Chapter IV Integration Theory

Chapter IV Integration Theory This chapter is devoted to the developement of integration theory. The main motivation is to extend the Riemann integral calculus to larger types of functions, thus leading to the Lebesgue integral calculus. The main advantages include a better approach to indefinite integrals, as well as nice convergence results. As the exposition suggests, integral calculus can be developed in a great deal of generality. 1. Construction of the integral In this section we construct the abstract integral. As a matter of terminology, we define a measure space as being a triple (, A, µ), where is some (non-empty) set, A is a σ-algebra on, and µ is a measure on A. The measure space (, A, µ) is said to be finite, if µ() <. Definition. Let (, A, µ) be a measure space, and let K be one of the fields R or R. A K-valued elementary µ-integrable function on (, A, µ) is an function f : K, with the following properties the range f() of f is a finite set; f 1 ({α}) A, and µ ( f 1 ({α}) ) <, for all α f() {0}. We denote by L 1 K,elem (, A, µ) the collection of all such functions. Remarks 1.1. Let (, A, µ) be a measure space. A. Every K-valued elementary µ-integrable function f on (, A), µ) is measurable, as a map f : (, A) ( K, Bor(K ). In fact, any such f can be written as f = α 1 κ A1 + + α n κ An, with α k K, A k A and µ(a k ) <, k = 1,..., n. Using the notations from III.1, we have the inclusion L 1 K,elem(, A, µ) A-Elem K (). B. If we consider the collection R = {A A : µ(a) < }, then R is a ring, and, we have the equality L 1 K,elem(, A, µ) = R-Elem K (). In particular, it follows that L 1 K,elem (, A, µ) is a K-vector space. The following result is the first step in the construction of the integral. Theorem 1.1. Let (, A, µ) be a measure space, and let K be one of the fields R or C. Then there exists a unique K-linear map I µ elem : L1 K,elem (, A, µ) K, such that (1) I µ elem (κ A) = µ(a), for all A A, with µ(a) <. 301

302 CHAPTER IV: INTEGRATION THEORY Proof. For every f L 1 K,elem (, A, µ), we define I µ elem (f) = α µ ( f 1 ({α}) ), α f() {0} with the convention that, when f() = {0} (which is the same as f = 0), we define I µ elem (f) = 0. It is obvious that Iµ elem satsifies the equality (1) for all A A with µ(a) <. One key feature we are going to use is the following. Claim 1: Whenever we have a finite pairwise disjoint sequence (A k ) n A, with µ(a k ) <, k = 1,..., n, one has the equality I µ elem (α 1κ A1 + + α n κ An ) = α 1 µ(a 1 ) + + α n µ(a n ), α 1,..., α n K. It is obvious that we can assume α j 0, j = 1,..., n. To prove the above equality, we consider the elementary µ-integrable function f = α 1 κ A1 + +α n κ An, and we observe that f() {0} = {α 1 } {α n }. It may be the case that some of the α s a equal. We list f() {0} = {β 1,..., β p }, with β j β k, for all j, k {1,..., p} with j k. For each k {1,..., p}, we define the set J k = { j {1,..., n} : α j = β k }. It is obvious that the sets (J k ) p are pairwise disjoint, and we have J 1 J p = {1,..., n}. Moreover, for each k {1,..., p}, one has the equality f 1 ({β k }) = j J k A j, so we get β k µ ( f 1 ({β k }) ) = β k µ(a j ) = α j µ(a j ), k {1,..., p}. j J k j J k By the definition of I µ elem we then get p I µ elem (f) = β k µ ( f 1 ({β k }) ) = p [ ] α j µ(a j ) = j J k α j µ(a j ). Claim 2: For every f L 1 K,elem (, A, µ), and every A A with µ(a) <, one has the equality (2) I µ elem (f + ακ A ) = Iµ elem (f) + αµ(a), α K. Write f = α 1 κ A1 + + α n κ An, with (A j ) n j=1 A pairwise disjoint, and µ(a j) <, j = 1,..., n. In order to prove (2), we are going to write the function f + ακ A in a similar way, and we are going to apply Claim 1. Consider the sets B 1, B 2,..., B 2n, B 2n+1 A defined by B 2n+1 = A (A 1 A n ), and B 2k 1 = A k A, B 2k = A k A, k = 1,..., n. It is obvious that the sets (B p ) 2n+1 p=1 are pairwise disjoint. Moreover, one has the equalities (3) B 2k 1 B 2k = A k, k {1,..., n}, as well as the equality (4) A = n+1 B 2k 1. j=1

