As usual, you've provided this poor algebraist with a couple of useful pieces of information that he's forgotten from his course in measure theory 30 years ago (or else never learned in the first place).
Thanks for the term "finitely additive integral." I felt a little defensive in my notes because I wasn't assuming countable additivity over disjoint sets, but apparently I don't need this for the Riemann integral.
Thanks also for the mention of the name Darboux. Someone else mentioned the name Daniel. I usually attribute the notion of the integral as a positive linear functional to Bourbaki, since that's where I learned it from. But Bourbaki wraps it in a formidible topological package which, at least for what I need, is quite unnecessary.
There is nothing in my notes <http://www.Math.Hawaii.Edu/~lee/ Integrals.pdf> that is in any way mathematically original. (Or at least if there is, it's probably wrong!) What's innovative in my notes is the observation that the Darboux/Daniel/Bourbaki approach to the integral can be used to handle most of the simple applications of integration taught in Calculus I without the need for explicit use of limits of sums. As my notes say, "Step functions are required for the proof, not for the actual calculation."
Secondly, what is primarily innovative in my notes and what required an enormous amount of effort from me is the explanation of this approach in terms that can be understood by a student with very little mathematical sophistication.
In my notes I have used the term "cumulative" rather than "finitely additive over disjoint intervals" and "stable" rather than "continuous." I have used the term "relationship" expressly because of the fact that it is not a generally used bit of mathematical jargon. Roughly speaking, in my notes it is used instead of "functional" or "mapping."
Finally, I have used a style that many mathematicians may find objectionable but which I find that students deal with much more easily. This style was common in books fifty or a hundred years ago. It emphasisizes explanations rather than proofs. These explanations are often given in terms of examples rather than a formidible array of subscripts and summation signs.
The notes contain an abundance of very simple diagrams (each of which cost me an enormous amount of effort to create in Latex). Most formulas are displayed but not labeled. When referring back to a previously given formula, I generally give the formula again rather than asking the reader to leaf back to an earlier page. All this is rather paper intensive. But I think that this forty-plus page paper reads a lot more quickly than the same material would if presented in 15 pages.
In article <email@example.com>, Herman Rubin <firstname.lastname@example.org> wrote: >In article <6b3q5e$b01@news.Hawaii.Edu>, Lee Lady <lady@Hawaii.Edu> wrote: >SNIP<
>>So here's the main thing I claim: > >>Theorem: Let Appl(f,I) be a mapping that associates a real number to >>each ordered pair (f,I) consisting of a Riemann-integrable function f >>defined on some reasonable subset of the real line and a compact >>subinterval I of the domain of f. Suppose further that Appl is >>additive over disjoint intervals (and over intervals which intersect >>only in a single point). Let G(x,y) be a continuous function of two >>variables. Suppose now that Appl(f,I) equals the integral over I >>of G(x,f(x)) whenever f is a constant function. Suppose also that >>the mapping Appl is monotonic, i.e. making f larger [resp. smaller] >>will always result in a larger [smaller] value for Appl(f,I). >>THEN: Appl(f,I) will equal the integral over I of G(x,f(x)) >>for all Riemann-integrable functions f. > >>Proof: Since both sides of the asserted equation are additive over >>disjoint intervals, the equation is valid not only for constant functions >>but for step functions. Now f can be obtained as a limit of step >>functions, and in fact there exist step functions s(x) and S(x) >>with s(x) <= f(x) <= S(x), and where s and S can be chosen so that >>the difference between Appl(s,I) and Appl(S,I) (as well as the >>difference between the corresponding integrals) is arbitrarily small.* >>The result now follows from the fact that both the mapping Appl and >>the integral are monotonic. (*I think that this follows from the >>hypotheses given. To say that s is a (uniformly) small function is to >>say that there exists a small real number c such that s(x) <= c for >>all x. It now follows from the stated hypotheses that Appl(s,I) will >>be small. However I need to deal with the case where there are two >>step functions s_1 and s_2 such that s_1 - s_2 is small. >>Unfortunately, I can't assume that Appl respects subtraction. So >>there might be a glitch here, as far as proving that Appl(s_1,I) - >>Appl(s_2,I) is small. Actually, it suffices to consider the case where >>s_1 and s_2 are constant functions. But maybe I need an additional >>hypothesis on the function G(x,y). )
Actually, no. I was too tired when I wrote the above. No additional hypothesis is required on G(x,y). The whole parenthetical comment beginning with the asterisk can be omitted. Since one is dealing with integrals of step functions, one really only needs to consider the case where s_1 and s_2 are constants and in this case the continuity of G(x,y) is all that is required.
>How does this really differ from the usual definition? It is >essentially the Darboux approach.
To repeat: I'm not defining a new kind of integral. (At least I hope not!) All I'm doing is justifying my claim that in setting up most common applications of integration, one does not need to go back to an explicit representation of the integral as a limit of sums. All one needs to do is to find an integral formula which gives the correct result in the case of constant functions.
>If the customary definition of the Riemann integral is modified >to not requiring closeness of the approximation for all sufficiently >fine partitions, but just the existence of a partition, it is the >statement that a function is Riemann integrable on an interval >if it can be squeezed between two step functions, the integrals >(already defined as the sum of products of length times height) >of which are close. It is this which, for arbitrary measures, >is the basis for the finitely additive integral.
>You are going through limits of sums.
Not explicitly. Not in the actual calculation. Only in the justification for the method.
> How much calculation is used >depends on the problem. Also, it is legitimate to use theorems to >reduce the amount of calculation; the Fundamental Theorem of Calculus >states that the antiderivative can be used to compute the integral, >and not just of real-valued functions.
Students (even mine!) are reasonably competent in calculating integrals by using anti-derivatives. Where they completely fail is in learning how to set integrals up in the first place.
I am suggesting the use of theory not merely to calculate the integral, but to set it up in the first place. As far as I can see, in terms of what one finds in existing calculus books, this is (alas!) extremely innovative.
I might also mention that the Fundamental Theorem can be very useful in setting integrals up in the first place. For instance, consider the example of arc length.
It is highly plausible to see arc length as determined by seeing the function in question as a limit of piecewise linear functions. However to justify this rigorously is beyond what one can do in a calculus book. The Fundamental Theorem, though, gives us an approach which is much more convincing.
Let s(x) by the length of the graph of the function f(x), measured from a fixed starting point a. To find an integral yielding s(x), we can consider the derivative ds/dx. Fix x. We are looking at a very small neighborhood of x, and when we zoom in on the corresponding tiny piece of the curve, we see that it looks like a straight line. (I am assuming, of course, that f(x) is a differentiable function.) It is now easy to see that ds/dx is given by on sec theta, where theta is the angle the curve makes to the horizontal. But of course we all know that f'(x), which is the slope of the curve, is given by tan theta. It then follows from the usual trig identity that ds/dx is the square root of 1 + f'(x)^2.
Therefore the formula for arc length is obtained by integrating the square root of 1 + f'(x)^2.
The interesting thing about this approach is that one finds a formula for arc length without ever needing a rigorous definition for arc length. I realize that to some mathematicians this will seem like heresy, but I prefer this to giving a definition of arc length that to many will seem arbitrary and not obviously in agreement with the intuitive notion of length.
-- Trying to understand learning by studying schooling is rather like trying to understand sexuality by studying bordellos. -- Mary Catherine Bateson, Peripheral Visions lady@Hawaii.Edu