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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 $E(\int_{0}^{t}X_udu|F_s)=E(\int_{0}^{s}X_udu|F_s)+E(\int_{s}^{t}X_udu|F_s)=E(\int_{0}^{s}X_udu|F_s)+(t-s)X_s$?

blah blah blah
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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

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

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saz
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    @BlueJay You are welcome. Note that we have to assume that the integrals are well-defined, i.e. $$\mathbb{E} \left( \int_0^t |X_u| , du \right)< \infty$$ for all $t \geq 0$. – saz Mar 30 '16 at 15:17
  • Actually, I do have a question. In this case, we aren't given that $E(\int_{s}^{t}|X_u|du)=\int_{s}^{t}E|X_u|du$. – blah blah blah Mar 30 '16 at 17:39
  • @BlueJay Well, if you want to calculate $$\mathbb{E} \left( \int_0^t X_u , \mathcal{F}_s \right),$$ then you need an additional assumption to ensure that this conditional expectation is well-defined (continuity of $(X_t)_t$ is not enough).... and $$\mathbb{E} \left( \int_0^t |X_u| , du \right)<\infty$$ is a pretty natural assumption to ensure this. – saz Mar 30 '16 at 17:57
  • I'm not sure I see what you're saying. In the link you sent me, they used that $E(\int_{s}^{t}|X_u|du)=\int_{s}^{t}E|X_u|du$. I'm not sure how $E(\int_{0}^{t}|X_u|du)<\infty$ implies this. – blah blah blah Mar 30 '16 at 18:02
  • @BlueJay By the monotonicity of the integral, we have $$\int_s^t |X_u| , du \leq \int_0^t |X_u| , du$$ for any $s \leq t$; hence, $$\mathbb{E} \left( \int_s^t |X_u| , du \right) \leq \mathbb{E}\left( \int_0^t |X_u| , du \right)< \infty.$$ – saz Mar 30 '16 at 18:04
  • I got that part. Is the point that because of integrability, you're able to use Fubini to show that $E(\int_{s}^{t}|X_u|du)=\int_{s}^{t}E|X_u|du$? Or is it always the case that we can bring the expected value inside an integral? – blah blah blah Mar 30 '16 at 18:06
  • @BlueJay No, actually the identity $\mathbb{E}(\int_s^t |X_u| , du) = \int_s^t \mathbb{E}(|X_u|) , du$ holds in any case because the integrand is non-negative (also known as Tonelli's theorem). The important point is that this integral is finite. The integrability is needed in equation $(2)$ (in the linked answer) in order to interchange the integrals with respect to $\mathbb{P}$ and $dr$. (And, as I mentioned in my previous comment, the integrability condition also ensures that all appearing integrals/expectations are well-defined.) – saz Mar 30 '16 at 18:10
  • ok thanks for all your help. – blah blah blah Mar 30 '16 at 18:13