[Math] Ito process is a martingale

Question

Let $(B_t)_{t\geq 0}$ be a Brownian motion. Is it true that if the generalized Ito process
$$
X_t=\int_0^tf(B_s)\;dB_s+\int_0^tg(B_s)\;ds,
$$
$f,g\in\mathcal{C}(\mathbb{R})$, is a martingale, then $g\equiv 0$? It is easy to see that $X_t\in L^2$ for any $t\geq 0$ and hence $(M_t)_{t\geq 0}:=(X^2_t-\langle X\rangle_t)_{t\geq 0}$, where $\langle X\rangle$, denotes the quadratic variation is a martingale. Hence, if I could manage to show that $(M_t)_t$ is of BV for $g\neq 0$, I would be done since any BV martingale is constant as. However, I am struggling a bit with the argument.

Answer 1

If $X$ is a martingale, then $\int_0^t g(B_s)\,ds = X_t-\int_0^t f(B_s)\,dB_s$ is a local martingale with paths of (locally) bounded variation. As you note, the only BV martingales are the constant ones, and this remains true for BV local martingales by the obvious localization argument.