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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?

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up vote 2 down vote accepted

I assume when you say "$f$ has a min ($m>0$)" we mean $f(x)>0$ on this set. Let $E_g$ and $E_f$ be the set of all points of discontinuity for $g$, $f$ respectively. (which have Lebesgue measure 0) Then let $$U=E_f^c\cap E_g^c.$$ Both $f,g$ must be continuous on $U$. Since $x^y$ is a continuous function from $(0,\infty)\times\mathbb{R}\rightarrow \mathbb{R}$ we see $f(x)^{g(x)}$ will be continuous on $U$ as the composition of continuous functions is continuous. (Note: For the range/domains to work out, we need $f(x)>0$). Now, as $m (U^c)=0$, that is the Lebesgue measure of $U^c$ is 0, we conclude $f(x)^{g(x)}$ is Riemannn integrable as desired. (The set of discontinuities is a subset of $U^c$)

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