Fubini's theorem and time integrals of stochastic processes Let $(X_t)$ be a real continuous stochastic process such that $E[ \int_0^1 (\frac{X_t}{1-t})^2 dt ] <\infty$. Let $f_t$ be the probability density function (PDF) of $(X_t)$.
Since $Y_t= \frac{X_t}{1-t}$ is $L^2$ in time and space, and that the measure of $\Omega \times [0,1]$ is finite, $(Y_t)$ is also $L^1$ in time and space, and hence,
$$I:=E\left[ \int_0^1 \frac{X_t}{1-t} dt \right] <\infty.$$
I would like to know if the following integral is equal to $I$:
$$\int_\mathbb R x \left(\int_0^1 \frac{f_t(x)}{1-t} dt\right) d x,$$
and if so, is $\int_0^1 \frac{f_t(x)}{1-t} dt$ the PDF of the random variable $\int_0^1 \frac{X_t}{1-t} dt$ ? Is the function $x \rightarrow x\int_0^1 \frac{f_t(x)}{1-t} dt$ guaranteed to be integrable in $x$ ?
I obtained this integral by following these steps:
$$I = \int_0^1 \frac{E[X_t]}{1-t} dt  = \int_0^1 \frac{ \int_\mathbb R x f_t(x) dx}{1-t} dt =\int_\mathbb R x \left(\int_0^1 \frac{f_t(x)}{1-t} dt\right) d x $$
I used Fubini two times, and I am not sure if knowing $I < \infty$ is enough to use Fubini in these ways.
 A: Your first application of Fubini's theorem is correct, but the second one does not seem to hold in general.
First application
Recall that to apply Fubini's theorem to $f$, you must have $f \in L^1( \Omega \times [0,1], \mathbb{P} \times \lambda)$. Under your assumptions, this holds, since $$\int_{\Omega \times [0,1]} \left| \frac{X_t}{1-t} \right| d (\mathbb{P} \times \lambda) \leq \int_{\Omega \times [0,1]} \left| \frac{X_t}{1-t} \right|^2 d (\mathbb{P} \times \lambda) = \mathbb{E} \int_0^1 \frac{X_t^2}{(1-t)^2}dt < \infty $$
where the inequality is Hölder's inequality and the equality is Tonelli's theorem (which allows you to exchange iterated integrals for positive functions). It follows that
$$\mathbb{E} \int_0^1 \frac{X_t}{1-t} dt = \int_0^1 \frac{\mathbb{E} (X_t)}{1-t} = \int_0^1 \frac{\int_{\mathbb{R} } x f_t(x) }{1-t}$$ by Fubini's theorem.
Second application
Your second application doesn't seem to hold. For a counterexample, consider the time-changed and time-reversed Brownian motion $X_t = B_{(1-t)^2}$, whose density is given by $$f_t(x) = \frac{1}{\sqrt{2 \pi} (1-t) } \exp \left( 
- \frac{x^2}{2(1-t)^2} \right)$$
Observe that your moment condition holds: $$\mathbb{E} \int_0^1 \frac{X_t^2}{(1-t)^2}dt = \int_0^1 \frac{\mathbb{E}X_t^2}{(1-t)^2}dt = \int_0^1 \frac{(1-t)^2}{(1-t)^2} = 1 < \infty$$
So Fubini's theorem tells us that:
$$\mathbb{E} \int_0^1 \frac{X_t}{1-t} dt = \int_0^1 \frac{\mathbb{E} X_t}{1-t} dt= 0$$
However, according to WolframAlpha, we get:
$$\int_0^1 \frac{f_t(x)}{1-t} = \sqrt{ \frac{\pi}{2} } \left( \frac{1}{x} - \mathrm{erf} \left( \frac{x}{\sqrt{2}} \right) \right)$$
whence,
$$\int_{\mathbb{R}} x \left(\sqrt{ \frac{\pi}{2} } \left( \frac{1}{x} - \mathrm{erf} \left( \frac{x}{\sqrt{2}} \right) \right) \right)dx $$
does not exist.
