Let $(\Omega,(\mathcal{F}_t),\mathbb{P})$ be a filtered probability space, and let $(W_t,\mathcal{F}_t)$ and $(B_t,\mathcal{F}_t)$ be two independent Brownian motions adapted to the filtration $\mathcal{F}_t$.
Now consider two $\mathcal{L}^2$ processes $f_t$ and $g_t$ adapted to the filtration $\mathcal{F}_t$, then the stochastic integral $M_t = \int_0^t f_s \,dW_s$ and $N_t =\int_0^t g_s\, dB_s$ are martingales and therefore the process defined by $$M_tN_t - \langle M,N \rangle_t, $$is a martingale, where $\langle M,N\rangle_t$ is the quadratic variation or the cross variation of the two processes. Now we also know that $$\langle M,N \rangle_t = \int_0^t f_sg_s\,d\langle W,B\rangle_s = 0, $$where the last inequality comes from the fact that $\langle W,B \rangle_t=0$ if $W_t$ and $B_t$ are independent Brownian motions. So I conclude that in this case the product of the stochastic integral $(M_tN_t,\mathcal{F}_t)$ is a martingale.
Now I'm not entirely confident about this statement, because using the notations above, both $W_t$ and $B_t$ are adapted to the filtration $\mathcal{F}_t$, but by independence $W_t$ is not adapted to the $\sigma$-algebra generated by $B_t$ denoted as $\mathcal{F}_t^B=\sigma{(B_s|s\leq t)}$. So if I take for instance $g_t = W_t$, is the process $$N_t = \int_0^tW_s\,dB_s, $$a martingale due to the independence of $W_t$ and $B_t$. Now assuming $(N_t=\int_0^tW_s\,dB_s,\mathcal{F}_t)$ is a martingale, and define $M_t=\int_0^tB_s\,dW_s$ symmetrically, is the process $M_tN_t - \langle M,N\rangle_t=M_tN_t - \int_0^t B_sW_s\,d\langle W_,B \rangle_t=M_tN_t-0=M_tN_t$ a martingale?
If this is again true, I believe this can be generalized to any two processes $M_t = \int_0^t f_s \,dW_s$ and $N_t =\int_0^t g_s\, dB_s$, where $f_t$ is an Ito process with local martingale part as an Ito integral of $B_t$ and $g_t$ is an Ito process with local martingale part as an Ito integral of $W_t$.
Can someone comment on these mathematics?