LEBESGUE INTEGATION EYE SJAMAA Supplementary notes Math 414, Spring 25 Introduction The following heuristic argument is at the basis of the denition of the Lebesgue integral. This argument will be imprecise, but it is meant to justify the conclusion, which is important. Let f be a function from to. Let us assume for the time being that f ; then we can think of the integral of f as the surface area between the graph of f and the x-axis. But a surface area is really a double integral, f(x) f(x) dx = dy dx, (1) and we may just as well reverse the order of integration. To do this eciently, we introduce the notion of the characteristic function of a subset A of a set. This is the function χ A : dened by { 1 if x A, χ A (x) = if x A. If = or 2, this function has the important property that χ A (x) dx = µ(a), (2) the measure (length or surface area) of A. For example, let A be the area (strictly) below the graph of a function f, which is given by the inequality y < f(x). This set can be described in three equivalent ways: A = { (x, y) 2 y < f(x) } = { (x, y) 2 y (, f(x)) } = { (x, y) 2 x f 1 (y, ) }. Date: April 25. 1
2 EYE SJAMAA This shows that χ A (x, y) = χ (,f(x)) (y) = χ f 1 (y, )(x). (3) We use (2) and (3) to change the order of integration in (1): f(x) dy dx = Plugging this into (1) we get f(x) dx = χ (,f(x)) (y) dy dx = χ f (y, )(x) dy dx 1 = χ f (y, )(x) dx dy = µ ( f 1 (y, ) ) dy. 1 µ ( f 1 (y, ) ) dy. (4) This calculation can be justied for suciently well-behaved functions. But Lebesgue's insight was that even for many badly behaved functions, for which the left-hand side of (4) does not make sense (as a iemann integral), the righthand side does make sense and can be taken as the denition of the integral. This Lebesgue integral turns out to be much more satisfactory than the iemann integral. Not only can we integrate far more functions than we used to, but more importantly the analytical properties of the Lebesgue integral are much better. For example, in the Lebesgue theory it is much easier to interchange its and integration, and multiple integrals are far less troublesome. There are many more advantages, some of which we will discuss if time allows. Finally, if a function is iemann integrable, it is Lebesgue integrable and the integrals are the same, so the new theory extends the old. Definition of the Lebesgue integral Let (, F, µ) be a measure space. A function f: is measurable if the set { x f(x) > y } = f 1 (y, ) is in F for all y. Observe that this denition does not involve the measure µ, but only the σ-algebra F. 1. Problem. Suppose f: is measurable. Using that F is a σ-algebra, show that (a) f 1 (, y], f 1 (, y) and f 1 [y, ) are in F for all y ; (b) f 1 (I) F for every interval in.
Conclude that f 1 (B) F for all Borel sets B. LEBESGUE INTEGATION 3 2. Problem. Suppose that = and F is the σ-algebra of Borel sets. If f: is continuous, then it is measurable. Nonnegative functions. We will start by dening the Lebesgue integral of nonnegative measurable functions. Let f: be measurable and f (i.e. f(x) for all x). Then we can dene a new function F f : [, ) [, ] by F f (y) = µ ( f 1 (y, ) ). 3. Definition. The (Lebesgue) integral of f is F f(y) dy. It is written as f dµ or f(x) dµ(x). The integral F f(y) dy is to be interpreted as an improper iemann integral. To justify this denition, we need to show that this iemann integral exists. First observe that F f and that F f is monotone nonincreasing: y 1 y 2 = F f (y 1 ) F f (y 2 ), because the measure µ is monotone. In particular, if F f (y) = for some y >, then F f (z) = for z [, y]. If this is the case, of course we dene F f(y) dy to be. If F f (y) is nite for all y >, we dene as usual F f (y) dy = a b b a F f (y) dy. (5) We claim that this is well-dened, either as a nite number or. This follows from the following fact. 4. Lemma. Let g be a bounded monotone nonincreasing function on [a, b], where < a b <. Then g is iemann integrable. Proof. Let P: a = x < x 1 < < x N = b be a partition of [a, b]. Then S + P (g) = N i=1 g(x i 1)(x i x i 1 ) and S P (g) = N i=1 g(x i)(x i x i 1 ) because g is monotone. Hence N S + P (g) S P (g) = ( g(xi 1 ) g(x i ) ) (x i x i 1 ) i=1 N ( g(xi 1 ) g(x i ) ) mesh P = ( g(a) g(b) ) mesh P. i=1 Therefore S + P (g) S P (g) as mesh P. (maximum interval length) of P. Here mesh P denotes the mesh
