Let $X$ be a stochastic process with every sample path RCLL(right continuous on $[0,\infty)$ and with left-hand finite limits on $(0,\infty)$). Let $A$ be an event that $X$ is continuous on $[0,t_0)$. Show $A\in\mathcal{F}_{t_0}^X$.

For $(t_n)_n$ a dense sequence contained in $(0,t_0)$, if we can show

$$A=\{\omega: \lim_{s\rightarrow t_n-}X_s(\omega)=X_{t_n}(\omega),\ \ \forall n\geq 1,\ \ t_n<t\}$$

then $A\in\mathcal{F}_{t_0}^X$ since every $\{\omega: \lim_{s\rightarrow t_n-}X_s(\omega)=X_{t_n}(\omega)\}\in\mathcal{F}_{t_0}^X$. Could someone show how to prove

$$A=\{\omega: \lim_{s\rightarrow t_n-}X_s(\omega)=X_{t_n}(\omega),\ \ \forall n\geq 1,\ \ t_n<t\}$$

  • 1
    $\begingroup$ I don't think your claim is true as written. If there's an $\omega$ such that $X(\omega)$ has a jump at some $t \notin \{t_n\}$ (which is entirely possible under the given assumptions, and could even happen with probability 1), then $\omega$ is in the set on the RHS, but not in $A$. $\endgroup$ Jun 23, 2013 at 15:14
  • $\begingroup$ However, as another approach to the measurability of $A$, you might use (after first verifying) the fact that an RCLL path is continuous on $[a,b]$ iff its restriction to $\mathbb{Q} \cap [a,b]$ (or your favorite dense subset) is uniformly continuous. $\endgroup$ Jun 23, 2013 at 15:18
  • $\begingroup$ Thanks a lot for your help! $\endgroup$
    – Higgs88
    Jun 23, 2013 at 16:10
  • 1
    $\begingroup$ If you get it worked out, you might like to post the solution here as an answer. $\endgroup$ Jun 23, 2013 at 18:01

1 Answer 1


Thanks to Nate Eldredge, I could have a solution here.

Set $\{t_n\}_{n\geq 1}=\mathbb{Q}\cap [0,t_0)$, then we firstly we prove $X$ is continuous on $[0,t_0)$ if and only if its restriction on $\{t_n\}_{n\geq 1}$ is uniformly continuous:

if $X$ is continuous on $[0,t_0)$, by definition we can extend $X$ on $[0,t_0]$ by defining $X_{t_0}:=\lim_{t\rightarrow t_0-}X_t$, then $X$ is continuous on $[0,t_0]$ and so it is uniformly continuous on $[0,t_0]$;

if $X$ is uniformly continous on $\{t_n\}_{n\geq 1}$, then by Cauchy's criterion, we can define $\widetilde{X}$ on $[0,t_0]$ by

$$\widetilde{X}_t:=\lim_{s\rightarrow t, s\in\{t_n\}_{n\geq 1}}X_s$$

Evidently $\widetilde{X}$ is continuous. For $X$ is right-continuous, we have necessarily $X=\widetilde{X}$, so $X$ is continuous on $[0,t_0)$.

Secondly by an argument largely used to describe the continuity of a function, we have

\begin{eqnarray} A&=&\{\omega: X(\omega) \text{ is uniformly continuous on } \{t_n\}_{n\geq 1}\}\\ &=&\bigcap_{m\geq 1}\bigcup_{k\geq 1}\bigcap_{n\geq 1}\{\omega: |X_t(\omega)-X_{t_n}(\omega)|<\frac{1}{m},\ \ \forall t\in \{t_l\}_{l\geq 1} \text{ with } |t-t_n|<\frac{1}{k}\} \end{eqnarray}

So $A$ is measuralbe $\clubsuit$


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