Why is the set of continous paths of a browian motion not measurable?
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2
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Øksendal states in his book "stochastic differential equations" (Defintion 2.2.1 iii), p.13) that
the set $H = { omega mid t → B_t (omega) text{is continuous}
}$ is not measurable with respect to the Borel $sigma$-algebra
$mathcal{B}$ on $(mathbb{R}^n)^{[0,infty)}$ (...) ($H$ involves an uncountable number of $t$'s),
where $B_t$ is a brownian motion and we identify $omega$ with the path of $B_t(omega)$.
Unfortunately I don't know much about $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ aside its defintion. According to this question a set $A$ is measurable iff there exists $Jsubseteq mathbb{R}$ with $|J|≤aleph_0$ and $Bin mathcal{B}(mathbb{R}^J)$ such that $A=B times mathbb{R}^{mathbb{R}setminus J}={f in mathbb{R}^mathbb{R} colon (f(j) colon t in J) in B}$.
I would appreciate it, if someone could provide me a reference for the statement above.
Edit: If I am not mistaken the product $sigma$-algebra $mathcal{B}((mathbb{R}^n))^{[0,infty)}$ is a true subset of the Borel-$sigma$-algebra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ and thus the statement above should be correct for the product, but not for the Borel algebra. Prior to the quote Øksendal writes (after Definition 2.1.4, p. 10):
$mathcal{B}$ [the algebra generated by cylindrical sets] is the same as the Borel $sigma$-algebra on $tilde{Omega}$ [$=(mathbb{R}^n)^T$] if $T = [0,infty)$ and $tilde{Omega}$ is given
the product topology
This should be false or am I missing something?
measure-theory stochastic-calculus brownian-motion stochastic-analysis borel-sets
asked Sep 19 at 14:03
Richard
1029
|
show 4 more comments
up vote
2
down vote
favorite
Øksendal states in his book "stochastic differential equations" (Defintion 2.2.1 iii), p.13) that
the set $H = { omega mid t → B_t (omega) text{is continuous}
}$ is not measurable with respect to the Borel $sigma$-algebra
$mathcal{B}$ on $(mathbb{R}^n)^{[0,infty)}$ (...) ($H$ involves an uncountable number of $t$'s),
where $B_t$ is a brownian motion and we identify $omega$ with the path of $B_t(omega)$.
Unfortunately I don't know much about $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ aside its defintion. According to this question a set $A$ is measurable iff there exists $Jsubseteq mathbb{R}$ with $|J|≤aleph_0$ and $Bin mathcal{B}(mathbb{R}^J)$ such that $A=B times mathbb{R}^{mathbb{R}setminus J}={f in mathbb{R}^mathbb{R} colon (f(j) colon t in J) in B}$.
I would appreciate it, if someone could provide me a reference for the statement above.
Edit: If I am not mistaken the product $sigma$-algebra $mathcal{B}((mathbb{R}^n))^{[0,infty)}$ is a true subset of the Borel-$sigma$-algebra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ and thus the statement above should be correct for the product, but not for the Borel algebra. Prior to the quote Øksendal writes (after Definition 2.1.4, p. 10):
$mathcal{B}$ [the algebra generated by cylindrical sets] is the same as the Borel $sigma$-algebra on $tilde{Omega}$ [$=(mathbb{R}^n)^T$] if $T = [0,infty)$ and $tilde{Omega}$ is given
the product topology
This should be false or am I missing something?
measure-theory stochastic-calculus brownian-motion stochastic-analysis borel-sets
asked Sep 19 at 14:03
Richard
1029
You might want to take a look at this question: math.stackexchange.com/q/508767/36150
– saz
Sep 19 at 16:23
Thank you for your reply. The linked question is asking about the "cylindrical $sigma$-algebra" (which should be $otimes_{t in [0,1]}mathcal{B}(mathbb{R})$?). Isn't the Borel $sigma$-algbra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ bigger than said product? Therefore I still don't understand why $H$ (or $C([0,1])$) could not be measurable with respect to $mathcal{B}((mathbb{R}^n)^{[0,infty)})$.
– Richard
Sep 19 at 18:01
1
What is your definition of the Borel $sigma$-algebra on $(mathbb{R}^n)^{[0,infty)}$? Typically, the Borel-$sigma$-algebra on this space is defined as the smallest $sigma$-algebra which makes all projections $pi_t$ measurable, and hence it equals the cyclindrical $sigma$-algebra.
