Question on measure theoretic conditional expectations. I have a question related to conditional expectations and  that is not really difficult per se, but I am having a harder time than I should to be rigorous about it.\
Assume that $\Omega =\{\omega: \omega \in [-1/2, 1/2]\} $ and for simplicity let's assume that the probability measure is the Lebesgue measure on ${\cal B}(\Omega), $ the Borel $\sigma$-algebra.  Let $X: \Omega \mapsto R $ be defined by $X(\omega) = \omega^2.  $ I would like to show that
$$ E[Y\mid \sigma(X)] = \frac{Y(\omega)}{2} + \frac{Y(-\omega)}{2}. $$ 
To prove this it is obviously enough to apply the definition, i.e.
prove that $Z := \frac{Y(\omega)}{2} + \frac{Y(-\omega)}{2} $ is 
$\sigma(X)$-measurable and that $\int_B Y \,\lambda(d\omega) = \int_B Z\,\lambda(d\omega) $ for every $B \in \sigma(X). $ 
I think that my difficulty stems from the fact that I am not seeing what $\sigma(X) $ should look like to pick the $B, $ even if I can show that $X $ given above is measurable with respect to ${\cal B}(\Omega). $ I wonder if anyone could throw in here some details.
Thank you.
 A: Random variable $X$ indentifies two points $\omega_1$ and $\omega_2$ if and only if $|\omega_1|=|\omega_2|,$ i.e. 
$$
X(\omega_1)=X(\omega_2)\Leftrightarrow |\omega_1|=|\omega_2|.
$$
It means that the pre-image $X^{-1}(B)$ of every Borel set $B\subset \mathbb{R}$ is  symmetric about the origin. This suggests that  $\sigma(X)$ consists of all symmetric about the origin Borel subsets of $[-\frac{1}{2},\frac{1}{2}].$ I will check that, indeed, 
$$
\sigma(X)=\bigg\{B\in \mathcal{B}\bigg(\bigg[-\frac{1}{2},\frac{1}{2}\bigg]\bigg):-B=B\bigg\}.
$$ 
Inclusion $\subset$ was discussed above. Conversely, assume that $B$ is a Borel subset of $[-\frac{1}{2},\frac{1}{2}]$ and $-B=B.$ Consider Borel set
$$
C=\bigg\{x^2:x\in B\bigg\}\subset \bigg[0,\frac{1}{4}\bigg]
$$
$C$ is Borel because it can be written as $\tilde{X}^{-1}(B\cap [0,\frac{1}{2}]),$ $\tilde{X}(\omega)=\sqrt{\omega}.$ It is easy to see that $B=X^{-1}(C)\in \sigma(X).$ The needed description of $\sigma(X)$ is obtained. Now, if $B\in \sigma(X),$ then
$$
\int_B Y(\omega)d\omega=\int_{-B} Y(-\omega)d\omega=\int_{B} Y(-\omega)d\omega
$$
So,
$$
\int_B Y(\omega)d\omega=\int_B \frac{Y(\omega)+Y(-\omega)}{2}d\omega=\int_B Z(\omega)d\omega
$$
