Do we have a formula (which maybe related to integral or probability) of $\mathbb{E}[X | \mathcal{G}] (\omega)$?

Question

My lecture note defines conditional expectation as follows:

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Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $\mathcal{G}$ be a sub-sigma field of $\mathcal{F}$. If $X$ is an integrable random variable, then the conditional expectation of $X$ given $\mathcal{G}$ is any integrable random variable $Z$ which satisfies the following two properties:

(CE1) $Z$ is $\mathcal{G}$-measurable.

(CE2) $$\forall \Lambda \in \mathcal{G}: \int_{\Lambda} Z \, d \mathbb{P}=\int_{\Lambda} X \, d \mathbb{P}$$

We denote $Z$ by $\mathbb{E}[X | \mathcal{G}]$.

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Let $X$ be an integrable random variable and $\mathcal G$ a sub-sigma field.

I would like to ask if we can deduce a formula (which maybe related to integral or probability) of $\mathbb{E}[X | \mathcal{G}] (\omega)$ from (CE1) and (CE2).

Thank you so much!

Answer 1

If $\mu(\cdot,\cdot)$ is the regular conditional distribution of $X$ given $\mathcal{G}$, then for an integrable function $g$, $$ \mathsf{E}[g(X)\mid \mathcal{G}](\omega)=\int_{\mathbb{R}} g(x)\mu(\omega,dx) \quad\text{a.s.} $$