Suppose a process $X_t$ is defined as follows for some $H(\omega,t)$:
$$X_t:=\int_{s=0}^{s=t}H(\omega,s)ds$$
Suppose we want to compute the expectation of $X_t$:
$$\mathbb{E}[X_t]=\int_{\Omega}\int_{s=0}^{s=t}H(\omega,s)dsd\mathbb{P}$$
One way to interpret Fubini's theorem is to argue that we can exchange the order of integration above as long as:
- (i) $\mathbb{E}[|H(\omega,t)|]<\infty$
- (ii) $H(\omega,t)$ is measurable
I would like to ask about condition (ii) (i.e. "measurability"): suppose we want to apply Fubuni to the following examples, how do we check or demonstrate the measurability condition in practice?
$$\tag{1} X_t:=\int_{s=0}^{s=t}W_s^2ds$$
$$\tag{2} X_t:=\int_{s=0}^{s=t}W_t^2ds$$
$$\tag{3} X_t:=\int_{s=0}^{s=t}\left(W_1(s)+W_2(s)\right)ds$$
Question (1): This expectation is easily computable ( $\mathbb{E}[X_t]=\int \mathbb{E}[W_s^2]ds = \int s ds=0.5s^2$), but how do I check the measurability condition if I have to? I suppose my entire probability space would be generated by the Weiener measure associated with $W_t$, so I would just have to "show" that $W_t$ is measurable w.r.t. the filtration $\mathcal{F}_t$ (which it actually generates): so could I just say that "by definition of the probability space, the integrand is measurable"?
Question (2): here the integrand is $W_t$ (i.e. the Wiener process at time $t$), whilst the integral runs from $s=0$ to $s=t$: is the integrand measurable (with respect to $\mathcal{F}_t$ ?) ?? How do I show it is / show it isn't? Can the expectation be put inside the integral?
Question (3): Again, how do I show the integrand is measurable? Similarly to (1), here the probability measure is generated by the set of two Brownian motions (correlated with $\rho t$), so simply by definition of the probability space, the integrand is measurable (with respect to $\mathcal{F}_t$ ? ) ?