Existence of measurable function for a special composition

Question

Let $Y$ be a $\left( \Omega\mathcal{,F} \right) \rightarrow \left( S,\mathcal{E} \right)$ random variable, and $X$ be a $\left( \Omega,\sigma\left( Y \right) \right) \rightarrow \left( \mathbb{R,}\mathcal{B}\left( \mathbb{R} \right) \right)$ random variable, where $\sigma\left( Y \right)$ is the $\sigma$-algebra generated by $Y$, then how do we prove there exists a $\left( S,\mathcal{E} \right) \rightarrow \left( \mathbb{R,}\mathcal{B}\left( \mathbb{R} \right) \right)$ measurable function $f$ such that $f \circ Y = X$?

Answer 1

Consider first the case when $X=I_A$ for some $A \in \sigma (Y)$. Any set in $\sigma (Y)$ is of the type $Y^{-1}(B)$ for some $B \in \mathcal E$. Hence $X=I_{Y^{-1}(B)}=I_B\circ Y$. This proves the result in this special case since $f=I_B$ is measurable. Now take linear combinations to show the same holds for simple functions $X$. For a non-negative measurable function $X$ choose simple functions $X_n$ converging to $X$ pointwise . If $X_n=f_n\circ Y$ for all $n$ then $X=f\circ Y$ where $f=\lim \sup f_n$ (or $f=\lim \inf f_n$). For the general case write $X$ as $X^{+}-X^{-1}$.