Let $B$ be a standard Brownian motion and
X_t=\int_0^t f_s ds+\int_0^t g_s dB_s,
where, $|f|$ and $|g|$ are both bounded, almost surely, by some positive constant $M$.
Is it true that $$ E\left(\int_0^t X_s ds\right)= \int_0^t \left(E X_s \right) ds\,? $$
As $X$ is not a martingale, I cannot define $\int X$ as an almost sure limit, and apply the method of this post.