The Lebesgue Measure and Integral


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1 The Lebesgue Measure and Integral Mike Klaas April 12, 2003 Introduction The Riemann integral, defined as the limit of upper and lower sums, is the first example of an integral, and still holds the honour of being the only integrating process taught before graduatelevel mathematics. It has several advantages, namely the relative simplicity behind the theory of the integral, its simplicity in calculation (particularly in numerical integration) and the fact that it works almost everywhere. From an analysis theory point of view, however, it is somewhat unsatisfactory. It can handle some discontinuity, but it is vulnerable to dense discontinuities (although countably discontinuous, nowhere dense functions can be Riemannintegrable). Perhaps more distressing is the strictness of the its convergence theorems. lim n fn = f when {f n } converges to f uniformly and in some other cases, but for many applications uniform convergence is simply not possible to achieve. The Lebesgue integral provides an alternative. While computationally unwieldy, it is a robust theoretic tool, equipped with powerful convergence theorems. It allows integrals to be found for many classes of functions on which the Riemann integral is not defined. Finally, it agrees with the Riemann integral when both exist, which soothes our conscience, and allows us to revert to the former when appropriate. 1 Measures A measure is a function that describes the nthdimensional size of a set. In R, this is the width of the set. In R 2, this is the area, in R 3, the volume, and soon in R n. The Riemann integral allows us to measure some sets in R 2, namely, the area of the set of points under a Riemannintegrable functions. It fails 1
2 for some bounded sets, however. Is the unmeasurability of these sets an inherent property of the sets or a failure of the Riemann integral? To answer this question it is instructive to examine the measure to which the Riemann integral is tied: the Jordan measure. 1.1 The Jordan Measure The outer Jordan measure of a set S R n is the infimum of the set of all possible coverings of S with a finite number of rectangles in R n. The Jordan measure of a set S R, for a bounded rectangle R R n is the common value of its outer measure and its inner measure, which is the outer measure of R\S minus the content of R (which is welldefined). Consider Q [0, 1] [0, 1] R. Given any ε, if Q [0, 1] is covered by a finite union of rectangles (closed intervals) of width ε, then the sum must be 1, since Q is dense in R (the rectangles are disjoint and must span [0, 1]). However, R\Q is also dense in R, so the inner measure is v([0, 1]) 1 = 1 1 = 0. Thus Q [0, 1] is not Jordan measurable. However, since Q is a countable set of singletons, it seems intuitively that it should have measure zero when compared to R. What if we first investigated the generic properties of a measure. If µ(e) : P(R n ) [0, ] can measure every subset of R n then we should expect it to be consistent with our idea of the measure of simple sets of spaces, such as intervals in R. We require, at least: 1. If E 1, E 2,... a possibly finite sequence of disjoint subsets of R n, then 2. If E F, then µ(e) = µ(f ) µ(e 1 E 2...) = µ(e 1 ) + µ(e 2 ) The unit cube in R n has measure 1. These seem like reasonable assumptions to make. However, it has been shown that they are mutually exclusive no such µ can exist! Relaxing our axioms doesn t help matters, either: even if only finite unions of sets are required to be additive (property 1), the schema breaks down for R n 3. The solution is to relax the requirement that all subsets of R n be measurable. That leads to a question: Can we develop a measure on which a larger class of sets is measurable compared to the Jordan measure? 2
3 1.2 Set Algebras Since there exist subsets of R n which stymy our ability to include them in a (seemingly reasonable) idea of a measure, we will concentrate on building families of subsets of R n which are consistent with the axioms we set forward. Fortunately, the family of subsets of R n which are consistent with these axioms encompasses virtually all subsets that are likely to be encountered in practice. Let X be a set, X. Any A P(X), A that is closed under finite unions and taking complements is an algebra of sets on X. If A is also closed under countable unions, it is called a σalgebra. Elements of algebras are denoted as E, F, etc., so we have E 1, E 2,..., E n A n E j A and E A X\E A. It is clear that algebras and σ algebras are closed under finite and countable intersections (respectively) as well, since if {E j } A, then {X\E j } A, and n (X\E j) A, so X\( n (X\E j) = n E j A. We also have = E X\E A, and X = E X\E A. Algebras are important in that they allow us means of classifying sets in hopes that we can find families of measurable sets. An important concept is the generation of an algebra from a set of subsets of X. First we need to verify that σalgebras are closed under intersection. Let M, N be σ algebras on X. Since all algebras contain X and, M N {X, } =, so the intersection cannot be empty (which would exclude it from being an algebra). Let E, {E j } M N. Then E, {E j } M, N, so X\E and n E j M, N, hence X\E, n E j M N. Thus M N is a σ algebra. Now, consider any A P(X). Let M(A) be the intersection of all σalgebras containing A. M(A) is unique and exists, since the intersection of σalgebras is never empty, and is the smallest σalgebra containing A. It is called the σalgebra generated by A. Up to this point we have resisted the urge to stray from generality and left X to be an arbitrary set. However, since our ultimate goal is compare the Lebesgue integral to the Riemann integral (which is defined on R n ), we can put some restrictions on X. If X is a topological space, we can consider the σalgebra generated by τ P(X) (the set of open subsets of X). This is the Borel σalgebra on X and is denoted B X. Since B X is closed under complementation, it not only contains but is generated by the set of closed subsets of X as well as τ. The Borel σalgebra on R denoted B R will be important when studying LebesgueStietjes measures later. 3
4 1.3 Definition of a Measure With the definition of a σalgebra in place, we can define a measure. Let X be a set and M a σalgebra on X. Let µ : M [0, ] be a function such that µ( ) = 0, and either or, If {E j } 1 M is a sequence of disjoint sets, then µ( E j) = µ(e j) If E 1, E 2,..., E n M are disjoint sets, then µ( n E j) = n µ(e j) If the former holds then µ is a countably additive measure on (X, M), and if the latter holds but not necessarily the former, µ is a finitely additive measure on (X, M). In either case, (X, M, µ) is called a measure space, M is the domain of µ, and the elements of M are measurable sets. Some properties of measures spaces that will be needed later are given in the following proposition. Theorem 1 Let (X, M, µ) be a measure space. 1. (Monotonicity) If E, F M and E F, then µ(e) µ(f ). 2. (Subadditivity) If {E j } 1 M, then µ( E j) µ(e j) 3. (Continuity from below) If {E j } 1 M and E 1 E 2, then µ( E j) = lim j µ(e j ) 4. (Continuity from above) If {E j } 1 M, E 1 E 2, and µ(e 1 ) <, then µ( E j) = lim j µ(e j ) Proof of 1. If E F, then F = E (F \E), so µ(f ) = µ(e (F \E)) = µ(e) + µ(f \E) (by the additivity of the measure) and this is µ(e) since µ(f \E) 0. One matter of interest to us in a measure space are its null sets, ie. sets E M s.t. µ(e) = 0 (sets of measure zero). A measure space whose σalgebra contains all its null sets is said to be complete. However, a noncomplete measure space (X, M, µ) can be extended in a straightforward manner to form a complete measure space in the following way: Proposition 1 Let N = {N M : µ(n) = 0}, and M = {E F : E M and F N, for some N N }. Then an extension µ of µ s.t. (X, M, µ) is a complete measure on X. (Proof skipped for brevity.) 4
5 1.4 The Lebesgue Measure Now that some of the concepts have been developed, it is possible to work toward the construction of the Lebesgue measure, upon which the Lebesgue integral is based. To accomplish this, we look at a certain family of continuous functions F (x) : R [0, ) which are defined as F (x) = µ((, x]), where µ is a Borel measure (which is simply a measure on R whose domain is B R ). F is increasing by the monotonicity of µ. Also, let {x j } be a strictly decreasing sequence in R with x j x. Then we have (, x 1 ] (, x 2 ] and µ((, x 1 ]) <. Since µ is continuous from above, µ((, x]) = µ( (, x j]) = lim j µ((, x j ]), hence F is right continuous. Also, if b > a, a, b R, then (, b] = (, a] (a, b], so µ((a, b]) = F (b) F (a). Since F can be so cleanly generated in the way, and from arbitrary µ, it is natural to wonder whether measures can be induces from functions satisfying these properties, namely being increasing and right continuous. As can be guessed by the wording of the last sentence, the answer is in the affirmative. The following proposition, which is given without proof due to its long and technical nature, provides the details. Theorem 2 If F : R R is any increasing, right continuous function, there is a unique Borel measure µ F on R s.t. µ F ((a, b]) = F (b) F (a) a, b R. Additionally, if the function G satisfies these properties as well, then µ F = µ G iff F (x) G(x) = c, a constant. The measure given by the theorem is actually slightly