– saz
Sep 19 at 18:42
Isn't the Borel $sigma$-algebra usually generated by the open sets. So if $tau$ is the product topology on $(mathbb{R}^n)^{[0,infty)}$, I would define $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ as $sigma(tau)$.
– Richard
Sep 19 at 19:50
I'm sorry but I still don't see it. So the product topology is the coarsest topology such that all projections are continuous. Let's consider $pi_i^{-1}({0})=mathbb{R}^{[0,i)}times {0}times mathbb{R}^{(i,infty)}$ for all $i in [0,infty)$ ($n=1$ for simplicity). This would be a closed set and the intersection ${0}timesdotstimes{0}$ would still be closed in $mathbb{R}^{[0,infty)}$ and therefore in $mathcal{B}((mathbb{R})^{[0,infty)})$. Couldn't elements of the $sigma$-algebra generated by cylindrical sets only depend on countable many restrictions?
– Richard
Sep 20 at 11:38
|
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Øksendal states in his book "stochastic differential equations" (Defintion 2.2.1 iii), p.13) that
the set $H = { omega mid t → B_t (omega) text{is continuous}
}$ is not measurable with respect to the Borel $sigma$-algebra
$mathcal{B}$ on $(mathbb{R}^n)^{[0,infty)}$ (...) ($H$ involves an uncountable number of $t$'s),
where $B_t$ is a brownian motion and we identify $omega$ with the path of $B_t(omega)$.
Unfortunately I don't know much about $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ aside its defintion. According to this question a set $A$ is measurable iff there exists $Jsubseteq mathbb{R}$ with $|J|≤aleph_0$ and $Bin mathcal{B}(mathbb{R}^J)$ such that $A=B times mathbb{R}^{mathbb{R}setminus J}={f in mathbb{R}^mathbb{R} colon (f(j) colon t in J) in B}$.
I would appreciate it, if someone could provide me a reference for the statement above.
Edit: If I am not mistaken the product $sigma$-algebra $mathcal{B}((mathbb{R}^n))^{[0,infty)}$ is a true subset of the Borel-$sigma$-algebra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ and thus the statement above should be correct for the product, but not for the Borel algebra. Prior to the quote Øksendal writes (after Definition 2.1.4, p. 10):
$mathcal{B}$ [the algebra generated by cylindrical sets] is the same as the Borel $sigma$-algebra on $tilde{Omega}$ [$=(mathbb{R}^n)^T$] if $T = [0,infty)$ and $tilde{Omega}$ is given
the product topology
This should be false or am I missing something?
measure-theory stochastic-calculus brownian-motion stochastic-analysis borel-sets
asked Sep 19 at 14:03
Richard
1029
Øksendal states in his book "stochastic differential equations" (Defintion 2.2.1 iii), p.13) that
the set $H = { omega mid t → B_t (omega) text{is continuous}
}$ is not measurable with respect to the Borel $sigma$-algebra
$mathcal{B}$ on $(mathbb{R}^n)^{[0,infty)}$ (...) ($H$ involves an uncountable number of $t$'s),
where $B_t$ is a brownian motion and we identify $omega$ with the path of $B_t(omega)$.
Unfortunately I don't know much about $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ aside its defintion. According to this question a set $A$ is measurable iff there exists $Jsubseteq mathbb{R}$ with $|J|≤aleph_0$ and $Bin mathcal{B}(mathbb{R}^J)$ such that $A=B times mathbb{R}^{mathbb{R}setminus J}={f in mathbb{R}^mathbb{R} colon (f(j) colon t in J) in B}$.
I would appreciate it, if someone could provide me a reference for the statement above.
Edit: If I am not mistaken the product $sigma$-algebra $mathcal{B}((mathbb{R}^n))^{[0,infty)}$ is a true subset of the Borel-$sigma$-algebra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ and thus the statement above should be correct for the product, but not for the Borel algebra. Prior to the quote Øksendal writes (after Definition 2.1.4, p. 10):
$mathcal{B}$ [the algebra generated by cylindrical sets] is the same as the Borel $sigma$-algebra on $tilde{Omega}$ [$=(mathbb{R}^n)^T$] if $T = [0,infty)$ and $tilde{Omega}$ is given
the product topology
This should be false or am I missing something?
