Let $(\Omega,\mathcal F,\mathbb P)$ is a standard Borel space (i.e. $\Omega$ is Polish and $\mathcal F = \mathcal B(\Omega)$).
Then $\mathcal F$ is separable and for every sub-sigma-algebra $\mathcal G \subset \mathcal F$, then there exists a regular conditional probability $\mathbb P^{\mathcal G} : \Omega \times \mathcal F \to [0,1]$ (i.e. satisfying a. for all $A \in \mathcal F$, $\mathbb P^{\mathcal G}(\cdot,A) : \Omega \to [0,1]$ is $\mathcal G$-measurable, b. for a.e. $\omega \in \Omega$, the map $\mathbb P^{\mathcal G}(\omega,\cdot) : \mathcal F \to [0,1]$ is a probability on $(\Omega,\mathcal F)$ and c. for all $A \in \mathcal F$ and $B \in \mathcal G$, we have $\mathbb P(A \cap B) = \int_\Omega 1_B(\omega) \mathbb P^{\mathcal G}(\omega, A) \mathbb P(d \omega)$).
I read here and there (Is the regular conditional probability of $\mathbb P$ dominated by $\mathbb P$ almost surely? ; https://mathoverflow.net/questions/354098/necessary-and-sufficient-conditions-for-almost-sure-absolute-continuity-of-regul ) that under this setting, for every $\mathcal G$, we have $\mathbb P^{\mathcal G}(\omega, \cdot) \ll \mathbb P$ almost surely (i.e. $\mathbb P({\omega : \mathbb P^{\mathcal G}(\omega, \cdot) \ll \mathbb P}) = 1$).
I must have a misunderstanding about r.c.p's because of the following:
take $(\Omega=[0,1],\mathcal F = \mathcal B([0,1]),\mathbb P = \mathrm{Leb})$ and $\mathcal G = \mathcal F = \mathcal B([0,1])$.
Then for $A \in \mathcal F$, we have $$\int_\Omega 1_A(\omega) P(d\omega) = \mathbb P(A) = \mathbb P(A \cap A) = \int_\Omega 1_A(\omega) \mathbb P^{\mathcal G}(\omega, A) \mathbb P(d\omega) \,,$$ which gives $1_A = 1_A \mathbb P^{\mathcal G}(\cdot, A)$, $\mathbb P$-almost surely. In addition, $$0 = \mathbb P(A \cap A^c) = \int_\Omega 1_{A^c}(\omega) \mathbb P^{\mathcal G}(\omega,A) \mathbb P(d\omega)\,,$$ thus $1_{A^c} \mathbb P^{\mathcal G}(\cdot,A) = 0$, $\mathbb P$-almost surely.
This implies that $\mathbb P^{\mathcal G}(\cdot,A) = 1_A$, $\mathbb P$-almost surely. (I can also directly check that $(\omega,A) \mapsto 1_A(\omega)$ satisfies a.,b.,c. from the definition of r.c.p above). Hence $\mathbb P^{\mathcal G}(\omega, \cdot) = \delta_\omega$.
Is there any reference for the fact that $\mathbb P(\omega : \mathbb P^{\mathcal G}(\omega,\cdot) \ll \mathbb P) = 1$ ? What is wrong with my above "counterexample" ?