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

First, I want to figure out why, for the simple case where $ g:[a,b]\rightarrow\mathbb{R} $ is bounded and continuous except at some point $ x_0\in[a,b], $ $ g$ is Riemann integrable on [a,b].

I know the Riemann integrability condition that there must exist some partition $ P $ for which $ U(P,f) - L(P,f)\leq\epsilon, \forall \epsilon>0 $.

For my attempt at a proof, I said:

Let $ P=\{x_0,x_1,\cdots,x_n:x_0=a<x_1<\cdots<x_n=b\} $ be a partition of $ [a,b] $. Let $ x_0\in[x_{k-1},x_k]. $ Then normally for an everywhere continuous function, since $ [a,b] $ is a compact set, the function is uniformly continuous and therefore $ \exists\delta(\epsilon):|x-y|<\delta\implies|f(x)-f(y)|<\epsilon,\forall\epsilon>0 $. Using this, we can make any of the subintervals in the partition arbitrarily small, and eventually make the difference between the upper and lower sums arbitrarily small by using the property that a continuous function will attain its maximum and minimum. But how do I do something similar for $ g $? Furthermore, how can I extend that to the case in which $ g $ is discontinuous at $ x_0, x_1, \cdots, x_n $?

EDIT: Using ncmathsadist's advice,

Let $\epsilon > 0$. Let $M = \sup_{x\in[a,b]} f$.

Let $ D=\{x_0,x_1,\cdots,x_{2n}:x_0<x_1<\cdots<x_{2n}\} $ be a subpartition containing all the points $ x_0,x_1,\cdots,x_n $ where $ g $ is discontinuous, such that $ \sum_{i=1}^{2n}(x_i-x_{i-1}) < \frac{\epsilon}{2M}$

Let $ C $ be a subpartition containing all the other points. By visiting the proof that a continuous function is Riemann integrable, I can construct a $ C $ so that:

$ U(C\bigcup D,f)-L(C\bigcup D,f)< \frac{\epsilon}{2M}\times M+\frac{\epsilon}{2}=\epsilon $

This is because $ g $ is bounded, and any contribution by $ g $ to the sum from the discontinuous point must be less than the maximum, $ M $.

share|cite|improve this question
This is looking good. To make things a bit clearer, let $K$ denote the union of the subintervals containing the discontinuities and $L$ denote the union of the subintervals that do not contain the discontinuities. You know $f$ is uniformly continuous on $L$. – ncmathsadist Jan 28 '12 at 13:07
Comment>How do we know that f is uniformly continuous on [a,b]\setminus {q|q is a discontinuity point of f in [a,b]} – user84029 Jun 26 '13 at 19:53
@prism: This is not an answer to the question, so it should not be left as one. (Rather it's a question, and the answer is: if there is at least one and only finitely many points of discontinuity, $f$ will not be uniformly continuous on the complement; if so it would extend continuously to all of $[a,b]$.) – Pete L. Clark Jun 26 '13 at 20:04
@PeteL.Clark: ooo.. that is right, but once I saw ' You know f is uniformly continuous on L.' and since I dont know I got excited to learn it, sorry for confusion. – user84029 Jun 26 '13 at 20:36
up vote 7 down vote accepted

Let $\epsilon > 0$. Put $M = \sup_{x\in[a,b]} f$. Now choose a partition so that the total length of the intervals containing the discontnuities of $f$ is smaller than $\epsilon/2M$. The contribution from the intervals with discontinuities of $f$ is smaller than $\epsilon/2$. Since $f$ is continuous on the rest, it is fairly easy to engineer an argument that supplies a partition whose upper and lower sums differ by less than $\epsilon/2$.

Assemble the pieces and you have Riemann integrability.

share|cite|improve this answer
Thanks. I added some steps using your advice. Am I heading in the right direction? – Shafat Arbaz Alam Jan 28 '12 at 2:23

$f$ is bounded $\Rightarrow \exists M \in R$ such that $f<|M|$. For all $\varepsilon>0$ :

Is $D=\{d_0,...,d_k\}$ the set of discontinuous points of $f$ and $P=\{x_0,...,x_n\}$ partition of $[a,b]$ containing $D$ such that:

  • the sum of intervals determined by $P$ such that one of the points is of discontinuous of $f$ is $\sum_{D}^{}(x_{i}-x_{j})<\frac{\varepsilon}{4M}$
  • $K=\{k_0,...,k_j\}$ is the set of other points of $P$ such that $\sum_{i=1}^j (M_i-m_i) (k_{i}-k_{i-1})<\frac{\varepsilon}{2}$ (in this points $f$ is continuous and we can make this choice).

So: $$ U(f, K \cup D)-L(f,K \cup D)=\sum_{i=1}^j (M_i-m_i) (k_{i}-k_{i-1})+\sum_{D}^{} (M_i-m_i) (x_{i}-x_{j}) $$ $$ <\sum_{i=1}^j (M_i-m_i) (k_{i}-k_{i-1})+2M \sum_{i=1}^j (y_{i}-y_{i-1})<\sum_{i=1}^j (M_i-m_i) (k_{i}-k_{i-1})+2M \frac{\varepsilon}{4M} $$

$$ =\sum_{i=1}^n (M_i-m_i) (k_{i}-k_{i-1})+\frac{\varepsilon}{2} <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} < \varepsilon $$

From that: $f$ is Darboux Integrable, consequently Riemann Integrable (and Cauchy Integrable)

share|cite|improve this answer
I'm trying to prove this but I'm not sure if I'm correct. Is little different than yours, someone can help me? – Felipe Jan 3 '14 at 3:23

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.