Chapter IV Integration Theory

Transcription

1 Chapter IV Itegratio Theory

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3 Lectures Costructio of the itegral I this sectio we costruct the abstract itegral. As a matter of termiology, we defie a measure space as beig a triple (, A, µ), where is some (o-empty) set, A is a σ-algebra o, ad µ is a measure o A. The measure space (, A, µ) is said to be fiite, if If µ() <. Defiitio. Let (, A, µ) be a measure space, ad let K be oe of the fields R or R. A K-valued elemetary µ-itegrable fuctio o (, A, µ) is a fuctio f : K, with the followig properties the rage f() of f is a fiite set; f 1 ({α}) A, ad µ ( f 1 ({α}) ) <, for all α f() {0}. We deote by L 1 K,elem (, A, µ) the collectio of all such fuctios. Remarks 1.1. Let (, A, µ) be a measure space. A. Every K-valued elemetary µ-itegrable fuctio f o (, A), µ) is measurable, as a map f : (, A) ( K, Bor(K ). I fact, ay such f ca be writte as f = α 1 κ A1 + + α κ A, with α k K, A k A ad µ(a k ) <, k = 1,...,. Usig the otatios from III.1, we have the iclusio L 1 K,elem(, A, µ) A-Elem K (). B. If we cosider the collectio R = {A A : µ(a) < }, the R is a rig, ad, we have the equality L 1 K,elem(, A, µ) = R-Elem K (). I particular, it follows that L 1 K,elem (, A, µ) is a K-vector space. The followig result is the first step i the costructio of the itegral. Theorem 1.1. Let (, A, µ) be a measure space, ad let K be oe of the fields R or C. The there exists a uique K-liear 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) <. Proof. For every f L 1 K,elem (, A, µ), we defie I µ elem (f) = α µ ( f 1 ({α}) ), α f() {0} 291

4 292 LECTURE with the covetio that, whe f() = {0} (which is the same as f = 0), we defie I µ elem (f) = 0. It is obvious that Iµ elem satsifies the equality (1) for all A A with µ(a) <. Oe key feature we are goig to use is the followig. Claim 1: Wheever we have a fiite pairwise disjoit sequece (A k ) A, with µ(a k ) <, k = 1,...,, oe has the equality I µ elem (α 1κ A1 + + α κ A ) = α 1 µ(a 1 ) + + α µ(a ), α 1,..., α K. It is obvious that we ca assume α j 0, j = 1,...,. To prove the above equality, we cosider the elemetary µ-itegrable fuctio f = α 1 κ A1 + +α κ A, ad we observe that f() {0} = {α 1 } {α }. 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 defie the set J k = { j {1,..., } : α j = β k }. It is obvious that the sets (J k ) p are pairwise disjoit, ad we have J 1 J p = {1,..., }. Moreover, for each k {1,..., p}, oe 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 defiitio of I µ elem we the 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), ad every A A with µ(a) <, oe has the equality (2) I µ elem (f + ακ A ) = Iµ elem (f) + αµ(a), α K. Write f = α 1 κ A1 + + α κ A, with (A j ) j=1 A pairwise disjoit, ad µ(a j) <, j = 1,...,. I order to prove (2), we are goig to write the fuctio f + ακ A i a similar way, ad we are goig to apply Claim 1. Cosider the sets B 1, B 2,..., B 2, B 2+1 A defied by B 2+1 = A (A 1 A ), ad B 2k 1 = A k A, B 2k = A k A, k = 1,...,. It is obvious that the sets (B p ) p=1 2+1 are pairwise disjoit. Moreover, oe has the equalities (3) B 2k 1 B 2k = A k, k {1,..., }, as well as the equality (4) A = +1 B 2k 1. Usig these equalities, ow we have f + ακ A = 2+1 p=1 β pκ Bp, where β 2+1 = α, ad β 2k = α k ad β 2k 1 = α k + α, k {1,..., }. Usig these equalities, j=1

