# Proof of Riemann Integral of an indicator function of an interval on Real

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.

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@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

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.