If and if is Riemann Integrable on both and , then is Riemann Integrable on .

**Proof:** Since we know is Riemann Integrable on and it follows that is bounded on those intervals as well. Let be the maximum of those bounds, then it follows that for every .

Let . Since we know is Riemann Integrable on and , and bounded on , it follows that there exists a and such that for a partition of with and a partition of with it follows that where and where .

Let . Let be a partition of such that . Then is in at most two intervals of . This would happen if is an endpoint of an interval. Suppose and for some . Consider the sum below. We wish to show that the following sum is less than .

Thus, is Riemann Integrable on .

**Reflection:** This is a type of proof that shows up a lot on the quals. The trick is that we know we can make most of the sum small since we know it’s RI on a part of the interval. Since our function is bounded this let’s us pick a delta that works nicely. Then we can make everything “small” and so it falls out from there.

Here’s a different perspective, using a slightly different definition (with, in my opinion, better notation).

Recall that the definition of Riemann integrability on an interval is that for any there is a such that any tagged partition with norm less than satisfies the following inequality:

for some fixed finite limit , with having intervals of length , and tags . This was standard notation in Dr. Shapiro’s analysis class, as the actual endpoints of the intervals are irrelevant unless the partition and function are explicitly specified.

So given Riemann integrability on both of the above intervals, we may assert the existence of two limits , such that for any epsilons , there exist corresponding , where for any tagged partitions with norms less than , respectively, the following inequalities hold:

with the number of intervals of their respective partitions. Without loss of generality we may let the norm of both be less than , and the inequality above still holds. Notice then, that is a tagged partition of with norm less than both of . Then the result follows as an application of the triangle inequality:

As are both arbitrary, given any we may let , then we see that the Riemann sum above is within , and the Riemann sum over the whole interval is .

correction: replace with , the values of the tags under . Otherwise the Riemann sum doesn’t make sense at all 🙂

And of course, if we don’t need to recall the definitions and notation, that cuts the proof shorter by three paragraphs. I was just trying to be overly pedantic.

When I attacked this problem, I took the same approach as Jeremy, which I thought was easier due to the notation. But then again, a function being Riemann Integrable and bounded should give us a hint to use the sup/inf definition of Riemann Integrable, which has proved to be very helpful in other problems.

Right, but as per my and Erin’s discussion, the problem statement never mentions boundedness. That comes as a result of Riemann integrability.