Properties of a cadlag continuous martingale

Question

So I get a little confused when it comes to trying to extend discrete time to continuous time martingales. For example I am unsure how to prove the following: If $D_n=\lbrace k2^{-n} \vert k \in Z\rbrace $ then for a cadlag martingale $(X_n)_{n \geq 0}$ the following is true $$\lim_{n \to \infty}\sup_{t \in D_n} \vert X_t \vert=\sup_{t\geq 0} \vert X_t\vert .$$ Do I have to use that $D= \bigcup_{n=1}^\infty D_n$ is dense in the real numbers? How do I proceed from here?

Answer 1

If you fix $\omega\in \Omega$, the sample path $t\mapsto X(t,\omega)$ is a cadlag function. So the problem reduces to show that for a cadlag function $f:\mathbb{R}_+\to \mathbb{R}$, $$ \lim_{n\to\infty}\sup_{t\in D_n}|f(t)|=\sup_{t\ge 0}|f(t)|. \quad (*) $$

First, by the denseness of $D$ and the cadlag property of $f$, for each $m\in \mathbb{N}$ we have

$$ \max_{t\in D_n\cap[0,m]}|f(t)|\to \sup_{t\in[0,m]}|f(t)|. \quad (**) $$

Then since both sides of $(**)$ are non-decreasing in $m$ and the LHS is non-decreasing in $n$ (for fixed $m$) we may send $m\to \infty$ to get $(*)$.