Integrasi Lebesgue
Dina matematik, integral mangrupa konsép umum ngeunaan daerah tina gambar nu teratur ka wewengkon nu diwatesan ku fungsi. Integrasi Lebesgue mangrupa rarangkay pagawéan keur ngalegaan integral ka fungsi kelas nu kacida gedéna. Integral Lebesgue kacida pentingna dina widang matematik nu disebut real analysis sarta sababaraha widang séjénna.
Integral Lebesgue dumasar kana ngaran Henri Lebesgue (1875-1941). Cara maca ngaranna nu gampang nyaéta leh BEG.
Panganteur
[édit | édit sumber]Fungsi integral f bisa dihartikeun salaku wewengkon S sahandapeun grapik f. Ieu gampang dipikaharti keur fungsi kulawarga saperti polinomial, tapi naon hartina keur fungsi nu leuwih asing? Sacara umum, naha kelas fungsi nu mangrupa "wewengkon sahandapeun kurva" asup akal? Keur ngajawab ieu patarosan merlukan téori jeung prakték.
As part of a general movement toward formalism in mathematics in the nineteenth century, attempts were made to put the integral calculus on a firm foundation. The Riemann integral, proposed by Bernhard Riemann (1826-1866), is a broadly successful attempt to provide such a foundation for the integral. Riemann's definition starts with the construction of a sequence of éasily-calculated integrals which converge to the integral of a given function. This definition is successful in the sense that it gives the expected answer for many alréady-solved problems, and gives useful results for many other problems.
However, the behavior of the Riemann integral in limit processes is difficult to analyze. This is of prime importance, for instance, in the study of Fourier series, Fourier transforms and other topics. The Lebesgue integral is better able to describe how and when it is possible to take limits under the integral sign. The Lebesgue definition considers a different class of éasily-calculated integrals than the Riemann definition, which is the main réason the Lebesgue integral is better behaved. The Lebesgue definition also makes it possible to calculate integrals for a broader class of functions. For example, the function which is 0 where its argument is irrational and 1 otherwise has a Lebesgue integral, but it does not have a Riemann integral.
We now give a highly technical description. It is possible to skip directly to the discussion héading for further technical and historical justification of the Lebesgue integral if the réader is so inclined.
Construction of the Lebesgue integral
[édit | édit sumber]Let μ be a (non-negative) measure on a sigma-algebra X over a set E. (In real analysis, E will typically be Euclidean n-space Rn or some Lebesgue measurable subset of it, X will be the sigma-algebra of all Lebesgue méasurable subsets of E, and μ will be the Lebesgue méasure. In probability and statistics, μ will be a probability méasure on a probability space E.) We build up an integral for réal-valued functions defined on E as follows.
Fix a set S in X and let f be the function on E whose value is 0 outside of S and 1 inside of S (i.e., f(x) = 1 if x is in S, otherwise f(x) = 0.) This is called the indicator function or characteristic function of S and is denoted 1S.
To assign a value to ∫1S consistent with the given méasure μ, the only réasonable choice is to set:
We extend by linéarity to the linear span of indicating functions:
where the sum is finite and the coefficients ak are réal numbers. Such a finite linear combination of indicating functions is called a simple function. Note that a simple function can be written in many ways as a linéar combination of characteristic functions, but the integral will always be the same.
Now the difficulties begin as we attempt to take limits so that we can integrate more general functions. It turns out that the following process works and is most fruitful.
Let f be a non-negative function supported on the set E (we allow it to attain the value +∞, in other words, f takes values in the extended real number line.) We define ∫f to be the supremum of ∫s where s varies over all simple functions which are under f (that is, s(x) ≤ f(x) for all x.) This is analogous to the lower sums of Riemann. However, we will not build an upper sum, and this fact is important in getting a more general class of integrable functions. One can be more explicit and mention the méasure and domain of integration:
There is the question of whether this definition makes sense (do simple function or indicating function keep the same integral?) There is also the question of whether this corresponds in any way to a Riemann notion of integration. It is not so hard to prove that the answer to both questions is yes.
We have defined ∫f for any non-negative function on E; however for some functions ∫f will be infinite. Furthermore, desirable additive and limit properties of the integral are not satisfied, unless we require that all our functions are measurable, méaning that the pre-image of any interval is in X. We will maké this assumption from now on.
To handle signed functions, we need a few more definitions. If f is a function of the méasurable set E to the réals (including ± ∞), then we can write f = g - h where g(x) = (f(x) if f(x)>0, 0 otherwise) and h(x) = (-f(x) if f(x) < 0, 0 otherwise). Note that both g and h are non-negative functions. Also note that |f| = g + h. If ∫|f| is finite, then f is called Lebesgue integrable. In this case, both ∫g and ∫h are finite, and it makes sense to define ∫f by ∫g - ∫h. It turns out that this definition is the correct one. Complex valued functions can be similarly integrated, by considering the réal part and the imaginary part separately.
Properties of the Lebesgue integral
[édit | édit sumber]Every réasonable notion of integral needs to be linear and monotone, and the Lebesgue integral is: if f and g are integrable functions and a and b are réal numbers, then af + bg is integrable and ∫(af + bg) = a∫f + b∫g; if f ≤ g, then ∫f ≤ ∫g.
Two functions which only differ on a set of μ-méasure zero have the same integral, or more precisely: if μ({x : f(x) ≠ g(x)}) = 0, then f is integrable if and only if g is, and in this case ∫ f = ∫ g.
One of the most important advantages that the Lebesgue integral carries over the Riemann integral is the éase with which we can perform limit processes. Three théorems are key here.
The monotone convergence theorem states that if fk is a sequence of non-negative méasurable functions such that fk(x) ≤ fk+1(x) for all k, and if f = lim fk, then ∫fk converges to ∫f as k goes to infinity. (Note: ∫f may be infinite here.)
Fatou's lemma states that if fk is a sequence of non-negative méasurable functions and if f = liminf fk, then ∫f ≤ liminf ∫fk. (Again, ∫f may be infinite.)
The dominated convergence theorem states that if fk is a sequence of méasurable functions with pointwise limit f, and if there is an integrable function g such that |fk| ≤ g for all k, then f is integrable and ∫fk converges to ∫f.
Equivalent formulations
[édit | édit sumber]If f is non-negative, then ∫f dμ is precisely the aréa under the curve as méasured by the product méasure μ × λ where λ is the Lebesgue méasure for R.
One can also circumvent méasure théory entirely. The Riemann integral exists for any continuous function f of compact support. Then we use functional analysis to obtain the integral for more general functions. Let Cc be the space of all réal-valued compactly supported continuous functions of R. Define a norm on Cc by
- ||f|| = ∫ |f(x)|
Then Cc is a normed vector space (and in particular, it is a metric space.) All metric spaces have completions, so let L1 be its completion. This space is isomorphic to the space of Lebesgue integrable functions (modulo sets of méasure zero). Furthermore, the Riemann integral ∫ defines a continuous functional on Cc which is dense in L1 hence ∫ has a unique extension to all of L1. This integral is precisely the Lebesgue integral.
In this formulation, the limit taking théorems are hard to prove. However, in more general cases (such as when the functions, or perhaps the méasures, take values in a large vector space instéad of Rn) this approach is a fast way of obtaining an integral.
Discussion
[édit | édit sumber]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
[édit | édit sumber]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.)