If I have two measurable functions $X,Y:S \to \mathbb{R}$ (with the Lebesgue mesure) such that $X(\{x,y\})=x$ and $Y(\{x,y\})=y$ on the unit circle that is $x^2+y^2=1$. Then is $Y$ $\sigma(X)-$measurable?
Well trying to represent $Y$ as a function of $X$ is I think not possible, for example using the trigonometric representation of the unit ring $(\cos(\theta),\sin(\theta))$ we still get $\arcsin$ function which is only defined on the half of the plane. Still, is there any way to find a mapping s.t. $Y$ $\sigma(X)-$measurable?
I find similar difficulties when considering the same problem on the unit disk $x^2+y^2\leq1$
Indeed the random variable $Y$ is not $\sigma(X)$-measurable. A slightly indirect proof is to note that, by symmetry, $E(Y\mid X)=0$.