Measurability question for càglàd process












3














I have encounter the following question, which is probably naive for probabilists, but let me still ask it:



Let ${X_t(omega): t in mathbb{R}}$ be a real valued càglàd process (that is ''continue à gauche et il existe limite à droite'' instead of the usual càdlàg assumption ), is the following function



$$ limsup_{t downarrow 0 } X_t(omega)$$



measurable? If this is the case, would you please tell me a reference for it, if this is not the case, then what kinds of conditions on the process will such a limit function measurable ?



Thanks a lot.






measure-theory stochastic-processes







share|cite|improve this question





























    3














    I have encounter the following question, which is probably naive for probabilists, but let me still ask it:



    Let ${X_t(omega): t in mathbb{R}}$ be a real valued càglàd process (that is ''continue à gauche et il existe limite à droite'' instead of the usual càdlàg assumption ), is the following function



    $$ limsup_{t downarrow 0 } X_t(omega)$$



    measurable? If this is the case, would you please tell me a reference for it, if this is not the case, then what kinds of conditions on the process will such a limit function measurable ?



    Thanks a lot.






    measure-theory stochastic-processes







    share|cite|improve this question



























      3












      3








      3


      1





      I have encounter the following question, which is probably naive for probabilists, but let me still ask it:



      Let ${X_t(omega): t in mathbb{R}}$ be a real valued càglàd process (that is ''continue à gauche et il existe limite à droite'' instead of the usual càdlàg assumption ), is the following function



      $$ limsup_{t downarrow 0 } X_t(omega)$$



      measurable? If this is the case, would you please tell me a reference for it, if this is not the case, then what kinds of conditions on the process will such a limit function measurable ?



      Thanks a lot.






      measure-theory stochastic-processes







      share|cite|improve this question















      I have encounter the following question, which is probably naive for probabilists, but let me still ask it:



      Let ${X_t(omega): t in mathbb{R}}$ be a real valued càglàd process (that is ''continue à gauche et il existe limite à droite'' instead of the usual càdlàg assumption ), is the following function



      $$ limsup_{t downarrow 0 } X_t(omega)$$



      measurable? If this is the case, would you please tell me a reference for it, if this is not the case, then what kinds of conditions on the process will such a limit function measurable ?



      Thanks a lot.






      measure-theory stochastic-processes




      measure-theory stochastic-processes






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 '18 at 16:59









      saz

      78.2k758123




      78.2k758123










      asked Jul 20 '14 at 7:53









      Yanqi QIU

      1214




      1214






















          1 Answer
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          6














          If $(X_t)_{t geq 0}$ is a càdlàg process, then it follows from the very definition of "càdlàg" that



          $$lim_{t downarrow 0} X_t(omega) = limsup_{t downarrow 0} X_t(omega) = X_0(omega).$$



          Consequently, the $limsup$ is measurable since $omega mapsto X_0(omega)$ is measurable (as $(X_t)_{t geq 0}$ is a stochastic process).




          Edit (càglàd case) If a function $f$ is right-continuous at some point $x in mathbb{R}$, then we have



          $$limsup_{y downarrow x} f(y) = limsup_{n to infty} f(y_n)$$



          for any sequence $y_n downarrow x$. Since the (pointwise) $limsup$ of a sequence of measurable functions is measurable, we conclude that



          $$limsup_{t downarrow 0} X_t(omega) = limsup_{n to infty} X_{t_n}(omega)$$



          is measurable whenever $(X_t)_{t geq 0}$ is a càglàd process. Here $t_n$ denotes an arbitrary sequence satisfying $t_n downarrow 0$.






          share|cite|improve this answer























          • Thanks a lot. And sorry, I made a mistake in my question, the process should be something like: caglad instead of cadlag
            – Yanqi QIU
            Jul 20 '14 at 9:16










          • I think one arbitrary $t_n$ does not work. But thanks to Saz's answer, now I think the right argument should be $$limsup_{t downarrow 0} X_t(omega) = lim_{n to infty} sup_{substack{0< q < 1/n\ q in mathbb{Q}} } X_q(omega).$$ The equality of course follows from the one-side continuity.
            – Yanqi QIU
            Jul 20 '14 at 14:33












          • @YanqiQIU Why does an arbitrary sequence in your oppinion not work? Sequential (right)continuity is equivalent to (right)continuity.
            – saz
            Jul 20 '14 at 15:03