measure-theory stochastic-calculus brownian-motion stochastic-analysis borel-sets
measure-theory stochastic-calculus brownian-motion stochastic-analysis borel-sets
asked Sep 19 at 14:03
Richard
1029
asked Sep 19 at 14:03
Richard
1029
edited Dec 5 at 23:30
asked Sep 19 at 14:03
Richard
1029
asked Sep 19 at 14:03
Richard
1029
asked Sep 19 at 14:03
Richard
1029
1029
You might want to take a look at this question: math.stackexchange.com/q/508767/36150
– saz
Sep 19 at 16:23
Thank you for your reply. The linked question is asking about the "cylindrical $sigma$-algebra" (which should be $otimes_{t in [0,1]}mathcal{B}(mathbb{R})$?). Isn't the Borel $sigma$-algbra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ bigger than said product? Therefore I still don't understand why $H$ (or $C([0,1])$) could not be measurable with respect to $mathcal{B}((mathbb{R}^n)^{[0,infty)})$.
– Richard
Sep 19 at 18:01
1
What is your definition of the Borel $sigma$-algebra on $(mathbb{R}^n)^{[0,infty)}$? Typically, the Borel-$sigma$-algebra on this space is defined as the smallest $sigma$-algebra which makes all projections $pi_t$ measurable, and hence it equals the cyclindrical $sigma$-algebra.
– saz
Sep 19 at 18:42
Isn't the Borel $sigma$-algebra usually generated by the open sets. So if $tau$ is the product topology on $(mathbb{R}^n)^{[0,infty)}$, I would define $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ as $sigma(tau)$.
– Richard
Sep 19 at 19:50
I'm sorry but I still don't see it. So the product topology is the coarsest topology such that all projections are continuous. Let's consider $pi_i^{-1}({0})=mathbb{R}^{[0,i)}times {0}times mathbb{R}^{(i,infty)}$ for all $i in [0,infty)$ ($n=1$ for simplicity). This would be a closed set and the intersection ${0}timesdotstimes{0}$ would still be closed in $mathbb{R}^{[0,infty)}$ and therefore in $mathcal{B}((mathbb{R})^{[0,infty)})$. Couldn't elements of the $sigma$-algebra generated by cylindrical sets only depend on countable many restrictions?
– Richard
Sep 20 at 11:38
|
show 4 more comments
You might want to take a look at this question: math.stackexchange.com/q/508767/36150
– saz
Sep 19 at 16:23
Thank you for your reply. The linked question is asking about the "cylindrical $sigma$-algebra" (which should be $otimes_{t in [0,1]}mathcal{B}(mathbb{R})$?). Isn't the Borel $sigma$-algbra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ bigger than said product? Therefore I still don't understand why $H$ (or $C([0,1])$) could not be measurable with respect to $mathcal{B}((mathbb{R}^n)^{[0,infty)})$.
– Richard
Sep 19 at 18:01
1
What is your definition of the Borel $sigma$-algebra on $(mathbb{R}^n)^{[0,infty)}$? Typically, the Borel-$sigma$-algebra on this space is defined as the smallest $sigma$-algebra which makes all projections $pi_t$ measurable, and hence it equals the cyclindrical $sigma$-algebra.
– saz
Sep 19 at 18:42
Isn't the Borel $sigma$-algebra usually generated by the open sets. So if $tau$ is the product topology on $(mathbb{R}^n)^{[0,infty)}$, I would define $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ as $sigma(tau)$.
– Richard
Sep 19 at 19:50
I'm sorry but I still don't see it. So the product topology is the coarsest topology such that all projections are continuous. Let's consider $pi_i^{-1}({0})=mathbb{R}^{[0,i)}times {0}times mathbb{R}^{(i,infty)}$ for all $i in [0,infty)$ ($n=1$ for simplicity). This would be a closed set and the intersection ${0}timesdotstimes{0}$ would still be closed in $mathbb{R}^{[0,infty)}$ and therefore in $mathcal{B}((mathbb{R})^{[0,infty)})$. Couldn't elements of the $sigma$-algebra generated by cylindrical sets only depend on countable many restrictions?
– Richard
Sep 20 at 11:38
You might want to take a look at this question: math.stackexchange.com/q/508767/36150
– saz
Sep 19 at 16:23
You might want to take a look at this question: math.stackexchange.com/q/508767/36150
– saz
Sep 19 at 16:23
Thank you for your reply. The linked question is asking about the "cylindrical $sigma$-algebra" (which should be $otimes_{t in [0,1]}mathcal{B}(mathbb{R})$?). Isn't the Borel $sigma$-algbra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ bigger than said product? Therefore I still don't understand why $H$ (or $C([0,1])$) could not be measurable with respect to $mathcal{B}((mathbb{R}^n)^{[0,infty)})$.
