If $X$ is a martingale, $X(0)=0$; $f$ left continuous, is $\int f X$ dt also a martingale? If $X(t)$ is a martingale, and $X(0) = 0$.
$f(t)$ is a left continuous function,
$$
g(t) = \int_0^t f(s) X(s) ds
$$
is $g(t)$ also a martingale?
I guess it shall be, but don't know how to prove that?
 A: Since $(X(t))_t$ is a martingale in the filtration $(\mathcal F_t)_t$, one knows that $\mathrm E(X(u)\mid\mathcal F_s)$ is $X(u)$ is $u\leqslant s$ and $X(s)$ if $u\geqslant s$. Hence, for every $s\leqslant t$,
$$
\mathrm E(g(t)\mid\mathcal F_s)=g(s)+\int_s^tf(u)\mathrm E\left(X(u)\mid F_s\right)\mathrm du=g(s)+X(s)\cdot \int_s^tf(u)\mathrm du.
$$
The only cases when $(g(t))_t$ is a martingale are when $f(t)X(s)=0$ almost surely, for almost every $s\leqslant t$.
Edit: Note however that $X(0)=0$ and that $(X(t))_t$ is a martingale hence $\mathrm E(X(t))=0$ for every $t$, and
$$
\mathrm E(g(t))=\int_0^tf(s)\mathrm E\left(X(s)\right)\mathrm ds=0.
$$
A: No, take $f(s) = s$ so
$$
g(t) = \int\limits_0^t s B_s\mathrm ds = \int\limits_0^t B_s\mathrm d\frac{s^2}{2} = \left.\frac{s^2}{2}B_s \right|_0^t - \frac12\int\limits_{0}^ts^2\mathrm dB_s = \frac12 t^2B_t - \frac12\int\limits_{0}^ts^2\mathrm dB_s.
$$
Note that $\frac12\int\limits_{0}^ts^2\mathrm dB_s$ is a martingale and if $g$ is a martingale then $\frac12 t^2 B_t$ has to be martingale, but:
$$
\mathsf E[t^2 B_t|\mathscr F_s] = t^2 B_s\neq s^2 B_s.
$$
