Regular Conditional Probability given $X$ In Williams' "Probability with Martingales" he defines a regular conditional probability given $\mathcal{G} \subset \mathcal{F}$ as the map
$$
\mathbf{P}(\cdot, \cdot) : \Omega \times \mathcal{F} \to [0,1)
$$
such that
\begin{align*}
 &1.\quad \forall F \in \mathcal{F}, \omega \mapsto \mathbf{P}(\omega,F) \text{ is a version of } \mathbf{P}(F \mid \mathcal{G}),\\
 &2.\quad\text{ for almost every } \omega \in \Omega, F \mapsto \mathbf{P}(\omega,F) \text{ is a probability measure on } (\Omega,\mathcal{F}).
\end{align*}
Property 1 tells us that for any $F \in \mathcal{F}$ and any $G \in \mathcal{G}$,
\begin{align*}
\int_G \mathbf{P}(\omega, F) \, \mathrm{d}P(\omega) & = \int_G \mathbf{P}(F \mid \mathcal{G})(\omega) \, \mathrm{d}P(\omega) \\
& = \int_G E(\mathbb{I}_F \mid \mathcal{G})(\omega) \, \mathrm{d}P(\omega) \\
& = \int_G \mathbb{I}_F(\omega) \, \mathrm{d}P(\omega) \\
& = P(F \cap G).
\end{align*}
Fair enough, but I'm trying to get a handle on the regular conditional probability given a random variable $X: \Omega \to \mathbb{R}$.  It seems like we should define it as the map
$$
\mathbf{P}(\cdot, \cdot) : \mathbb{R} \times \mathcal{F} \to [0,1)
$$
such that
\begin{align*}
 &1.\quad \forall F \in \mathcal{F}, x \mapsto \mathbf{P}(x,F) \text{ is a version of } \mathbf{P}(F \mid \sigma(X)),\\
 &2.\quad\text{ for almost every } x \in \mathbb{R}, F \mapsto \mathbf{P}(x,F) \text{ is a probability measure on } (\Omega, \mathcal{F}).
\end{align*}
However, I'm not sure how to get the "defining equation" as above, but here's my shot.  Let $F \in \mathcal{F}$ and $C \in \sigma(X) \subset \mathcal{F}$.  Note $C = X^{-1}(B)$ for some $B \in \mathcal{B}$.  Letting $\Lambda_X$ be the distribution of $X$, by property 1 we get
\begin{align*}
\int_B \mathbf{P}(x,F) \, \mathrm{d}\Lambda_X(x) & = \int_C \mathbf{P}(X(\omega), F) \, \mathrm{d} P(\omega) \\
& = \int_C \mathbf{P}(F \mid \sigma(X))(\omega) \, \mathrm{d} P(\omega) \\
& = \int_C E(\mathbb{I}_F \mid \sigma(X))(\omega) \, \mathrm{d} P(\omega) \\
& = \int_C \mathbb{I}_F(\omega) \, \mathrm{d} P(\omega) \\
& = P(F \cap \{X \in B\}).
\end{align*}
Does this seem correct?
Update Based on Conrado Costa's answer, I think the proper definition for the regular conditional probability given $X$ should be as follows.  Below, let $X: (\Omega, \mathcal{F}) \to (E, \mathcal{E})$.
Assume the regular conditional probability given $\sigma(X)$ exists.  That is, there exists a function $\mathbf{P}^2 : \Omega \times \mathcal{F} \to [0,1]$ such that


*

*$\forall F \in \mathcal{F}$, $\omega \mapsto \mathbf{P}^2(\omega,F)$ is a version of $\mathbf{P}(F \mid \sigma({X}))$,

*$\forall \omega \in \Omega$, $F \mapsto \mathbf{P}^2(\omega, F)$ is a probability measure on $(\Omega, \mathcal{F})$.


In particular, property 1 tells us that $\forall F \in \mathcal{F}$, $\omega \mapsto \mathbf{P}^2(\omega,F)$ is $(\sigma(X), \mathcal{B}[0,1])$-measurable.  Then, by Theorem 18 listed below in Conrado Costa's answer, there exists a function $\mathbf{P}^1: E \times \mathcal{F} \to [0,1]$ that is $(\mathcal{E}, \mathcal{B}[0,1])$-measurable such that $\forall F \in \mathcal{F}$, $\omega \mapsto \mathbf{P}^2(\omega,F) = \mathbf{P}^1(X(\omega),F)$.
So, for any $F \in \mathcal{F}$ and any $C \in \sigma(X)$, by the change of variables theorem we get
\begin{align*}
\int_{X(C)} \mathbf{P}^1(x,F) \, d\Lambda_X(x) & = \int_C \mathbf{P}^2(\omega,F) \, dP(\omega) \\
& = \int_C \mathbf{P}(F \mid \sigma(X))(\omega) \, dP(\omega) \\
& = \int_C E(\mathbb{I}_F \mid \sigma(X))(\omega) \, dP(\omega) \\
& = \int_C \mathbb{I}_F(\omega) \, dP(\omega) \\
& = P(F \cap \{X \in B\}).
\end{align*}
 A: This is a subtle question.
you wrote:
\begin{align*}
 &1.\quad \forall F \in \mathcal{F}, x \mapsto \mathbf{P}(x,F) \text{ is a version of } \mathbf{P}(F \mid \sigma(X)),\\
 &2.\quad\text{ for almost every } x \in \mathbb{R}, F \mapsto \mathbf{P}(x,F) \text{ is a probability measure on } (\Omega, \mathcal{F}).
\end{align*}
for 1 It is important to note that the first $\mathbf{P}$ the one atributed to $\mathbf{P}(x,F)$ is a function from $\Bbb{R} \times \mathcal{F}$ into $[0,1]$, the second $\mathbf{P}$ (atributed to $\mathbf{P}(F \mid \sigma(X))$ is a function from $\Omega \times \mathcal{F}$ into $[0,1]$.
So Let's give them different names, call the first one $\mathbf{P}^1(x,F)$ and the second one $\mathbf{P}^2(\omega, F)$
The second one is readily defined from your previous consideration as you already know how to deal with regular conditional probability distributions. 
The first one needs the following theorem: (taken from Meyers Probability and potentials, page 10)

It tells us that $\mathbf{P}^2(\omega,F) = P^1(X(\omega),F)$ Use this $\mathbf{P}^1(x,F)$ as your object and then you can safely say that
\begin{align*}
\int_B \mathbf{P}^1(x,F) \, \mathrm{d}\Lambda_X(x) & = \int_C \mathbf{P}^1(X(\omega), F) \, \mathrm{d} P(\omega) = \int_C\mathbf{P}^2(\omega, F) \, \mathrm{d} P(\omega) \; (\ldots)\\
\end{align*}