5 CHAPTER IV: INTEGRATION THEORY 293 combied with Claim 1, ad (3) ad (4), we ow get 2+1 I µ elem (f + ακ A ) = = αµ(b 2+1 ) + p=1 β p µ(b p ) = [ (αk + α)µ(b 2k 1 ) + α k µ(b 2k ) ] = [ +1 ] = α µ(b 2k 1 ) = αµ ( +1 = αµ(a) + B 2k 1 ) ) + [ + 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), ad the Claim is prove. We ow prove that I µ elem is liear. The equality I µ elem (f + g) = Iµ elem (f) + Iµ elem (g), f, g L1 K,elem(, A, µ) follows from Claim 2, usig a obvious iductive argumet. The equality I µ elem (αf) = αiµ elem (f), α K, f L1 K,elem(, A, µ). is also pretty obvious, from the defiitio. The uiqueess is also clear. Defiitio. With the otatios above, the liear map is called the elemetary µ-itegral. I µ elem : L1 K,elem(, A, µ) K I what follows we are goig to ecouter also situatios whe certai relatios amog measurable fuctios hold almost everywhere. We are goig to use the followig. Covetio. Let T be oe of the spaces [, ] or C, ad let r be some relatio o T (i our case r will be either =, or, or, o [, ]). Give a measurable space (, A, µ), ad two measurable fuctios 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) } belogs to A, ad it has µ-ull complemet i, i.e. µ( A) = 0. (If r is oe of the relatios listed above, the set A automatically belogs to A.) The abreviatio µ-a.e. stads 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. The f L 1 K,elem (, A, µ), ad Iµ elem (f) = 0. Ideed, if we defie the set N = {x : f(x) 0},

6 294 LECTURE the N A ad µ(n) = 0. ice f 1 ({α}) N, α f() {0}, it follows that µ ( f 1 ({α}) ) = 0, α f() {0}, ad the by the defiitio of the elemetary µ-itegral, we get I µ elem (f) = 0. Oe useful property of elemetary itegrable fuctios is the followig. Propositio 1.1. Let (, A, µ) be a measure space, let f, g L 1 R,elem (, A, µ), ad let h A-Elem R () be such that The h L 1 R (, A, µ), ad f h g, µ-a.e. (5) I µ elem (f) Iµ elem (h) Iµ elem (h). Proof. Cosider the sets A = {x : f(x) > h(x)} ad B = {x : h(x) > g(x)}, which both belog to A, ad have µ(a) = µ(b) = 0. The set M = A B also belogs to A ad has µ(m) = 0. Defie the fuctios f 0 = f(1 κ M ), g 0 = g(1 κ M ), ad h 0 = h(1 κ M ). It is clear that f 0, g 0, ad h 0 are all i A-Elem R (). Moreover, we have the equalities f 0 = f, µ-a.e., g 0 = g, µ-a.e., ad h 0 = h, µ-a.e., so by Remark??, combied with Theorem 1.1, the fuctios f 0 = f + (f 0 f) ad g 0 = (g 0 g) + g both belog to L 1 R (, A, µ), ad we have the equalities (6) I µ elem (f 0) = I µ elem (f) ad Iµ elem (g 0) = I µ elem (g). Notice ow that we have the (absolute) iequality f 0 h 0 g 0. Let us show that h 0 is elemetary itegrable. tart with some α h 0 () {0}. If α > 0, the, usig the iequality 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, the, usig the iequality h 0 f 0, we get h 1 0 ( ) 1 ({α}) f0 (, 0) f0 1 ({λ}), λ f 0() {0} which proves agai that µ ( h 1 0 ({α})) <. Havig show that h 0 is elemetary itegrable, we ow compare the umbers I µ elem (f), Iµ elem (h 0), ad I µ (g). Defie the fuctios f 1 = h 0 f 0, ad g 1 = g 0 h 0. By Theorem 1.1, we kow that f 1, g 1 L 1 R,elem (, A, µ). ice f 1, g 1 0, we have f 1 (), g 1 () [0, ), so it follows immediately that I µ elem (f 1) 0 ad I µ elem (g 1) 0. Now, agai usig Theorem 1.1, ad (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). ice h = h 0, µ-a.e., by the above Remark it follows that h L 1 R,elem (, A, µ), ad I µ elem (h) = Iµ elem (h 0), so the desired iequality (5) follows immediately. We ow defie aother type of itegral.

