Baxter & Rennie at pag. 162 state the following theorem.

Let $W$ be an $n$-dimensional $\mathbb Q$-Brownian motion and let $M_t=(M_1(t),...,M_n(t))$ be an $n$-dimensional $\mathbb Q$-martingale process, which has volatility matrix $(\sigma_{ij}(t))$, in that $dM_j(t) = \Sigma_i \sigma_{ij}(t) dW_i(t)$ and the matrix is non singular (for all $t$ with probability 1). Let $N_t$ be any one-dimensional $\mathbb Q$-martingale.

Then there exists an $n$-dimensional $F$-previsible process $\phi_t = (\phi_1(t), ... \phi_n(t))$ such that $\int^T_0 (\Sigma_j \sigma_{ij}(t) d\phi_j(t))^2 dt <\infty$, and the martingale $N$ can be written as $N_t = N_0 + \Sigma_j \int_0^T \phi_j(s) dM_j(s)$. Further $\phi$ is (essentially) unique.

My question is: because each of the $M_j(t)$ is a $\mathbb Q$-martingale, why can't I chose the $\phi_j$ to write $N_t=N_0 + \int_0^T \phi_j(s) dM_j(s)$ $\forall j$? In other words, what is the idea behind uniqueness in the $n$-dimensional case ($n>1$)?

measure, without emphasizing that the filtration really plays a role. That seems unfortunate. Maybe there is a standing assumption that the probability space comes equipped with a Brownian motion $W$ and the filtration is the one generated by $W$, which would at least make the theorem correct but seems very awkward. I don't have the full book to check. $\endgroup$ – Nate Eldredge Oct 16 '18 at 23:42