I am particularly interested in the proof of the Début Theorem as it is formulated in George Lowther's blog. There is one detail that escapes me.
Let $\sigma$ be a stopping time, $t\geq 0$ a positive real number, $a\in\mathbb{R}$ a constant and $X$ a right-continuous adapted process. The proof implicitly uses this identity
$$ \bigcup_{\sigma\leq s\leq t}\{X_s>a\} = \bigcup_{q\in\mathbb{Q}\bigcap [0,1]}\{\sigma\leq q\,t\}\bigcap\{X_{q\,t}>a\} $$
which follows from the right-continuity of the paths of $X$. It is easy to prove the inclusion $\supseteq$, that is if there exists $q\in\mathbb{Q}\bigcap [0,1]$ such that $X_{q\,t}>a$ and $\sigma\leq q\,t$ then we have immediately that $\exists s$ with $\sigma\leq s$ such that $X_s>a$ (and we don't need here right-continuity). The other inclusion is less evident. Suppose that $\exists s$ such that $\sigma\leq s\leq t$ such that $X_s>a$. How do I find the $q\in\mathbb{Q}\bigcap [0,1]$ that I need? I have somehow to use right-continuity, but how?
Let $\omega \in $ lhs. Then, there exists $s\in [0,t]$ s.t. $\sigma(\omega)\leq s$ and $X_s(\omega)>a$ or equivalently $u \in [0,1]$ s.t. $\sigma(\omega)\leq ut$ and $X_{ut}(\omega)>a$.
There exist $q\in \mathbb{Q}\cap[0,1]$ s.t. $\sigma(\omega)\leq ut\leq qt$. Consider $X_{qt}(\omega)$ for all such $q$. Since $t\mapsto X_t(\omega)$ is right continuous, for any $\varepsilon >0$ there exists a $q$ among those s.t. $X_{qt}(\omega)\in (X_{ut}(\omega)-\varepsilon,X_{ut}(\omega)+\varepsilon)$. Choose $\varepsilon >0$ s.t. the following holds: $$a<X_{ut}(\omega)-\varepsilon<X_{ut}(\omega)<X_{ut}(\omega)+\varepsilon$$ Then there exists $q$ s.t. $\sigma(\omega)\leq ut\leq qt$ and $X_{qt}(\omega)>a$. So $$\omega \in \bigcup_{q \in \mathbb{Q}\cap [0,1]}\{\sigma \leq qt\}\cap \{X_{qt}>a\}$$