4 EYE SJAMAA This implies that for all measurable f the iemann integral G(a, b) = b a F f(y) dy is well-dened. Now we need to show that the iterated it a G(a, b) b exists. We will do this using a monotonicity argument that is quite easy, but will recur frequently in Lebesgue theory, so let's discuss it in some detail. Note that G has the properties a 1 a 2 = G(a 1, b) G(a 2, b) for all b; b 1 b 2 = G(a, b 1 ) G(a, b 2 ) for all a; i.e. G is monotone nondecreasing in a and monotone nonincreasing in b. A comment on double sequences. First we consider double sequences {x mn } of real numbers. 5. Problem. Suppose x mn is of the form f n (x m ), where {f n } n is a sequence of continuous real-valued functions on converging uniformly to a function f and {x m } m is a sequence of reals converging to x. Prove that x mn = x mn = f(x). m m n n This is frankly not too useful in Lebesgue's theory of integration, but the next result is. 6. Theorem. Suppose {x mn } m,n is monotone nondecreasing in each variable separately, i.e. m 1 m 2 implies x m1,n x m2,n for all n and n 1 n 2 implies x m,n1 x m,n2 for all m. Let L = sup{ x mn m, n } { }. Then n x mn = x mn = x mn = L. m (m,n) (, ) m n Proof. For each n let L n = sup{ x mn m } = m x mn. If n 1 n 1 then x m,n1 x m,n2 and thus L n1 = m x mn 1 m x mn 2 = L n2, so {L n } is monotone nondecreasing. Let M = sup{ L n n } = n L n. Then for all m and n, x mn L n M, so M is an upper bound for x mn, so M L. Let ɛ >. Choose n such that M L n < ɛ/2 and choose m such that L n x mn < ɛ/2; then M x mn < ɛ, so M L. This proves n x mn = L n = M = L. m n (6)
LEBESGUE INTEGATION 5 By interchanging the roles of m and n we see that m n x mn is also equal to L. To nish the proof it is enough to show that (m,n) (, ) x mn = L. Let ɛ >. Choose N 1 such that L L n < ɛ/2 for n N 1. Now choose N 2 such that L N1 x mn1 < ɛ/2 for m N 2. Put N = max{n 1, N 2 } and let m, n N. Then x mn x N2 N 1, so L x mn L x N2 N 1 = L L N1 + L N1 x N2 N 1 < ɛ 2 + ɛ 2 = ɛ, so (m,n) (, ) x mn = L. 7. Problem. (a) Let φ: (, x ) (, y ) (, ] be a function, where x, y (, ]. Assume that φ is monotone nondecreasing in each variable in the sense that x 1 x 2 implies φ(x 1, y) φ(x 2, y) for all y and y 1 y 2 implies φ(x, y 1 ) φ(x, y 2 ) for all x. Show that x x φ(x, y) = φ(x, y) = y y (x,y) (x,y ) y y φ(x, y) x x in the sense that all of these its exist and have the same value. (They may be innite.) There are no continuity assumptions on φ! (Try to mimick the proof of theorem 6.) (b) Suppose that g is iemann integrable on [a, b] for all < a b <. Let G(a, b) be the iemann integral b g(y) dy. Show that a a G(a, b) = b G(a, b) = a,b b a G(a, b). (c) Applying this result to (5), conclude that every nonnegative measurable function has a well-dened Lebesgue integral, and that does not matter in which order we take the it in (5). 8. Problem. Let A be a subset of. ecall from the introduction that the characteristic function of A is dened by χ A (x) = 1 if x A and χ A (x) = if x A. Prove that χ A is measurable if and only if A is measurable (i.e. A F). Compute χ A dµ directly from the denition. 9. Problem. A function f: is simple if it is measurable and takes on only nitely many values. Prove that a simple function f can be written as a linear combination, f = n i=1 c iχ Ai, where the A i are disjoint measurable subsets and c i. Now assume f is nonnegative, i.e. all c i are nonnegative. Compute the integral of f directly from the denition. (You are not allowed to use linearity of the integral. At this point we have not yet proved that the Lebesgue integral is linear.)
