In basic calculus: 'Fubini's theorem' allows us to switch order of integration in double integrals without changing the bounds provided we are integrating over a rectangle. From here:
If the area is not a rectangle, we will very likely have to change the bounds.
In stochastic calculus:
Let $T > 0$. Given a stochastic process $\{ X_t \}_{t \in [0,T]}$ on $(\Omega, \mathscr{F}, \mathbb{P})$, we can say by 'Fubini's theorem' that
$$\mathbb{E} [\int_0^T X_t^2 dt] = \int_0^T \mathbb{E}[X_t^2] dt$$
or, if I'm not mistaken
$$\int_{\Omega} \int_0^T X_t^2 dt d\mathbb{P} = \int_0^T \int_{\Omega} X_t^2 d\mathbb{P} dt$$
The left hand side of the first equation comes from here: Itô isometry, so all the assumptions apply.
Question: Why is it that the bounds do not change?
Well obviously we can just look at the link earlier, but how can this be put in terms of 'Fubini's theorem' in basic calculus? I'm looking for an explanation for a beginning stochastic calculus student who knows only some basics of measure theory. As far as a beginning stochastic Calculus student knows, 'Fubini's theorem' is that proposition from basic Calculus which is for Riemann integrals and assumes rectangular area of integration. So how is that applicable here? Is such an area $\Omega \times [0, T]$ rectangular?
Best Answer
Based on saz' comments:
'if we integrate over the whole space, then we don't have to change the bounds of integration'
$$\int_{\mathbb R} \int_{\mathbb R} f(x,y) dx dy = \int_{\mathbb R} \int_{\mathbb R} f(x,y) dy dx$$
for applicable f. Also,
$$\int_{\Omega} \int_{\mathbb R} f(t,\omega) dt d\mathbb P = \int_{\mathbb R} \int_{\Omega} f(\omega,t) d\mathbb P dt$$
for applicable f. Now apply: $$f(t, \omega) := X_t^2(\omega)1_{[0,T]}(t)$$
A notable difference b/w the 2 Fubini's Thms is that in basic calculus, f is required to be continuous. In stochastic calculus, $f$ is not required to be continuous, but $f$ is required to be measurable.