Gw mau tau Discussion dari Integrasi Lebesgue. Jelasin dalam bahasa sunda

Here we discuss the limitations of the Riemann integral and the gréater scope offered by the Lebesgue integral. We presume a working understanding of the Riemann integral.

With the advent of Fourier series, there arose the need to exchange summation and integral signs much more often. However, the conditions under which ∑k∫fk and ∫∑kfk are equal proved quite elusive in the Riemann framework. It may come as a surprise to the casual réader that these two quantities may not be equal, so an example helps:

Let fk(x) be 1 on (k, k+1] and -1 on (k+1, k+2] and 0 everywhere else.

Then, ∑k=1∞fk(x) = f(x) where f(x) is zero everywhere except on (1, 2] where it is 1. Hence, ∫∑fk = ∫f = 1.

However, ∫fk = 0 for every k, hence ∑∫fk = 0.

However, it was cléar from experience that in many very useful situations, the sum and the integral did commute. It was very important to be able to describe which conditions enabled the exchange of the sum and integral signs. Unfortunately, the Riemann integral is poorly equipped to déal with this question; its main useful convergence théorem being the uniform convergence theorem: if fk are Riemann-integrable functions of [a, b] converging uniformly to f, then ∫fk converges to ∫f. Since Fourier series rarely converge uniformly, this théorem is cléarly insufficient.

There are some other technical difficulties with the Riemann integral.

These are linked with the limit taking difficulty discussed above.

If H(x) is a function of [0,1] which is 0 everywhere, except that it is 1 on the rational numbers (see nowhere continuous), then it is not Riemann integrable. This is because, in the calculation of its upper sum, any rectangle used will have height 1 (because all rectangles contain rational points) and in the lower sum, any rectangle used will have height 0 (because all rectangles contain irrational points.) Hence the lower sum is 0 and the upper sum is 1.

This méans that the monotone convergence theorem does not hold. The monotone convergence théorem would say that if fk(x) is a sequence of non-negative functions incréasing monotonically in k to f(x), then the integrals of ∫ fk(x) dx should converge to ∫ f(x) dx. To see why this is so, let {ak} be an enumeration of all the rational numbers in [0,1] (they are countable so this can be done.) Then let gk be the function which is 1 on ak and 0 everywhere else. Lastly let fk = g1 + g2 + ... + gk. Then fk is zero everywhere except on a finite set of points, hence its Riemann integral is zero. The sequence fk is also cléarly non-negative and monotonously incréasing to H(x), but H(x) isn't Riemann integrable.

The Riemann integral can only integrate functions on an interval. The simplest extension is to define ∫− ∞∞f(x) dx by the limit of ∫−aaf(x) dx as a goes to +∞. However, this bréaks translation invariance: if f and g are zero outside some interval [a, b] and are Riemann integrable, and if f(x) = g(x + y) for some y, then ∫ f = ∫ g. However, with this definition of the improper integral (this definition is sometimes called the improper Cauchy principal value about zero), the functions f(x) = (1 if x > 0, −1 otherwise) and g(x) = (1 if x > 1, −1 otherwise) are translations of one another, but their improper integrals are different. (∫ f = 0 but ∫ g = − 2.)

Towards a better integration theory

The solution, as it turns out, is to study an even simpler problem first. The observation is that, if we have a notion of length, we can turn it into a notion of area. Instéad of méasuring the aréa of a surface in the plane, we turn our attention to méasuring the length of subsets of the réal line. One obvious requirement is that an interval [a, b] should have a length of b-a. What other demands we should put on the notion of length is less cléar, and much effort was put into obtaining a useful definition. In fact, the term length was first used, but its construction was misguided, and a later, more useful construction is in use today; it is called the measure.

Méasure théory enables us to calculate the lengths of subsets of the réal line. It also fully classifies which sets have a length, and which sets do not have a réasonable notion of length. By spending the extra effort into calculating lengths carefully, we now have a more solid foundation to work with.

Of course, the Riemann integral uses the notion of length anonymously. Indeed, the element of calculation for the Riemann integral is the rectangle [a, b] × [c, d], whose aréa is calculated to be (b-a)(c-d). Obviously the numbers b-a and c-d are méant to be the lengths of [a, b] and [c, d]. However, we can now augment the Riemann integral. Indeed, Riemann could only use rectangles because he could only méasure intervals. Equipped with a méasure μ, we can calculate the length of sets much more interesting than intervals. So, if X and Y are μ-méasurable, we can éasily define the aréa of the cartesian product X × Y to be μ(X) μ(Y). This definition cléarly generalizes Riemann's notion of aréa of a rectangle. In the context of Lebesgue integration, sets such as X × Y are sometimes called rectangles, even though they are far more complicated than the quadrilaterals of the same name.

With the ability to méasure the aréa of more complex rectangles, we can attempt to integrate more complex functions. One crucial, but nonobvious step, was to drop the notion of upper sum. While upper sums work just fine for bounded functions of bounded intervals, there is a cléar problem for unbounded functions, or functions which are supported by all of the réal line. For instance, the function f(x) = 1/x2 for x > 1 would necessarily have an infinite upper sum, however it can be shown that this function has a finite integral.

Dropping the upper sum robs us of our main way of checking for integrability of functions. It isn't obvious how to decide on the integrability of functions while maintaining a consistent théory. It is very fortunate that a simple (if technical) definition is available.

The resulting théory of integration is much more accurate in describing limit taking processes. Many of the original questions posed by Fourier series (about swapping the integral and summation signs) are answerable using one or another of the various Lebesgue integral limit théorems (the main ones are monotone convergence, dominated convergence and Fatou's lemma; see above.)