# How is $P(|A)$ defined from $P(|\mathcal{B})$?

Let $(\Omega, \mathcal A, \operatorname{P})$ be a probability space, with a real random variable $X$ and a sub-σ-algebra $\mathcal B \subseteq \mathcal A$

In probability theory with measure theoretical treatment, conditional expectation $E(X| \mathcal{B})$ is defined first, and then conditional probability $P(A| \mathcal{B}), A \in \mathcal{A}$ is defined as $E(I_A| \mathcal{B})$.

There is another commonly used conditional probability given an event $P(A|C):=\frac{P(A \cap C)}{P(C)}, A, C \in \mathcal{A}, P(C) >0$. Can it be defined from $E(X| \mathcal{B})$ or $P(A| \mathcal{B})$?

For example, let $C \in \mathcal{A}, P(C) > 0$, and $\sigma(C):=\{\emptyset, C, \bar{C}, \Omega\}$.

is $P( | \sigma(C) )$ regular? That is, is $P( | \sigma(C) )(\omega)$ a probability measure?

is $P( | \sigma(C) )(\omega)$ same for all $\omega \in C$, and same for all $\omega \in \bar{C}$?

is $\frac{P(A \cap C)}{P(C)} = P( A | \sigma(C) )(\omega)$ for all $\omega \in C$ and $A \in \mathcal A$?

Thanks!

Whenever $\mathsf P(C)>0$ you have $$\mathsf P(A|\sigma(C))(\omega) = 1_C(\omega) \mathsf P(A|C) + 1_{\bar C}(\omega)\mathsf P(A|\bar C).$$ which you can easily check by the definition of the conditional expectation. Thus, for any $\omega$ you have $\mathsf P(\cdot|\sigma(C))(\omega)$ is either of two probability measures $\mathsf P(\cdot|C)$ or $\mathsf P(\cdot|\bar C)$ which positively answers your question.