1. Construction of the integral 303 Using these equalities, now we have f + ακ A = 2n+1 p=1 β pκ Bp, where β 2n+1 = α, and β 2k = α k and β 2k 1 = α k + α, k {1,..., n}. Using these equalities, combined with Claim 1, and (3) and (4), we now get 2n+1 I µ elem (f + ακ A ) = = αµ(b 2n+1 ) + p=1 β p µ(b p ) = [ (αk + α)µ(b 2k 1 ) + α k µ(b 2k ) ] = [ n+1 ] = α µ(b 2k 1 ) = αµ ( n+1 = αµ(a) + B 2k 1 ) ) + [ n + α k [ µ(b2k 1 ) + µ(b 2k ) ]] = α k µ(b 2k 1 B 2k ) = α k µ(a k ) = αµ(a) + I µ elem (f), and the Claim is proven. We now prove that I µ elem is linear. The equality I µ elem (f + g) = Iµ elem (f) + Iµ elem (g), f, g L1 K,elem(, A, µ) follows from Claim 2, using an obvious inductive argument. The equality I µ elem (αf) = αiµ elem (f), α K, f L1 K,elem(, A, µ). is also pretty obvious, from the definition. The uniqueness is also clear. Definition. With the notations above, the linear map is called the elementary µ-integral. I µ elem : L1 K,elem(, A, µ) K In what follows we are going to encounter also situations when certain relations among measurable functions hold almost everywhere. We are going to use the following. Convention. Let T be one of the spaces [, ] or C, and let r be some relation on T (in our case r will be either =, or, or, on [, ]). Given a measurable space (, A, µ), and two measurable functions f 1, f 2 : T, if the set f 1 r f 2, µ-a.e. A = { x : f 1 (x) r f 2 (x) } belongs to A, and it has µ-null complement in, i.e. µ( A) = 0. (If r is one of the relations listed above, the set A automatically belongs to A.) The abreviation µ-a.e. stands for µ-almost everywhere. Remark 1.2. Let (, A, µ) be a measure space, let f A-Elem K () be such that f = 0, µ-a.e.

304 CHAPTER IV: INTEGRATION THEORY Then f L 1 K,elem (, A, µ), and Iµ elem (f) = 0. Indeed, if we define the set N = {x : f(x) 0}, then N A and µ(n) = 0. ince f 1 ({α}) N, α f() {0}, it follows that µ ( f 1 ({α}) ) = 0, α f() {0}, and then by the definition of the elementary µ-integral, we get I µ elem (f) = 0. One useful property of elementary integrable functions is the following. Proposition 1.1. Let (, A, µ) be a measure space, let f, g L 1 R,elem (, A, µ), and let h A-Elem R () be such that Then h L 1 R,elem (, A, µ), and f h g, µ-a.e. (5) I µ elem (f) Iµ elem (h) Iµ elem (h). Proof. Consider the sets A = {x : f(x) > h(x)} and B = {x : h(x) > g(x)}, which both belong to A, and have µ(a) = µ(b) = 0. The set M = A B also belongs to A and has µ(m) = 0. Define the functions f 0 = f(1 κ M ), g 0 = g(1 κ M ), and h 0 = h(1 κ M ). It is clear that f 0, g 0, and h 0 are all in A-Elem R (). Moreover, we have the equalities f 0 = f, µ-a.e., g 0 = g, µ-a.e., and h 0 = h, µ-a.e., so by Remark??, combined with Theorem 1.1, the functions f 0 = f + (f 0 f) and g 0 = (g 0 g) + g both belong to L 1 R,elem (, A, µ), and we have the equalities (6) I µ elem (f 0) = I µ elem (f) and Iµ elem (g 0) = I µ elem (g). Notice now that we have the (absolute) inequality f 0 h 0 g 0. Let us show that h 0 is elementary integrable. tart with some α h 0 () {0}. If α > 0, then, using the inequality h 0 g 0, we get h 1 ( ) 0 ({α}) g 1 0 (0, ) g0 1 ({λ}), λ g 0() {0} which proves that µ ( h 1 0 ({α})) <. Likewise, if α < 0, then, using the inequality h 0 f 0, we get h 1 0 ( ) 1 ({α}) f0 (, 0) f0 1 ({λ}), λ f 0() {0} which proves again that µ ( h 1 0 ({α})) <. Having shown that h 0 is elementary integrable, we now compare the numbers I µ elem (f 0), I µ elem (h 0), and I µ elem (g 0). Define the functions f 1 = h 0 f 0, and g 1 = g 0 h 0. By Theorem 1.1, we know that f 1, g 1 L 1 R,elem (, A, µ). ince f 1, g 1 0, we have f 1 (), g 1 () [0, ), so it follows immediately that I µ elem (f 1) 0 and I µ elem (g 1) 0. Now, again using Theorem 1.1, and (6), we get I µ elem (h 0) = I µ elem (f 0 + f 1 ) = I µ elem (f 0) + I µ elem (f 1) I µ elem (f 0) = I µ elem (f); I µ elem (h 0) = I µ elem (g 0 g 1 ) = I µ elem (g 0) I µ elem (g 1) I µ elem (g 0) = I µ elem (g).