– Richard
Sep 19 at 18:01
Thank you for your reply. The linked question is asking about the "cylindrical $sigma$-algebra" (which should be $otimes_{t in [0,1]}mathcal{B}(mathbb{R})$?). Isn't the Borel $sigma$-algbra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ bigger than said product? Therefore I still don't understand why $H$ (or $C([0,1])$) could not be measurable with respect to $mathcal{B}((mathbb{R}^n)^{[0,infty)})$.
– Richard
Sep 19 at 18:01
1
1
What is your definition of the Borel $sigma$-algebra on $(mathbb{R}^n)^{[0,infty)}$? Typically, the Borel-$sigma$-algebra on this space is defined as the smallest $sigma$-algebra which makes all projections $pi_t$ measurable, and hence it equals the cyclindrical $sigma$-algebra.
– saz
Sep 19 at 18:42
What is your definition of the Borel $sigma$-algebra on $(mathbb{R}^n)^{[0,infty)}$? Typically, the Borel-$sigma$-algebra on this space is defined as the smallest $sigma$-algebra which makes all projections $pi_t$ measurable, and hence it equals the cyclindrical $sigma$-algebra.
– saz
Sep 19 at 18:42
Isn't the Borel $sigma$-algebra usually generated by the open sets. So if $tau$ is the product topology on $(mathbb{R}^n)^{[0,infty)}$, I would define $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ as $sigma(tau)$.
– Richard
Sep 19 at 19:50
Isn't the Borel $sigma$-algebra usually generated by the open sets. So if $tau$ is the product topology on $(mathbb{R}^n)^{[0,infty)}$, I would define $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ as $sigma(tau)$.
– Richard
Sep 19 at 19:50
I'm sorry but I still don't see it. So the product topology is the coarsest topology such that all projections are continuous. Let's consider $pi_i^{-1}({0})=mathbb{R}^{[0,i)}times {0}times mathbb{R}^{(i,infty)}$ for all $i in [0,infty)$ ($n=1$ for simplicity). This would be a closed set and the intersection ${0}timesdotstimes{0}$ would still be closed in $mathbb{R}^{[0,infty)}$ and therefore in $mathcal{B}((mathbb{R})^{[0,infty)})$. Couldn't elements of the $sigma$-algebra generated by cylindrical sets only depend on countable many restrictions?
– Richard
Sep 20 at 11:38
I'm sorry but I still don't see it. So the product topology is the coarsest topology such that all projections are continuous. Let's consider $pi_i^{-1}({0})=mathbb{R}^{[0,i)}times {0}times mathbb{R}^{(i,infty)}$ for all $i in [0,infty)$ ($n=1$ for simplicity). This would be a closed set and the intersection ${0}timesdotstimes{0}$ would still be closed in $mathbb{R}^{[0,infty)}$ and therefore in $mathcal{B}((mathbb{R})^{[0,infty)})$. Couldn't elements of the $sigma$-algebra generated by cylindrical sets only depend on countable many restrictions?
– Richard
Sep 20 at 11:38
|
show 4 more comments
1 Answer
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(I'm going to assume that $Omega=Bbb R^{[0,infty)}$.) As Øksendal notes, if $H$ were measurable then there would be a countable set $(t_n)subset[0,infty)$ and a Borel set $Binmathcal B({Bbb R}^{Bbb N})$ such that $H={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$. In particular, this would mean that if $omega$ and $omega^*$ were two elements of $Omega$ with $omega(t_n)=omega^*(t_n)$ for all $ninBbb N$, and if $omegain H$, then so too $omega^*in H$. But clearly if we take any element $omegain H$ (that is, a continuous path) and define $omega^*(t):=omega(t)$ for all $t$ except one point $t^*in [0,infty)setminus{t_1,t_2,ldots}$ and set $omega^*(t^*)=omega(t^*)+17$, then $omega^*$ passes the test (involving $(t_n)$ and $B$) for inclusion in $H$ but $omega^*$ is not continuous.
answered Sep 21 at 19:38
John Dawkins
13k11017
Could you provide me the page number on which Øksendal writes that? Thank you :) Also I'm not entirely sure whether this is correct or not. If I am not mistaken $I := {{0}timesdotstimes{0}}$ should be in $mathcal{B}((mathbb{R})^{[0,infty)})$, but I can't display said set $I$ as $I={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$, since $(t_n)subset[0,infty)$ is supposed to be countable.
