In the following let $f : [a,b] \to \mathbb R$ be bounded functions.

For a regulated function, the integral could be written as the limit $$ \int_a^b f(x) dx = \lim_{n\to \infty} \int_a^b \varphi_n(x) dx $$ of a sequence $\varphi_n$ of step functions converging uniformly to $f$. Similar if $f$ is measurable, then there exists a sequence of simple functions $\varphi_n$ such that its Lebesgue-Integral could be written as $$ \int_a^b f(x) dx = \lim_{n\to \infty} \int_a^b \varphi_n(x) dx. $$

Now is there a similar convergence characterisation in terms of simpler functions for the Riemann-Integral? The closest I could think of is that $f$ is Riemann-integrable iff there exists sequences $\varphi_n, \psi_n$ of step functions such that $$ \varphi_n \le f \le \psi_n \quad \mbox{and} \quad \int (\psi_n - \varphi_n) d x < \frac{1}{n}. $$

Are there other characterisations or approximation formulations in terms of simpler functions in case $f$ is Riemann-integrable?

  • $\begingroup$ I imagine that you know Lebesgue characterization of Riemann-integrable functions? $\endgroup$ – mathcounterexamples.net Jun 26 '15 at 16:53
  • $\begingroup$ Yes, a function is Riemann-integrable iff its set of discontinuities has measure zero, but I am looking more for some approximation by sequences. $\endgroup$ – StefanH Jun 26 '15 at 16:57

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