Let us start from a one-dimensional Brownian Motion $B(l)$, $l=0<\ldots<s<\ldots<t$.
If we want to compute $\mathbb{E}\left[B(s)B(t)\right]$ by applying Fubini's theorem, knowing that for Fubini's theorem to be applied it is necessary that integrand is integrable, is there any easy way to show that: $$B(s)B(t)\text{ is integrable}$$ that is: $$\mathbb{E}\{B(s)B(t)\}<\infty? $$
Or, use $|B(s)B(t)|\le (1/2)[B(s)^2+B(t)^2]$ and the square integrability of $B(s)$ and $B(t)$ noted already by Kolmo.