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 $.

  • $\begingroup$ 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 $L$. $\endgroup$ Jan 28, 2012 at 13:07
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    $\begingroup$ 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]} $\endgroup$
    – user84029
    Jun 26, 2013 at 19:53
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    $\begingroup$ @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 $[a,b]$.) $\endgroup$ Jun 26, 2013 at 20:04
  • $\begingroup$ @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. $\endgroup$
    – user84029
    Jun 26, 2013 at 20:36

2 Answers 2


$f$ 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)


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.

  • $\begingroup$ Thanks. I added some steps using your advice. Am I heading in the right direction? $\endgroup$ Jan 28, 2012 at 2:23
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    $\begingroup$ @ncmathsadist you are using fact that function is continuous on rest set, hence it will be Riemann integrable, but for that set has to compact. but that won't happen. Where I am getting confused? $\endgroup$
    – MeetR
    Aug 21, 2017 at 13:52

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