How do I prove that this conditional expectation equates to $\frac{1}{2}$? I am tasked with answering the following question:

Idea:
I understand what the sets in $\mathscr{G}$ look like (i.e. potentially a series of sets on one side of $\frac{1}{2}$ and the corresponding symmetrical representation on the right). Also, my understanding is that $\mathbb{E}[X | \mathscr{G}]$ describes what can be deduced about the value of $X$ given this symmetry about $\frac{1}{2}$.
Would I be correct to somehow show that: $$\int_G X(\omega) \ d\mathbb{P} = \int_G \frac{1}{2} \ d\mathbb{P} ,\ \ \forall G \in \mathscr{G}$$ Because then it would surely follow that $\mathbb{E}[X | \mathscr{G}] = \frac{1}{2}$, right? The issue is I'm not sure if this approach is correct or how to actually do that if it is.
Thank you in advance for any help you may be able to provide!
 A: For $B=A\cup(1-A) \in \mathcal G$, then by 
\begin{align}\mathbb E[X\mathbf1_{B}]&=\int_{B}X(ω)d\mathbb P=\int_{A}X(ω)d\mathbb P+\int_{1-A}X(ω)d\mathbb P=\int_{A}ω\;d\mathbb P+\int_{1-A}ω\;d\mathbb P\\[0.3cm]&=\int_{A}ω\;d\mathbb P+\int_{A}1-ω\;d\mathbb P=\int_{A}d\mathbb P=\mathbb P(A)=\frac12\mathbb P(B)\end{align} Since $ξ=\mathbb E[X\mid \mathcal G]$ is by definition $\mathcal G-$ measurable and such that $$\mathbb E[X\mathbf 1_B]=\mathbb E[ξ\mathbf 1_B]$$ for all $B\in \mathcal G$, you get by the above that $$\frac12\mathbb P(B)=\mathbb E[ξ\mathbf 1_B]=ξ\mathbb E[\mathbf 1_B]=ξ\mathbb P(B)$$ 
A: Indeed you are right to show this equality, because (as per the definition of conditional expectation), you are supposed to show
$\mathbb E[XZ] = \mathbb E[\frac 1 2 Z]$ for all $\mathcal G$-measurable functions $Z$. Your equality of integrals is a special case where $Z = \Bbb 1_G$ where $G\in \mathcal G$. To extend to general $Z$ you can apply the functional monotone class theorem : http://math.ucsd.edu/~pfitz/downloads/courses/spring05/math280c/mct.pdf .
Now to prove your equality, try to do it yourself using the definition of $G$. It is very straightforward.
