Mixing
Is $Eni xmapsto(x,x)$ only measurable if $E$ is a Polish space?
$begingroup$
Let $(E,mathcal E)$ be a measurable space and $$f:Eto Etimes E;,;;;xmapsto(x,x).$$ I've read that $f$ is $(mathcal E,mathcal Eotimesmathcal E)$-measurable, if $E$ is a Polish space and $mathcal E=mathcal B(E)$. I don't get why $E$ needs to be a Polish space. If $pi_i:Etimes Eto E$ denotes the projection onto the $i$th coordinate, then $pi_icirc f=text{id}_E$ is obviously $(mathcal E,mathcal E)$-measurable. So, $f$ should be $(mathcal E,mathcal Eotimesmathcal E)$-measurable. What am I missing?
measure-theory
asked Jan 1 at 21:52
0xbadf00d0xbadf00d
1,81241430
$endgroup$
|
show 7 more comments
$begingroup$
Let $(E,mathcal E)$ be a measurable space and $$f:Eto Etimes E;,;;;xmapsto(x,x).$$ I've read that $f$ is $(mathcal E,mathcal Eotimesmathcal E)$-measurable, if $E$ is a Polish space and $mathcal E=mathcal B(E)$. I don't get why $E$ needs to be a Polish space. If $pi_i:Etimes Eto E$ denotes the projection onto the $i$th coordinate, then $pi_icirc f=text{id}_E$ is obviously $(mathcal E,mathcal E)$-measurable. So, $f$ should be $(mathcal E,mathcal Eotimesmathcal E)$-measurable. What am I missing?
measure-theory
asked Jan 1 at 21:52
0xbadf00d0xbadf00d
1,81241430
$endgroup$
$begingroup$
Can you justify the "So,"
$endgroup$
– mathworker21
Jan 1 at 21:54
1
$begingroup$
@mathworker21 That's a more general property. $f:Eto F_1times F_2$ is $(mathcal E,mathcal F_1otimesmathcal F_2)$ measurable if and only if $pi_icirc f$ is $(mathcal E,mathcal F_i)$-measurable.
$endgroup$
– 0xbadf00d
Jan 1 at 21:55
$begingroup$
Are you proposing $E$ can be an arbitrary topological space? You need to be able to talk about $mathcal{B}(E)$.
$endgroup$
– mathworker21
Jan 1 at 21:55
$begingroup$
@mathworker21 I'm claiming that $E$ doesn't need to be a topological space at all. $f$ should be measurable for any measurable space $(E,mathcal E)$.
$endgroup$
– 0xbadf00d
Jan 1 at 21:56
$begingroup$
@mathworker21 $mathcal B(E)$ is the Borel $sigma$-algebra on $E$ if $E$ is a topological space. Read the question: I've said that I read that the measurability would follow in the special case where $E$ is Polish and $mathcal E=mathcal B(E)$. I claim that this assumption is not needed, i.e. $(E,mathcal E)$ can be an arbitary measurable space.
$endgroup$
– 0xbadf00d
Jan 1 at 21:57
|
show 7 more comments
$begingroup$
Let $(E,mathcal E)$ be a measurable space and $$f:Eto Etimes E;,;;;xmapsto(x,x).$$ I've read that $f$ is $(mathcal E,mathcal Eotimesmathcal E)$-measurable, if $E$ is a Polish space and $mathcal E=mathcal B(E)$. I don't get why $E$ needs to be a Polish space. If $pi_i:Etimes Eto E$ denotes the projection onto the $i$th coordinate, then $pi_icirc f=text{id}_E$ is obviously $(mathcal E,mathcal E)$-measurable. So, $f$ should be $(mathcal E,mathcal Eotimesmathcal E)$-measurable. What am I missing?
