Let $X$ be a process with every sample path left-continuous on $(0,\infty)$ and with right-hand limits on $[0,\infty)$, and let A be the event that $X$ is continuous on $[0, t_0]$. Let $X$ be adapted to a right-continuous filtration $\{\mathscr{F}_t\}$. Show that $A\in\mathscr{F}_{t_0}.$

A stochastic process is adapted to a filtration if for all $t \ge 0$ it is $\mathscr{F}_t$-measurable. The event $A$ can be written as:

$$A ~=~\bigcap_{l>1} ~~ \bigcup_{k>1} \bigcap_{\left|s-r\right|< {1\over k}} \left\{\omega:|X_s-X_r|< {1\over l}\wedge (s,r)\in [0,t_0]^2\right\} $$

A right continuous filtration is defined as:

$$ \mathscr{F}_t = \bigcap_{h>0} \mathscr{F}_{t+h} $$

I think that I need to show that $X$ has an RCLL (right continuous left-hand limits) modification, $Y$. I am not sure how to tackle this problem, is the definition of $A$ above relevant?

  • 2
    $\begingroup$ Why do you think you need to show that $X$ has a RCLL modification? As far as I can see, $X$ is RCLL by assumption. Moreover, $A \in \mathcal{F}_{t_0}$ follows directly from the representation you mentioned in your question (it is a countable union of $\mathcal{F}_{t_0}$-measurable sets). $\endgroup$
    – saz
    Dec 29, 2014 at 15:40
  • $\begingroup$ The process is LCRL for all sample paths and adapted to a right-continuous filtration, how does this imply that it is RCLL? Also $A \in \mathscr{F}_{t_0}$? How is it significant that the filtration is RC, I cant see how the different parts of the hypothesis are required nor that I have solved the problem. $\endgroup$
    – shilov
    Dec 30, 2014 at 18:38
  • 1
    $\begingroup$ I see your point. See my answer below. $\endgroup$
    – saz
    Dec 30, 2014 at 19:25

1 Answer 1


Lemma: Let $f:[0,\infty) \to \mathbb{R}$ be left-continuous on $(0,\infty)$ and with right-hand limits on $[0,\infty)$. Then: $f$ is continuous on $[0,t_0)$ if, and only if, $$\forall \epsilon>0 \exists \delta>0: |f(s)-f(r)| \leq \epsilon \quad \text{for all} \, |s-r| \leq \delta, s,r \in [0,t_0) \cap \mathbb{Q}. \tag{1}$$

Proof: "$\Rightarrow$": Obvious. "$\Leftarrow$": Since $f$ is left-continuous and has right-hand limits, $f$ is continuous at $t$ if $$f(t) = f(t+) = \lim_{s \downarrow t} f(s).$$ It follows from $(1)$ and the left-continuity that $$\lim_{s \downarrow t, s\in \mathbb{Q}} f(s)= f(t).$$ Since $$\lim_{s \downarrow t} f(s) = \lim_{s \downarrow t, s \in \mathbb{Q}} f(s)$$ the claim follows.

If we rephrase $(1)$ for the stochastic process $X$, then it becomes

$$A_1 := \{X \, \text{continuous on $[0,t_0)$}\} = \bigcap_{\ell>1} \bigcup_{k \geq 1} \bigcap_{\substack{|s-r| \leq \frac{1}{k} \\ s,r \in \mathbb{Q} \cap [0,t_0)}} \left\{|X_s-X_r| \leq \frac{1}{\ell} \right\}.$$

Since $X$ is adapted, we get $A_1 \in \mathcal{F}_{t_0}$ (as a countable union of measurable sets).

Note however, that $(1)$ does not give continuty at $t=t_0$; that is, we need an extra condition to ensure continuity at $t=t_0$ - and here comes the right-continuity of the filtration into play.

$f$ is continuous in $t=t_0$ if and only if for any $\gamma>0$ and $\epsilon>0$ we can find $\delta>0$ such that $$|f(t_0)-f(s)| \leq \epsilon \quad \text{for all $s \in [t_0-\delta,t_0+(\delta \wedge \gamma)] \cap \mathbb{Q}$.}$$

Roughly speaking: In order to decide whether $f$ is continuous at $t=t_0$ it does not suffices to know $f$ up to time $t_0$, we need some information about the future. Consequently,

$$A_2 := \{X \, \text{continuous at $t=t_0$}\} = \bigcap_{m \geq 1} \underbrace{\bigcap_{\ell \geq 1} \bigcap_{\substack{|s-t_0| \leq \frac{1}{k} \\ s \in \mathbb{Q} \cap [0,t_0+\frac{1}{m})}} \left\{|X_s-X_r| \leq \frac{1}{\ell} \right\}}_{\in \mathcal{F}_{t_0+1/m}} \in \mathcal{F}_{t_0+} = \mathcal{F}_{t_0}.$$

Hence, $A= A_1 \cap A_2 \in \mathcal{F}_{t_0}.$

  • $\begingroup$ Would the downvoter care to comment....? $\endgroup$
    – saz
    Jan 2, 2017 at 16:47
  • $\begingroup$ Could you give me a more detailed explanation on $\lim _{s \downarrow t, s \in \mathbb{Q}} f(s)=f(t)$ in Lemma, please? I think $\lim _{s \uparrow t, s \in \mathbb{Q}} f(s)=f(t)$ follows from the left continuity, not $s \downarrow t$. @saz $\endgroup$
    – mathmrk
    Aug 20, 2021 at 2:23

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