Expectation of a stochastic integral Let $M$ be a right-continuous local martingale, $s,t$ two times (stopping times, if you like). Under what conditions does the following hold:
$$E\left(\int_s^t X \, dM\mid\mathcal{F}_s\right)\le 0$$
My guess is:
It holds whenever: $E(\int_s^t X \, dM\mid\mathcal{F}_s)> -\infty$
But I have no idea where to start...
[This is no homework]
 A: If $M$ is a martingale and $X$ is properly integrable, then
$$I_t := \int_s^t X(r) \, dM_r, \qquad t \geq s$$
defines a martingale and, consequently,
$$\mathbb{E} \left( \int_s^t X(r) \, dM_r \mid \mathcal{F}_s \right) = \mathbb{E}(I_t \mid \mathcal{F}_s) = I_s =0$$
for any $t \geq s$. Now, if $M$ is a local martingale and $X$ locally integrable, then there exists a sequence of stopping times $(\tau_n)_n$ such that
$$I_{t}^{\tau_n} := \int_s^{t \wedge \tau_n} X(r) \, dM_r$$
is a martingale. Thus,
$$\mathbb{E} \left( \int_s^{t \wedge \tau_n} X(r) \, dM_r \mid \mathcal{F}_s \right) = 0.$$
As $\lim_{n \to \infty} I_t^{\tau_n} = I_t$, this leaves us with the question whether we may apply some "Fatou-like" argumentation, i.e. whether
$$\mathbb{E} \left( \int_s^{t} X(r) \, dM_r \mid \mathcal{F}_s \right) \leq \limsup_{n \to \infty} \mathbb{E} \left( \int_s^{t \wedge \tau_n} X(r) \, dM_r \mid \mathcal{F}_s \right) = 0.$$
Obviously, this works out if the dominated convergence theorem is applicable; that is
$$\sup_{n \in \mathbb{N}} \left| \int_s^{t \wedge \tau_n} X(r) \, dM_r \right| \in L^1(\mathbb{P}).$$

Note that we can in general not expect that the stated inequality holds without further assumptions on $X$ or $M$. Consider for example $X=1$, then
$$\mathbb{E}\left( \int_s^t X \, dM \mid \mathcal{F}_s \right) \leq 0 \Leftrightarrow \mathbb{E}(M_t \mid \mathcal{F}_s) \leq M_s$$
i.e. the inequality holds if $M$ is a supermartingale.
