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.
measure-theory reference-request
asked Jan 2 at 0:19
0xbadf00d0xbadf00d
1,88741530
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add a comment |
$begingroup$
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.
measure-theory reference-request
asked Jan 2 at 0:19
0xbadf00d0xbadf00d
1,88741530
$endgroup$
$begingroup$
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 $xin E$.
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– SmileyCraft
Jan 2 at 0:29
add a comment |
$begingroup$
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.
measure-theory reference-request
asked Jan 2 at 0:19
0xbadf00d0xbadf00d
1,88741530
$endgroup$
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
measure-theory reference-request
asked Jan 2 at 0:19
0xbadf00d0xbadf00d
1,88741530
asked Jan 2 at 0:19
0xbadf00d0xbadf00d
1,88741530
asked Jan 2 at 0:19
0xbadf00d0xbadf00d
1,88741530
asked Jan 2 at 0:19
0xbadf00d0xbadf00d
1,88741530
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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votes
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True for the Borel sigma algebra of any second countable space (in partular separable metric space).
answered Jan 2 at 8:25
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
$begingroup$
Do you have a reference at hand?
$endgroup$
– 0xbadf00d
Jan 2 at 10:15
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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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
True for the Borel sigma algebra of any second countable space (in partular separable metric space).
answered Jan 2 at 8:25
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
True for the Borel sigma algebra of any second countable space (in partular separable metric space).
answered Jan 2 at 8:25
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
True for the Borel sigma algebra of any second countable space (in partular separable metric space).
answered Jan 2 at 8:25
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
True for the Borel sigma algebra of any second countable space (in partular separable metric space).
answered Jan 2 at 8:25
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Jan 2 at 8:25
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Jan 2 at 8:25
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
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
add a comment |
$begingroup$
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
add a comment |
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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