– Richard
Sep 22 at 20:20
Sorry, I misquoted. What I wrote is a rephrasing of of your own statement just before the reference "According to this question". In any event, your example $I$ is not a subset of $mathcal B(Bbb R)^{[0,infty)}$.
– John Dawkins
Sep 23 at 16:56
Yes, that's correct. However the "According to this question"-statement seems wrong. I am still unsure whether it is in $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ (not the product) or not.
– Richard
Sep 23 at 20:06
John, why do you say « if H were measurable then there would be a countable set $(t_n)subset [0,infty)$ and a Borel set B» ?
– An old man in the sea.
Oct 1 at 17:00
(I take $n=1$) A set $HsubsetOmega:=Bbb R^{[0,infty)}$ is measurable with respect to the product $sigma$-field $mathcal F:=mathcal B(Bbb R)^{[0,infty)}$ if and only if it has the form ${omega: (omega(t_1),omega(t_2),ldots)in A}$. This is an old result, due to Doob. Here's a sketch. Let $mathcal G$ be the collection of subsets of $Omega$ of the prescribed form. Clearly $mathcal Gsubsetmathcal F$. Also, $mathcal G$ is a $sigma$-field containing all the generators of $mathcal F$. From this it follows that $mathcal Fsubsetmathcal G$.
– John Dawkins
Oct 2 at 21:17
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(I'm going to assume that $Omega=Bbb R^{[0,infty)}$.) As Øksendal notes, if $H$ were measurable then there would be a countable set $(t_n)subset[0,infty)$ and a Borel set $Binmathcal B({Bbb R}^{Bbb N})$ such that $H={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$. In particular, this would mean that if $omega$ and $omega^*$ were two elements of $Omega$ with $omega(t_n)=omega^*(t_n)$ for all $ninBbb N$, and if $omegain H$, then so too $omega^*in H$. But clearly if we take any element $omegain H$ (that is, a continuous path) and define $omega^*(t):=omega(t)$ for all $t$ except one point $t^*in [0,infty)setminus{t_1,t_2,ldots}$ and set $omega^*(t^*)=omega(t^*)+17$, then $omega^*$ passes the test (involving $(t_n)$ and $B$) for inclusion in $H$ but $omega^*$ is not continuous.
answered Sep 21 at 19:38
John Dawkins
13k11017
Could you provide me the page number on which Øksendal writes that? Thank you :) Also I'm not entirely sure whether this is correct or not. If I am not mistaken $I := {{0}timesdotstimes{0}}$ should be in $mathcal{B}((mathbb{R})^{[0,infty)})$, but I can't display said set $I$ as $I={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$, since $(t_n)subset[0,infty)$ is supposed to be countable.
– Richard
Sep 22 at 20:20
Sorry, I misquoted. What I wrote is a rephrasing of of your own statement just before the reference "According to this question". In any event, your example $I$ is not a subset of $mathcal B(Bbb R)^{[0,infty)}$.
– John Dawkins
Sep 23 at 16:56
Yes, that's correct. However the "According to this question"-statement seems wrong. I am still unsure whether it is in $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ (not the product) or not.
– Richard
Sep 23 at 20:06
John, why do you say « if H were measurable then there would be a countable set $(t_n)subset [0,infty)$ and a Borel set B» ?
– An old man in the sea.
Oct 1 at 17:00
(I take $n=1$) A set $HsubsetOmega:=Bbb R^{[0,infty)}$ is measurable with respect to the product $sigma$-field $mathcal F:=mathcal B(Bbb R)^{[0,infty)}$ if and only if it has the form ${omega: (omega(t_1),omega(t_2),ldots)in A}$. This is an old result, due to Doob. Here's a sketch. Let $mathcal G$ be the collection of subsets of $Omega$ of the prescribed form. Clearly $mathcal Gsubsetmathcal F$. Also, $mathcal G$ is a $sigma$-field containing all the generators of $mathcal F$. From this it follows that $mathcal Fsubsetmathcal G$.