measure-theory
asked Jan 1 at 21:52
0xbadf00d0xbadf00d
1,81241430
$endgroup$
Let $(E,mathcal E)$ be a measurable space and $$f:Eto Etimes E;,;;;xmapsto(x,x).$$ I've read that $f$ is $(mathcal E,mathcal Eotimesmathcal E)$-measurable, if $E$ is a Polish space and $mathcal E=mathcal B(E)$. I don't get why $E$ needs to be a Polish space. If $pi_i:Etimes Eto E$ denotes the projection onto the $i$th coordinate, then $pi_icirc f=text{id}_E$ is obviously $(mathcal E,mathcal E)$-measurable. So, $f$ should be $(mathcal E,mathcal Eotimesmathcal E)$-measurable. What am I missing?
measure-theory
measure-theory
asked Jan 1 at 21:52
0xbadf00d0xbadf00d
1,81241430
asked Jan 1 at 21:52
0xbadf00d0xbadf00d
1,81241430
asked Jan 1 at 21:52
0xbadf00d0xbadf00d
1,81241430
asked Jan 1 at 21:52
0xbadf00d0xbadf00d
1,81241430
asked Jan 1 at 21:52
0xbadf00d0xbadf00d
1,81241430
1,81241430
$begingroup$
Can you justify the "So,"
$endgroup$
– mathworker21
Jan 1 at 21:54
1
$begingroup$
@mathworker21 That's a more general property. $f:Eto F_1times F_2$ is $(mathcal E,mathcal F_1otimesmathcal F_2)$ measurable if and only if $pi_icirc f$ is $(mathcal E,mathcal F_i)$-measurable.
$endgroup$
– 0xbadf00d
Jan 1 at 21:55
$begingroup$
Are you proposing $E$ can be an arbitrary topological space? You need to be able to talk about $mathcal{B}(E)$.
$endgroup$
– mathworker21
Jan 1 at 21:55
$begingroup$
@mathworker21 I'm claiming that $E$ doesn't need to be a topological space at all. $f$ should be measurable for any measurable space $(E,mathcal E)$.
$endgroup$
– 0xbadf00d
Jan 1 at 21:56
$begingroup$
@mathworker21 $mathcal B(E)$ is the Borel $sigma$-algebra on $E$ if $E$ is a topological space. Read the question: I've said that I read that the measurability would follow in the special case where $E$ is Polish and $mathcal E=mathcal B(E)$. I claim that this assumption is not needed, i.e. $(E,mathcal E)$ can be an arbitary measurable space.
$endgroup$
– 0xbadf00d
Jan 1 at 21:57
|
show 7 more comments
$begingroup$
Can you justify the "So,"
$endgroup$
– mathworker21
Jan 1 at 21:54
1
$begingroup$
@mathworker21 That's a more general property. $f:Eto F_1times F_2$ is $(mathcal E,mathcal F_1otimesmathcal F_2)$ measurable if and only if $pi_icirc f$ is $(mathcal E,mathcal F_i)$-measurable.
$endgroup$
– 0xbadf00d
Jan 1 at 21:55
$begingroup$
Are you proposing $E$ can be an arbitrary topological space? You need to be able to talk about $mathcal{B}(E)$.
$endgroup$
– mathworker21
Jan 1 at 21:55
$begingroup$
@mathworker21 I'm claiming that $E$ doesn't need to be a topological space at all. $f$ should be measurable for any measurable space $(E,mathcal E)$.
$endgroup$
– 0xbadf00d
Jan 1 at 21:56
$begingroup$
@mathworker21 $mathcal B(E)$ is the Borel $sigma$-algebra on $E$ if $E$ is a topological space. Read the question: I've said that I read that the measurability would follow in the special case where $E$ is Polish and $mathcal E=mathcal B(E)$. I claim that this assumption is not needed, i.e. $(E,mathcal E)$ can be an arbitary measurable space.