1. Construction of the integral 305 ince h = h 0, µ-a.e., by the above Remark it follows that h L 1 R,elem (, A, µ), and I µ elem (h) = Iµ elem (h 0), so the desired inequality (5) follows immediately. We now define another type of integral. Definition. Let (, A, µ) be a measure space. A measurable function f : [0, ] is said to be µ-integrable, if (a) every h A-Elem R (), with 0 h f, is elementary µ-integrable; (b) sup { I µ elem (h) : h A-Elem R(), 0 h f } <. If this is the case, the above supremum is denoted by I µ +(f). The space of all such functions is denoted by L 1 +(, A, µ). The map is called the positive µ-integral. I µ + : L 1 +(, A, µ) [0, ) The first (legitimate) question is whether there is an overlap between the two notions of integrability. This is anwered by the following. Proposition 1.2. Let (, A, µ) be a measure space, and let f A-Elem R () be a function with f 0. The following are equivalent (i) f L 1 +(, A, µ); (ii) f L 1 R,elem (, A, µ). Moreover, if f is as above, then I µ elem (f) = Iµ +(f). Proof. The implication (i) (ii) is trivial. To prove the implication (ii) (i) we start with an arbitrary elementary h A-Elem R (), with 0 h f. Using Proposition 1.1, we clearly get (a) h L 1 R,elem (, A, µ); (b) I µ elem (h) Iµ elem (f). Using these two facts, it follows that f L 1 +(, A, µ), as well as the equality sup { I µ elem (h) : h A-Elem R(), h f } = I µ elem (f), which gives I µ +(f) = I µ elem (f). We now examine properties of the positive integral, which are similar to those of the elementary integral. The following is an analogue of Proposition 1.1. Proposition 1.3. Let (, A, µ) be a measure space, let f L 1 +(, A, µ), and let g : [0, ] be a measurable function, such that g f, µ-a.e., then g L 1 +(, A, µ), and I µ +(g) I µ +(f). Proof. tart with some elementary function h A-Elem R (), with 0 h g. Consider the sets M = {x : h(x) > f(x)} and N = {x : g(x) > f(x)}, which obviously belong to A. ince M N, and µ(n) = 0, we have µ(m) = 0. If we define the elementary function h 0 = h(1 κ M ), then we have h = h 0, µ-a.e., and 0 h 0 f, so it follows that h 0 L 1 R,elem (, A, µ), and Iµ elem (h 0) I +(f). µ ince h = h 0, µ-a.e., by Proposition 1.1., it follows that h L 1 R,elem (, A, µ), and I µ elem (h) = Iµ elem (h 0) I +(f). µ By definition, this gives g L 1 +(, A, µ) and I +(g) µ I +(f). µ

306 CHAPTER IV: INTEGRATION THEORY Remark 1.3. Let (, A, µ) be a measure space, and let f L 1 +(, A, µ). Although f is allowed to take the value, it turns out that this is inessential. More precisely one has µ ( f 1 ({ }) ) = 0. This is in fact a consequence of the equality (7) lim t µ ( f 1 ([t, ]) ) = 0. Indeed, if we define, for each t (0, ), the set A t = f 1 ([t, ]) A, then we have 0 tκ At f. This forces the functions tκ At, t (0, ) to be elementary integrable, and µ(a t ) Iµ +(f), t (0, ). t This forces lim t µ(a t ) = 0. The next result explains the fact that positive integrability is a decomposable property. Proposition 1.4. Let (, A, µ) be a measure space. uppose (A k ) n A is a pairwise disjoint finite sequence, with A 1 A n =. For a measurable function f : [0, ], the following are equivalent. (i) f L 1 +(, A, µ); (ii) fκ Ak L 1 +(, A, µ), k = 1,..., n. Moreover, if f satisfies these equivalent conditions, one has I +(f) µ = I +(fκ µ Ak ). Proof. The implication (i) (ii) is trivial, since we have 0 fκ Ak f, so we can apply Proposition 1.3. To prove the implication (ii) (i), start by assuming that f satisfies condition (ii). We first observe that every elementary function h A-Elem R (), with 0 h f, has the properties: (a) h L 1 R,elem (, A, µ); (b) I µ elem (h) n Iµ +(fκ Ak ). This is immediate from the fact that we have the equality h = n hκ A k, and all function hκ Ak are elementary, and satisfy 0 hκ Ak fκ Ak, and then everything follows from Theorem 1.1 and the definition of the positive integral which gives I µ elem (hκ A k ) I +(fκ µ Ak ). Of course, the properties (a) and (b) above prove that f L 1 +(, A, µ), as well as the inequality I +(f) µ I +(fκ µ Ak ). To prove that we have in fact equality, we start with some ε > 0, and we choose, for each k {1,..., n}, a function h k L 1 R,elem (, A, µ), such that 0 h k fκ Ak, and I µ elem (h k) I +(fκ µ Ak ) ε n. By Theorem 1.1, the function h = h 1 + + h n belongs to L 1 R,elem (, A, µ), and has n (8) I µ elem (h) = I µ elem (h k) ( n I +(fκ µ Ak ) ) ε.