– John Dawkins
Oct 2 at 21:17
add a comment |
up vote
1
down vote
(I'm going to assume that $Omega=Bbb R^{[0,infty)}$.) As Øksendal notes, if $H$ were measurable then there would be a countable set $(t_n)subset[0,infty)$ and a Borel set $Binmathcal B({Bbb R}^{Bbb N})$ such that $H={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$. In particular, this would mean that if $omega$ and $omega^*$ were two elements of $Omega$ with $omega(t_n)=omega^*(t_n)$ for all $ninBbb N$, and if $omegain H$, then so too $omega^*in H$. But clearly if we take any element $omegain H$ (that is, a continuous path) and define $omega^*(t):=omega(t)$ for all $t$ except one point $t^*in [0,infty)setminus{t_1,t_2,ldots}$ and set $omega^*(t^*)=omega(t^*)+17$, then $omega^*$ passes the test (involving $(t_n)$ and $B$) for inclusion in $H$ but $omega^*$ is not continuous.
answered Sep 21 at 19:38
John Dawkins
13k11017
Could you provide me the page number on which Øksendal writes that? Thank you :) Also I'm not entirely sure whether this is correct or not. If I am not mistaken $I := {{0}timesdotstimes{0}}$ should be in $mathcal{B}((mathbb{R})^{[0,infty)})$, but I can't display said set $I$ as $I={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$, since $(t_n)subset[0,infty)$ is supposed to be countable.
– Richard
Sep 22 at 20:20
Sorry, I misquoted. What I wrote is a rephrasing of of your own statement just before the reference "According to this question". In any event, your example $I$ is not a subset of $mathcal B(Bbb R)^{[0,infty)}$.
– John Dawkins
Sep 23 at 16:56
Yes, that's correct. However the "According to this question"-statement seems wrong. I am still unsure whether it is in $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ (not the product) or not.
– Richard
Sep 23 at 20:06
John, why do you say « if H were measurable then there would be a countable set $(t_n)subset [0,infty)$ and a Borel set B» ?
– An old man in the sea.
Oct 1 at 17:00
(I take $n=1$) A set $HsubsetOmega:=Bbb R^{[0,infty)}$ is measurable with respect to the product $sigma$-field $mathcal F:=mathcal B(Bbb R)^{[0,infty)}$ if and only if it has the form ${omega: (omega(t_1),omega(t_2),ldots)in A}$. This is an old result, due to Doob. Here's a sketch. Let $mathcal G$ be the collection of subsets of $Omega$ of the prescribed form. Clearly $mathcal Gsubsetmathcal F$. Also, $mathcal G$ is a $sigma$-field containing all the generators of $mathcal F$. From this it follows that $mathcal Fsubsetmathcal G$.
– John Dawkins
Oct 2 at 21:17
add a comment |
up vote
1
down vote
up vote
1
down vote
(I'm going to assume that $Omega=Bbb R^{[0,infty)}$.) As Øksendal notes, if $H$ were measurable then there would be a countable set $(t_n)subset[0,infty)$ and a Borel set $Binmathcal B({Bbb R}^{Bbb N})$ such that $H={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$. In particular, this would mean that if $omega$ and $omega^*$ were two elements of $Omega$ with $omega(t_n)=omega^*(t_n)$ for all $ninBbb N$, and if $omegain H$, then so too $omega^*in H$. But clearly if we take any element $omegain H$ (that is, a continuous path) and define $omega^*(t):=omega(t)$ for all $t$ except one point $t^*in [0,infty)setminus{t_1,t_2,ldots}$ and set $omega^*(t^*)=omega(t^*)+17$, then $omega^*$ passes the test (involving $(t_n)$ and $B$) for inclusion in $H$ but $omega^*$ is not continuous.
answered Sep 21 at 19:38
John Dawkins
13k11017
(I'm going to assume that $Omega=Bbb R^{[0,infty)}$.) As Øksendal notes, if $H$ were measurable then there would be a countable set $(t_n)subset[0,infty)$ and a Borel set $Binmathcal B({Bbb R}^{Bbb N})$ such that $H={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$. In particular, this would mean that if $omega$ and $omega^*$ were two elements of $Omega$ with $omega(t_n)=omega^*(t_n)$ for all $ninBbb N$, and if $omegain H$, then so too $omega^*in H$. But clearly if we take any element $omegain H$ (that is, a continuous path) and define $omega^*(t):=omega(t)$ for all $t$ except one point $t^*in [0,infty)setminus{t_1,t_2,ldots}$ and set $omega^*(t^*)=omega(t^*)+17$, then $omega^*$ passes the test (involving $(t_n)$ and $B$) for inclusion in $H$ but $omega^*$ is not continuous.
answered Sep 21 at 19:38
John Dawkins
13k11017
answered Sep 21 at 19:38
John Dawkins
13k11017
answered Sep 21 at 19:38
John Dawkins
13k11017
answered Sep 21 at 19:38
John Dawkins
13k11017
13k11017
Could you provide me the page number on which Øksendal writes that? Thank you :) Also I'm not entirely sure whether this is correct or not. If I am not mistaken $I := {{0}timesdotstimes{0}}$ should be in $mathcal{B}((mathbb{R})^{[0,infty)})$, but I can't display said set $I$ as $I={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$, since $(t_n)subset[0,infty)$ is supposed to be countable.
