I've come across this problem in my studies, where the book (Real Analysis (4th Edition) by Royden) gives an example of two measurable functions whose composition is nonmeasurable. The two functions are constructed as follows:
1) Let $\Phi$ be the Cantor-Lebesgue function, and define the function $\Psi\ R \rightarrow R$ by $\Psi(x)$=$\Phi(x)+x\ \forall x \in R$ which is continuous and strictly incerasing
2) $\chi_A$ which is the characteristic function of a measurable set $A$, while $\Psi(A)$ is nonmeasurable.
First I am confused why $\Psi^{-1}$ is measurable knowing that $\Psi(A)$ is nonmeasurable. Is continuity sufficient for it to be measurable?!
Secondly, I cannot prove that $f=\chi_A \circ \Psi^{-1}$ is nonmeasurable.
Thanks.