Cadlag process and measurability. Let $(\Omega,(\mathcal{F_t})_{t\geq0},P)$ be a filtered probability space and $X=(X_t)_{t\geq0}$ a real-valued adapted cadlag process. Let $A\subset\Omega$ (resp. $B\subset\Omega$) be the event that $X$ is continuous (resp right-continuous) on $[0,t)$.
Show that $A$, $B\in\mathcal{F_t}$.
I am not to show how to show this for $A$ but is it not trivial for $B$ as since $X$ is cadlag, we must then have $B=\Omega$??
Any help is most welcomed and needed. Thanks.
 A: If $(X_t)_{t \geq 0}$ is càdlàg, then it follows from the very definition that $B=\Omega$, so there is nothing to show.
In order to show $A \in \mathcal{F}_t$, we note that
$$\begin{align*} A &= \bigcap_{n \in \mathbb{N}} \underbrace{\{\omega; [0,t-1/n] \ni s \mapsto X_s(\omega) \, \text{is (uniformly) cts.}\}}_{=:A_n}, \end{align*}$$
i.e. it suffices to show $A_n \in \mathcal{F}_t$. As $(X_t)_{t \geq 0}$ is càdlàg, we know that the limits
$$\begin{align*} X(t-) &:= \lim_{s \uparrow t} X(s)  = \lim_{\mathbb{Q} \ni s \uparrow t} X(s) \tag{1} \\ X(t) &= \lim_{s \downarrow t} X(s) = \lim_{\mathbb{Q} \ni s \downarrow t} X(s) \tag{2} \end{align*}$$
exist for all $t \geq 0$. Therefore,
$$A_n = \{\omega; \forall s \in [0,t-1/n]: X(s-,\omega) = X(s,\omega)\}.$$
By $(1)$ and $(2)$, we have
$$\begin{align*} &\forall s \in [0,t-1/n]: X(s-,\omega) = X(s,\omega)\\  &\iff  \{\forall \epsilon>0 \exists \delta>0: r,s \in [0,t-1/n] \cap \mathbb{Q}, |r-s| \leq \delta \Rightarrow |X(r,\omega)-X(s,\omega)| \leq \epsilon\} \end{align*}$$
Consequently,
$$A_n = \bigcap_{i \geq 1} \bigcup_{j \geq 1} \bigcap_{\substack{r,s \in [0,t-1/n] \cap \mathbb{Q} \\ |r-s| \leq \frac{1}{j}}} \left\{ |X(r)-X(s)| \leq \frac{1}{i} \right\} \in \mathcal{F}_{t- \frac{1}{n}} \subseteq \mathcal{F}_t.$$
