TITLE: Exchange integral and conditional expectation
QUESTION [8 upvotes]: I know that if we have $E[\int_0^1 |X_t|dt] < \infty$ we may apply Fubini's theorem and compute $E[\int_0^1 X_tdt] = \int_0^1 E[X_t]dt$. Is there a similar version that allows the exchange of integral and conditional expectation? I want to say $E[\int_0^1 X_tdt|F_s] = \int_0^1 E[X_t|F_s]dt$ but I don't know why it should be true.

REPLY [5 votes]: Let $(\Omega,\cal A,\mu)$ be a probability space, and assume that $X\colon [0,1]\times\Omega\to \Bbb R$ is measurable for the product $\sigma$-algebra and is integrable with respect to the product measure. In particular, we don't need the assumption of path-continuity. 
We can approximate $X$ in $L^1(\lambda \otimes\mu)$ and almost everywhere by simple function, i.e. finite linear combinations of elements of the form $\chi_A$, where $A\in \mathcal{A}\otimes \mathcal{B}([0,1])$. As such an element can be approximated in $L^1(\lambda\otimes\mu)$ by a finite linear combination of elements of the form $\chi_A\chi_B$, $A\in\cal A$, $B\in\mathcal B[0,1]$, we can consider linear combinations $\sum_kc_k\chi_{A_k}\chi_{B_k}$. We will denote $\{X_n\}$ this approximating sequence, which can be chosen in order to make $|X_n|\leqslant |X|$ almost everywhere.
Using dominated convergence theorem for conditional expectation, we just have see prove the equality for $X_n$ instead of $X$. We can indeed take $E(|X(t,\cdot)| \mid\mathcal F)$ as a dominating function, since $|X_n|\leqslant X$.
Note that by dominated convergence, $$\int_0^1E[X(t,\cdot)\mid\mathcal F]dt=\lim_{n\to +\infty}\int_0^1E[X_n(t,\cdot)\mid\mathcal F]dt,$$
and $\int_0^1E[X_n(t,\cdot)\mid\mathcal F]dt$ is $\mathcal F$-measurable. 
By linearity of conditional expectation, we will be done once we show that for any
$A\in\cal A,$ $B\in\mathcal B[0,1]$, we have 
$$E\left[\int_0^1\chi_A(\cdot)\chi_B(t)dt\mid \mathcal F\right]=\int_0^1E\left[\chi_A(\cdot)\chi_B(t)\mid \mathcal F\right]dt.$$
It's the case since $\chi_B(t)$ doesn't depend on $\omega$.