Re: Why do they teach Riemannian sums?

Matthias says:

For example, the function f(x) that is equal to (-1^k)/k on each
interval [k,k+1) (for positive integer k) is such that the limit as b
goes to infinity of int^b_0 f(x) dx exists, and thus the naive improper
Riemann integral exists. But clearly there are integrals over
subsets of the domain that diverge, so monotonicity fails.

I say:

As far as my analysis book goes, a function can not be improperly integrable unless it is first locally integrable, that is intregable on any closed interval inside the original integral. So, as far as WHAT YOU'RE SAYING, it seems to me that the function you give is not at all improperly integrable. Therefore it is not Riemann integrable with an endpoint of Infinity if what you say is true. You say the "naive improper Riemann integral" exists. Score 1 for the naive improper Riemann integral and 0 for Matthias... unless I'm just not understanding what you're saying here.

However, I don't see that what you're saying is true. Would this be something where the right endpoint is an integer such as the integral from 1 to 2? It seems to me that even that would be integrable. However, you tell us that it is clear that there are places where in the domain where it is not integrable so if you would please explain that to me, that would be great. Thanks. It appears to me, that from the definition of integral given in Wade's "Real Analysis" that this is in actuality integrable on every closed interval.

Take the integral from 1 to 2 of your function as an example. The integral is defined to be the value of the upper and lower integrals if they are in fact equal. The upper integral is defined as the infimum of the set of all upper Riemann sums of the function and any partition of the interval. The lower integral is defined as the supremum of the set of all lower Riemann sums of the function and any partition of the interval.

I say it's clear that the lowest possible upper sum will be the same as the highest possible lower sum because the only problem would be at one point, the endpoint of 2 but in the end, the partition that maximizes the lower sum and the partition that minimizes the upper sum will be one where each subinterval is infinitely small so that one point doesn't make a difference. Therefore, unless I've made a mistake (and I certainly haven't rigorously proved the integral exists) that integral exists. And in general, it's not too hard to see that all of the integrals on the domain exist. I admit that I am not great at this so if I'm wrong, please show me. But if what I said is correct, then it's not at all naive to say that this is also improperly integrable.

Also, I believe you mean the function ((-1)^k)/k as the function you give is in reality just -1/k.

Matthias says:

I would say the only reason that my definition isn't part of the
textbook definition is that nobody cares about Riemann integrals any
more, so they are frozen with the definitions they had in 1880. In my
opinion, monotonicity is too important a property of integrability to
say that functions such as f(x) above are integrable. Whether the
naive improper Riemann integral is useful for the physical sciences is
another matter.

I say:

Also, monotonicity is about being always nondecreasing or always nonincreasing, at least after a certain point. The way you use it suggests one of three things to me, 1. You don't know what monotonicity means, 2. You're using a correct definition that I don't know, 3. You're using the definition I know in a right way and I don't get it. This function is clearly not monotonic, but neither is sine. When you say, "In my opinion, monotonicity is too important a property of integrability to say that functions such as f(x) above are integrable", are you suggesting sine should not be integrable since it is not monotonic?
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