I would like to prove or give a counterexample to the following. Thanks for your help!
Let $f: \mathbb A \rightarrow \mathbb R$ be a measurable function. While $ (\Omega, \mathbb A) $ is a Measure-Space.
$f^{n}(x)$ := $f(x)^{n}$
a) Does it follow that $f^{3}$ is Lebesgue measurable?
b) Does it also follow that $f^{2}$ is Lebesgue measurable, if not give a counterexample?
If $\psi:\Bbb R\to \Bbb R$ continuous then for every measurable function $f:\Bbb A\to \Bbb R$ we have, $\psi\circ f:\Bbb A\to \Bbb R$ is measurable. To prove this, let $c\in \Bbb R$ the, $$(\psi\circ f)^{-1}(-\infty,c)=f^{-1}\big(\psi^{-1}(-\infty,c)\big)=f^{-1}(U),$$ where $U$ is an open subset of $\Bbb R$ as $\psi$ is continuous. Since $f$ is measurable we have, $f^{-1}(U)\in \Omega$. So $\psi\circ f$ is measurable.