Suppose, $f:\mathbb{R} \to \mathbb{R}$ is Borel-measurable. Define $A$ to be the smallest $\sigma-$algebra containing the sets $ \{\ x: f(x)> a \}\ $ for every $a\in \mathbb{R}$. Suppose $g:\mathbb{R}\to \mathbb{R}$ is measurable with respect to this $\sigma-$ algebra $A$. Prove that there exists a Borel-measurable function $h:\mathbb{R} \to \mathbb{R}$ such that $g=h\circ f$.
Now, what I understand is the the new $\sigma-$algebra is contained in the Borel-sigma algebra. Hence g is Borel-measurable. But how do we construuct such an h?
Thanks in advance!!
Start with the case in which $g$ is the characteristic function of a set $E=\{x: f(x)>a\}$. In this case $h=\chi_{(a,\infty)}$. Then take $g=\chi_E$ with $E$ first countable union of sets $\{x: f(x)>a\}$, then $E$ an arbitrary element of $A$. Then mover to $g$ a simple function, then $g$ pointwise limit of simple functions.