Regular conditional probability for the 0-1 sigma-algebra

Question

Let $(\Omega, \mathcal A, P)$ be a probability space and $\mathcal F := \{A \in \mathcal A \mid P(A) \in \{0,1 \}\}$. Show that $\mathcal F$ is a $\sigma$-algebra and find a regular conditional probability $P(\cdot\mid\mathcal F)(\cdot)$.

The proof of $\mathcal F$ being a $\sigma$-algebra is not the problem. But for the regular conditional probability, I know, that \begin{align} E(X\mid\mathcal F)=E(X) \end{align} for $X:(\Omega, \mathcal A, P) \rightarrow (E, \mathcal E)$.

I think because of \begin{align} P(A\mid\mathcal F)(\omega):=E(\mathbb{1}_{A}\mid\mathcal F)(\omega) \end{align} for $A \in \mathcal A$, I can deduce \begin{align} P(X \in B\mid \mathcal F)=P(X \in B) \end{align} for $B \in \mathcal E$.

If $P(X \in B)$ would be a markov kernel, I would have found my regular conditional probability here. So now I have to somehow get $P(X\in B)$ to be a markov kernel. But how to do so?

Are my thoughts going in the right direction? Maybe someone can help me. I don't think it is that difficult, but I don't see it right now...

Answer 1

An RCP must satisfy:

1. For each $A\in \mathcal{A},$ $\mathsf{P}(A\mid\mathcal{F})(\cdot)=\mathsf{E}[1_A\mid\mathcal{F}](\cdot)$, $\mathsf{P}\mid_{\mathcal{F}}-$a.s., where $\mathsf{P}(A\mid\mathcal{F})(\cdot)$ is measurable for $\mathcal{F}$;
2. For every $\omega\in\Omega$, $\mathsf{P}(\cdot\mid\mathcal{F})(\omega)$ is a probability measure on $\mathcal{A}$.

Since, for $A\in \mathcal{A}$ and any $F\in\mathcal{F}$, $\int 1_{F\cap A}d\mathsf{P}=\int\mathsf{P}(A)1_Fd\mathsf{P}$, $\mathsf{P}(A)$ is a version of $\mathsf{E}[1_A\mid \mathcal{F}]$. Moreover, since $\mathsf{P}(A)$ is a constant, it is measurable for $\mathcal{F}$ and $\mathsf{P}$ is trivially a probability measure on $\mathcal{A}$ for each $\omega\in \Omega$.