          • Dear saz, since the process in question is not right continuous but only right continuous. (I am sorry that the very first version of my question confused you.)
            – Yanqi QIU
            Jul 20 '14 at 22:03










          • @YanqiQIU Well, yes, of course. But to any caglad function $g$ we can associate a cadlag function $g$ by setting $$g(t) = lim_{s downarrow t} f(s).$$ And this shows in particular (now using the right-continuity of $g$) that $$limsup_{n to infty} g(t_n) = limsup_{s downarrow t} g(s) = g(t)=f(t+).$$
            – saz
            Jul 21 '14 at 5:40











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          active

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          6














          If $(X_t)_{t geq 0}$ is a càdlàg process, then it follows from the very definition of "càdlàg" that



          $$lim_{t downarrow 0} X_t(omega) = limsup_{t downarrow 0} X_t(omega) = X_0(omega).$$



          Consequently, the $limsup$ is measurable since $omega mapsto X_0(omega)$ is measurable (as $(X_t)_{t geq 0}$ is a stochastic process).




          Edit (càglàd case) If a function $f$ is right-continuous at some point $x in mathbb{R}$, then we have



          $$limsup_{y downarrow x} f(y) = limsup_{n to infty} f(y_n)$$



          for any sequence $y_n downarrow x$. Since the (pointwise) $limsup$ of a sequence of measurable functions is measurable, we conclude that



          $$limsup_{t downarrow 0} X_t(omega) = limsup_{n to infty} X_{t_n}(omega)$$



          is measurable whenever $(X_t)_{t geq 0}$ is a càglàd process. Here $t_n$ denotes an arbitrary sequence satisfying $t_n downarrow 0$.






          share|cite|improve this answer























          • Thanks a lot. And sorry, I made a mistake in my question, the process should be something like: caglad instead of cadlag
            – Yanqi QIU
            Jul 20 '14 at 9:16










          • I think one arbitrary $t_n$ does not work. But thanks to Saz's answer, now I think the right argument should be $$limsup_{t downarrow 0} X_t(omega) = lim_{n to infty} sup_{substack{0< q < 1/n\ q in mathbb{Q}} } X_q(omega).$$ The equality of course follows from the one-side continuity.
            – Yanqi QIU
            Jul 20 '14 at 14:33












          • @YanqiQIU Why does an arbitrary sequence in your oppinion not work? Sequential (right)continuity is equivalent to (right)continuity.
            – saz
            Jul 20 '14 at 15:03










          • Dear saz, since the process in question is not right continuous but only right continuous. (I am sorry that the very first version of my question confused you.)
            – Yanqi QIU
            Jul 20 '14 at 22:03










          • @YanqiQIU Well, yes, of course. But to any caglad function $g$ we can associate a cadlag function $g$ by setting $$g(t) = lim_{s downarrow t} f(s).$$ And this shows in particular (now using the right-continuity of $g$) that $$limsup_{n to infty} g(t_n) = limsup_{s downarrow t} g(s) = g(t)=f(t+).$$
            – saz
            Jul 21 '14 at 5:40
















          6














          If $(X_t)_{t geq 0}$ is a càdlàg process, then it follows from the very definition of "càdlàg" that



          $$lim_{t downarrow 0} X_t(omega) = limsup_{t downarrow 0} X_t(omega) = X_0(omega).$$



          Consequently, the $limsup$ is measurable since $omega mapsto X_0(omega)$ is measurable (as $(X_t)_{t geq 0}$ is a stochastic process).




          Edit (càglàd case) If a function $f$ is right-continuous at some point $x in mathbb{R}$, then we have



          $$limsup_{y downarrow x} f(y) = limsup_{n to infty} f(y_n)$$



          for any sequence $y_n downarrow x$. Since the (pointwise) $limsup$ of a sequence of measurable functions is measurable, we conclude that



          $$limsup_{t downarrow 0} X_t(omega) = limsup_{n to infty} X_{t_n}(omega)$$



          is measurable whenever $(X_t)_{t geq 0}$ is a càglàd process. Here $t_n$ denotes an arbitrary sequence satisfying $t_n downarrow 0$.






          share|cite|improve this answer























          • Thanks a lot. And sorry, I made a mistake in my question, the process should be something like: caglad instead of cadlag
            – Yanqi QIU
            Jul 20 '14 at 9:16










