U(f,P) - L(f,P) < epsilon proof

[Buri]

I won't state the theorem in full but showing that a function is integrable if and only if for any epsilon there exists a partition P such that U(f,P) - L(f,P) < epsilon.

I'm finding the proof in the forward direction a bit confusing. If you assume the function is integrable that means that the lower and upper integrals are equal. Now the proof goes on to say that there exists a partition P' such that L(f,P') is within a distance of epsilon/2 of the integral. But how is this true!? The lower integral is defined to be the sup{L(f,P)} where P ranges over all partitions. Now let's say that sup{L(f,P)} is actually not included in the set of all lower sums, then I know that there must exist a partition such that the lower sum using this partition will lie really close to sup{L(f,P} as if there weren't it would contradict that sup{L(f,P} is indeed the supremum of all lower sums. But if it IS included in the set of all lower sums then its not necessarily true that I can find a partition really close to it as the set of all lower sums *could be* a set of discrete points. I guess to eliminate the discrete set of lower sums, L(f,P) would have to be continuous in P, is this true? Otherwise, I just don't see how there exists a P' such that L(f,P') is within epsilon/2 of the integral. Neither my professor nor Munkres in Analysis on Manifolds justify the existence of such a P'.

Any clarification? Thanks!

 

[Petek]

I suggest that you create a simple example, and calculate all the sums and such. Perhaps that will help. For example, let

f(x) = 1 if 0 $\leq$ x $\leq$ 1/2, and
f(x) = 2 if 1/2 < x $\leq$ 1

Set up a couple of partitions and see what you get,

 

[Landau]

I haven't given this much thought, but isn't that just the definition of the supremum? The supremum of a set of real numbers is a limit point of that set, hence can be approximated arbitrarily close.

If the supremum happens to be in the set itself, i.e. the supremum is in fact the maximum, then the partition P for which it occurs satisfies L(f,P)=sup{L(f,P'): P' a partition}, so L(f,P) is just equal to the lower integral, hence has distance zero.

You will agree that zero is smaller than epsilon/2...?

 

[Buri]

I haven't given this much thought, but isn't that just the definition of the supremum? The supremum of a set of real numbers is a limit point of that set, hence can be approximated arbitrarily close.

If the supremum happens to be in the set itself, i.e. the supremum is in fact the maximum, then the partition P for which it occurs satisfies L(f,P)=sup{L(f,P'): P' a partition}, so L(f,P) is just equal to the lower integral, hence has distance zero.

You will agree that zero is smaller than epsilon/2...?

"The supremum of a set of real numbers is a limit point of that set, hence can be approximated arbitrarily close."

I've never actually seen this definition. And as a matter of fact, I've always thought that that ONLY worked with open sets because there I must be able to approximate it with something different than the supremum but when closed I hadn't realized that I can take the supremum itself. If I had known all this I wouldn't have gotten all confused lol

Thanks for the help guys!

 

[Landau]

"The supremum of a set of real numbers is a limit point of that set, hence can be approximated arbitrarily close."

I've never actually seen this definition.

It's not a definition, but a result/theorem.

Thanks for the help guys!

You're welcome :)