Why can you simplify $E[\int_s^t X_u du |F_s] = X_s \int_s^t du$?

Question

As per the question, I am trying to simplify the expression $$E[\int_0^t X_u du | Fs]$$ where $X_u$ is standard Brownian motion, and $F_s$ is the filtration of the process until time $s$. Now you can split the integral into $0$ to $s$ and from $s$ to $t$, but why can you simplify $E[\int_s^t X_u du |F_s] = X_s \int_s^t du = X_s(t-s)$?

Thanks a lot in advance!

Answer 1

$E[\int_s^t X_u du |F_s] =E[\int_s^t [X_u-X_s] du |F_s]+E[\int_s^t X_s du |F_s]$. The second term is, of course, $X_s(t-s)$. Now the integral in the first term can be written as a limit of Riemann sums and this shows that the integral is independent of $F_s$. Hence, the first term is $0$ (since $E(X_u-X_s)=0$ for all $u$).

Answer 2

Using Fubini's Theorem, we can write $$\mathbb{E} \left[ \int_s^t X_u du \middle| \mathcal{F}_s \right] = \int_s^t \mathbb{E}\left[ X_u | \mathcal{F}_s \right] du = \int_s^t X_s du = X_s(t-s).$$ Basically, here I'm just swapping the expectation and integral (you can verify the details, but this only requires that the Brownian motion is a square integrable martingale and the bounds of the integral are finite, so all should be good).