I know that for being able to apply Fubini I have before to prove the measurability and the integrability condition. In particular (for what concerne the integrability) I have to prove that: $$E\bigg[\int_0^t\big| \phi_s \big|ds \bigg]<\infty$$ and only if this is true I can interchange the two integrals.
Now lets take two example involving the Brownian motion ($B_t$) as integrand:
$1)$ $E\bigg[\int_0^t B_sds \bigg]$
$2)$ $E\bigg[\int_0^t sgn(B_s)ds \bigg]$ where $sgn(x)=1$ if $x \ge 0$ and $sgn(x)=-1$ if $x < 0$.
Is the integrability condition in these cases satisfied? In particular:
Does the continuity of $B_t$ imply that $E\bigg[\int_0^t |B_s |ds \bigg] <\infty?$
Does the fact that $\int_0^t sgn(B_s)ds$ can assume as maximum value $t$ imply that $E\bigg[\int_0^t |sgn(B_s)|ds \bigg]$ is finite?
I would like to understand which are the most common methods for observing that the integrability condition is satisfied.
$E\int_0^{t}|B_s|ds=\int_0^{t}E|B_s|ds$ by Tonelli's Theorem. Since $B_s$ has same distribution as $\sqrt s X$ where $X$ has standard normal distribution we get $E|B_s|=\sqrt sE|X|$.Finally observe that $\int_0^{t} \sqrt s ds <\infty$.
The second is simpler. Use Tonelli's Theorem again and just note that $|sign (B_s)|=1$.