# Stochastic Processes - Bass exercise 1.6

I'm working on this problem similar to one from Bass' book on stochastic processes and I cannot for the life of me make any progress.

Let $X$ be a stochastic process and suppose that each sample path of $X$ is cadlag. Put $\mathcal{F}^X_t = \sigma(X_s:s \leq t)$, $\Delta X_t = X_{t} - X_{t^-}$, $\mathcal{F}_{\infty} = \sigma(\mathcal{F}^X_t:t\geq 0 )$, and $A_c = \{\omega \in \Omega: \text{ for some }t > 0\text{, } \Delta X_t(\omega) > c\}$.

Show that $A_c \in \mathcal{F}_{\infty}$.

So far I know that for $t>0$, $\Delta X_t(\omega) > c$ if and only if there exists $K \in \mathbb{N}$ such that for each $n \in \mathbb{N}$, there exists $m \in \mathbb{N}$ such that for each $t_1 \in (0,t) \cap \mathbb{Q}, t_2 \in (t,\infty)\cap \mathbb{Q}$, if $|t_1 - t_2| < 1/m$, then $X_{t_2}(\omega) - X_{t_1}(\omega)>c + 1/K -1/n$.

If I put $B_t = \{\omega \in \Omega: \Delta X_t(\omega) > c\}$, $R_{t_1, t_2, m} = \{\omega \in \Omega: |t_1 - t_2| < 1/m \}$, $S_{K, t_1,t_2,n} = \{\omega \in \Omega: X_{t_2}(\omega) - X_{t_1}(\omega) > c +1/K - 1/n\}$, then $R_{t_1, t_2, m}$, $S_{K, t_1,t_2,n} \in \mathcal{F}_\infty$. So $B_t = \bigcup\limits_{K \in \mathbb{N}}\bigcap\limits_{n \in \mathbb{N}} \bigcup\limits_{m \in \mathbb{N}} \bigcap\limits_{t_1 \in (0,t)\cap \mathbb{Q}} \bigcap\limits_{t_2 \in (t, \infty)\cap \mathbb{Q}} R_{t_1, t_2, m}^c \cup S_{K, t_1, t_2, n}$ and hence $B_t \in \mathcal{F}_{\infty}$.

Now I'm stuck trying to answer the question since $A_c = \bigcup\limits_{t>0}B_t$ and I cannot figure out how to get a countable union.

I thought a little about how cadlag functions have at most countably many discontinuities, but I couldn't figure a way to work that in.

I'm sort of expecting this entire process to be an unhelpful direction to begin at, but it was the best that I could come up with. The question reminds me of exercise 1.7 in Karatzas and Shreve. However in that problem I was able to get a uniform characterization of continuity to be able to get a countable union/intersection. With this one, I'm not sure if that's possible, so feel free to offer any other approach to solving the problem. Thanks everyone.

• I think there is something off with your characterization of $\Delta X_t(\omega)>c$. For instance the function $f(x) := 1_{[1,\infty)}(x)$ satisfies for $t:=1$ and $c=1$ your second condition, but clearly $\Delta f(1)=1$ is not strictly larger than $c=1$. I think you are actually characterizing $\Delta X_t(\omega) \color{red}{\geq} c$. – saz May 15 '18 at 18:59
• Regarding your original problem: Take a look at the lemma in this answer. – saz May 15 '18 at 19:08
• saz, you are correct. I will fix that. I'll need to look at that other answer you referenced. Thank you. – Ceeerson May 15 '18 at 19:30
• saz, I just read the lemma you referenced and you just answered my question. I was stuck on this problem for a while and am still getting used to these new techniques. So thank you very much, this was very helpful! :) – Ceeerson May 15 '18 at 19:41
• That $B_t$ is $\mathcal{F}_t$-measurable is not difficult: For, notice that both $X(t,\cdot)$ and $X(t-,\cdot)$ are both $\mathcal{F}_t$-measurable. – Danny Pak-Keung Chan May 15 '18 at 20:06

Define $$A^{n,k}_{p,q}$$ for $$p,q\in\mathbb{Q}$$ with $$p, $$n\in\mathbb{N}$$, $$0\leq k<2^n$$ as the set of all $$\omega\in\Omega$$ such that:
1. $$X_{p+(q-p)i/2^n}(\omega) for all $$i,j$$ with $$0\leq i\leq k < j\leq 2^n$$,
2. $$|X_{p+(q-p)i/2^n}(\omega)-X_{p+(q-p)j/2^n}(\omega)| for all $$i,j$$ with $$0\leq i\leq j\leq k$$,
3. $$|X_{p+(q-p)i/2^n}(\omega)-X_{p+(q-p)j/2^n}(\omega)| for all $$i,j$$ with $$k.
This represents a subdivision of the interval $$(p,q)$$ where the $$X$$-value at the first $$k+1$$ points are more than $$c$$ smaller than the $$X$$-values at the remaining points. Define $$A_{p,q}^n$$ by the (disjoint) union $$A_{p,q}^n:=\bigcup_{k=0}^{2^n-1} A^{n,k}_{p,q}$$ and $$A_{p,q}$$ by $$A_{p,q} := \bigcup_{m=1}^\infty \bigcap_{n\geq m} A_{p,q}^n.$$ Here $$A^n_{p,q}$$ allows $$k$$ to occur at any of the subdivision points except the last, and $$A_{p,q}$$ requires $$\omega$$ to eventually exist in $$A_{p,q}^n$$ for all sufficiently large $$n$$. Then $$A_c=\bigcup_{\substack{p,q\in\mathbb{Q}\\p Hopefully this is clear.