If $f,g$ are Riemann-Integrable on $[a,b]$ then $h(x)=f(x)^{g(x)}$ is Riemann-Integrable

Question

If $f,g$ are Riemann-Integrable on $[a,b]$, and there is $m$ such that $0<m\leq{f(x)}$ for every $x$ in $[a,b]$ then $h(x)= f(x)^{g(x)}$ is Riemann-Integrable on $[a,b]$. I want to use Lebesgue's Theorem, but I don't know how to prove that the set of discontinuities of $h$ have measure zero.

Answer 1

The function $h$ is bounded. And whenever $f$ and $g$ are continuous, $h$ is continuous too. So, since the sets of points of discontinuity of both $f$ and $g$ have Lebesgue measure $0$, the set of points of discontinuity of $h$ have Lebesgue measure $0$ too). See, you can indeed apply Lebesgue's theorem.