1. Construction of the integral 307 We obviously have h = h k fκ Ak = f, so we get I µ elem (h) Iµ +(f), thus the inequality (8) gives I +(f) µ ( n I +(fκ µ Ak ) ) ε. ince this inequality holds for all ε > 0, we get I +(f) µ n Iµ +(fκ Ak ), and we are done. Remark 1.4. Let (, A, µ) be a measure space, and let A. Define A = {A : A A} = {A A : A }, so that A A is a σ-algebra on. The restriction of µ to A will be denoted by µ. With these notations, (, A, µ ) is a measure space. It is not hard to see that for a measurable function f : [0, ], the conditions fκ L 1 +(, A, µ), f L 1 +(, A, µ ) are equivalent. Moreover, in this case one has the equality I µ +(fκ ) = I µ + (f ). This is a consequence of the fact that these two conditions are equivalent if f is elementary, combined with the fact that the restriction map h h establishes a bijection between the sets { h A-ElemR () : 0 h fκ }, { k A -Elem R () : 0 k f }. The following is an analogue of Theorem 1.1 for the positive integral. Theorem 1.2. Let (, A, µ) be a measure space. (i) If f L 1 +(, A, µ), and α [0, ), then 1 αf L 1 +(, A, µ), and one has the equality I µ +(αf) = αi µ +(f). (ii) If f 1, f 2 L 1 +(, A, µ), then f 1 + f 2 L 1 +(, A, µ), and one has the equality I µ +(f 1 + f 2 ) = I µ +(f 1 ) + I µ +(f 2 ). Proof. (i). This part is obvious. (ii). Fix f 1, f 2 L 1 +(, A, µ), and define f = f 1 + f 2. Obviously, f is measurable, and with values in [0, ). In order to prove that f is integrable, we define the sets A = {x : f 1 (x) f 2 (x)} and B = A. On the one hand, we clearly have 0 fκ A 2f 2 κ A 2f 2, with 2f 2 L 1 +(, A, µ), by part (i), so by Proposition 1.3 it follows that fκ A L 1 +(, A, µ). On the other hand, we have fκ B 2f 1 κ B 2f 1, and again, as above, it follows that fκ B L 1 +(, A, µ). Now we have both fκ A and fκ B in L 1 +(, A, µ), with A, B A disjoint such that A B =, so by 1 Here we use the convention that when α = 0, we take αf = 0.

308 CHAPTER IV: INTEGRATION THEORY Proposition 1.4, it follows that f indeed belongs to L 1 +(, A, µ). We still have to prove the equality (9) I µ +(f) = I µ +(f 1 ) + I µ +(f 2 ). To prove this equality, we fix for the moment some integer n 1, and we define the sets A 0 = f1 1 ({0}) and A k = { k 1 x : n f(x) < f 1(x) k n f(x)}, k = 1,..., n. ince 0 f 1 f, we have the equality A 0 A 1 A n =. It is also clear that the sets (A k ) n k=0 are pairwise disjoint. By construction we have f 1κ A0 = 0, and the inequalities k 1 (10) n fκ A k f 1 κ Ak k n fκ A k, k {1,..., n}. The function f 2 = f f 1 will then satisfy f 2 κ A0 = fκ A0, and the inequalities n k (11) n fκ A k f 2 κ Ak n k + 1 fκ n Ak, k {1,..., n}. We apply now I µ + to both sides of (10) and (11), and using part (i) and Proposition 1.3 we get k 1 n Iµ +(fκ Ak ) I +(f µ 1 κ Ak ) k n Iµ +(fκ Ak ), n k n Iµ +(fκ Ak ) I +(f µ 2 κ Ak ) n k + 1 I µ n +(fκ Ak ), so if we add up we get n 1 (12) n Iµ +(fκ Ak ) I +(f µ 1 κ Ak ) + I +(f µ 2 κ Ak ) n + 1 n Iµ +(fκ Ak ), for all k {1,..., n}. ince we also have the equality I µ +(f 1 κ A0 ) + I µ +(f 2 κ A0 ) = I µ +(fκ A0 ), when we add it and the inequalities (12), we get n 1 I µ n +(fκ Ak ) [I +(f µ 1 κ Ak ) + I +(f µ 2 κ Ak )] n + 1 n k=0 k=0 I +(fκ µ Ak ), which by Proposition 1.4 reads n 1 (13) n Iµ +(f) I +(f µ 1 ) + I +(f µ 2 )] n + 1 n Iµ +(f). ince the inequalities (13) hold for all integers n 1, they will clearly force the equality (9). Definitions. Let (, A, µ) be a measure space. Denote the extended real line [, ] by R. A measurable function f : R is said to be µ-integrable, if there exist functions f 1, f 2 L 1 +(, A, µ), such that (14) f(x) = f 1 (x) f 2 (x), x [ f1 1 1 ({ }) f2 ({ })]. By Remark 1.3, we know that the sets f 1 k ({ }), k = 1, 2, have measure zero. The equality (14) gives then the fact f = f 1 f 2, µ-a.e. We define k=0 L 1 R(, A, µ) = { f : R : f µ-integrable }.

1. Construction of the integral 309 We also define the space of honest real-valued µ-integrable functions, as L 1 R(, A, µ) = { f L 1 R(, A, µ) : < f(x) <, x }. Finally, we define the space of complex-valued µ-integrable functions as L 1 C(, A, µ) = { f : C : Re f, Im f L 1 R(, A, µ) }. The next result collects the basic properties of L 1 R. states that it is an almost vector space. Among other things, it Theorem 1.3. Let (, A, µ) be a measure space. (i) For a measurable function f : R, the following are equivalent: (a) f L 1 R(, A, µ); (b) f L 1 +(, A, µ). (ii) If f, g L 1 R(, A, µ), and if h : R is a measurable function, such that h(x) = f(x) + g(x), x [ f 1 ({, }) g 1 ({, }) ], then h L 1 R(, A, µ). (iii) If f L 1 R(, A, µ), and α R, and if g : R is a measurable function, such that then g L 1 R(, A, µ). (iv) One has the inclusion g(x) = αf(x), x f 1 ({, }), L 1 R,elem(, A, µ) L 1 +(, A, µ) L 1 R(, A, µ). Proof. (i). Consider the functions measurable functions f ± : [0, ] defined as f + = max{f, 0} and f = max{ f, 0}. To prove the implication (a) (b), assume f L 1 R(, A, µ), which means there exist f 1, f 2 L 1 +(, A, µ), such that f(x) = f 1 (x) f 2 (x), x [ f1 1 1 ({ }) f2 ({ })]. Notice that we have the inequalities (15) (16) f + f 1, µ-a.e., f f 2, µ-a.e.. Indeed, if we put N = f1 1 1 ({ }) f2 ({ }), then µ(n) = 0, and if we start with some x N, we either have f 1 (x) f 2 (x) 0, in which case we get f + (x) = f(x) = f 1 (x) f 2 (x) f 1 (x), f (x) = 0 f 2 (x), or we have f 1 (x) f 2 (x), in which case we get In other words, we have f + (x) = 0 f 1 (x), f (x) = f(x) = f 2 (x) f 1 (x) f 2 (x). f + (x) f 1 (x) and f (x) f 2 (x), x N,

