Consider the probability space $\Omega = \{-1, 0, 1\}$ with the $\sigma$-algebra of all possible events and a probability measure $P$. Consider also the smaller $\sigma$-algebra
$$F = \{\emptyset, \{0\}, \{1, -1\}, \Omega\}$$
Prove that, for every function $f : \Omega \mapsto \mathbb{R}$,
$$E(f | F)(\omega) = \frac{f(\omega) + f(-\omega)}{2}$$
Let $B_1, B_2$ be the partition of $F$ with $B_1 = \{0\}$ and $B_2 = \{1, -1\}$. If, for example, $\omega \in B_1$, then;
$$E(f | F)(\omega) = \frac{1}{P(B_1)}(f(-1)P(-1|B_1) + f(1)P(1|B_1) = f(1) + f(-1)$$
which is obviously not the required answer. What am I doing wrong?
By definition, $\mathbb E(f\mid\mathcal F)$ is $\mathcal F$ measurable, hence it has the form $a\chi_{\{0\}}+b\chi_{\{-1,1\}}$. Integrating this function over $\{0\}$ and $\{-1,1\}$, we get the values of $a$ and $b$.