Conditional Expectation of Martingale

Question

Letting $(X_t, F_t)_{t \in \mathbb{R}}$ be a martingale with continuous realizations and $0 \leq s \leq t$, I want to find $E(\int_{0}^{t}X_udu|F_s)$. I understand that $E(X_u|F_s)=X_s$ for $u\geq s$. Does this imply that $E(\int_{0}^{t}X_udu|F_s)=E(\int_{0}^{s}X_udu|F_s)+E(\int_{s}^{t}X_udu|F_s)=E(\int_{0}^{s}X_udu|F_s)+(t-s)X_s$?

Answer 1

Yes, you have to use Fubini's theorem for conditional expectations, see this question. Moreover, note that

$$\mathbb{E} \left( \int_0^s X_u \, du \mid \mathcal{F}_s \right) = \int_0^s X_u \, du.$$

Adding all up, we get

$$\mathbb{E} \left( \int_0^t X_u \, du \mid \mathcal{F}_s \right) = \int_0^s X_u \, du + (t-s) X_s.$$