stronger than is indicated. µ F is not complete, generally, but can by made into a complete measure µ F using the construction outlined in Proposition 1, whose domain contains B R but is a strict superset of it. The complete measure is called the LebesgueStietjes measure associated with F, labelled µ when F is clear. µ s domain is denoted M µ. The simplest function F that induces a useful measure is F (x) = x. Let m be the complete measure associated with F ; this is the Lebesgue measure. M m, the domain of m, is denoted by L. E is Lebesgue measurable if E L. To perform calculations of the Lebesgue measure of sets, one more tool is required. This is the outer measure. Informally, the outer measure is a means of approximating a measure on a larger set by using increasingly fine coverings of sets of a type we already know how to measure. The clearest analogy is calculating the area of an arbitrary bounded figure B in R 2. Since B is bounded, we can find a rectangle X such that B X R 2. We can 5
6 cover B with a rectangle mesh which approximates its area, and do likewise for X\B. As the mesh gets finer, the limit of these two areas converges (for Jordan measurable B) and this is the area of B. Carathéory was instrumental in advancing this area of measure theory. The important result for our purposes is the following (a premeasure is a measure whose domain is an algebra rather than a σalgebra): Proposition 2 Let µ 0 be a premeasure on A P(X), with A, X A. For E X, define µ (E) = inf µ 0 (A j ) : A j A and E A j Then µ is an outer measure, and µ A = µ 0. The last assertion in the theorem simply confirms that the outer measure constructed is equivalent to the premeasure we started out with when restricted to the original set. When applied to the Lebesgue measure, we get: m(e) = inf [F (b j ) F (a j )] : E (a j, b j ] (1) m(e) = inf (b j a j ) : E (a j, b j ] (2) m(e) = inf (b j a j ) : E (a j, b j ) (3) (3) is due to a Lemma. It is a convenient form for calculating the Lebesgue measure of a subset of R. For instance, singletons in R have Lebesgue measure zero since {a} can be covered by an open set (a ε, a +ε) for arbitrarily small ε. Every countable union of null sets are themselves null, meaning Q has measure zero. The ability to measure Q (a dense null set) where the Jordan measure fails is essential to the difference between the Riemann and Lebesgue integral. Fortunately, however, for Jordan measurable sets, the Lebesgue measure agrees with the Jordan measure. 6
7 2 Integration The machinery developed in the previous section was quite general, applicable to measures on generic sets. For instance, a measure could be defined on a set of events X with µ(e M) the probability of the event set E X, and µ(x) = 1. However, in this section we will restrict our attention to integration using the Lebesgue measure. 2.1 Simple and Measurable Functions Lebesgue integration is based on the general principle of examining the sets of xvalues on which a function takes a given yvalue. These sets are also used to determine whether the function is measurable. Given f : X Y, E Y, there is a mapping f 1 (E) = {x X : f(x) E}. It is easy to see that {f 1 (E) : E N } is a σalgebra if N is a σalgebra. Thus, if (X, M), (Y, N ) are measurable spaces, then we call f a measurable function if f 1 (E) M for all E N. The measurability of a function f is directly related to its integrability. First, though, consider the sets f 1 (E) X above. If A is an arbitrary subset of X, we define { 1 if x A, χ A (x) = 0 otherwise. Let f : X C be a measurable function with a finite subset {a 1, a 2,..., a n } of C as its range. Then f is called a simple function, and f = n a j χ Aj, where A j = f 1 ({a j }), j = 1... n Lebesgue integration is defined in terms of simple functions. Thus, we need to approximate general functions using simply functions. First we present the Lebesgue integral for simple functions. 2.2 Lebesgue Integration of Simple Functions Let s(x) = n a jχ Aj be simple, positive, and m(a j ) <, 1 j n. The Lebesgue Integral of s(x) over X is s(x) = X s(x)dm = 7 n a j m(a j )
