After having read about Doob's inequalities in discrete time, I am trying to understand the move to continuous time. I know that the Martingale regularization theorem tells me that there is (under the usual hypotheses) a càdlàg version of my martingale. I also know that it can be shown that the sample paths of càdlàg processes only have countably many discontinuities. What I fail to understand so far, is how this comes into play in moving from discrete to continuous.
Specifically I'm looking at a proof of Doob's maximal inequality from Alan Sola's Part III Probability notes at Cambridge (unfortunately he seems to have moved and the pdf isn't online anymore). So let's assume that $(X_t)_{t\geq 0}$ is a càdlàg martingale and let $\lambda\geq 0$. Then he considers the usual dyadic discretization of the interval $[0,t]$:
$$ D_n = \left\{kt2^{-n}:k=0,...,2^n\right\}$$ and defines discrete martingales $X^n$ and filtrations $\mathcal{F}^{n}$ by setting (for $k=0,\ldots,2^n$)
$$X^n_k=X_{\dfrac{kt}{2^n}}$$ $$\mathcal{F}^n_k=\mathcal{F}_{\dfrac{kt}{2^n}}$$
Then the usual maximal inequality in discrete time gives me that $$\lambda P\left(\max_{k\leq 2^n}|X^n_k| >\lambda\right)\leq E|X_t|$$
Next, we define for $n=1,2,\ldots$
$$A_n=\left\{\sup_{s\in D_n}|X_s| > \lambda\right\}$$
and note that these events are increasing in $n$, that is $A_n \subset A_{n+1}$.
The next step is the one I don't understand: Because $X$ càdlàg,
$$ \left\{\sup_{s\leq t}|X_s| > \lambda \right\} = \bigcup_{n=1}^\infty A_n$$
Could someone please point me in a direction or explain to me how cadlag comes into play here?
Thanks!
Assume $\sup_{s<t} |X_s|> \lambda$ therefore, there is an $t^*\in [0,t)$ with $|X(t^*)|> \lambda$ or $|X(t)|> \lambda$.
$\{|X(t)| > \lambda\} \subset \bigcup_{n=1}^\infty A_n$
If there is an $t^*\in [0,t)$ with $|X(t^*)|> \lambda$ then there is a sequence a dyadic $s_n = \frac{\lfloor{2^n t}\rfloor + 1}{2^n} \in D_n$ $s_n \downarrow t^*$ and right continuity gives us that there is $n \in \Bbb{N}$ for which $|X(s_n)|> \lambda$ and therefore
$\{\exists t^* \in [0,t) ,|X(t^*)|> \lambda\} \subset \cup_{n=1}^\infty A_n$
So we can conclude that $$ \bigg\{\sup_{s\leq t}|X_s| > \lambda\bigg\} \subset \bigcup_{n=1}^\infty A_n$$
the other inclusion $$ \bigg\{\sup_{s\leq t}|X_s| > \lambda\bigg\} \supset \bigcup_{n=1}^\infty A_n $$
Follows from the fact that
$$\bigg\{\sup_{s\leq t}|X_s| > \lambda\bigg\} \supset A_n \quad \forall n$$