          • I think one arbitrary $t_n$ does not work. But thanks to Saz's answer, now I think the right argument should be $$limsup_{t downarrow 0} X_t(omega) = lim_{n to infty} sup_{substack{0< q < 1/n\ q in mathbb{Q}} } X_q(omega).$$ The equality of course follows from the one-side continuity.
            – Yanqi QIU
            Jul 20 '14 at 14:33












          • @YanqiQIU Why does an arbitrary sequence in your oppinion not work? Sequential (right)continuity is equivalent to (right)continuity.
            – saz
            Jul 20 '14 at 15:03










          • Dear saz, since the process in question is not right continuous but only right continuous. (I am sorry that the very first version of my question confused you.)
            – Yanqi QIU
            Jul 20 '14 at 22:03










          • @YanqiQIU Well, yes, of course. But to any caglad function $g$ we can associate a cadlag function $g$ by setting $$g(t) = lim_{s downarrow t} f(s).$$ And this shows in particular (now using the right-continuity of $g$) that $$limsup_{n to infty} g(t_n) = limsup_{s downarrow t} g(s) = g(t)=f(t+).$$
            – saz
            Jul 21 '14 at 5:40














          6












          6








          6






          If $(X_t)_{t geq 0}$ is a càdlàg process, then it follows from the very definition of "càdlàg" that



          $$lim_{t downarrow 0} X_t(omega) = limsup_{t downarrow 0} X_t(omega) = X_0(omega).$$



          Consequently, the $limsup$ is measurable since $omega mapsto X_0(omega)$ is measurable (as $(X_t)_{t geq 0}$ is a stochastic process).




          Edit (càglàd case) If a function $f$ is right-continuous at some point $x in mathbb{R}$, then we have



          $$limsup_{y downarrow x} f(y) = limsup_{n to infty} f(y_n)$$



          for any sequence $y_n downarrow x$. Since the (pointwise) $limsup$ of a sequence of measurable functions is measurable, we conclude that



          $$limsup_{t downarrow 0} X_t(omega) = limsup_{n to infty} X_{t_n}(omega)$$



          is measurable whenever $(X_t)_{t geq 0}$ is a càglàd process. Here $t_n$ denotes an arbitrary sequence satisfying $t_n downarrow 0$.






          share|cite|improve this answer














          If $(X_t)_{t geq 0}$ is a càdlàg process, then it follows from the very definition of "càdlàg" that



          $$lim_{t downarrow 0} X_t(omega) = limsup_{t downarrow 0} X_t(omega) = X_0(omega).$$



          Consequently, the $limsup$ is measurable since $omega mapsto X_0(omega)$ is measurable (as $(X_t)_{t geq 0}$ is a stochastic process).




          Edit (càglàd case) If a function $f$ is right-continuous at some point $x in mathbb{R}$, then we have



          $$limsup_{y downarrow x} f(y) = limsup_{n to infty} f(y_n)$$



          for any sequence $y_n downarrow x$. Since the (pointwise) $limsup$ of a sequence of measurable functions is measurable, we conclude that



          $$limsup_{t downarrow 0} X_t(omega) = limsup_{n to infty} X_{t_n}(omega)$$



          is measurable whenever $(X_t)_{t geq 0}$ is a càglàd process. Here $t_n$ denotes an arbitrary sequence satisfying $t_n downarrow 0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 20 '14 at 12:12

























          answered Jul 20 '14 at 7:57









          saz

          78.2k758123




          78.2k758123












          • Thanks a lot. And sorry, I made a mistake in my question, the process should be something like: caglad instead of cadlag
            – Yanqi QIU
            Jul 20 '14 at 9:16










          • I think one arbitrary $t_n$ does not work. But thanks to Saz's answer, now I think the right argument should be $$limsup_{t downarrow 0} X_t(omega) = lim_{n to infty} sup_{substack{0< q < 1/n\ q in mathbb{Q}} } X_q(omega).$$ The equality of course follows from the one-side continuity.
            – Yanqi QIU
            Jul 20 '14 at 14:33












          • @YanqiQIU Why does an arbitrary sequence in your oppinion not work? Sequential (right)continuity is equivalent to (right)continuity.
            – saz
            Jul 20 '14 at 15:03










          • Dear saz, since the process in question is not right continuous but only right continuous. (I am sorry that the very first version of my question confused you.)
            – Yanqi QIU
            Jul 20 '14 at 22:03










