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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.

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    $\begingroup$ Won't $h$ be continuous at any point where $f$ and $g$ both are? $\endgroup$ – Angina Seng Sep 8 at 15:50
  • $\begingroup$ @AnginaSeng, yes, but what implication does this have in the points of discontinuity? $\endgroup$ – Mateo Soto Arango Sep 8 at 15:59
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    $\begingroup$ 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. $\endgroup$ – peek-a-boo Sep 8 at 16:04
  • $\begingroup$ @peek-a-boo i got it. The problem was that i thought that perhaps $h$ could have discontinuities other than $f$ and $g$. $\endgroup$ – Mateo Soto Arango Sep 8 at 16:09
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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.

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  • $\begingroup$ I don't see why the set of points of discontinuity of h is countable, that is because h depends of f and g? And f,g have countable points of discontinuity? $\endgroup$ – Mateo Soto Arango Sep 8 at 15:58
  • $\begingroup$ @MateoSotoArango I have edited my answer. What do you think now? $\endgroup$ – José Carlos Santos Sep 8 at 16:09
  • $\begingroup$ 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. $\endgroup$ – Mateo Soto Arango Sep 8 at 16:15
  • $\begingroup$ No, it cannot, since $h=f^g=\exp(g\log(f))$. So, $h$ can be obtained from $f$ and $g$ through composition. $\endgroup$ – José Carlos Santos Sep 8 at 16:54

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