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.

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

How accurate is the following assertion?

A function is Riemann-integrable (necessarily over an interval) iff.:

(1) it is defined on the interval, except perhaps at the end-points, and

(2) it is bounded on the interval, and

(3) it is continuous on the interval, except at a finite number of points.

EDIT: Revised assertion based on the answers given:

A (bounded) function is Riemann-integrable (over a closed interval) if:

it is continuous on the interval, except at a finite number of points.

share|cite|improve this question
No, any monotone increasing function is Riemann-integrable, and there are monotone-increasing functions with infinitely many discontinuities. – Thomas Andrews Feb 15 '13 at 20:23
@FlybyNight I think he means the function is defined at all points, not the integral. He is writing conditions for when the integral exists, so he wouldn't be setting conditions that it exists if it exists. – Thomas Andrews Feb 15 '13 at 20:24
@ThomasAndrews I think you're right. The post to which I first replied has since been edited to reflect that point. – Fly by Night Feb 15 '13 at 20:28
Yes, it is true for "if." The "only if" part is the problem. – Thomas Andrews Feb 15 '13 at 20:34
@ThomasAndrews Thanks Thomas. – Ryan Feb 15 '13 at 20:42
up vote 2 down vote accepted

Your 'iff' assertion is incorrect.

A bounded function on an interval is Riemann integrable iff the set of discontinuities has measure zero.

share|cite|improve this answer
Thanks. Can you consider my newest edit and tell me if the claim is now true IF the "iff." is edited to a mere "if"? – Ryan Feb 15 '13 at 20:31
Yes, it follows from the statement above (a finite set has zero Lebesgue measure). – copper.hat Feb 15 '13 at 20:34
Thanks for your succinct answer! Unfortunately as I don't know measure theory, your answer currently flies over my head. – Ryan Feb 15 '13 at 20:50
A set $A \subset \mathbb{R}$ has measure zero iff for any $\epsilon>0$, you can find a countable collection of intervals $I_k$ such that $\sum_k l(I_k) < \epsilon$ and $A \subset \cup_k I_k$. – copper.hat Feb 15 '13 at 21:10

Even if we are talking about the Riemann integral, the claim is still false (the if and only if part is wrong). You can have an infinite number of discontinuities in an interval and the Riemann integral would still exist.

Example, consider this

$$\int_0^1 \frac{1}{\lfloor1/x\rfloor} dx=\frac{\pi^2}{6}-1$$

which has a jump discontinuity on each of the unit fractions $x=1/2,1/3,1/4,...$ and it is constant otherwise. So we have an infinite number of rectangles with decreasing area. You find the area of each rectangle and add them all up. The infinite sum does converge. I believe the criteria is that as long as the set of all discontinuities has Lebesgue measure zero, the Riemann integral will exist. This was one of the big questions Lebesgue finally answered. What exactly are the necessary and sufficient conditions for the Riemann integral to exist.

share|cite|improve this answer
Yes, with "if" instead of "iff" the above statement is true. – Fixed Point Feb 15 '13 at 20:32
No the assertion isn't true as it currently stands. – Fixed Point Feb 15 '13 at 20:34
Essentially, just delete "only if" from the assertion in question above, and it becomes true, right? – Ryan Feb 15 '13 at 20:40
Yes that would work. The measure theory part is saying something like "infinite discontinuities are okay but it has to be a small enough infinity. Yes there are different 'sizes' of infinities so if the infinity is 'too big' the integral won't exist." – Fixed Point Feb 15 '13 at 20:43
Thanks! How should I edit the conditions in my assertion if I want to retain the "iff."? – Ryan Feb 15 '13 at 20:49

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.