– Richard
Sep 22 at 20:20
Sorry, I misquoted. What I wrote is a rephrasing of of your own statement just before the reference "According to this question". In any event, your example $I$ is not a subset of $mathcal B(Bbb R)^{[0,infty)}$.
– John Dawkins
Sep 23 at 16:56
Yes, that's correct. However the "According to this question"-statement seems wrong. I am still unsure whether it is in $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ (not the product) or not.
– Richard
Sep 23 at 20:06
John, why do you say « if H were measurable then there would be a countable set $(t_n)subset [0,infty)$ and a Borel set B» ?
– An old man in the sea.
Oct 1 at 17:00
(I take $n=1$) A set $HsubsetOmega:=Bbb R^{[0,infty)}$ is measurable with respect to the product $sigma$-field $mathcal F:=mathcal B(Bbb R)^{[0,infty)}$ if and only if it has the form ${omega: (omega(t_1),omega(t_2),ldots)in A}$. This is an old result, due to Doob. Here's a sketch. Let $mathcal G$ be the collection of subsets of $Omega$ of the prescribed form. Clearly $mathcal Gsubsetmathcal F$. Also, $mathcal G$ is a $sigma$-field containing all the generators of $mathcal F$. From this it follows that $mathcal Fsubsetmathcal G$.
– John Dawkins
Oct 2 at 21:17
add a comment |
Could you provide me the page number on which Øksendal writes that? Thank you :) Also I'm not entirely sure whether this is correct or not. If I am not mistaken $I := {{0}timesdotstimes{0}}$ should be in $mathcal{B}((mathbb{R})^{[0,infty)})$, but I can't display said set $I$ as $I={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$, since $(t_n)subset[0,infty)$ is supposed to be countable.
– Richard
Sep 22 at 20:20
Sorry, I misquoted. What I wrote is a rephrasing of of your own statement just before the reference "According to this question". In any event, your example $I$ is not a subset of $mathcal B(Bbb R)^{[0,infty)}$.
– John Dawkins
Sep 23 at 16:56
Yes, that's correct. However the "According to this question"-statement seems wrong. I am still unsure whether it is in $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ (not the product) or not.
– Richard
Sep 23 at 20:06
John, why do you say « if H were measurable then there would be a countable set $(t_n)subset [0,infty)$ and a Borel set B» ?
– An old man in the sea.
Oct 1 at 17:00
(I take $n=1$) A set $HsubsetOmega:=Bbb R^{[0,infty)}$ is measurable with respect to the product $sigma$-field $mathcal F:=mathcal B(Bbb R)^{[0,infty)}$ if and only if it has the form ${omega: (omega(t_1),omega(t_2),ldots)in A}$. This is an old result, due to Doob. Here's a sketch. Let $mathcal G$ be the collection of subsets of $Omega$ of the prescribed form. Clearly $mathcal Gsubsetmathcal F$. Also, $mathcal G$ is a $sigma$-field containing all the generators of $mathcal F$. From this it follows that $mathcal Fsubsetmathcal G$.
– John Dawkins
Oct 2 at 21:17
Could you provide me the page number on which Øksendal writes that? Thank you :) Also I'm not entirely sure whether this is correct or not. If I am not mistaken $I := {{0}timesdotstimes{0}}$ should be in $mathcal{B}((mathbb{R})^{[0,infty)})$, but I can't display said set $I$ as $I={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$, since $(t_n)subset[0,infty)$ is supposed to be countable.
– Richard
Sep 22 at 20:20
Could you provide me the page number on which Øksendal writes that? Thank you :) Also I'm not entirely sure whether this is correct or not. If I am not mistaken $I := {{0}timesdotstimes{0}}$ should be in $mathcal{B}((mathbb{R})^{[0,infty)})$, but I can't display said set $I$ as $I={omegainOmega: (omega(t_1),omega(t_2),ldots)in B}$, since $(t_n)subset[0,infty)$ is supposed to be countable.