$endgroup$
– 0xbadf00d
Jan 1 at 21:57
$begingroup$
Can you justify the "So,"
$endgroup$
– mathworker21
Jan 1 at 21:54
$begingroup$
Can you justify the "So,"
$endgroup$
– mathworker21
Jan 1 at 21:54
1
1
$begingroup$
@mathworker21 That's a more general property. $f:Eto F_1times F_2$ is $(mathcal E,mathcal F_1otimesmathcal F_2)$ measurable if and only if $pi_icirc f$ is $(mathcal E,mathcal F_i)$-measurable.
$endgroup$
– 0xbadf00d
Jan 1 at 21:55
$begingroup$
@mathworker21 That's a more general property. $f:Eto F_1times F_2$ is $(mathcal E,mathcal F_1otimesmathcal F_2)$ measurable if and only if $pi_icirc f$ is $(mathcal E,mathcal F_i)$-measurable.
$endgroup$
– 0xbadf00d
Jan 1 at 21:55
$begingroup$
Are you proposing $E$ can be an arbitrary topological space? You need to be able to talk about $mathcal{B}(E)$.
$endgroup$
– mathworker21
Jan 1 at 21:55
$begingroup$
Are you proposing $E$ can be an arbitrary topological space? You need to be able to talk about $mathcal{B}(E)$.
$endgroup$
– mathworker21
Jan 1 at 21:55
$begingroup$
@mathworker21 I'm claiming that $E$ doesn't need to be a topological space at all. $f$ should be measurable for any measurable space $(E,mathcal E)$.
$endgroup$
– 0xbadf00d
Jan 1 at 21:56
$begingroup$
@mathworker21 I'm claiming that $E$ doesn't need to be a topological space at all. $f$ should be measurable for any measurable space $(E,mathcal E)$.
$endgroup$
– 0xbadf00d
Jan 1 at 21:56
$begingroup$
@mathworker21 $mathcal B(E)$ is the Borel $sigma$-algebra on $E$ if $E$ is a topological space. Read the question: I've said that I read that the measurability would follow in the special case where $E$ is Polish and $mathcal E=mathcal B(E)$. I claim that this assumption is not needed, i.e. $(E,mathcal E)$ can be an arbitary measurable space.
$endgroup$
– 0xbadf00d
Jan 1 at 21:57
$begingroup$
@mathworker21 $mathcal B(E)$ is the Borel $sigma$-algebra on $E$ if $E$ is a topological space. Read the question: I've said that I read that the measurability would follow in the special case where $E$ is Polish and $mathcal E=mathcal B(E)$. I claim that this assumption is not needed, i.e. $(E,mathcal E)$ can be an arbitary measurable space.
$endgroup$
– 0xbadf00d
Jan 1 at 21:57
|
show 7 more comments
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$begingroup$
Can you justify the "So,"
$endgroup$
– mathworker21
Jan 1 at 21:54
1
$begingroup$
@mathworker21 That's a more general property. $f:Eto F_1times F_2$ is $(mathcal E,mathcal F_1otimesmathcal F_2)$ measurable if and only if $pi_icirc f$ is $(mathcal E,mathcal F_i)$-measurable.
$endgroup$
– 0xbadf00d
Jan 1 at 21:55
$begingroup$
Are you proposing $E$ can be an arbitrary topological space? You need to be able to talk about $mathcal{B}(E)$.
$endgroup$
– mathworker21
Jan 1 at 21:55
$begingroup$
@mathworker21 I'm claiming that $E$ doesn't need to be a topological space at all. $f$ should be measurable for any measurable space $(E,mathcal E)$.
$endgroup$
– 0xbadf00d
Jan 1 at 21:56
$begingroup$
@mathworker21 $mathcal B(E)$ is the Borel $sigma$-algebra on $E$ if $E$ is a topological space. Read the question: I've said that I read that the measurability would follow in the special case where $E$ is Polish and $mathcal E=mathcal B(E)$. I claim that this assumption is not needed, i.e. $(E,mathcal E)$ can be an arbitary measurable space.
$endgroup$
– 0xbadf00d
Jan 1 at 21:57