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I'm try to solving the following exercise:

Let $X=(X_t)_{t\geq0}$ be a real valued stochastic process such that, for all $\omega\in \Omega$ the map $t\rightarrow X_t(\omega)$ is right continuous and for every $t>0$ the left limit

$$\lim_{s\uparrow t}X_s (\omega)$$ exists. Consider the set, for $T>0$

$$C=\{\omega \in \Omega:t\rightarrow X_t(\omega) \text{ is continuous on } [0,T]\}$$ Show that it belongs to $$\mathcal{F}_T^X:=\sigma(X_t:t\in[0,T]).$$

I know that we have a clear solution in (Show that a certain set is measurable) but can someone help me to keep going with this solution:

From right continuity we have

$C=\{\omega \in \Omega:\lim_{s\uparrow t}X_s (\omega)=X_t(\omega) ,\forall t\in[0,T]\}=\bigcap_{t\in[0,T]}\{\omega \in \Omega:\lim_{s\uparrow t}X_s (\omega)=X_t(\omega)\}$

now I want to have in the last equality a countable intersection but I'm not able to do that. I know that for a cadlag process holds that the set of discontinuity points is a countable set but in my opinion this set could depend from the choice of $\omega$ and so I don't know how to conclude.

foubw
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  • thanks for the solution. So my approach, that is completely different from the solution is wrong or not?? – foubw Nov 12 '15 at 19:31
  • In the solution you say that a right continuous function $f$ is uniformly continuous on $[0,t]$ if and only if $f$ is uniformly continuous on $\mathbb{Q}\cap[0,t]$. Why does it hold? – foubw Nov 12 '15 at 19:37
  • See this: http://math.stackexchange.com/questions/76854/extending-a-function-by-continuity-from-a-dense-subset-of-a-space –  Nov 12 '15 at 19:49
  • but if I take a function like that: $f(x)=1$ if $x<\sqrt{2}$ and $f(x)=0$ if $x\geq \sqrt{2}$ for $x\in [0,2]$ f is $cadlag$, continous in $\mathbb{Q}$, but not in $[0,2]$ – foubw Nov 12 '15 at 20:04
  • The function $f$ in your comment is not uniformly continuous. –  Nov 12 '15 at 20:12
  • Oh sorry, now I understand where was my mistake. Do you have any suggestion for my first question? – foubw Nov 12 '15 at 20:26

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