Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm quite stuck on how to prove that this function: $$ f(x) = \begin{cases} 1 & x \in [0,{1\over 2}) \\ x - {1\over 2} & x \in [{1 \over 2}, 1] \end{cases} $$ is Riemann integrable. I've tried setting the partition $P_1 = \{0,{1\over 2} - \delta, {1\over 2} + \delta, 2\}$, but that turned out not to work, and I've tried a partition $P_N$ of $2N$ evenly spaced points, but it seems that for $f(x) = x - {1\over 2}$, then $U_{P}$ and $L_P$ will always differ by ${1\over 2N}$ for $N$ intervals, i.e., they will always differ by ${1\over 2}$ so I'm slightly stuck. thanks for any tips.

share|cite|improve this question
Notice that the mesh of the partition you have is also getting finer. – user39431 Nov 7 '13 at 23:16
up vote 1 down vote accepted

Notice that, a function $f$ is Riemann integrable on $[a,b]$ if

i) it is bounded,

ii) it has a countable number of discontinuities ( or it is discontinuous on a set of measure zero. )

share|cite|improve this answer
I understand, however, I am trying to prove this Riemann integrability from first principles, and so would like to find a way that involves a choice of partition rather than appeal to another theorem – Moderat Nov 8 '13 at 2:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.