6 EYE SJAMAA 1. Problem. Let be an arbitrary set and F = P(), the σ-algebra of all subsets of. Show that every function f: is measurable. Let µ be the counting measure dened by µ(a) = number of elements of A (= if A is innite). What is f dµ for f: [, )? Justify your answer. 11. Problem. Let be an arbitrary set and F = P(). Let x and let δ be the Dirac measure based at x. This is dened by δ (A) = 1 if x A and δ (A) = if x A. What is f dδ for f: [, )? Justify your answer. (Physicists and engineers often write this integral as f(x)δ (x) dx.) Infinite values. It is often useful to be able to integrate functions which take on innite values. (Think of vertical asymptotes.) This is now straightforward: a function f: [, ] is measurable if the set {x f(x) > y} = f 1 ((y, ]) = f 1 (y, ) f 1 ( ) is in F for all y. If f: [, ] is measurable and nonnegative, we dene, just as before, F f (y) = µ ( {x f(x) > y} ) and f dµ = F f(y) dy. 12. Problem. If µ ( f 1 ( ) ), then f dµ =. We say that f g if f(x) g(x) for all x, and that a sequence of functions is {f i } monotone nondecreasing if f 1 f 2 f 3. Such a sequence has the property that for each x the sequence {f i (x)} is a monotone nondecreasing sequence of reals and so has a well-dened it i f i (x) = sup{ f i (x) i }. Let us call this it f(x). The function f: [, ] thus dened is the pointwise it of the f i. We can now state one of Lebesgue's most famous results. 13. Theorem (monotone convergence theorem). Let {f i } be a monotone nondecreasing sequence of nonnegative measurable functions f i : [, ], and let f: [, ] be the pointwise it. Then f is nonnegative and measurable and f dµ = i f i dµ (where both sides may be innite ). (This theorem fails miserably for iemann integrals! The reason is that the pointwise it of a monotone nondecreasing sequence of nonnegative iemann integrable functions on an interval [a, b] is not necessarily iemann integrable.) Proof. It is obvious that f is nonnegative. For each y the sets A i (y) = {x f i (x) > y} are measurable and satisfy A 1 (y) A 2 (y). Put A(y) = {x f(x) > y}. By denition x A(y) if and only if f(x) = i f i (x) > y. This is
LEBESGUE INTEGATION 7 the case if and only if f i (x) > y, i.e. x A i (y), for i suciently large. In other words A(y) = i=1 A i(y). Being a countable union of measurable sets, A(y) is measurable. Therefore f is measurable. Now consider the functions F fi (y) = µ ( A i (y) ) and F f (y) = µ ( A(y) ). We already know that each of these functions is monotone nonin creasing and hence iemann integrable by Lemma 4. Moreover, the sequence {F fi } is monotone nondecreasing and F f (y) = i F fi (y) because µ is monotone and continuous from below. Exercise 14 below now implies that F f i (y) dy F f(y) as i. 14. Problem. Let {g i } be a monotone nondecreasing sequence of nonnegative functions from [, ) to [, ]. Let g = i g i be the pointwise it. Assume that each individual g i is a monotone nondecreasing function (or nonincreasing). (a) Show that g is monotone nondecreasing (resp. nonincreasing). (Hence the iemann integral g(y) dy is well-dened.) (b) Prove that g(y) = i g i(y) dy. (First prove this for the integrals over [a, b] for < a b < and then let a and b. You will get an iterated it and will need to change the order in which you take the its. Use Exercise 7(a) to justify this.) We shall see that every nonnegative measurable function can be written as the pointwise it of a monotone nondecreasing sequence of nonnegative simple functions. Together with the monotone convergence theorem we can then show that our denition of the Lebesgue integral is equivalent to the one used in the book. Other applications of the monotone convergence theorem include the Fatou theorem and the dominated convergence theorem of Lebesgue. These can be found in the book and will be covered in class. Arbitrary functions. Now that we have dened the Lebesgue integral of a nonnegative function, it is straightforward to dene the Lebesgue integral of an arbitrary measurable function f. (As before we will allow functions f: with values in the \extended" real numbers = {, } = [, ].) We use the old trick of writing f = f + f as the dierence of two nonnegative functions, where f + = max(f, ) and f = min(f, ). It is not hard to show that both f + and f are measurable. (See Problem 15 below.) Now put f dµ = f + dµ f dµ.
8 EYE SJAMAA However, for this to make sense we need to assume that both f + dµ and f dµ are nite. If this is the case we call f (Lebesgue) integrable. (Note that for f dµ to be well-dened it is enough to require at least one of the integrals f + dµ and f dµ to be nite. Nevertheless, we will call f integrable only if both are nite.) Observe that f = f + + f. Hence, if f is Lebesgue integrable, then so is f. In other words, every Lebesgue integrable function is absolutely integrable. 15. Problem. Suppose f, g: are measurable. Show that max{f, g} and min{f, g} are measurable. Conclude that f +, f and f are measurable. 16. Problem. Let = {1, 2, 3,... }, F = P(), µ = counting measure. Then f(n) = ( 1) n /n is not Lebesgue integrable. However, it is improperly Lebesgue integrable in the sense that it is Lebesgue integrable on {1, 2,..., N} for all N > and N f dµ exists. {1,2,...,N} 17. Problem. Let = [, ), F the Borel σ-algebra on, µ = Lebesgue measure. Then f(x) = (sin x)/x is not Lebesgue integrable. However, it is improperly Lebesgue integrable in the sense that it is Lebesgue integrable on [, N] for all N > and N f dµ exists. [,N]