If $f$ is finite and Lebesgue measurable on $R^n$ and $\phi$ is Borel measurable on $R^1$, then $\phi∘f$ is Lebesgue measurable.
For any open set $G,$ we have that $\phi^{-1}(G)$ is borel measurable. Then how we can do next. I think the proof is similar to continuous version, but I stuke here. What should I do next?
For any open set $G$, $$ (\phi\circ f)^{-1}(G)=f^{-1}(\phi^{-1}(G)). $$ Given that $\phi$ is Borel, $\phi^{-1}(G)$ is Borel, and given that $f$ is Lebesgue measurable, $f^{-1}(\phi^{-1}(G))$ is measurable, so $\phi\circ f$ is Lebesgue measurable.