          • @YanqiQIU Well, yes, of course. But to any caglad function $g$ we can associate a cadlag function $g$ by setting $$g(t) = lim_{s downarrow t} f(s).$$ And this shows in particular (now using the right-continuity of $g$) that $$limsup_{n to infty} g(t_n) = limsup_{s downarrow t} g(s) = g(t)=f(t+).$$
            – saz
            Jul 21 '14 at 5:40


















          • Thanks a lot. And sorry, I made a mistake in my question, the process should be something like: caglad instead of cadlag
            – Yanqi QIU
            Jul 20 '14 at 9:16










          • I think one arbitrary $t_n$ does not work. But thanks to Saz's answer, now I think the right argument should be $$limsup_{t downarrow 0} X_t(omega) = lim_{n to infty} sup_{substack{0< q < 1/n\ q in mathbb{Q}} } X_q(omega).$$ The equality of course follows from the one-side continuity.
            – Yanqi QIU
            Jul 20 '14 at 14:33












          • @YanqiQIU Why does an arbitrary sequence in your oppinion not work? Sequential (right)continuity is equivalent to (right)continuity.
            – saz
            Jul 20 '14 at 15:03










          • Dear saz, since the process in question is not right continuous but only right continuous. (I am sorry that the very first version of my question confused you.)
            – Yanqi QIU
            Jul 20 '14 at 22:03










          • @YanqiQIU Well, yes, of course. But to any caglad function $g$ we can associate a cadlag function $g$ by setting $$g(t) = lim_{s downarrow t} f(s).$$ And this shows in particular (now using the right-continuity of $g$) that $$limsup_{n to infty} g(t_n) = limsup_{s downarrow t} g(s) = g(t)=f(t+).$$
            – saz
            Jul 21 '14 at 5:40
















          Thanks a lot. And sorry, I made a mistake in my question, the process should be something like: caglad instead of cadlag
          – Yanqi QIU
          Jul 20 '14 at 9:16




          Thanks a lot. And sorry, I made a mistake in my question, the process should be something like: caglad instead of cadlag
          – Yanqi QIU
          Jul 20 '14 at 9:16












          I think one arbitrary $t_n$ does not work. But thanks to Saz's answer, now I think the right argument should be $$limsup_{t downarrow 0} X_t(omega) = lim_{n to infty} sup_{substack{0< q < 1/n\ q in mathbb{Q}} } X_q(omega).$$ The equality of course follows from the one-side continuity.
          – Yanqi QIU
          Jul 20 '14 at 14:33






          I think one arbitrary $t_n$ does not work. But thanks to Saz's answer, now I think the right argument should be $$limsup_{t downarrow 0} X_t(omega) = lim_{n to infty} sup_{substack{0< q < 1/n\ q in mathbb{Q}} } X_q(omega).$$ The equality of course follows from the one-side continuity.
          – Yanqi QIU
          Jul 20 '14 at 14:33














          @YanqiQIU Why does an arbitrary sequence in your oppinion not work? Sequential (right)continuity is equivalent to (right)continuity.
          – saz
          Jul 20 '14 at 15:03




          @YanqiQIU Why does an arbitrary sequence in your oppinion not work? Sequential (right)continuity is equivalent to (right)continuity.
          – saz
          Jul 20 '14 at 15:03












          Dear saz, since the process in question is not right continuous but only right continuous. (I am sorry that the very first version of my question confused you.)
          – Yanqi QIU
          Jul 20 '14 at 22:03




          Dear saz, since the process in question is not right continuous but only right continuous. (I am sorry that the very first version of my question confused you.)
          – Yanqi QIU
          Jul 20 '14 at 22:03












          @YanqiQIU Well, yes, of course. But to any caglad function $g$ we can associate a cadlag function $g$ by setting $$g(t) = lim_{s downarrow t} f(s).$$ And this shows in particular (now using the right-continuity of $g$) that $$limsup_{n to infty} g(t_n) = limsup_{s downarrow t} g(s) = g(t)=f(t+).$$
          – saz
          Jul 21 '14 at 5:40




          @YanqiQIU Well, yes, of course. But to any caglad function $g$ we can associate a cadlag function $g$ by setting $$g(t) = lim_{s downarrow t} f(s).$$ And this shows in particular (now using the right-continuity of $g$) that $$limsup_{n to infty} g(t_n) = limsup_{s downarrow t} g(s) = g(t)=f(t+).$$
          – saz
          Jul 21 '14 at 5:40


















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