When ${X_t=Y_t, text{for all $t$}}$ is measurable or not? [closed]











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Let $(X_t)$ and $(Y_t)$ two stochastic process on $(Omega ,mathcal F,mathbb P)$. I heard that ${X_t=Y_ttext{ for all }t}$ or $sup_{tin mathbb R}X_t$ are not always measurable but if $X_t$ and $Y_t$ are both a.s. continuous, then there are measurable.



1) Why is it the case ?



2) Could someone give an example where there are not measurable ?










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closed as off-topic by Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin Dec 4 at 12:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
    – Surb
    Dec 3 at 12:58












  • @Surb: yes I am.
    – NewMath
    Dec 3 at 13:06










  • Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
    – Surb
    Dec 3 at 13:08












  • Could you explain why ? @Surb
    – NewMath
    Dec 3 at 13:08










  • Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
    – Surb
    Dec 3 at 13:09















up vote
-1
down vote

favorite












Let $(X_t)$ and $(Y_t)$ two stochastic process on $(Omega ,mathcal F,mathbb P)$. I heard that ${X_t=Y_ttext{ for all }t}$ or $sup_{tin mathbb R}X_t$ are not always measurable but if $X_t$ and $Y_t$ are both a.s. continuous, then there are measurable.



1) Why is it the case ?



2) Could someone give an example where there are not measurable ?










share|cite|improve this question













closed as off-topic by Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin Dec 4 at 12:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
    – Surb
    Dec 3 at 12:58












  • @Surb: yes I am.
    – NewMath
    Dec 3 at 13:06










  • Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
    – Surb
    Dec 3 at 13:08












  • Could you explain why ? @Surb
    – NewMath
    Dec 3 at 13:08










  • Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
    – Surb
    Dec 3 at 13:09













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Let $(X_t)$ and $(Y_t)$ two stochastic process on $(Omega ,mathcal F,mathbb P)$. I heard that ${X_t=Y_ttext{ for all }t}$ or $sup_{tin mathbb R}X_t$ are not always measurable but if $X_t$ and $Y_t$ are both a.s. continuous, then there are measurable.



1) Why is it the case ?



2) Could someone give an example where there are not measurable ?










share|cite|improve this question













Let $(X_t)$ and $(Y_t)$ two stochastic process on $(Omega ,mathcal F,mathbb P)$. I heard that ${X_t=Y_ttext{ for all }t}$ or $sup_{tin mathbb R}X_t$ are not always measurable but if $X_t$ and $Y_t$ are both a.s. continuous, then there are measurable.



1) Why is it the case ?



2) Could someone give an example where there are not measurable ?







probability measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 at 12:51









NewMath

936




936




closed as off-topic by Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin Dec 4 at 12:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin Dec 4 at 12:57


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
    – Surb
    Dec 3 at 12:58












  • @Surb: yes I am.
    – NewMath
    Dec 3 at 13:06










  • Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
    – Surb
    Dec 3 at 13:08












  • Could you explain why ? @Surb
    – NewMath
    Dec 3 at 13:08










  • Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
    – Surb
    Dec 3 at 13:09


















  • Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
    – Surb
    Dec 3 at 12:58












  • @Surb: yes I am.
    – NewMath
    Dec 3 at 13:06










  • Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
    – Surb
    Dec 3 at 13:08












  • Could you explain why ? @Surb
    – NewMath
    Dec 3 at 13:08










  • Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
    – Surb
    Dec 3 at 13:09
















Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
– Surb
Dec 3 at 12:58






Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
– Surb
Dec 3 at 12:58














@Surb: yes I am.
– NewMath
Dec 3 at 13:06




@Surb: yes I am.
– NewMath
Dec 3 at 13:06












Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
– Surb
Dec 3 at 13:08






Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
– Surb
Dec 3 at 13:08














Could you explain why ? @Surb
– NewMath
Dec 3 at 13:08




Could you explain why ? @Surb
– NewMath
Dec 3 at 13:08












Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
– Surb
Dec 3 at 13:09




Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
– Surb
Dec 3 at 13:09















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