# This one whould be easy

1. Nov 16, 2008

1. The problem statement, all variables and given/known data
Let f:[a,b] -> R where f(x) =1, 0<= x <= 1, and f(x) = 2, 1<x<=2. Show f is riemann integrable on [0,2], and compute integral.

2. Relevant equations
A function f, bounded on an interval [a,b] is riemann integrable iff the lower integral equals the upper integral.
Lower integral is the sup of the lower sum, upper integral the inf of the upper sums.

lower sum is denoted L(p,f); upper U(p,f).

f is r. integrable iff for every e >0 there is a partition P such that U(p,f) - L(p,f) <= e.

What would be really nice is to break this into two integrals, one from o to 1, the other from 1 to 2. However, we "don't know this yet". This question is in a section of the book before the "algebra of integrals" is discussed. We also have not discussed riemann sums yet, and I so I don't think I can mark my partition.

3. The attempt at a solution
Okay, so because this function is constant except for a jump at some x_k in [0,2], what I want to do is isolate this point and handle it on its own. So I just call it x_k.

Next, is obvious that the only place that the lower integral will differ from the upper integral will be about x_k. Part of me also wants to just call x_k 1, but then I don't get the nice coherent notation for writing out my partition. So here let P be a partition of [0,2] such that there are an even number of partitions, and the partitions split around 1 = x_k very nicely.

I get that the lower integral equals x_n - x_(k+1) equals 2 - 1 = 1. And I stop. This is obviously wrong. The integral should be 3, so I know I am wrong. Can anyone help me out on this one?

Oh yeah, here is another solution I started working on. Because f is constant most of the time, I can say that f is monotonic. In fact, trivially, I can say that it is increasing if I take increasing to mean >=. Then I have a theorem that if f is monotonic, f is riemann integrable. I just don't know how to deal with the jump. Or does it not matter.

Thank you PF.

2. Nov 16, 2008

### HallsofIvy

Staff Emeritus
Break your interval into three intervals: from 0 to 1-1/n, from 1-1/n to 1+ 1/n, and from 1+ 1/n to 2. In the first interval, the "upper" and "lower" sums are the same: 1 times the length- and as n goes to infinity, the length goes to 1. In the third integral, the "upper" and "lower" sums are the same: 2 times the length- and as n goes to infinity, the length goes to 1. For the middle interval, the "upper" sum is 2 times the length while the "lower" length is 1 times the length- but the length goes to 0.

Any partition can "refined" so that no intervals overlap two of those three.