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There is a real value function $f$ for $x\in [0,\infty]$.

$f$ is bounded. $0 \leq f(x) \leq 1$.

And $f$ is non-decreasing. $f(x) \leq f(x+\epsilon)$ for all $\epsilon >0$.

Then, is there any example of $f$ that we cannot use (or define) Riemann integral for $f$?

martian03
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  • The question is ambiguous. Do you want to know whether the restriction of $f$ to each interval $[0,a]$ is Riemann-integrable? Or whether the improper Riemann integral $\int_0^\infty f(x),\mathrm dx$ converges? – José Carlos Santos Jan 21 '22 at 16:06
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    Just for the notice, $f\left(x\right)$ is a real, $f$ is the function. For example $f$ is bounded, not $f\left(x\right)$ – Atmos Jan 21 '22 at 16:06
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    See for example https://math.stackexchange.com/q/2989242/42969 or https://math.stackexchange.com/q/1035758/42969 or https://math.stackexchange.com/q/920833/42969 – Martin R Jan 21 '22 at 16:14
  • @JoséCarlosSantos Interval is not important. I just want to know an example of a function that is improper for Riemann integral. (Honestly I am not sure the exact definition of 'imporper') – martian03 Jan 21 '22 at 17:27
  • @Atmos Thank you. I edited. – martian03 Jan 21 '22 at 17:27
  • @MartinR wow. Thank you. It seems that if a real valued, monotone, closed function is always Riemann integrable. – martian03 Jan 21 '22 at 17:30
  • @martian03: Sorry, I do not have a reference. I just found those other questions with a Google search for “Increasing function is Riemann integrable”. – Martin R Jan 21 '22 at 17:37
  • @MartinR Thank you very much. You gave me the right answer for my question. You may post this answer and I will select it as the correct answer. – martian03 Jan 21 '22 at 17:42
  • @martian03: For proper Riemann integrals $\int_a^b f(x) dx$ this has been answered before, there is no need to add another answer. For improper integrals such as $\int_0^\infty f(x) dx$ it is wrong, simply choose $f(x) = 1$ as a counterexample. – Martin R Jan 21 '22 at 19:39

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