Example of almost surely continuous stochastic process $(X_t)$ with ${omega mid tmapsto X_t(omega )text{...












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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.










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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.










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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.










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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






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      edited Dec 3 '18 at 19:18









      Did

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      asked Dec 3 '18 at 19:15









      NewMathNewMath

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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.






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            $begingroup$

            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.






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              $begingroup$

              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.






              share|cite|improve this answer









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                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







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                share|cite|improve this answer










                answered Dec 3 '18 at 19:18









                SurbSurb

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                37.6k94375






























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