8 and for measurable E X, the Lebesgue Integral of s(x) over E is n χ E (x)s(x)dm = a j m(a j ) E Even with this restricted integral, we can produce some interesting results. Consider the Dirichlet function on X = [0, 1] given by { 1 if x Q f : [0, 1] [0, ], f = 0 otherwise f is not Riemannintegrable. However, f = 2 a jχ Aj with a 1 = 0, A 1 = [0, 1]\Q, a 2 = 1, A 2 = [0, 1] Q, so f fulfils the above criteria for integrability. We can compute the integral [0,1] f = n a jm(a j ) = 1m([0, 1] Q) = 1 0 = The Integral for Arbitrary Measurable f First, define L + to be the vector space of all Lebesgue measurable functions from X to [0, ]. We will approximate f L + using simple functions. Proposition 3 Let f L +. Then a sequence {φ n } of simple functions such that 0 φ 1 φ 2 f, φ n f pointwise, and φ n f uniformly on any set on which f is bounded. Proof. For n = 0, 1, 2,... and 0 k 2 2n 1, let (( k En k = f 1 2 n, k + 1 ]) 2 n and F n = f 1 ((2 n, ]) and define φ n = 2 2n 1 k=0 k 2 n χ E k n + 2n χ Fn Consider an En k in φ n. Clearly, En k En+1 2k E2k+1 n+1, so φ n φ n+1. Also, on the x set where f 2 n, at any x, 0 f(x) φ n (x) 1 2. If this is R (ie., f n is bounded), then φ n f uniformly. Otherwise, convergence is pointwise, since we can choose n large enough such that 2 n > f(x). Now we can define the integral of f L +, namely { } f = fdm = sup φdm : 0 φ f, φ simple X X 8
9 For other f, we decompose the function into a linear combination of functions in L + by defining f + = max{f(x), 0}, f = max{ f(x), 0}, f = f + f Given arbitrary f : X C, set g = R(f), h = I(f), and all of g +, g, h +, h L +, so there exist sequences of simple functions ψ +, ψ, ζ +, ζ that converge to g +, g, h +, h, respectively, and (g f = + g + i(h + h ) ) = g + ( g + i h + h ) f is integrable if all of g +, g, h +, h are. 3 Summary and Conclusion 3.1 Convergence We had said that the Lebesgue integral is a powerful theoretic alternative to the Riemann integral in the converge theorems it boasts. One example is the following: Theorem 3 (Monotone Convergence Theorem) If {f n } is a sequence in L + such that f j f j+1 for all j, and f = lim n f n, then f = lim n fn This theorem has a plethora of uses, and demonstrates the flexibility of the Lebesgue integral. In particular, we assumed its truth in the previous section when we only required the existence of one sequence of simple functions that increased to f to find the integral of f, instead of a supremum over an infinite family of functions. 3.2 Other Properties of the Lebesgue Integral When considering the Lebesgue integral vis a vis the Riemann integral, it s worth pondering if any useful properties are lost. Almost every familiar property of the integral is preserved, as the following proposition demonstrates. Proposition 4 Let φ, ψ be simple functions in L +, f L 1 realvalued. Then 9
10 1. If c 0, cφ = c φ. 2. (φ + ψ) = φ + ψ. 3. If φ ψ, then φ ψ. 4. f f. Proof. Let n a jχ Ej and m k=1 b kχ Fk be the standard representations of φ and ψ, respectively. cφ = c n a jχ Ej = n c a jχ Ej = n c a jm(e j ) = c n a jm(e j ) = c n a jχ Ej = c φ which proves (1). For (2), we know that n E j = m k=1 F k = X, thus we can write E j = E j X = E j m k=1 F k = m k=1 (E j F k ) and analogously F k = n (F k E j ). These unions are disjoint since {E j } and {F k } are disjoint. Therefore we can write n φ + ψ = a j m(e j ) + = = m b k m(f k ) k=1 n m a j m( (E j F k )) + n k=1 k=1 m a j m(e j F k ) + (since m is finitely additive) m n b k m( (F k E j )) k=1 m k=1 = (a j + b k )m(e j F k ) j,k = (φ + ψ), by similar reasoning. n b j m(f k E j ) For (3), using the same construction, observe that if φ ψ, then a j b k on E j F k, giving φ = a j m(e j F k ) b k m(e j F k ) = ψ. j,k j,k To prove (4), simply observe that if f = f + f, then f = f + + f, so f = f + f f + + f = f. Also, parts (1) to (3) hold for arbitrary f L 1 by the Monotone Convergence Theorem. 10
11 This doesn t mean that all properties of the Riemann integral are present in the Lebesgue integral, however. Integration by parts is one example. 3.3 Conclusion The Lebesgue integral and measure provide a substantial improvement to the Riemann integral (resp. Jordan measure). Not only does it agree with the Riemann integral everywhere but it allows much more esoteric subsets of R to be measured. In fact, it has been proven that the Axiom of Choice must be assumed true to construct a nonlebesguemeasurable set. The Lebesgue integral also enjoys a more robust theoretic background, exhibiting a much richer array of convergence theorems with weaker conditions than their Riemann equivalents, if they exist. It strengthens the properties of some metric spaces that have integraldefined metrics (M can be such that it is not complete using the Riemann integral, but is complete when using the Lebesgue integral). It is, however, quite technical and requires a significant amount of measure theoretic background to use in practice, since the sets f 1 (E) can be quite wild even for continuous f. 11
12 References djao. Planet Math: Lebesgue Integral. From: Follard, Gerald B. Real Analysis: Modern Techniques and Their Applications. John Wiley & Sons, Inc., New York, nd Ed. Wachsmuth, Bert G. 7.4: Lebesgue Integral. From: Weisstein, Eric W. Jordan Measure. From: 12
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