Here are two related exercise from Karatzas and Shreve

Let $X$ be a process, every sample path of which is right continuous with left limits. Let $A$ be the event that $X$ is continuous on $[0,t_0)$. Show that $A\in\mathcal{F}_{t_0}^X$ (that is the $\sigma$-algebra generated by $X$)

Embarassingly, I am not sure how to do this. If I write down

$\bigcap_{\epsilon\in \{1,1/2,1/3,...\}}\bigcup_{q\in \mathbb{Q}^+}\bigcap_{p\in\mathbb{Q},0<t-p\leq q} \{|X_t-X_p|<\epsilon\} $

Is this equivalent to $X$ is continuous at $t$? (If so, this is measurable w.r.t $\mathcal{F}^X_t$) but this wouldn't be helpful, because, we want left continuity on an interval, and still it wouldn't be a countable union of events which are measurable.

If $X$ is only almost surely RCLL, then $A$ can fail to be in $\mathcal{F}_{t_0}^X$ but if $\mathcal{F}_t$ is a filtration such that $\mathcal{F}_{t_0}^X\subset\mathcal{F}_t$, $t\geq 0$, and $\mathcal{F}_{t_0}$ contains all null sets, then $A\in \mathcal{F}_{t_0}$

I assume this depend on the previous part. Will skip for the time being.

  • 3
    $\begingroup$ A suggestion: an RCLL path is continuous on $[a,b]$ iff its restriction to $[a,b] \cap \mathbb{Q}$ is uniformly continuous. (1: a function on the rationals has at most one RCLL extension to the reals. 2: a function on $[a,b] \cap \mathbb{Q}$ has a continuous extension to $[a,b]$ iff it is uniformly continuous.) $\endgroup$ Oct 25, 2013 at 22:51
  • $\begingroup$ @NateEldredge thank you. i will give this some thoughts. I get the feeling this works. $\endgroup$
    – Lost1
    Oct 25, 2013 at 23:23

1 Answer 1


The set you wrote is the event that $X_t$ is left-continuous at $t > 0$. However, as you commented, it wouldn't work because you need one more intersection over $t \in [0, t_0)$ to say $X_t$ is left-continuous for all $t \in [0, t_0)$ but it may not be expressed as a countable intersection. This is overcome by employing uniform continuity instead, where RCLL property plays a key role.

So let's first show the following lemma:

An RCLL function $f \colon [0, +\infty) \to \mathbb{R}$ is continuous on $[0, t_0)$ if and only if the following holds: $$ \forall \varepsilon > 0, \ \exists \delta > 0, \ \forall q_1, q_2 \in \mathbb{Q} \cap [0, t_0), \ \lvert q_1 - q_2 \rvert < \delta \implies \lvert f(q_1) - f(q_2) \rvert < \varepsilon \qquad \text{(1)} $$

Once this lemma is proved, it's easy to see that $$ A = \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}} \bigcap_{\substack{q_1, q_2 \in \mathbb{Q} \cap [0, t_0)\\ \lvert q_1 - q_2 \rvert < 1/m}} \biggl\{ \lvert X_{q_1} - X_{q_2} \rvert < \frac{1}{n} \biggr\} $$ and, hence, $A$ is $\mathcal{F}_{t_0}^X$-measurable. To prove the lemma, let's define $\widetilde{f} \colon [0, t_0] \to \mathbb{R}$ by $$ \widetilde{f}(t) = \begin{cases} f(t) & \text{if}\ t \in [0, t_0), \\ \displaystyle\lim_{s \uparrow t_0} f(s) & \text{if}\ t = t_0. \end{cases} $$ This $\widetilde{f}$ is well-defined because $f$ has finite left-hand limits.

First, suppose that $f$ is continuous on $[0, t_0)$. Then, obviously $\widetilde{f}$ is continuous on $[0, t_0]$ but, since $[0, t_0]$ is compact, it's also uniformly continuous, i.e.: $$ \forall \varepsilon > 0, \ \exists \delta > 0, \ \forall s, t \in [0, t_0], \ \lvert s - t \rvert < \delta \implies \lvert \widetilde{f}(s) - \widetilde{f}(t) \rvert < \varepsilon. $$ Condition $(1)$ immediately follows from this.

