Multiplication rule and regular conditional probability I've been studying the conditions of existence of the regular conditional probability and have a question about it. Let's $(\Omega, \mathcal{B}, P)$ be a product probability space, and let's say the regular conditional probability exists :
\begin{align}
P_{\times}(A\times B) = \int_B P(A|\omega)dP_0(\omega)
\end{align}
with $P_0$ being the marginal of $P_{\times}$.
I was wondering how to derive from this the well known multiplication rule for measurable sets :
\begin{align}
P(A\cap B) = P(A|B)P(B)
\end{align}
those two results seems pretty general to me and are somewhat related (I think). I just want to wrap my head around the two equations. If we assume the sets to be independents I managed to arrive to the same result $P(A\cap B) = P(A)P(B)$. So how theses equations relate ?
 A: Assume you have $(\Omega, \mathcal{B},\Bbb{P}_0)$ be a probability space where you define $X,Y$ two random variables. Let $\sigma(Y)$ be the sigma-algebra generated by $Y$. Assume that a regular conditional probability for $\Bbb{P}_0$ given $\sigma(Y)$ exists, that is assume there is a function $K_Y: \mathcal{B}\times \Omega \to [0,1]$ with the following properties:
1) $A \mapsto K_Y (A,\omega)$ is a probability for every $\omega \in \Omega$
2) $\omega \mapsto K_Y(A,\omega)$ is $\sigma(Y)$- measurable for every $A \in \mathcal{B}$
3) $K_Y(A,\omega) = \Bbb{E}[1_A \vert \sigma(Y)]$
Now note that
$$\Bbb{P}_0(X \in A, Y \in B) = \Bbb{E}[\Bbb{E}[1_{\{X \in A\}} 1_{\{Y \in B\}} \vert \sigma(Y)]] = \Bbb{E}[1_{\{Y \in B\}}\Bbb{E}[1_{\{X \in A\}}  \vert \sigma(Y)]]  = \Bbb{E}[1_{\{Y \in B\}}K(\{X \in A\},\omega)] = \int_B K_Y(\{X \in A\},\omega) \, d\Bbb{P}_0$$
So now calculate
$$\Bbb{P}_0(X \in A \vert Y \in B) = \frac{\Bbb{P}_0(X \in A, Y \in B)}{\Bbb{P}_0( Y \in B)} = \int_B K_Y(\{X \in A\},\omega) \, \frac{d\Bbb{P}_0}{\Bbb{P}_0( Y \in B)}$$
The probability of $[A] = \{X \in A\}$ given $[B] = \{Y \in B\}$ is therefore the mean of the conditional probabilities of $[A]$ given $\omega \in [B]$.
Note finally that $K(A,\omega)$ gives you a more general notion of conditional probability, once it allows you to speak of the conditional probability of an event given an outcome $\omega$ (even if $\Bbb{P}_0(\{\omega\}) = 0$)
