Prove E(A|G) = $1/2(1_A(\omega) + 1_A(-\omega))$, where G = $\sigma(x)$ for $x = \omega^2$

Question

I am extremely stuck on this problem, no idea how to even get started. Any help appreciated.

Let $(\Omega, F, P) = ([0, 1], B([0, 1]), \operatorname{Leb}/2)$, where Leb refers to the Lebesgue measure. Show that for the random variable $X(\omega) = \omega^2$, and $\mathcal G = \sigma(X)$, that for $A \in \mathcal F$...

$\mathbb E(A|\mathcal G) = E(1_A|\mathcal G) = \frac{1}{2}(1_A(\omega) + 1_A(-\omega))$.

EDIT: Image of the question.

Answer 1

You cannot replace the interval $[-1,1]$ in the original question to $[0,1]$. (If $A \in \mathcal F$ then $A$ is a subset of $[0,1]$ so $I_A(-\omega)=0$ for all $\omega \in [0,1]$). If this question has to make sense you have change the interval to $[-1,1]$.

Hints for solution. We can describe $\mathcal G$ as $\{B\cup -B:B \, \text {is Borel in} [0,1]$. Once you prove that you can easily verify the identity $E(A|\mathcal G)=\frac 1 2 (I_A+I_{-A})$ by applying the definition of definition of conditional expectation and using the property $ Leb (-C) =Leb (C)$ for any Borel set $C$. To show that $\mathcal G$ has above form note first that $\mathcal G=\{X^{-1}(B): B \,\text {Borel}\}$. Then note that $B\cup -B=X^{-1}(B)$ for any Borel set $B$ in $[0,1]$. Hence sets of the form $B\cup -B=X^{-1}(B)$ belong to $\mathcal G$. The converse is very easy and I leave that to you.