Let $\Omega = [0,1]$ be our probability space with sigma algebra of borel sets on $[0,1]$ and Lebesgue measure on $[0,1]$. Let Y be a random variable such that $Y(\omega) = Y(1-\omega)$ for every $\omega$ in $\Omega$.
Assume that $Y_{[0, \frac{1}{2}]}$ is a homeomorphism on $Y([0, \frac{1}{2}])$.
What does $\sigma-$algebra generated by $Y$ look like?
The information that $Y_{[0, \frac{1}{2}]}$ is a homeomorphism tells us that the image of every borel set in $[0, \frac{1}{2}]$ is a borel set. Is that correct? What does that tell us about $\sigma(Y)?$
I need this in order to prove that for any random variable $X$ and for random variable $Y$ described above we have $$\mathbb{E}(X|Y)(\omega) = \frac{X(\omega) + X(1- \omega)}{2}$$
When I integrate $\mathbb{E}(X|Y)$ over a $\sigma(Y)-$measurable set $B$ I get $$\int_B\mathbb{E}(X|Y) dP = \frac{1}{2}\int_B X(\omega) dP(\omega) + \frac{1}{2}\int_B X(1-\omega) dP(\omega)$$
Now we know that $Y(\omega) = Y(1-\omega)$ but we do not know that about $X$. Anyway, shouldn't $\mathbb{E}(X|Y)$ be a measurable function of $Y$?