I am taking issue with the Riemann Integral

1. Jul 4, 2010

Poopsilon

Basically from what I understand the integral of a function, say ∫x^2dx from say 0 to 1, can be represented as the supremum/infimum of the function values within each of a countably infinite number of vanishingly small intervals in the domain created by a countably infinite number of partition points, each multiplied against the corresponding length of the interval within which each supremum/infimum resides.
By The Nested Interval Theorem (I think!) we know each of these intervals has only a single value as n->∞, which means that the infimum=supremum, the upper and lower integrals are equal, thus the integral exists etc.
My concern is that each of these intervals has length 1/n as n->∞, which means they can be put into a one-to-one correspondence with the natural numbers, meaning the number of heights at which the integral is evaluated compose a countably infinite set. Yet! the domain of the original function is the real numbers, which are uncountably infinite! I'm still a novice in the areas of higher mathematics but to my mathematical intuition this implies that we are using methods that are only precise to the level of a countably infinite set to approximate the area under the curve of a function which has an uncountably infinite domain. Does this not entail that we are not taking into account all possible inputs that the function can take? Is it not possible that this function is behaving in strange ways, possibly where L(ƒ,Pn)≠U(ƒ,Pn), if we were to zoom infinitely in on some point of the curve?

Hopefully some of you more experienced analysts can shed some light on this for me, and any other words of wisdom about analysis in general would be great to.

2. Jul 4, 2010

Dickfore

You are absolutely right. Consider the Dirichlet function:

$$f(x) = \left\{\begin{array}{l} 1, \ x \in \mathbb{Q} \\ 0, \ x \notin \mathbb{Q} \end{array}\right.$$

It is not Riemann integrable.

3. Jul 4, 2010

Poopsilon

I guess my confusion comes from the fact that the Archimedean Sequence is designed such that each interval in the partition of the function's domain collapses onto a single point, yet it has a length of 1/n. So there are only a total of n of them which means we have essentially evaluated the function at n points, but we have neglected all those points on the real number line (an infinite number of them) that if entered into the function would provide a unique output.

If I understand what you mean by bringing up the Dirichlet Function, you're saying that we have functions with an interval domain that do act strangely at a level only the density of the real numbers can accommodate, but it seems that it is possible that even continuous functions of a real variable may act strangely at a level that we are blind to if we can only discuss it within the context of sequences and their limits.

4. Jul 4, 2010

Dickfore

Continuity of a function implies Riemann integrability.

5. Jul 12, 2010

lunde

I believe that it's actually the least upper bound property and not the nested interval property. In terms of your main question, the rational numbers form a countable dense subset of R, which allows us to use Riemann integrals on it. It is always possible to find a rational number (and thus from a countable subset) arbitrarily close to any real number which places us within good striking distance of making sure the irrationals don't do anything too crazy. There are a lot of examples like this where the rationals serve as an adequate countable skeleton that allows us to do many things within an uncountable set that we couldn't do otherwise. Also, you might want to look up nets on wikipedia, I personally perfer a net of partitions partially ordered by fineness over the clunky upper and lower riemann integral set up.