Independence of a function and integral of a function I have a probability space $(\Omega, M, P)$ and non-negative integrable functions on $\Omega \times [0,1]$, $F_1(\omega, t)$ and $F_2(\omega, t)$. For each $t \in [0,1]$, we have that $F_1(\omega, t)$ and $F_2(\omega, s)$ are independent for any $s \in [0,1]$. Also $ \int_{\Omega} F_n(\omega, t) dP(\omega) = 1  $ for each $t$.
Does it follow that $X_1(\omega) = \int_a^b F_1(t,\omega) \ dt$, $0 \leq a< b \leq 1$
and $F_2(s,\omega)$ are independent for any fixed $s$?
I would appreciate any hint, solutions, comments! 
Thanks!
 A: First: In probability it is better to forget the $\omega$, the basic objects are the random elements (variables, processes or measures) and not particular events. 
Have that been said, the hypotheses tell us that for every $t,s \in [0,1]$ we have pairwise independence for $F_1(t)$ and $F_2(s)$. As Did remarks in the comments this is not enough to conclude the independence of the sigma algebras $\mathcal{F} := \sigma(F_1(t): t \in [0,1])$ and $\mathcal{G} := \sigma(F_2(t): t \in [0,1])$. To achieve this we need the stronger condition that for $0 \leq t_1 < \ldots < t_k \leq 1$ and $0 \leq s_1 < \ldots < s_l \leq 1$ the following identity holds
$$ P( F_1(t_i) \in D_i, \, 1\leq i \leq k, \, F_2(s_j) \in E_j, 1 \leq j \leq l) =\\P( F_1(t_i) \in D_i, 1\leq i \leq k)\cdot P(F_2(s_j) \in E_j, 1 \leq j \leq l) $$
where the $D$'s and $E$'s are borelian sets.
Under these conditions the result follows inmediatly since $X_1$ is a pointwise limit of $\mathcal{F}$-measurables random variables thus it is itself $\mathcal{F}$-measurable. On the other hand every $F_2(s), s \in [0,1]$ is $\mathcal{G}$-measurable, and so they are independent random variables.
