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

If $$f,g$$ are Riemann-Integrable on $$[a,b]$$, and there is $$m$$ such that $$0 for every $$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.

• Won't $h$ be continuous at any point where $f$ and $g$ both are? – Angina Seng Sep 8 at 15:50
• @AnginaSeng, yes, but what implication does this have in the points of discontinuity? – Mateo Soto Arango Sep 8 at 15:59
• you know that if $f$ and $g$ are continuous at $x$ then $h$ is. So, look at this in the contrapositive manner: if $h$ is not continuous at $x$ then either $f$ isn't continuous at $x$ or $g$ isn't continuous at $x$. i.e $D_h\subseteq D_f \cup D_g$, where $D_f$ represents the set of discontinuities of $f$. Since the two sets on the right have (Lebesgue) measure zero, so does their union, and hence the set on the left also has measure zero. – peek-a-boo Sep 8 at 16:04
• @peek-a-boo i got it. The problem was that i thought that perhaps $h$ could have discontinuities other than $f$ and $g$. – Mateo Soto Arango Sep 8 at 16:09

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.
• I think that's right, The problem was that i thought that perhaps $h$ could have discontinuities other than $f$ and $g$. Thanks for answering. – Mateo Soto Arango Sep 8 at 16:15
• No, it cannot, since $h=f^g=\exp(g\log(f))$. So, $h$ can be obtained from $f$ and $g$ through composition. – José Carlos Santos Sep 8 at 16:54