Let $X$ be a set, $(Y, \mathbb A)$ a measurable space and $g: X \to Y$ a surjective function. We consider on $X$ the $\sigma$-algebra induced by the function $g$ ($g^{-1}(\mathbb A))$. If the function $f: X \to \mathbb R$ and $\mathbb B(\mathbb R)$ is the Borel $\sigma$-algebra on $\mathbb R$ generated by the Euclidean topology. Why does it hold that $f$ is measurable from $(X, g^{-1}(\mathbb A))$ to $(\mathbb R, \mathbb B (\mathbb R))$ if and only if there exists a measurable function $h: (Y, \mathbb A) \to (\mathbb R, \mathbb B (\mathbb R))$ such that $f = h \circ g$ ?
I think that one can apply here the factorization lemma. But I'm not sure how to handle the functions...may anybody help me?
One way is obvious: if $f$ has the form $h\circ g$ then it is measurable. For the converse start with $f$ of the form $I_C$ where $C \in g^{-1}(\mathcal A)$. Then $C=g^{-1}(A)$ for some $A \in \mathcal A$. Hence $f=I_C=I_A\circ g$. Now extend this to non-negative simple functions in an obvious way. If $f$ is a non-negative measurable function w.r.t. $\mathcal A$ then there exist non-negative simple functions $f_n$ increasing to $f$. If $f_n=h_n\circ g$ then $f=\lim\sup h_n \circ g$ and $\lim\sup h_n$ is measurable. [We can also use $\lim \inf$]. Now use $f=f^{+}-f^{-}$ to prove the same for any measurable $f$.