310 CHAPTER IV: INTEGRATION THEORY so we indeed get (15) and (16). Using these inequalities, and Proposition 1.3, it follows that f ± L 1 +(, A, µ), so by Theorem 1.2, it follows that f + + f = f also belongs to L 1 +(, A, µ). To prove the implication (b) (a), start by assuming that f L 1 +(, A, µ). Then, since we obviously have the inequalities 0 f ± f, again by Proposition 1.3, it follows that f ± L 1 +(, A, µ). ince we obviously have f(x) = f + (x) f (x), x f 1 ({, }), it follows that f indeed belongs to f ± L 1 R(, A, µ). (ii). Assume f, g, and h are as in (ii). By (i), both functions f and g are in L 1 +(, A, µ). By Theorem 1.2, it follows that the function k = f + g also belongs to L 1 +(, A, µ). Notice that we have the equality so the hypothesis on h reads which then gives f 1 ({, }) g 1 ({, }) = k 1 ({ }), h(x) = f(x) + g(x), x k 1 ({ }), h(x) = f(x) + g(x) f(x) + g(x), x k 1 ({ }). Of course, since µ ( k 1 ({ }) ) = 0, this gives h k, µ-a.e., and using (i) it follows that h indeed belongs to L 1 R(, A, µ). (iii). Assume f, α, and g are as in (iii). Exactly as above, we have g = α f, µ-a.e., and then by Theorem 1.2 it follows that g L 1 +(, A, µ). (iv). The inclusion L 1 +(, A, µ) L 1 R(, A, µ) is trivial. To prove the inclusion L 1 R,elem (, A, µ) L1 R(, A, µ), we use parts (ii) and (iii) to reduce this to the fact that κ A L 1 R(, A, µ), for all A A, with µ(a) <. But this fact is now obvious, because any such function belongs to L 1 +(, A, µ) L 1 R(, A, µ). Corollary 1.1. Let (, A, µ) be a measure space, and let K be one of the fields R or C. (i) For a K-valued measurable function f : K, the following are equivalent: (a) f L 1 K (, A, µ); (b) f L 1 +(, A, µ). (ii) When equipped with the pointwise addition and scalar multiplication, the space L 1 K (, A, µ) becomes a K-vector space. Proof. (i). The case K = R is immediate from Theorem 1.3 In the case when K = C, we use the obvious inequalities (17) max { Re f, Im f } f Re f + Im f. If f L 1 C (, A, µ), then both Re f and Im f belong to L1 R (, A, µ), so by Theorem 1.3, both Re f and Im f belong to L 1 +(, A, µ). By Theorem 1.2, the function g = Re f + Im f belongs to L 1 +(, A, µ), and then using the second inequality in (17), it follows that f belongs to L 1 +(, A, µ).

1. Construction of the integral 311 Conversely, if f belongs to L 1 +(, A, µ), then using the first inequality in (17), it follows that both Re f and Im f belong to L 1 +(, A, µ), so by Theorem 1.3, both Re f and Im f belong to L 1 R (, A, µ), i.e. f belongs to L1 C (, A, µ). (ii). This part is pretty clear. If f, g L 1 K (, A, µ), then by (i) both f and g belong to L 1 +(, A, µ), and by Theorem 1.2, the function f + g will also belong to L 1 +(, A, µ). ince f + g f + g, it follows that f + g itself belongs to L 1 +(, A, µ), so using (i) again, it follows that f + g indeed belongs to L 1 K (, A, µ). If f L1 K (, A, µ) and α K, then f belongs to L1 +(, A, µ), so αf = α f again belongs to L 1 +(, A, µ), which by (i) gives the fact that αf belongs to L 1 K (, A, µ). (18) (19) Remark 1.5. Let (, A, µ) be a measure space. Then one has the equalities L 1 +(, A, µ) = { f L 1 R(, A, µ) : f() [0, ] } ; L 1 K,elem(, A, µ) = L 1 K(, A, µ) A-Elem K (). Indeed, by Theorem 1.3 that we have the inclusion L 1 +(, A, µ) { f L 1 R(, A, µ) : f() [0, ] }. The inclusion in the other direction follows again from Theorem 1.3, since any function that belongs to the right hand side of (18) satisfies f = f. The inclusion L 1 K,elem(, A, µ) L 1 K(, A, µ) A-Elem K () is again contained in Theorem 1.3. To prove the inclusion in the other direction, it suffices to consider the case K = R. tart with h L 1 R (, A, µ) A-Elem R(), which gives h L 1 +(, A, µ). The function h is obviously in A-Elem R (), so we get h L 1 R,elem (, A, µ). ince L1 R,elem (, A, µ) is a vector space, it will also contain the function h. The fact that h itself belongs to L 1 R,elem (, A, µ) then follows from Proposition 1.1, combined with the obvious inequalities h h h. The following result deals with the construction of the integral. Theorem 1.4. Let (, A, µ) be a measure space. There exists a unique map I µ R(, A, µ) R, with the following properties: (i) Whenever f, g, h L 1 R(, A, µ) are such that h(x) = f(x) + g(x), x [ f 1 ({, }) g 1 ({, }) ], it follows that I µ R(h) = I µ R(f) + I µ R(g). (ii) Whenever f, g L 1 R(, A, µ) and α R are such that g(x) = αf(x), x f 1 ({, }), it follows that I µ R(g) = αi µ R(f). (iii) I µ R(f) = I µ +(f), f L 1 +(, A, µ). Proof. Let us first show the existence. tart with some f L 1 R(, A, µ), and define the functions f ± : [0, ] by f + = max{f, 0} and f = max{ f, 0} so that f = f + f, and f +, f L 1 +(, A, µ). We then define I µ R(f) = I µ +(f + ) I µ +(f ). It is obvious that I µ R satisfies condition (iii).

