Is there a "First Principles" for integration?

[mathwonk]

One more comment on when functions are not integrable even using lebesgue integration.
Intuitively, a function is lebesgue integrable if it is not too big and not too rough. Since the lebesgue integral is a monotone function of its integrand, a lebesgue integrable function cannot be approximated from below by integrable functions whose integrals are unbounded above. E.g. the function f(x) = 1/x on [0,1], with say f(0) = 0, is too big in this sense, because the usual approximating Riemann integrable functions obtained by restricting this function to the subintervals [e,1], as e—>0+, have unbounded integrals. So this function is not only not riemann integrable, but also not lebesgue integrable.

A function is too rough, if it is not the pointwise limit a.e. of any sequence of riemann integrable functions. It takes more sophistication to produce a function that is too rough, but the axiom of choice allows such a construction. Define an equivalence relation on the interval [0,1], so that two numbers are equivalent iff their difference is rational, and then choose exactly one element from each equivalence class to form a subset S of [0,1].

Then each number in [0,1] is equivalent to exactly one element of S. Thus [0,1] can be covered by a sequence of disjoint rational translates of S. (Given a rational number, add it to each element of S, and if the sum is > 1, chop it off at 1 and slide it back to start at 0, i.e. just subtract 1 from it. Or if you prefer, consider [0,1] essentially as the quotient space R/Z where rational translation is defined.] Now write this countable collection of translates of S as a sequence: S1, S2, S3,….and define functions as follows: gj = 1 at exactly all points of Sj, and zero elsewhere. Then f 1 = g1, f2 = g1+g2, f3 = g1+g2+g3,…..

Now the sequence fj converges upward pointwise everywhere to the constant function 1 on [0,1]. These functions fj are all bounded between 0 and 1, hence not too big. But 1 is integrable with integral 1 on [0,1]. So if the functions fj are integrable, they must have integrals bounded by 1, and their sequence of integrals must converge upward to 1. Now lebesgue measure is translation - invariant, so since the sets Sj are all essentially translations of S, the functions gj all must have the same integral, equal to some number between zero and 1. But what is the value of that integral? It cannot be zero since then the integral of all the functions fj would also be 0, and then 1 would be the limit of a sequence of zeroes, contradiction. But also the integral of the gj cannot be equal to a positive number e>0, since then the integral of fn would equal the sum of the integrals of g1,…,gn, hence equal to ne, which eventually becomes larger than 1, also a contradiction. So the functions gj and fj, although not too big, are too “rough” to be lebesgue integrable.

I'll probably quit here. Thanks for your indulgence. I myself think I learned something by doing and summarizing the reading needed for this discussion.

One more comment gleaned from that long mathoverflow discussion linked above. Someone there claimed that a great Harvard mathematician, maybe Brauer, remarked that after one reads the abstract book Measure Theory, by Halmos, one will know very well how to integrate the function 1, but will not know anything about how to integrate the function x.

 

[FactChecker]

One more comment on when functions are not integrable even using lebesgue integration.
Intuitively, a function is lebesgue integrable if it is not too big and not too rough. Since the lebesgue integral is a monotone function of its integrand, a lebesgue integrable function cannot be approximated from below by integrable functions whose integrals are unbounded above. E.g. the function f(x) = 1/x on [0,1], with say f(0) = 0, is too big in this sense, because the usual approximating Riemann integrable functions obtained by restricting this function to the subintervals [e,1], as e—>0+, have unbounded integrals. So this function is not only not riemann integrable, but also not lebesgue integrable.

The function is both Riemann and Lebesgue integrable. The definite integral of both on [0,1] are infinite.

 

[mathwonk]

Forgive me, but of course my assertions use the definitions I gave, and my definition of lebesgue integrable includes that it has finite integral. and my definition of riemann integrable (which is the one given by riemann) requires the function to be bounded, hence the riemann integral is also finite. I know some people admit infinity as a value of an integral, but the sources I use, such as Berberian and Lang define "integrability" as meaning the integral is finite. Cf. page 165 of Berberian, or page 237 of Lang Analysis II. (The same is true in Rudin, Riesz-Nagy, and Fleming, hence I suspect this was probably Lebegue's own usage, or he may have used the word "summable" or a French version of that.) It may be a little confusing since Rudin e.g. does write down integrals which are infinite, but he calls a function "Lebesgue integrable" only when its integral is finite. Fleming also writes such integrals, calling them "divergent". So in these books a function whose integral is infinite is not "integrable".

