$E\int_0^T|X(t)|^2dt=0$ then $X=0 dP\times dt$

Question

$X:\Omega\times [0,T]\to \mathbb{R}$ - process measurable with respect to the product sigma field such that $E\int_0^T|X(t)|^2dt=0$. Prove that then $X=0 dP\times dt$ Can anyone prove it?

Answer 1

Let $A=\{(\omega,t): X(t))(\omega)=0\}$. By hypothesis we have $\int_0^{T} |X(t)|^{2}dt=0$ for almost all $\omega$. And this implies that For almost all $\omega$, $X(t))(\omega)=0$ for almost all $t$. Now apply Fubini's Theorem to $(P\times m)(A^{c})=\int I_{A^{c}} dP\times dt$. We get $\int \int I_{A^{C}} dt dP=\int 0dP=0$.