Doob–Dynkin lemma: let $( X,Y : [0,1] \to [0,1] )$ be two Borel-measurable functions, where $[0,1]$ is equipped with the Borel $\sigma$-algebra. Then $Y$ is $\sigma(X)$-measurable if and only if $Y=g(X)$ for some Borel-measurable function $( g : [0,1] \to [0,1] )$.
Does anyone have an "example" of a Borel-measurable function $( X : [0,1] \to [0,1] )$ and of a function $( g : [0,1] \to [0,1] )$ (not Borel-measurable) such that $Y=g(X)$ , $Y$ is Borel-measurable and $Y$ is not $\sigma(X)$-measurable?
Thank you