Interpretation of conditional expectation as a random variable I have a couple problems understanding the conditional expectation as a random variable. Consider the fair dice roll as a random variable $X$. Let $C$ be the event that the dice shows a one and consider $A=\sigma(C)$. We have $E[X|C]=1, E[X|C^c]=4$ and thus one version of the expected value of $X$ conditioned on $A$ is $E[X|A]=1\cdot \chi_C+4\cdot\chi_{C^c}$. Now this is a function from $\Omega$ to $\mathbb{R}$. I can't grasp how to interpret $E[X|A](\omega)$, since in general $E[X|A]$ is characterized by having certain integrals over the $\sigma$-field, which defines it up to null sets. Of course, in this case there is no ambiguity, but in a countable setting $E[X|A]$ is an equivalence class. I do understand that $E[X|A]$ is basically averaging of $X$ over $C$ and $C^c$. But how do I interpret $E[X|A](\omega)$ for a specific $\omega$.
 A: In short, $E(X|\mathcal{A})(\omega)$ is the value of the random variable $E(X|\mathcal{A})$ when outcome $\omega$ occurs.
It sounds like you are specifically looking at sub-$\sigma$-algebras that are generated by partitions of the entire collection of outcomes: $\Omega=\cup A_i$ where the $A_i$ are pair-wise disjoint and $\mathcal{A}=\sigma(A_1,A_2,...)$. To interpret $E(X|\mathcal{A})(\omega)$ in this setting, you simply find the element of the partition $B$ such that $\omega\in B$. Note that there is only one such $B$. Then we get that $E(X|\mathcal{A})(\omega)=E(X|B)$. 
Let $\mathcal{A}=\sigma(C)$ for some outcome $C\subset\Omega$. For your example, either $\omega\in C$ or $\omega\in C^c$ since $\Omega=C\cup C^c$. Therefore 
$$E(X|\mathcal{A})(\omega)=\left\{
\begin{aligned}
&E(X|C)&\text{if} &\quad \omega\in C\\
&E(X|C^c) &\text{if} &\quad \omega\in C^c.
\end{aligned}
\right. $$
One way to think about it is that, as an observer, you cannot distinguish between the outcomes in event $C$. If $\omega_1$ and $\omega_2$ are both in $C$, then to you, they look identical to you even if $X(\omega_1)\neq X(\omega_2)$. If I tell you that the outcome of the experiment is a particular $\tilde{\omega}\in C$, then what you hear me say is that "event $C$ occurred". Even though I'm speaking the specific outcome name $ `` \tilde{\omega} " $, you hear only $ `` C " $. So you calculate the expectation of $X$ given the information you heard. You know for certain event $C$ has occurred, therefore you calculate $E(X|C)$. 
$$E(X|C)=\frac{E(X1_C)}{P(C)}=\frac{1}{P(C)}\int_C X P(d\omega) \quad \text{so long as}\quad P(C)\neq0$$
Example:
Let $X$ be uniform on $\{1,2,3\}$, and 
$$\mathcal{A}=\sigma(\{1\})=\{\emptyset,\Omega,\{1\},\{2,3\}\}.$$ 
(Aside: Just to be careful, we have to remember that $\{1\}=\{\omega\in\Omega : X(\omega)=1\}$.)
We want to examine the random variable $E(X|\mathcal{A})$. Think of this random variable as representing your best estimate of $X$ when your own ability to distinguish between different outcomes of the experiment of checking the (random) value of $X$ is limited by the events in $\mathcal{A}$. 
If the value of $X$ is $1$, you can see it perfectly, and you output $$E(X|\mathcal{A})(1)=E(X|\{1\})=1$$ because $\{1\}\in\mathcal{A}$. However, if $X$ takes on the value $2$ or $3$, you cannot distinguish between them so $$E(X|\mathcal{A})(2)=E(X|\mathcal{A})(3)=E(X|\{2,3\})=2.5.$$
Note that $E(X|\mathcal{A})(\omega)= E(X|\{2,3\})$ if $\omega\in\{2\}$ or if $\omega\in\{3\}$.
