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I am re-reading lecture notes and I am having a few issues. Okay, I understand the concept of a partition. You basically have this bounded function $f:[a,b] \rightarrow \mathbb{R}$ and we are creating the idea of the integral out of scratch in order to do calculus with it. Then he introduced the concept of a lower and upper darboux sum, where you split your partition into partition intervals divided between partition points. So a Lower Darboux Sum, for example, is the summation of the infimums of each partition interval multiplied by the length of the partition interval.

However, the following page it begins to talk about "lower and upper integrals" and one of the very first explanations here I am having trouble understanding what it "looks like."

It says the lower integral of f from a to b is equal to the supremum{LowerSum(f,Partition) | P a partition of [a,b]}. What does this mean at all?

The way I understand it right now is basically: If you were to add all the blue squares together, that's your "Lower Darboux Sum" but what is the jump to having the sup of this "Lower Sum". To me, the Lower Sum is just a number, so what is the meaning behind supremum-ing it?

Also, there was a proof that I was supposed to do that said verbatim "Supposed that the bounded function f:[a,b] to R has the property that $f(x) \geq 0$ for all x \in [a,b]. Prove that the lower integral of f is always greater than or equal to zero. Geometrically, it seems obvious, and I basically just explained that the LowerSum is equal to the summation of each individual infimum multiplied by the difference in length of the interval in question, but I was wondering if there was some way of more clearly (at least to me) stating that "well, it seems geometrically equivalent, and because your function is always positive, it implies that this area you are calculating will also be positive."

Here is a photo of my textbook with what I'm talking about in the beginning of this post, the proof is not as important presently:

I appreciate any insight.

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@TuckerRapu There was no need for the drastic changes you introduced. I have reverted them. In general, edits should attempt not to change the style of the original poster. – Andrés E. Caicedo Mar 5 '14 at 7:15
@AndresCaicedo I didn't know the changes were so drastic. Example. I thought it's better to link to the picture in the beginning of the post. I was fazed the picture for the beginning is at the end. And I rectified some latex. Do you think these are vindicated? – Tucker Rapu Mar 5 '14 at 7:20
@TuckerRapu Let's leave the post as it is. – Andrés E. Caicedo Mar 5 '14 at 7:21
@AndresCaicedo My suggested edit didn't aid? or is there another reason – Tucker Rapu Mar 5 '14 at 12:41
@AndresCaicedo Just to let you know, there's a meta question about this edit. – user61527 Mar 5 '14 at 17:02
up vote 3 down vote accepted

Your textbook should have a proof of the following fact:

If $P'$ is a refinement of the partition $P$ of $[a,b]$, then $L(f,P') \ge L(f,P)$.

Let's assume there is some kind of "true" integral that we have defined already, and that the integral exists. Each lower sum approximates the true integral, but will be a bit too small (unless your function is constant). The above fact states that by adding points to your partition, the lower sum gets larger and a bit closer to the true value of the integral. Take a look at the diagrams on this page. You can see that as we increase the number of points in our partition, the lower sum will not shoot off to infinity but will instead converge to some value. We define the (lower) integral this way - as the "limit" of $L(f,P)$ as the number of points in $P$ goes to infinity, assuming that the limit exists. But this is vague, because I could add a lot of points to one specific area of the interval $[a,b]$ and ignore other areas. That's why we take the $\sup$ over all possible partitions.

For your proof: I'm assuming you have the following definition: $$L(f,P)=\sum_{i=0}^n \inf \{f(x) \mid x \in [x_{i-1},x_i]\} (x_i - x_{i-1}).$$ The only thing in this expression that could possibly be negative is the $\inf$, but $f$ is nonnegative. So every $L(f,P)$ is nonnegative, which means that $\sup L(f,P)$ is nonnegative.

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