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. 
 A: Define $A^{n,k}_{p,q}$ for $p,q\in\mathbb{Q}$ with $p<q$, $n\in\mathbb{N}$, $0\leq k<2^n$ as the set of all $\omega\in\Omega$ such that:


*

*$X_{p+(q-p)i/2^n}(\omega)<X_{p+(q-p)j/2^n}(\omega)-c$ for all $i,j$ with $0\leq i\leq k < j\leq 2^n$,

*$|X_{p+(q-p)i/2^n}(\omega)-X_{p+(q-p)j/2^n}(\omega)|<c/4$ for all $i,j$ with $0\leq i\leq j\leq k$,

*$|X_{p+(q-p)i/2^n}(\omega)-X_{p+(q-p)j/2^n}(\omega)|<c/4$ for all $i,j$ with $k<i\leq j\leq 2^n$.
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 $$A_c=\bigcup_{\substack{p,q\in\mathbb{Q}\\p<q}} A_{p,q}.$$
Hopefully this is clear.
