Given the r.v. $\mathbb E(X \mid \mathcal D)(\omega)$, how do I find the conditional expectation w.r.t. an event $D \in \mathcal D$? Given a conditional expectation random variable $\mathbb E(X \mid \mathcal D)(\omega)$, how do I find the conditional expectation $X$ w.r.t. an event $D \in \mathcal D$? I'm not sure when we can evaluate $\mathbb E(X \mid \mathcal D)(\omega)$ at some special $\omega_0$ to yield some sort of explicit number, and when we have to do something different. 
For example, in the example where $X$ is an r.v. taking on values $\{1,2,3,4,5,6\}$ each with probability $1/6$ (e.g. a dice roll), then $\mathbb E(X\mid X \text{ is odd}) = 3$ and $\mathbb E(X\mid X \text{ is even}) = 4$. So now taking $\mathcal D$ to be the sigma-algebra $\{\varnothing, X^{-1}(\{1,3,5\}), X^{-1}(\{2,4,6\}), \Omega\}$, we get that $\mathbb E(X\mid\mathcal D)$ evaluated at any $\omega \in X^{-1}(\{1,3,5\})$ yields $3$, and for any $\omega \in X^{-1}(\{2,4,6\})$ yields $4$. However, this is a simple example with no guarantee that every time I want to find explicitly $\mathbb E(X\mid D)$ I just choose some $\omega \in D$ to evaluate $\mathbb E(X\mid\mathcal D)$ on.
Would it be $\mathbb E[\mathbb E(X\mid\mathcal D) \cdot 1_D]$? If so, wouldn't that be equal to $\mathbb E[X \cdot 1_D]$ by the definition of conditional expectation w.r.t a sigma-algebra? If that were so, what's the point of going to all the trouble of defining $\mathbb E(X\mid\mathcal D)(\omega)$ if we were just going to use $\mathbb E[X \cdot 1_D]$?
 A: By definition Conditional_expectation_with_respect_to_an_event,
 the conditional expectation of $X$ given the event $A$ is
$$E(X\mid A)=\frac{E(X1_A)}{P(A)}$$ 
For example
$$E(X\mid X \text{ is even})=\frac{E(X1_{ \{2,4,6\} })}{P(\{2,4,6\})}
=\frac{\frac{1}{6}(2+4+6)}{\frac{3}{6}}=\frac{12}{3}=4$$
Relation with sigma field:
lets $\sigma=\sigma\left(\{A, A^c \} \right)=\{\emptyset ,\Omega , A, A^c \}$
$$E(X|\sigma)=E(X|A)1_A + E(X|A^c)1_{A^c}$$ 
if you know $E(X|\sigma)(\omega)$ you know $E(X|1_A)$ since if 
$\omega\in A$
$$E(X|\sigma)(\omega)=E(X|A)$$
A little more general let $\sigma=\sigma\left(\{A_1,\cdots, A_n \} \right)=\{\emptyset ,\Omega , all \ A_i, all A_i\cup A_j ,\cdots, \}$
Where $A_i$ are disjoint and are a partition of $\Omega$.
In this case 
$E(X|\sigma)=E(X|A_1)1_{A_1}+E(X|A_2)1_{A_2}+\cdots+E(X|A_n)1_{A_n}$
For Example $\Omega=\{1,2,3,4,5,6\}$ and $A_1=\{1,2 \}$ ,$A_2=\{3,4 \}$, and 
$A_3=\{5,6 \}$. Similarly If you know $E(X|\sigma)$ you already know $E(X|A_j)$. So you can calculate any $E(X|D)$ where $D\in \sigma$ like $$E(X|A_1\cup A_3)=\frac{E(X1_{A_1\cup A_3})}{P(A_1\cup A_3)}=\frac{E(X1_{A_1})+E(X1_{A_3})}{P(A_1)+ P(A_3)}
=\frac{E(X|A_1)P(A_1)+E(X|A_3)P(A_3)}{P(A_1)+ P(A_3)}$$
since you know all $E(X|A_i)$.
In general case, By definition of $E(X|\sigma)$, for all $D\in \sigma$
$$E(X1_D)=E( E(X|\sigma)1_D).$$
So if $P(D)> 0$
$$E(X|D)=\frac{E(X1_D)}{P(D)}=\frac{E( E(X|\sigma)1_D)}{P(D)}$$
If we come back to $E(X|A_1\cup A_3)$ so  
$$E(X|A_1\cup A_3)=\frac{E( E(X|\sigma)1_D)}{P(D)}$$
$$=\frac{E( E(X|A_1)1_{A_1}(\omega) + E(X|A_3)1_{A_3}(\omega))}{P(A_1 \cup A_3)}=\frac{E(X|A_1)P(A_1)+E(X|A_3)P(A_3)}{P(A_1 \cup A_3)}$$ that is same with previous.
