Expectation of Integrals of Brownian Motion Hello I am not a native english speaker so please let me know if something does not make sense. I am interested in computing the following:
$$E\int_0^T(B_s(\omega,t))^4dt$$
Or at least showing it is finite because I want to prove that $(B_s(\omega,t))^2\in\mathcal H[0,T]$. Thank you.

Hint Apply Tonelli's theorem and use that $\mathbb{E}(B_t^4)=3t^2$ as $B_t$ is Gaussian with mean $0$ and variance $t$.

Remark Please note that the expression $B_s(\omega,t)$ does not make sense at all. The Brownian motion depends on the time (usually denoted by $t$) and the "random" $\omega$. So, in your case, it should read 
$$\mathbb{E} \int_0^T B_t^4 \, dt \qquad \text{or} \qquad \mathbb{E}\int_0^T B(t)^4 \, dt$$ 
or
$$\int\!\!\! \int_0^T B_t(\omega)^4 \, dt \, d\mathbb{P}(\omega) \qquad \text{or} \int\!\!\! \int_0^T B(t,\omega)^4 \, dt \, d\mathbb{P}(\omega)$$ 
(Mind that $\mathbb{E}X = \int X(\omega) \, d\mathbb{P}(\omega)$; it is not correct to write $\mathbb{E}X(\omega)$ for some random variable $X$.)