312 CHAPTER IV: INTEGRATION THEORY The key fact that we need is contained in the following. Claim: Whenever f L 1 R(, A, µ), and f 1, f 2 L 1 +(, A, µ) are such that f(x) = f 1 (x) f 2 (x), x [ f1 1 1 ({ }) f2 ({ })], it follows that we have the equality I µ R(f) = I µ +(f 1 ) I µ +(f 2 ). Indeed, since we have f = f + f, it follows immediately that we have the equality f 2 (x) + f + (x) = f 1 (x) + f (x), x [ f1 1 1 ({ }) f2 ({ })], which gives f 2 + f + = f 1 + f, µ-a.e. By Theorem 1.2, this immediately gives which then gives I µ +(f 2 ) + I µ +(f + ) = I µ +(f 1 ) + I µ +(f ), I µ +(f 1 ) I µ +(f 2 ) = I µ +(f + ) I µ +(f ) = I µ R(f). Having proven the above Claim, let us show now that I µ R has properties (i) and (ii). Assume f, g and h are as in (i). Notice that if we define h 1 = f + + g + and h 2 = f + g, then we clearly have 0 h 1 f + g and 0 h 2 f + g, so h 1 and h 2 both belng to L 1 +(, A, µ). By Theorem 1.2, we then have (20) I µ +(h 1 ) = I µ +(f + ) + I µ +(g + ) and I µ +(h 2 ) = I µ +(f ) + I µ +(g ). Notice also that, because of the equalities h 1 1 ({ }) = f 1 ({ } g 1 ({ }) and h2 1 ({ }) = f 1 ({ } g 1 ({ }), we have h = h 1 (x) h 2 (x), x [ h 1 1 ({ }) h 1 2 ({ })], so by the above Claim, combined with (20), we get I µ R(h) = I +(h µ 1 ) I +(h µ 2 ) = I +(f µ + ) + I +(g µ + ) I +(f µ ) I +(g µ ) = I µ R(f) + I µ R(g). Property (ii) is pretty obvious. The uniqueness is also obvious. If we start with a map J : L 1 R(, A, µ) R with properties (i)-(iii), then for every f L 1 R(, A, µ), we must have J(f) = J(f + ) J(f ) = I µ +(f + ) I µ +(f ). (For the second equality we use condition (iii), combined with the fact that both f + and f belong to L 1 +(, A, µ).) Corollary 1.2. Let (, A, µ) be a measure space, and let K be either R or C. There exists a unique linear map I µ K (, A, µ) K, such that I µ K (f) = Iµ +(f), f L 1 +(, A, µ) L 1 K(, A, µ). Proof. Let us start with the case K = R. In this case, we have the inclusion L 1 R(, A, µ) L 1 R(, A, µ), so we can define I µ R as the restriction of Iµ R to L 1 R (, A, µ). The uniqueness is again clear, because of the equalities I µ R (f) = Iµ R (f + ) I µ R (f ) = I µ +(f + ) I µ +(f ).

