# Cadlag and bounded martingale is constant?

Let $M=(M_t)_{t\in[0,1]}$ be a cadlag (right continuous with left limit) and bounded martingale w.r.t. right continuous filtration $(\mathcal F_t)_{t\in [0,1]}$. Assume that $\|M_t-M_s\|_\infty \to 0$ as $t\downarrow s$, can we conclude that $M$ is constant a.s.? The infinite norm here is $\|X\|_\infty :=\mbox{ess} \sup_{\omega \in \Omega}|X(\omega)|$.

My attempt was to prove $M$ is continous and has finite variation. If it's true then we are done, however I'm stuck.

No, $M$ does not need to be constant.

(Counter)Example: Let $X \in L^1(\mathbb{P})$ be a bounded random variable with mean $0$. Define

$$M_t(\omega) := \begin{cases} 0, & t \in [0,1/2), \\ X, & t \in [1/2,1] \end{cases} \qquad \mathcal{F}_t := \begin{cases} \{\emptyset,\Omega\}, & t \in [0,1/2), \\ \sigma(X), & t \in [1/2,1]. \end{cases}$$

Clearly, $(M_t)_{t \geq 0}$ is bounded, has càdlàg sample paths and satisfies

$$\lim_{t \downarrow s} \|M_t-M_s\|_{\infty} = 0.$$

Moreover, it is not difficult to see that $(M_t,\mathcal{F}_t)_{t \geq 0}$ is a martingale.

• Ah, I see, thanks a lot. I'm thinking if we replace $t\downarrow s$ by $t\to s$, then whether one gets the finite variation of $M$ from $\lim_{t\to s}\|M_t-M_s\|_\infty =0$?
– ntt
Jun 10, 2017 at 18:48
• @ntt I don't think so. Obviously you get continuity, but, as far as I can see, not finite variation.
– saz
Jun 11, 2017 at 4:25