Example of almost surely continuous stochastic process $(X_t)$ with ${omega mid tmapsto X_t(omega )text{...
Does someone has an example of stochastic process $(X_t)$ that is almost surely continuous but ${omega mid tmapsto X_t(omega )text{ continuous}}$ is not measurable ? It look strange for me. Because if $(X_t)$ is a.s. continuous then ${omega mid tmapsto X_t(omega )text{ continuous}}^c$ has measure $0$ and thus is measurable.
probability-theory measure-theory
edited Dec 3 '18 at 19:18
Did
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asked Dec 3 '18 at 19:15
NewMathNewMath
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Does someone has an example of stochastic process $(X_t)$ that is almost surely continuous but ${omega mid tmapsto X_t(omega )text{ continuous}}$ is not measurable ? It look strange for me. Because if $(X_t)$ is a.s. continuous then ${omega mid tmapsto X_t(omega )text{ continuous}}^c$ has measure $0$ and thus is measurable.
probability-theory measure-theory
edited Dec 3 '18 at 19:18
Did
246k23221456
asked Dec 3 '18 at 19:15
NewMathNewMath
3808
add a comment |
Does someone has an example of stochastic process $(X_t)$ that is almost surely continuous but ${omega mid tmapsto X_t(omega )text{ continuous}}$ is not measurable ? It look strange for me. Because if $(X_t)$ is a.s. continuous then ${omega mid tmapsto X_t(omega )text{ continuous}}^c$ has measure $0$ and thus is measurable.
probability-theory measure-theory
edited Dec 3 '18 at 19:18
Did
246k23221456
asked Dec 3 '18 at 19:15
NewMathNewMath
3808
Does someone has an example of stochastic process $(X_t)$ that is almost surely continuous but ${omega mid tmapsto X_t(omega )text{ continuous}}$ is not measurable ? It look strange for me. Because if $(X_t)$ is a.s. continuous then ${omega mid tmapsto X_t(omega )text{ continuous}}^c$ has measure $0$ and thus is measurable.
probability-theory measure-theory
probability-theory measure-theory
edited Dec 3 '18 at 19:18
Did
246k23221456
asked Dec 3 '18 at 19:15
NewMathNewMath
3808
edited Dec 3 '18 at 19:18
Did
246k23221456
asked Dec 3 '18 at 19:15
NewMathNewMath
3808
edited Dec 3 '18 at 19:18
Did
246k23221456
edited Dec 3 '18 at 19:18
Did
246k23221456
edited Dec 3 '18 at 19:18
Did
246k23221456
246k23221456
asked Dec 3 '18 at 19:15
NewMathNewMath
3808
asked Dec 3 '18 at 19:15
NewMathNewMath
3808
asked Dec 3 '18 at 19:15
NewMathNewMath
3808
3808
add a comment |
add a comment |
1 Answer
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Null sets are measurables in complete measurable spaces only. If it's not complete, there are null sets that are not measurable. Take for example $Omega =[0,1]$, $mathcal F=mathcal B([0,1])$ the Borel set of $[0,1]$ and $mathbb P$ the Lebesgue measure. Let $N$ a null set that is not a Borel set (such set exist). Then $$X_t(omega )=begin{cases}t&omega notin N\ boldsymbol 1_{mathbb Qcap [0,1]}(t)&omega in Nend{cases},$$
is such an example. If you measure space is complete (i.e. null set are measurable), then indeed, ${omega mid tmapsto X_t(omega )}$ is measurable for the reason you said.
answered Dec 3 '18 at 19:18
SurbSurb
37.4k94375
add a comment |
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1 Answer
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1 Answer
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Null sets are measurables in complete measurable spaces only. If it's not complete, there are null sets that are not measurable. Take for example $Omega =[0,1]$, $mathcal F=mathcal B([0,1])$ the Borel set of $[0,1]$ and $mathbb P$ the Lebesgue measure. Let $N$ a null set that is not a Borel set (such set exist). Then $$X_t(omega )=begin{cases}t&omega notin N\ boldsymbol 1_{mathbb Qcap [0,1]}(t)&omega in Nend{cases},$$
is such an example. If you measure space is complete (i.e. null set are measurable), then indeed, ${omega mid tmapsto X_t(omega )}$ is measurable for the reason you said.
answered Dec 3 '18 at 19:18
SurbSurb
37.4k94375
add a comment |
Null sets are measurables in complete measurable spaces only. If it's not complete, there are null sets that are not measurable. Take for example $Omega =[0,1]$, $mathcal F=mathcal B([0,1])$ the Borel set of $[0,1]$ and $mathbb P$ the Lebesgue measure. Let $N$ a null set that is not a Borel set (such set exist). Then $$X_t(omega )=begin{cases}t&omega notin N\ boldsymbol 1_{mathbb Qcap [0,1]}(t)&omega in Nend{cases},$$
is such an example. If you measure space is complete (i.e. null set are measurable), then indeed, ${omega mid tmapsto X_t(omega )}$ is measurable for the reason you said.
answered Dec 3 '18 at 19:18
SurbSurb
37.4k94375
add a comment |
Null sets are measurables in complete measurable spaces only. If it's not complete, there are null sets that are not measurable. Take for example $Omega =[0,1]$, $mathcal F=mathcal B([0,1])$ the Borel set of $[0,1]$ and $mathbb P$ the Lebesgue measure. Let $N$ a null set that is not a Borel set (such set exist). Then $$X_t(omega )=begin{cases}t&omega notin N\ boldsymbol 1_{mathbb Qcap [0,1]}(t)&omega in Nend{cases},$$
is such an example. If you measure space is complete (i.e. null set are measurable), then indeed, ${omega mid tmapsto X_t(omega )}$ is measurable for the reason you said.
answered Dec 3 '18 at 19:18
SurbSurb
37.4k94375
Null sets are measurables in complete measurable spaces only. If it's not complete, there are null sets that are not measurable. Take for example $Omega =[0,1]$, $mathcal F=mathcal B([0,1])$ the Borel set of $[0,1]$ and $mathbb P$ the Lebesgue measure. Let $N$ a null set that is not a Borel set (such set exist). Then $$X_t(omega )=begin{cases}t&omega notin N\ boldsymbol 1_{mathbb Qcap [0,1]}(t)&omega in Nend{cases},$$
is such an example. If you measure space is complete (i.e. null set are measurable), then indeed, ${omega mid tmapsto X_t(omega )}$ is measurable for the reason you said.
answered Dec 3 '18 at 19:18
SurbSurb
37.4k94375
answered Dec 3 '18 at 19:18
SurbSurb
37.4k94375
answered Dec 3 '18 at 19:18
SurbSurb
37.4k94375
answered Dec 3 '18 at 19:18
SurbSurb
37.4k94375
37.4k94375
add a comment |
add a comment |
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