Let $\Omega=\{a,b,c\}$, $\mathcal{F}=2^{\Omega}$ and $\mathbb{P}(a)=\mathbb{P}(b)=\mathbb{P}(c)=1/3$, so $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space.
Let $X$ be a random variable as follows: $$ X(\omega)= \begin{cases} 1, & \text{ $\omega=a$ } \\ 2, & \text{ $\omega=b$ } \\ 3, & \text{ $\omega=c$ } \end{cases} $$
Let $\mathcal{G}= \sigma(\{a\}) \ \subset \mathcal{F}$ .
Let $Y=\mathbb{E}(X|\mathcal{G})$ be the conditional expectation of $X$ given $\mathcal{G}$, that is:
(i) $Y$ is $\mathcal{G}$-measurable;
(ii) for all $G \in \mathcal{G}$, $\mathbb{E}(X; G)=\mathbb{E}(Y; G)$ .
I saw in some solution that:
$$ Y(\omega)= \begin{cases} 1, & \text{ $\omega=a$ } \\ 2.5, & \text{ $\omega=b$ } \\ 2.5, & \text{ $\omega=c$ } \end{cases} $$
Intuitively, I can understand why this is the result, since $\{a\} \in \mathcal{G}$, so $Y$ should get the same value in $a$ as $X$, and $b$ and $c$ have the same probability so each of them gets half of their original sum, but I wasn't able to prove it formally using the definition above and the properties of conditional expectation dervied from it.
Thanks!
Edit:
The example above was given as a solution to this question:
And this is the suggested solution:
