Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The theorem is as follow:

Let $a<b$ and let $c,d \in [a,b]$ with $c<d$. Then $1_{[c,d)}$ is Riemann Integrable over $[a,b]$ and $$\int_{a}^{b} 1_{[c,d)} dx = d-c$$

I am using Shroeder's Mathematical Analysis and I came across this proof, but I think the proof in the book is sort of unclear and therefore I am seeking if there is any other source that has a proof on this theorem. I tried google but they all concern Lebesgue integration of Dirchlet function which is not what I want.

share|improve this question
@Jasper Loy Choosing norm of partition less than $\delta=min\{\frac{\epsilon}{2},\frac{d-c}{3}\}$. Yes I will go over it myself couple more times, just need an additional resource to refer. –  Daniel Jul 11 '12 at 8:11

1 Answer 1

up vote 1 down vote accepted

Divide $[a,b]$ into $n$ intervals, each of length $\frac{b-a}{n}$. Since $c,d$ lie in at least one of these intervals, we know that $[c,d)$ contains at least $(d-c) \frac{n}{b-a} -2$ of these intervals, and that $[c,d)$ is contained entirely in at most $(d-c) \frac{n}{b-a}+2$ of these intervals. Thus we have that the lower ($L$) and upper ($U$) Riemann sums are bounded by $$\frac{b-a}{n} ((d-c) \frac{n}{b-a} -2) \leq L \leq U \leq \frac{b-a}{n} ((d-c) \frac{n}{b-a} +2).$$ Letiing $n\to \infty$ yields the desired result.

share|improve this answer

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.