1. Construction of the integral 313 In the case K = C, we define I µ C (f) = Iµ R (Re f) + iiµ R (Im f). The linearity is obvious. The uniqueness is also clear, because the restriction of I µ C to L 1 R (, A, µ) must agree with Iµ R. Definition. Let (, A, µ) be a measure space, and let K be one of the symbols R, R, or C. For any f L 1 K (, A, µ), the number Iµ K (f) (which is real, if K = R or R, and is complex if K = C) will be denoted by f dµ, and is called the µ-integral of f. This notation is unambiguous, because if f L 1 R (, A, µ), then we have Iµ R(f) = I µ C (f) = Iµ R (f). Remark 1.6. If (, A, µ) is a measure space, then for every A A, with µ(a) <, using the above Corollary, we get κ A dµ = I +(κ µ A ) = µ(a). By linearity, if K = R, C, one has then the equality h dµ = I µ elem (h), h L1 K,elem(, A, µ). To make the exposition a bit easier, it will adopt the following. Convention. If (, A, µ) is a measure space, and if f : [0, ] is a measurable function, which does not belong to L 1 +(, A, µ), then we define f dµ =. Remarks 1.7. Let (, A, µ) be a measure space. A. Using the above convention, when h A-Elem R () is a function with h() [0, ), the condition h dµ = is equivalent to the existence of some α h() {0}, with µ ( h 1 ({α}) ) =. B. Using the above convention, for every measurable function f : [0, ], one has the equality { } f dµ = sup h dµ : h A-Elem R (), 0 h f. C. If f, g : [0, ] are measurable, then one has the equalities (f + g) dµ = f dµ + g dµ, (αf) dµ = α f dµ, α [0, ), even in the case when some term is infinite. (We use the convention + t =, t [0, ], as well as α =, α (0, ), and 0 = 0.) D. If f, g : [0, ] are measurable, and f g, µ-a.e., then (using B) one has the inequality f dµ g dµ,

314 CHAPTER IV: INTEGRATION THEORY even if one side (or both) is infinite. E. Let K be one of the symbols R, R, or C, and let f : K be a measurable function. Then the function f : [0, ] is measurable. Using the above convention, the condition that f belongs to L 1 K (, A, µ) is equivalent to the inequality f dµ <. In the remainder of this section we discuss several properties of integration that are analoguous to those of the positive/elementary integration. We begin with a useful estimate Proposition 1.5. Let (, A, µ) be a measure space, and let K be one of the symbols R, R, or C. For every function f L 1 K (, A, µ), one has the inequality f dµ f dµ. Proof. Let us first examine the case when K = R, R. In this case we define f + = max{f, 0} and f = max{ f, 0}, so we have f = f + f, as well as f = f + + f. Using the inequalities I +(f µ ± ) 0, we have f dµ = I +(f µ + ) I +(f µ ) I +(f µ + ) + I +(f µ ) = f dµ; f dµ = I +(f µ + ) + I +(f µ ) I +(f µ + ) + I +(f µ ) = f dµ. In other words, we have ± f dµ f dµ, and the desired inequality immediately follows. Let us consider now the case K = C. Consider the number λ = f dµ, and let us choose some complex number α C, with α = 1, and αλ = λ. (If λ 0, we take α = λ 1 λ ; otherwise we take α = 1.) Consider the measurable function g = αf. Notice now that ( ) ( ) Re g dµ + i Im g dµ = g dµ = α f dµ = αλ = λ 0, so in particular we get λ = If we apply the real case, we then get (21) λ Re g dµ. Re g dµ. Notice now that, we have the inequality Re g g = f, which gives I +( µ ) Re g I µ ( f ) = f dµ, so the inequality (21) immediately gives f dµ = λ f dµ.

1. Construction of the integral 315 Corollary 1.3. Let (, A, µ) be a measure space, and let K be one of the symbols R, R, or C. If a measurable function f : K satisfies f = 0, µ-a.e, then f L 1 K (, A, µ), and f dµ = 0. Proof. Consider the measurable function f : [0, ], which satisfies f = 0, µ-a.e. By Proposition 1.3, it follows that f L 1 +(, A, µ), hence f L 1 K (, A, µ), and f dµ = 0. Of course, the last equality forces f dµ = 0. Corollary 1.4. Let K be either R or C. If (, A, µ) is a finite measure space, then every bounded measurable function f : K belongs to L 1 K (, A, µ), and satisfies f dµ µ() sup f(x). x Proof. If we put β = sup x f(x), then we clearly have f βκ, which shows that f L 1 +(, A, µ), and also gives f dµ βκ dµ = µ() β. Then everything follows from Proposition 1.6. Comment. The introduction of the space L 1 R(, A, µ), of extended real-valued µ-integrable functions, is useful mostly for technical reasons. In effect, everything can be reduced to the case when only honest real-valued functions are involved. The following result clarifies this matter. Lemma 1.1. Let (, A, µ) be a measure space, and let f : R be a measurable function. The following ar equivalent (i) f L 1 R(, A, µ); (ii) there exists g L 1 R (, A, µ), such that g = f, µ-a.e. Moreover, if f satisfies these equivalent conditions, then any function g, satisfying (ii), also has the property f dµ = g dµ. Proof. Consider the set F = {x : < f(x) < }, which belongs to A. We obviously have the equality F = f 1 ({ }). (i) (ii). Assume f L 1 R(, A, µ), which means that f L 1 +(, A, µ). In particular, we get µ( F ) = 0. Define the measurable function g = fκ F. On the one hand, it is clear, by construction, that we have < g(x) <, x. On the other hand, it is clear that g F = f F, so using µ( F ) = 0, we get the fact that f = g, µ-a.e. Finally, the inequality 0 g f, combined with Proposition 1.3, gives g L 1 +(, A, µ), so g indeed belongs to L 1 R (, A, µ). (ii) (i). uppose there exists g L 1 R (, A, µ), with f = g, µ-a.e., and let us prove that (a) f L 1 R(, A, µ); (b) f dµ = g dµ. The first assertion is clear, because by Proposition 1.3, the equality f = g, µ-a.e., combined with g L 1 +(, A, µ), forces f L 1 +(, A, µ), i.e. f L 1 R(, A, µ). To prove (b), we consider the difference h = f g, which is a measurable function h : R, and satisfies h = 0, µ-a.e. By Corollary 1.3, we know that h L 1 R(, A, µ), and h dµ = 0. By Theorem 1.3, we get f dµ = g dµ + h dµ = g dµ. x

