Can someone explain why the piecewise constant function $f:[0,1]\mapsto \mathbb{R}$, defined by

$f(t)=\left\{ \begin{array}{ll} 0 & t\in[0,1/2] \\ 1 & t\in(1/2,1] \\ \end{array} \right. $

is Riemann integrable?

I have shown that the sequence $f_n\in C_\mathbb{R}[0,1]$, defined by

$f_n(t)=\left\{ \begin{array}{ll} 0, & t\in[0,1/2] \\ n(t-1/2), & t\in[1/2,1/2 + 1/n] \\ 1, & t\in[1/2 +1/n,1] \\ \end{array} \right. $

tends pointwise to $f(t)$ in $[0,1]$ as $n\rightarrow \infty$, and now I want an explanation why $f$ is Riemann integrable.

  • $\begingroup$ Does the question require you to use the fact about $f_n$ is integrable? I think it is easier to show $f$ is Darboux integrable directly. $\endgroup$ – Empiricist Mar 10 '15 at 3:42
  • $\begingroup$ No it doesn't, but I've said what I've done before in case it is needed. What do you mean by ''Darboux integrable''? I haven't heard it before. $\endgroup$ – P.D. Mar 10 '15 at 3:50
  • $\begingroup$ I think this time the sequence may not be very useful indeed because we do not have uniform convergence. Darboux integrability deals with the so-called upper integral and the lower integral and it is equivalent to Riemann integrability. $\endgroup$ – Empiricist Mar 10 '15 at 3:55
  • $\begingroup$ Yeah I know it is not very useful.. but if you think of something please feel free to comment it. $\endgroup$ – P.D. Mar 10 '15 at 4:06
  • $\begingroup$ You can prove that a bounded function discontinuous at most countably many points is Riemann integrable. $\endgroup$ – Ink Mar 10 '15 at 5:38

The answer to your question about Riemann integrability is Lebesgue criteria for Riemann integrability:

A function is Riemann integrable if and only if it is bounded and discontinuous on a set of measure $0$.

  • $\begingroup$ So the explanation to this is that $f$ is discontinuous and bounded? $\endgroup$ – P.D. Mar 10 '15 at 5:28
  • $\begingroup$ @P.D.: your function is Riemann integrable since it is bounded and discontinuous on a set of measure $0$ (discontinuous at the point $x=1/2$). $\endgroup$ – science Mar 10 '15 at 5:29
  • $\begingroup$ what do you mean by a set of measure 0? $\endgroup$ – P.D. Mar 10 '15 at 5:31
  • $\begingroup$ @P.D.: This comes from measure theory which I do not think you have studied it yet. However your set of discontinuity $\left\{ 0\right\}$ has measure $0$ since it s countable. But at least you have studied that if the function is bounded and has a finite number of discontinuities then it is Riemann integrable, $\endgroup$ – science Mar 10 '15 at 5:34
  • $\begingroup$ I think it is even more advanced than Darboux integrability. Probably OP just knows the concept and basic properties of Riemann integrability. $\endgroup$ – Empiricist Mar 10 '15 at 7:30

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