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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.

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From Itō isometry and boundedness of $f_s, g_s$, we can get a uniform $L^2(\Bbb{P})$ bound for $X=(X_s,0\leq s\leq t)$. This implies uniform $L^1(\Bbb{P})$-boundedness of $X$. Thus $$\int_{0}^{t}\Bbb{E}\left|X_s\right|\,ds$$ converges and we may apply Fubini's theorem. – Sangchul Lee Nov 13 '12 at 18:28
@sos440 Excellent answer. Thank you. – Nicolas Essis-Breton Nov 13 '12 at 18:33

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