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Let $f$ be a Riemann-integrable function on a closed interval $[a,b] \subset \mathbb{R}$. Let g be a function on $\mathbb{R}$. What conditions must g satisfy so that $g \circ f$ is also Riemann-integrable ? Thank you!

marked as duplicate by KReiser, Brahadeesh, José Carlos Santos calculus Dec 7 at 8:38

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  • @WilliamSun The answer to the linked question shows that if $g$ is Riemann-integrable then it is not necessary that $g \circ f$ is Riemann-integrable. The question here is different, so it is not a duplicate. I am voting to close it as off-topic, though, because of the lack of context. – Brahadeesh Dec 7 at 8:24
  • @Brahadeesh Originally I planed to put the post mathoverflow.net/questions/20045/… in my answer below as a duplicate of the question. But the link I gave in my comment mentioned the post below in the comment section, which I think will provide the op with other useful informations. Another problem is I attempted to change the link of duplicate but I haven’t figure out how to do it. – William Sun Dec 7 at 8:30
  • @WilliamSun I don't think it's possible to change the link of duplicate once you've voted. You've written a good answer, and I've upvoted it. But sadly this post will probably be closed as off-topic unless the OP improves it by providing more context. – Brahadeesh Dec 7 at 8:33

See this post Riemann Integrability of Compositions

If $f$ is a Riemann integrable function defined on $[a,b],\ g$ is a differentiable function with non-zero continuous derivative on $[c,d]$ and the range of $g$ is contained in $[a,b]$, then $f\circ g$ is Riemann integrable on $[c,d]$.

Quote from Is the composite function integrable? See the last result mentioned in the paper.

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