316 CHAPTER IV: INTEGRATION THEORY The following result is an analogue of Proposition 1.1 (see also Proposition 1.3). Proposition 1.6. Let (, A, µ), and let f 1, f 2 L 1 R(, A, µ). uppose f : R is a measurable function, such that f 1 f f 2, µ-a.e. Then f L 1 R(, A, µ), and one has the inequality f 1 dµ f dµ f 2 dµ. Proof. First of all, since f 1 and f 2 belong to L 1 R(, A, µ), it follows that f 1 and f 2, hence also f 1 + f 2, belong tp L 1 +(, A, µ). econd, since we have f 2 f 2 f 1 + f 2 and f 1 f 1 f 1 f 2 (everyhwere!), the inequalities f 1 f f 2, µ-a.e., give which reads f 1 f 2 f f 1 + f 2, µ-a.e., f f 1 + f 2, µ-a.e. ince f 1 + f 2 L 1 +(, A, µ), by Proposition 1.3., we get f L 1 +(, A, µ), so f indeed belongs to L 1 R(, A, µ). To prove the inequality for integrals, we use Lemma 1.2, to find functions g 1, g 2, g L 1 R (, A, µ), such that f 1 = g 1, µ-a.e., f 2 = g 2, µ-a.e., and f = g, µ-a.e. Lemma 1.2 also gives the equalities f 1 dµ = g 1 dµ, f 2 dµ = g 2 dµ, and f dµ = g dµ, so what we need to prove are the inequalities (22) g 1 dµ g dµ g 2 dµ. Of course, we have g 1 g g 2, µ-a.e. To prove the first inequality in (22), we consider the function h = g g 1 L 1 R (, A, µ), and we prove that h dµ 0. But this is quite clear, because we have h 0, µ-a.e., which means that h = h, µ-a.e., so by Lemma 1.2, we get h dµ = h dµ = I +( h ) µ 0. The second inequality in (22) is prove the exact same way. The next result is an analogue of Proposition 1.4. Proposition 1.7. Let (, A, µ) be a measure space, and let K be one of the symbols R, R, or C. uppose (A k ) n A is a pairwise disjoint finite sequence, with A 1 A n =. For a measurable function f : K, the following are equivalent. (i) f L 1 K (, A, µ); (ii) fκ Ak L 1 K (, A, µ), k = 1,..., n. Moreover, if f satisfies these equivalent conditions, one has (23) f dµ = fκ Ak dµ.

1. Construction of the integral 317 Proof. It is fairly obvious that fκ Ak = f κ Ak. Then the equivalence (i) (ii) follows from Proposition 1.4 applied to the function f : [0, ]. In the cases when K = R, C, the equality (23) follows immediately from linearity, and the obvious equality f = n fκ A k. In the case when K = R, we take g L 1 R (, A, µ), such that f = g, µ-a.e. Then we obviously have fκ A k = gκ Ak, µ-a.e., for all k = 1,..., n, and the equality (23) follows from the corresponding equality that holds for g. Remark 1.8. The equality (23) also holds for arbitrary measurable functions f : [0, ], if we use the convention that preceded Remarks 1.7. This is an immediate consequence of Proposition 1.4, because the left hand side is infinite, if an only if one of the terms in the right hand side is infinite. The following is an obvious extension of Remark 1.4. Remark 1.9. Let K be one of the symbols R, R, or C, let (, A, µ) be a measure space. For a set A, and a measurable function f : K, one has the equivalence fκ L 1 K(, A, µ) f L 1 K(, A, µ ). If this is the case, one has the equality (24) fκ dµ = f dµ. The above equality also holds for arbitrary measurable functions f : [0, ], again using the convention that preceded Remarks 1.7. Notation. The above remark states that, whenver the quantities in (24) are defined, they are equal. (This only requires the fact that f is measurable, and either f L 1 K(, A, µ ), or f() [0, ].) In this case, the equal quantities in (24) will be simply denoted by f dµ. Exercise 1. Let I be some non-empty set. Consider the σ-algebra P(I), of all subsets of I, equipped with the counting measure { Card A if A is finite µ(a) = if A is infinite Prove that L 1 R(I, P(I), µ) = L 1 R (I, P(I), µ). Prove that, if K is either R or C, then L 1 K(I, P(I), µ) = l 1 K(I), the Banach space discussed in II.2 and II.3. Exercise 2. There is an instance when the entire theory developped here is essentially vacuous. Let be a non-empty set, and let A be a σ-algebra on. For a measure µ on A, prove that the following are equivalent (i) L 1 +(, A, µ) = { f : [0, ] : f measurable, and f = 0, µ-a.e. } ; (ii) for every A A, one has µ(a) {0, }. A measure space (, A, µ), with property (ii), is said to be degenerate. Exercise 3. Let (, A, µ) be a measure space, and let f : [0, ] be a measurable function, with f dµ = 0. Prove that f = 0, µ-a.e. Hint: Define the measurable sets A n = {x : f(x) 1 }, and analyze the relationship n between f and κ An.