Consequence of Doob-Meyer decomposition

Question

Let $\mathcal{H}^2_0$ be the space of all $L^2$ bounded RCLL martingales null at zero. As a consequence of Doob-Meyer we know:

For every $M\in \mathcal{H}^2_0$ there exist a unique adapted, increasing RCLL process $[M]$ null at zero with $\Delta [M]=(\Delta M)^2$ such that $M^2-[M]$ is a uniformly integrable martingale null at zero.

Suppose $L\in \mathcal{H}^{2,c}_0$ (the $c$ indicates that the martingale is continuous) and $M\in \mathcal{H}^{2,d}_0$, where $\mathcal{H}^{2,d}_0$ is the orthogonal complement in the Hilbert space $\mathcal{H}^2$ with the inner product $(L,M):=E[M_\infty L_\infty]$

I should show that $[L,M]=0$. The hint was to prove two things:

1. $LM$ is a uniformly integrable martingale

2. $[L,M]$ is continuous.

I was able to prove $1$. However why do I have to prove 2? By the theorem above we know there is an unique adapted RCLL process $[L,M]$ null at zero, of finite variation such that $\Delta [M,L]=\Delta L \Delta M$ and $LM-[L,M]$ is a uniformly integrable martingale. The process identically to zero is RCLL, adapted and of finite variation. Moreover by 1, we know $LM-0$ is auniformly integrable martingale. By uniqueness this would imply $[L,M]=0$. Am I wrong? I do not see why I have to show that $[L,M]$ is continuous. Even more, why should it be? Of course it is RCLL (due to polarization), but in general not continuous.

Thanks in advance!

Answer 1

You are right that the process $0$ is an adapted, RCLL and increasing process. You have shown that $LM$ is a uniformly integrable martingale, so in order to conclude that $[L,M]=0$, you just have to show that $\Delta [L,M]=\Delta L\Delta M$. I believe this is why they want you to show that this particular quadratic variation process, $[L,M]=0$, is continuous (they are not asking you to show that every quadratic variation process is continuous because, as you said yourself, this is not the case). Now that $[L,M]=0$ is continuous then $\Delta [L,M]=\Delta L\cdot \Delta M$ is obvious.