7 CHAPTER IV: INTEGRATION THEORY 295 Defiitio. Let (, A, µ) be a measure space. A measurable fuctio f : [0, ] is said to be µ-itegrable, if (a) every h A-Elem R (), with 0 h f, is elemetary µ-itegrable; (b) sup { I µ elem (h) : h A-Elem R(), 0 h f } <. If this is the case, the above supremum is deoted by I µ +(f). The space of all such fuctios is deoted by L 1 +(, A, µ). The map is called the positive µ-itegral. I µ + : L 1 +(, A, µ) [0, ) The first (legitimate) questio is whether there is a overlap betwee the two defiitios. This is awered by the followig. Propositio 1.2. Let (, A, µ) be a measure space, ad let f A-Elem R () be a fuctio with f 0. The followig are equivalet (i) f L 1 +(, A, µ); (ii) f L 1 R,elem (, A, µ). Moreover, if f is as above, the I µ elem (f) = Iµ +(f). Proof. The implicatio (i) (ii) is trivial. To prove the implicatio (ii) (i) we start with a arbitrary elemetary h A-Elem R (), with 0 h f. Usig Propositio 1.1, we clearly get (a) h L 1 R,elem (, A, µ); (b) I µ elem (h) Iµ elem (f). Usig 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 ow examie properties of the positive itegral, which are similar to those of the elemetary itegral. The followig is a aalogue of Propositio 1.1. Propositio 1.3. Let (, A, µ) be a measure space, let f L 1 +(, A, µ), ad let g : [0, ] be a measurable fuctio, such that g f, µ-a.e., the g L 1 +(, A, µ), ad I µ +(g) I µ +(f). Proof. tart with some elemetary fuctio h A-Elem R (), with 0 h g. Cosider the sets M = {x : h(x) > f(x)} ad N = {x : g(x) > f(x)}, which obviously belog to A. ice N N, ad µ(n) = 0, we have µ(m) = 0. If we defie the elemetary fuctio h 0 = h(1 κ M ), the we have h = h 0, µ-a.e., ad 0 h 0 f, so it follows that h 0 L 1 R,elem (, A, µ), ad Iµ elem (h 0) I µ (f). ice h = h 0, µ-a.e., by Propositio 1.1., it follows that h L 1 R,elem (, A, µ), ad I µ elem (h) = Iµ elem (h 0) I +(f). µ By defiitio, this gives g L 1 +(, A, µ) ad I +(g) µ I +(f). µ Remark 1.3. Let (, A, µ) be a measure space, ad let f L 1 +(, A, µ). Although f is allowed to take the value, it turs out that this is iessetial. More precisely oe has µ ( f 1 ({ }) ) = 0.

8 296 LECTURE This is i fact a cosequece of the equality (7) lim t µ ( f 1 ([t, ]) ) = 0. Ideed, if we defie, for each t (0, ), the set A t = f 1 ([t, ]) A, the we have 0 tκ At f. This forces the fuctios tκ At, t (0, ) to be elemetary itegrable, ad µ(a t ) Iµ +(f), t (0, ). t This forces lim t µ(a t ) = 0. The ext result explais the fact that positive itegrability is a decomposable property. Propositio 1.4. Let (, A, µ) be a measure space. uppose (A k ) A is a pairwise disjoit fiite sequece, with A 1 A =. For a measurable fuctio f : [0, ], the followig are equivalet. (i) f L 1 +(, A, µ); (ii) fκ Ak L 1 +(, A, µ), k = 1,...,. Moreover, if f satisfies these equivalet coditios, oe has I +(f) µ = I +(fκ µ Ak ). Proof. The implicatio (i) (ii) is trivial, sice we have 0 fκ Ak f, so we ca apply Propositio 1.3. To prove the implicatio (ii) (i), start by assumig that f satisfies coditio (ii). We first observe that every elemetary fuctio h A-Elem R (), with 0 h f, has the properties: (a) h L 1 R,elem (, A, µ); (b) I µ elem (h) Iµ +(fκ Ak ). This is immediate from the fact that we have the equality h = hκ A k, ad all fuctio hκ Ak are elemetary, ad satisfy 0 hκ Ak fκ Ak, ad the everythig follows from Theorem 1.1 ad the defiitio of the positive itegral which gives I µ elem (hκ A k ) I +(fκ µ Ak ). Of course, the properties (a) ad (b) above prove that f L 1 +(, A, µ), as well as the iequality I +(f) µ I +(fκ µ Ak ). To prove that we have i fact equality, we start with some ε > 0, ad we choose, for each k {1,..., }, a fuctio h k L 1 R,elem (, A, µ), such that 0 h k fκ Ak, ad I µ elem (h k) I +(fκ µ Ak ) ε. By Theorem 1.1, the fuctio h = h h belogs to L 1 R,elem (, A, µ), ad has (8) I µ elem (h) = I µ elem (h k) ( I +(fκ µ Ak ) ) ε. We obviously have h = h k fκ Ak = f,