Of course this all depends on your definition of "integrable". Fortunately I gave my definition, and it included that the approximating sequence of riemann integrable funcions is cauchy in the sense of the integral norm. Thus, with my definition of " lebesgue integrable" it is a consequence that the lebesgue integral is finite.
...
I have had some difficulty tracking down a copy of Lebesgue's own work, but finally found an online copy of his "Lecons sur l'integration" where on p. 115, he defines an unbounded function to be "sommable" if the infinite series of its approximating sums is convergent, hence finite. He of course does not use the term "Lebesgue integrable". Interestingly, he states repeatedly that he is unaware of any examples of bounded functions which are not summable, or measurable, apparently being unaware (in 1903) of the example in post #31, using the axiom of choice.

 

[FactChecker]

Forgive me, but of course my assertions use the definitions I gave, and my definition of lebesgue integrable includes that it has finite integral. and my definition of riemann integrable (which is with the one given by riemann) requires the function to be bounded, hence the riemann integral is also finite. I know some people admit infinity as a value of an integral, but the sources I use, such as Berberian and Lang define "integrability" as meaning the integral is finite. Cf. page 165 of Berberian, or page 237 of Lang Analysis II.

Of course this all depends on your definition of "integrable". Fortunately I gave my definition, and it included that the approximating sequence of riemann integrable funcions is cauchy in the sense of the integral norm. Thus, with my definition of " lebesgue integrable" it is a consequence that the lebesgue integral is finite.

Ok. I don't like it, but I guess I can accept it.

 

[mathwonk]

I don't claim these are the only possible defitinions, (even if, in the case of riemann's, they have the historical weight of being riemann's own). You are perfectly free to give, or cite, different ones, but these are the ones that appear in the sources I am aware of. I will try to find Lebesgue's own definition.

Or you are free to consider my discussion as a characterization of functions that are not only integrable, but that also have finite integral. As such the discusion should have value to you even if you choose to use the word "integrable" in a different sense.

Here's a potential fix. Introduce the word "summable" to mean integrable with finite integral, and then interpret this whole discussion as about summable functions.

And thank you for reminding me of the possible confusion in terminology.

 

[mathwonk]

After some thought, I can imagine cases where one would like to distinguish between integrals that seem to have infinite value, as opposed to those to which no value can be assigned, finite or infinite. I can imagine proceeding as follows: Separate the function f into the difference of two non negative functions f+ and f-, by separating the parts of the graph that are above the x-axis from the part below it. I.e. f+(x) = max{f(x), 0}, and f-(x) = max{-f(x),0}.

Then say a non negative function f is integrable if it is the pointwise limit from below a.e. by non negative step functions, and if so, that its integral is the sup of the integrals of those step functions. Then the integral is finite if the integrals of those step functions can be taken to be bounded above. Then, one can define the integral of f by subtracting the integrals of f+ and f-, provided at least one of f+ or f-, has finite integral.

So one loses in this theory some familiar properties, such as the ability to add integrals, unless they are of the same sign, or at least one is finite. Fubini's theorem is also somewhat more complicated. Even if one deals entirely with positive valued functions, the key question to ask, regardless of the terminology, seems to be whether the integral is finite or not. When functions have both positive and negative values, here is a simple example from Fleming showing how Fubini can go wrong: the function f(x,y) = (1/y).cos(x), on the square with both x and y in the interval [0,π]. If y is constant the x -integrals are all zero, but if x is constant, the y integrals are almost all infinite, so the double integral does not exist even though one of the repeated ones does exist and is finite.

Then in this more general terminology, a function f would fail to be integrable if either it is too rough, or both f+ and f- are too big. So the essential issues are the same.