Let $(\Sigma, F, P)$ be a probability space, let $X$ be an $F$-random variable, let $G$ be a sub-sigma algebra of $F$, let $\omega\in\Omega$, let $A$ be the intersection of all the sets in $G$ that contain $\omega$, and suppose that $A\in F$. Then is it true that $E(X|G)(\omega) = E(X|A)$?
I just wrote this statement down based on my intuition of what conditional expectation means, and I want to verify if it’s actually true. Note that sigma algebras need not be closed under uncountable intersections, so $A$ need not be an element of $G$. But I assumed $A$ is at least an element of $F$ so that $E(X|A)$ is meaningful.
No, when you condition on a sigma field, there is little connection to the basic definition of conditionings. The only result I know is when you condition on a discrete random variable. In other words, let $Z$ be an integrable random variable and let $X$ be a discrete random variable. Suppose all the values $X$ take with positive probability are $(x_k).$ Then $$ \mathbf{E}(Z \mid X) = \sum_k \mathbf{E}(Z \mid X = x_k) \mathbf{1}_{\{X = x_k\}} $$ and here $\mathbf{E}(Z \mid X = x_k)$ is the usual conditional expectation over an event $A$ with positive probabiltiy $\mathbf{E}(Z \mid A) = \dfrac{1}{\mathbf{P}(A)} \int\limits_A Z d\mathbf{P}.$ (Note that when $Z = \mathbf{1}_{\{Y \in B\}}$ then you have $$ \mathbf{E}(Z \mid A) = \dfrac{P(Y \in B, X = x_k)}{P(X = x_k)} = P(Y \in B \mid X = x_k).) $$