I have the following process where $B$ is a Brownian motion
$$\int_0^T \frac{B(T)-B(s)}{T-s}ds$$
I want to verify that the expectation exists, but to do so mandates exchange of the integrals via Fubini. How do I prove that
$$\int_{\Omega\times[0,T]}\frac{B(T,\omega)-B(s,\omega)}{T-s}d(\omega,s)<\infty$$
What are the general techniques for proving such results and exactly how do I think about this integral that is also over the space $\Omega$?
By Tonelli's Theorem we can write $E\int_0^{T} |\frac {B(T)-B(s)} {T-s}| ds=\int_0^{T} E|\frac {B(T)-B(s)} {T-s}| ds$. To compute $E|\frac {B(T)-B(s)} {T-s}|$ observe that $Y=\frac {B(T)-B(s)} {\sqrt{T-s}}$ has standard normal distribution. Hence $E|\frac {B(T)-B(s)} {T-s}|=\frac a {\sqrt {T-s}}$ where $a=E|Y|$. Finally $E\int_0^{T} |\frac {B(T)-B(s)} {T-s}| ds=\int_0^{T} \frac a {\sqrt {T-s}}ds$ which is fiite. Now we can apply Fubini's Theorem since $\frac {B(T)-B(s)} {T-s}$ is integrable on the product space.