We are given two random variables defined on $[0,1]$:
$$X(\omega) = 2 \omega -1 + |2 \omega -1|$$ $$Y(\omega) = 1-|2 \omega^2 -1|$$
I am supposed to find $\mathbb{E}(X|Y)$ which by definition is a $\sigma (Y)-$measurable random variable and $\forall B \in \sigma (Y): \ \int_B XdP = \int_B \mathbb{E}(X|Y)dP$. Here $P$ denotes Lebesgue measure on $[0,1]$.
I think that the sigma algebra generated by $Y$ is $$\sigma(Y) = \{B \in \mathcal{B}[0,1] : \ \forall x \in B: \sqrt{1-x^2} \in B\}$$
Then $B = [a,b] \cup [\sqrt{1-b^2}, \sqrt{1-a^2}], \ a \le b \le \frac{\sqrt{2}}{2}$, would be a generator of $\sigma(Y) $.
So I computed $$\int_B XdP = \int_{1/2}^b (4x-2)dx + \int_{\sqrt{1-b^2}}^{\sqrt{1-a^2}} (4x-2)dx = 4b^2 - 2a^2 + 2 (\sqrt{1-b^2} - \sqrt{1-a^2}) -2b + \frac{1}{2}$$
My problem is that $X=0$ on $[0, \frac{1}{2}]$ and none of the sets contained in $[0, \frac{1}{2}]$ is $\sigma(Y)-$measurable . I don't know how to construct $\mathbb{E}(X|Y)$ in order for the integrals to be equal. I suppose we have to take $\sqrt{1- \frac{Y}{2}}$ for $\omega \in [\frac{\sqrt{2}}{2},1]$ and $\sqrt{\frac{Y}{2}}$ on $[0,\frac{\sqrt{2}}{2}]$ but it's not obvious to me how.
Could you help me with that?