$If p(\omega, A) = P(A|\sigma)(\omega)$ is a regular version of conditional probability, then for every $X \in L^1 $
$E(X|\sigma)(\omega) = \int X(x)p(\omega, dx)$
I have to show this. I have seen the exercise was asked before but with $E(f(X)|\sigma)(\omega)$ And it was proven by using indicator functions. So my question is can i do it her too?