I know that the functions $f$ and $g$ are Riemann-integrable on some closed set. Also, $f$ has a min (m>0) on the closed set.
Let $h(x) = f(x)^{g(x)}$
I want to prove that $h$ is Riemann-integrable.
I've been trying to use the Lebesgue's Theorem but I'm not sure how to prove the set of discontinuities of $h$ is measure zero without assuming it is Riemann-integrable.
I've tried breaking up the integral of $f$ into it's continuous sections, then the discontinuities of $f^g$ for each section is equal to the discontinuities of $g$. Hence, we have a countable union of measure zero sets, which is measure zero. But, am I allowed to assume the integral of $f^g$ exists for each section?