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$\phi: K \to R$

K compact, $\phi$ is $C^0$

$f: [a,b] \to K$

Then $\phi \ o \ f$ is RI.

I do know that the composition of two integrable and continuous functions is integrable. In this problem, I know $\phi$ is RI because it is continuous. However, all I know about $f$ is that it maps a compact interval $[a,b]$ to a compact space $K$. I am not told it is continuous nor that it is RI. So I believe my question is:

Is a continuous function (therefore RI function) composed with a not necessarily continuous nor RI function still RI? And how would I prove that? Any help is appreciated! Thanks!

Nolan P
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    Of course not. Composing any non-Riemann Integrable function with the identity function is the same non-Riemann Integrable function. – balddraz Dec 13 '20 at 15:53
  • Hmm. Am I interpreting the question wrong then? I checked my notes this is the one my professor gave us. It doesn't seem like it is true then. – Nolan P Dec 13 '20 at 16:12
  • It does not hold as pointed out if $f$ is not Riemann integrable. If it is then the proof is given here. – RRL Dec 13 '20 at 19:56

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