In the book "Theory of Stochastic Processes with Applications to Financial Mathematics and Risk Theory" (D. Gusak, A. Kukush, etc.) there is a problem 3.33:
Let $\{ X(t), t \in \mathbb{R} \}$ be a separable stochastic process taking its values in complete metric space. Prove that if the process $X$ has a cadlag modification then there exists a set $\tilde{N} \in \mathcal{F}$ with $P(\tilde{N}) =0$ such that the trajectory $X( \cdot, \omega)$ is cadlag for any $\omega \notin \tilde{N}$.
I can prove that if conditions of the problem hold true then there exists a set $\tilde{N}$ such that $\forall \omega \notin \tilde{N}$ $\forall t_0 \in \mathbb{R}$ $ \exists A(t_0, \omega):=\lim_{t \to t_0 - 0} X(t, \omega)$, $ \exists B(t_0, \omega):= \lim_{t \to t_0 + 0} X(t, \omega)$ and one of the following conditions holds: either $X(t_o, \omega) = A(t_0, \omega)$ or $X(t_o, \omega) = B(t_0, \omega)$. But I don't know how to prove, that $X(t_o, \omega) = B(t_0, \omega)$.
Moreover, I think that the problem is incorrect, because, if I am not mistaken, there is a counterexample. Consider $\Omega=[0,1]$, $\mathcal{F} = \mathcal{B}[0,1]$ and Lebesgue measure on it. Consider $X(t, \omega) = I(t > \omega)$, $\tilde{X}(t, \omega) = I(t \ge \omega)$. Then $\tilde{X}$ is a cadlag modification of $X$, $X$ is a separable process, but $X$ is not cadlag.
Is my counterexample correct? If it is correct, then how we may reformulate the problem in order to make it correct? And if my counterexample is not correct, then what is wrong with it and how we can solve the problem?
The definition I use is the following one:
Definition: A random function $X$ is called separable if there exist a countable dense subset $T_0 \subset T$ and a set $N \in \mathcal{F}$ with $P(N)=0$ such that for every open set $G \subset T$ and closed set $F \subset \mathbb{X}$, $$\{ \omega| X(t, \omega) \in F, t \in G \} \triangle \{ \omega| X(t, \omega) \in F, t \in G \cap T_0\} \subset N.$$