Q:

Is this martingale constant 0?

I have a martingale X where $X_0 = 0$ a.s.
And for each $\omega$, the path $f(t)=X_t(\omega)$ is of bounded variation in the classical sense. That $K(\omega)=\sup\{\Sigma_j|X_{t_{j+1}}(\omega)-X_{t_j}| : \{t_j\}\text{is a partition of }$[0,T]$\}<\infty$.
Is it then true that for each t $M_t(\omega)=0$ a.s.?
What I tried to do is this:
I create an arbitrary parition where the norm converges to 0. For each $\omega$ I have that $\Sigma_{j}|X_{t_{j+1}}-X_{t_j}|$ is bounded, and I have that for each $\omega$, $\Sigma_{j}|X_{t_{j+1}}-X_{t_j}|^2$ converges to 0.
If I can show that $E[X_t^2]=0$ for any t, I will be done.
What I have is that:
$E[X_t^2]=E[(\Sigma_j\{X_{t_{j+1}}-X_{t_j}\})^2]$. I can multiply this out and using that the cross-terms disappear beacuse of the martingale property of constant expectation. So I get $E[X_t^2]=E[\Sigma_j\{X_{t_{j+1}}-X_{t_j}\}^2]$.
I know that what is inside the last expectation goes pointwise to 0 as the partitions get smaller. But in order to finish the argument I need to be able to take the limit outside the expectation, can this be done?
Or is it false that this is true?

A:

Example: Let $N(t)$ be a unit rate Poisson process. Then $M(t):=N(t)-t$, $t\ge 0$, is a martingale with paths of finite variation (on each finite time interval) and $M(0)=0$.

