product sigma filed.
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 11:40
















1












$begingroup$

True for the Borel sigma algebra of any second countable space (in partular separable metric space).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you have a reference at hand?
    $endgroup$
    – 0xbadf00d
    Jan 2 at 10:15










  • $begingroup$
    It is easy to show that the product sigma field is nothing but the Borel sigma field of the product space under the assumption of second countability. Since the Diagonal is a closed set its complement is open, hence Borel in the product, hence belongs to the product sigma filed.
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 11:40














1












1








1





$begingroup$

True for the Borel sigma algebra of any second countable space (in partular separable metric space).






share|cite|improve this answer









$endgroup$



True for the Borel sigma algebra of any second countable space (in partular separable metric space).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 8:25









Kavi Rama MurthyKavi Rama Murthy

55.4k42057




55.4k42057












  • $begingroup$
    Do you have a reference at hand?
    $endgroup$
    – 0xbadf00d
    Jan 2 at 10:15










  • $begingroup$
    It is easy to show that the product sigma field is nothing but the Borel sigma field of the product space under the assumption of second countability. Since the Diagonal is a closed set its complement is open, hence Borel in the product, hence belongs to the product sigma filed.
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 11:40


















  • $begingroup$
    Do you have a reference at hand?
    $endgroup$
    – 0xbadf00d
    Jan 2 at 10:15










  • $begingroup$
    It is easy to show that the product sigma field is nothing but the Borel sigma field of the product space under the assumption of second countability. Since the Diagonal is a closed set its complement is open, hence Borel in the product, hence belongs to the product sigma filed.
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 11:40
















$begingroup$
Do you have a reference at hand?
$endgroup$
– 0xbadf00d
Jan 2 at 10:15




$begingroup$
Do you have a reference at hand?
$endgroup$
– 0xbadf00d
Jan 2 at 10:15












$begingroup$
It is easy to show that the product sigma field is nothing but the Borel sigma field of the product space under the assumption of second countability. Since the Diagonal is a closed set its complement is open, hence Borel in the product, hence belongs to the product sigma filed.
$endgroup$
– Kavi Rama Murthy
Jan 2 at 11:40




$begingroup$
It is easy to show that the product sigma field is nothing but the Borel sigma field of the product space under the assumption of second countability. Since the Diagonal is a closed set its complement is open, hence Borel in the product, hence belongs to the product sigma filed.
$endgroup$
– Kavi Rama Murthy
Jan 2 at 11:40


















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