– Richard
Sep 22 at 20:20
Sorry, I misquoted. What I wrote is a rephrasing of of your own statement just before the reference "According to this question". In any event, your example $I$ is not a subset of $mathcal B(Bbb R)^{[0,infty)}$.
– John Dawkins
Sep 23 at 16:56
Sorry, I misquoted. What I wrote is a rephrasing of of your own statement just before the reference "According to this question". In any event, your example $I$ is not a subset of $mathcal B(Bbb R)^{[0,infty)}$.
– John Dawkins
Sep 23 at 16:56
Yes, that's correct. However the "According to this question"-statement seems wrong. I am still unsure whether it is in $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ (not the product) or not.
– Richard
Sep 23 at 20:06
Yes, that's correct. However the "According to this question"-statement seems wrong. I am still unsure whether it is in $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ (not the product) or not.
– Richard
Sep 23 at 20:06
John, why do you say « if H were measurable then there would be a countable set $(t_n)subset [0,infty)$ and a Borel set B» ?
– An old man in the sea.
Oct 1 at 17:00
John, why do you say « if H were measurable then there would be a countable set $(t_n)subset [0,infty)$ and a Borel set B» ?
– An old man in the sea.
Oct 1 at 17:00
(I take $n=1$) A set $HsubsetOmega:=Bbb R^{[0,infty)}$ is measurable with respect to the product $sigma$-field $mathcal F:=mathcal B(Bbb R)^{[0,infty)}$ if and only if it has the form ${omega: (omega(t_1),omega(t_2),ldots)in A}$. This is an old result, due to Doob. Here's a sketch. Let $mathcal G$ be the collection of subsets of $Omega$ of the prescribed form. Clearly $mathcal Gsubsetmathcal F$. Also, $mathcal G$ is a $sigma$-field containing all the generators of $mathcal F$. From this it follows that $mathcal Fsubsetmathcal G$.
– John Dawkins
Oct 2 at 21:17
(I take $n=1$) A set $HsubsetOmega:=Bbb R^{[0,infty)}$ is measurable with respect to the product $sigma$-field $mathcal F:=mathcal B(Bbb R)^{[0,infty)}$ if and only if it has the form ${omega: (omega(t_1),omega(t_2),ldots)in A}$. This is an old result, due to Doob. Here's a sketch. Let $mathcal G$ be the collection of subsets of $Omega$ of the prescribed form. Clearly $mathcal Gsubsetmathcal F$. Also, $mathcal G$ is a $sigma$-field containing all the generators of $mathcal F$. From this it follows that $mathcal Fsubsetmathcal G$.
– John Dawkins
Oct 2 at 21:17
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You might want to take a look at this question: math.stackexchange.com/q/508767/36150
– saz
Sep 19 at 16:23
Thank you for your reply. The linked question is asking about the "cylindrical $sigma$-algebra" (which should be $otimes_{t in [0,1]}mathcal{B}(mathbb{R})$?). Isn't the Borel $sigma$-algbra $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ bigger than said product? Therefore I still don't understand why $H$ (or $C([0,1])$) could not be measurable with respect to $mathcal{B}((mathbb{R}^n)^{[0,infty)})$.
– Richard
Sep 19 at 18:01
1
What is your definition of the Borel $sigma$-algebra on $(mathbb{R}^n)^{[0,infty)}$? Typically, the Borel-$sigma$-algebra on this space is defined as the smallest $sigma$-algebra which makes all projections $pi_t$ measurable, and hence it equals the cyclindrical $sigma$-algebra.
– saz
Sep 19 at 18:42
Isn't the Borel $sigma$-algebra usually generated by the open sets. So if $tau$ is the product topology on $(mathbb{R}^n)^{[0,infty)}$, I would define $mathcal{B}((mathbb{R}^n)^{[0,infty)})$ as $sigma(tau)$.
– Richard
Sep 19 at 19:50
I'm sorry but I still don't see it. So the product topology is the coarsest topology such that all projections are continuous. Let's consider $pi_i^{-1}({0})=mathbb{R}^{[0,i)}times {0}times mathbb{R}^{(i,infty)}$ for all $i in [0,infty)$ ($n=1$ for simplicity). This would be a closed set and the intersection ${0}timesdotstimes{0}$ would still be closed in $mathbb{R}^{[0,infty)}$ and therefore in $mathcal{B}((mathbb{R})^{[0,infty)})$. Couldn't elements of the $sigma$-algebra generated by cylindrical sets only depend on countable many restrictions?
– Richard
Sep 20 at 11:38