Conversely, suppose that condition $(1)$ holds. We shall show that $\widetilde{f}$ is uniformly continuous on $[0, t_0]$. For any fixed $\varepsilon > 0$, by condition $(1)$ and the definition of $\widetilde{f}$, we can pick $\delta_1, \delta_2 > 0$ so that $$ \forall q_1, q_2 \in \mathbb{Q} \cap [0, t_0), \ \lvert q_1 - q_2 \rvert < \delta_1 \implies \lvert \widetilde{f}(q_1) - \widetilde{f}(q_2) \rvert < \frac{\varepsilon}{3} $$ and $$ \forall q \in \mathbb{Q} \cap [0, t_0], \ \lvert q - t_0 \rvert < \delta_2 \implies \lvert \widetilde{f}(q) - \widetilde{f}(t_0) \rvert < \frac{2\varepsilon}{3}, $$ respectively. Then, by choosing $\delta = \min\{\delta_1, \delta_2\}$, we shall show that, for all $s, t \in [0, t_0]$, $\lvert s - t \rvert < \delta$ implies $\lvert \widetilde{f}(s) - \widetilde{f}(t) \rvert < \varepsilon$. Since it obviously holds when $s = t$, without loss of generality, we may assume that $s < t$. Let's consider two cases: $t < t_0$ and $t = t_0$. If $t < t_0$, the right continuity of $f$ guarantees the existence of $q_1 \in \mathbb{Q} \cap (s, t)$ and $q_2 \in \mathbb{Q} \cap (t, t_0)$ such that $$ s < q_1 < s + \delta, \qquad \lvert \widetilde{f}(s) - \widetilde{f}(q_1) \rvert < \frac{\varepsilon}{3}, \\ t < q_2 < t + q_1 - s, \qquad \lvert \widetilde{f}(t) - \widetilde{f}(q_2) \rvert < \frac{\varepsilon}{3}. $$ Then, noting that $\lvert q_1 - q_2 \rvert < \lvert s - t \rvert < \delta \le \delta_1$, we have $\lvert \widetilde{f}(q_1) - \widetilde{f}(q_2) \rvert < \varepsilon / 3$ by the selection of $\delta_1$. Hence, $$ \lvert \widetilde{f}(s) - \widetilde{f}(t) \rvert \le \lvert \widetilde{f}(s) - \widetilde{f}(q_1) \rvert + \lvert \widetilde{f}(q_1) - \widetilde{f}(q_2) \rvert + \lvert \widetilde{f}(q_2) - \widetilde{f}(t) \rvert < \varepsilon. $$ For the case where $t = t_0$, pick $q \in \mathbb{Q} \cap (s, t_0)$ in the same way as $q_1$ above. Then, by noting that $\lvert q - t_0 \rvert < \lvert s - t_0 \rvert < \delta \le \delta_2$ and by the selection of $\delta_2$, we have $$ \lvert \widetilde{f}(s) - \widetilde{f}(t_0) \rvert \le \lvert \widetilde{f}(s) - \widetilde{f}(q) \rvert + \lvert \widetilde{f}(q) - \widetilde{f}(t_0) \rvert < \varepsilon. $$ We have shown that $\widetilde{f}$ is uniformly continuous on $[0, t_0]$, which implies the continuity of $f$ on $[0, t_0)$.   Q.E.D.

Please see p.38 for the answer to the latter question.

  • $\begingroup$ jeeze man, must have taken you so long to write all this. thanks pal. $\endgroup$
    – Lost1
    Dec 12, 2013 at 18:16
  • $\begingroup$ I figured that it'd follow from the comment, but I didn't bother proving it. Thank you so much for this and it is always good to see a probablist on this site. $\endgroup$
    – Lost1
    Dec 12, 2013 at 18:18

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