# why the function f is Riemann integrable

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.

• 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. – Empiricist Mar 10 '15 at 3:42
• 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. – P.D. Mar 10 '15 at 3:50
• 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. – Empiricist Mar 10 '15 at 3:55
• Yeah I know it is not very useful.. but if you think of something please feel free to comment it. – P.D. Mar 10 '15 at 4:06
• You can prove that a bounded function discontinuous at most countably many points is Riemann integrable. – Ink Mar 10 '15 at 5:38

A function is Riemann integrable if and only if it is bounded and discontinuous on a set of measure $0$.
• So the explanation to this is that $f$ is discontinuous and bounded? – P.D. Mar 10 '15 at 5:28
• @P.D.: your function is Riemann integrable since it is bounded and discontinuous on a set of measure $0$ (discontinuous at the point $x=1/2$). – science Mar 10 '15 at 5:29
• @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, – science Mar 10 '15 at 5:34