9 CHAPTER IV: INTEGRATION THEORY 297 so we get I µ elem (h) Iµ (f), thus the iequality (8) gives I µ (f) ( I +(fκ µ Ak ) ) ε. ice this iequality holds for all ε > 0, we get I µ (f) Iµ +(fκ Ak ), ad we are doe. Remark 1.4. Let (, A, µ) be a measure space, ad let A. We ca A = {A : A A} = {A A : A }, so that A A is a σ-algebra o. The restrictio of µ to A will be deoted by µ. With these otatios, (, A, µ ) is a measure space. It is ot hard to see that for a measurable fuctio f : [0, ], the coditios fκ L 1 +(, A, µ), f L 1 +(, A, µ ) are equivalet. Moreover, i this case oe has the equality I µ +(fκ ) = I µ + (f ). This is a cosequece of the fact that these two coditios are equivalet if f is elemetary, combied with the fact that the restrictio map h h establishes a bijectio betwee the sets { h A-ElemR () : 0 h fκ }, { k A -Elem R () : 0 k f }. The ext result gives a alterative defiitio of the positive itegral, for fuctios that are domiated by elemetary itegrable oes. Propositio 1.5. Let (, A, µ) be a measure space, let f : [0, ] be a measurable fuctio. Assume there exists h 0 L 1 R,elem (, A, µ), with h 0 f. The f L 1 +(, A, µ), ad oe has the equality (9) I µ +(f) = if { I µ elem (h) : h L1 R,elem(, A, µ), h f }. Proof. ice h 0 0, by Propositio 1.2, we kow that h 0 L 1 +(, A, µ). The fact that f L 1 +(, A, µ) the follows from Propositio 1.3, combied with the iequality h 0 f. More geerally, agai by Propositios 1.2 ad 1.3, we kow that for ay h L 1 R,elem (, A, µ), with h f, we have h L1 +(, A, µ), as well as the iequality I µ +(f) I µ +(h) = I µ elem (h). o, if we deote the right had side of (9) by J(f), we have I +(f) µ J(f) I µ elem (h 0). We ow prove the other iequality I +(f) µ J(f). If h 0 = 0, there is othig to prove. Assume h 0 is ot idetically zero. Without ay loss of geerality, we ca assume that h 0 = βκ B, for some β (0, ) ad B A with µ(b) <. (If we defie B = h 1 0 ( (0, ) ) = α h 0() {0} h 1 0 ({α}), ad if we set β = max h 0(), the we clearly have µ(b) <, ad h 0 βκ B.) For every iteger 1, we defie the sets A 1,..., A A by A k = f 1( ( (k 1)β, kβ ]), k = 1,...,,

10 298 LECTURE ad we defie the elemetary fuctios (k 1)β g = κ A k ad h = kβ κ A k. The mai features of these costructios are collected i the followig. Claim: For every 1, the fuctios g ad h are elemetary itegrable, ad satisfy the iequalities 0 g f h h 0, as well as I µ elem (h ) I µ elem (g ) + βµ(b). To prove this fact, we fix 1, ad we first remark that the sets (A k ) are pairwise disjoit. ice 0 f h 0 = βκ B, we have A 1 A = f 1( (0, β] ) B. I particular, if we defie A = A 1 A B, we have h = β κ A k = βκ A βκ B. kβ κ A k Let us prove the iequalities g f h. tart with some arbitrary poit x, ad let us show that g (x) f(x) h (x). If f(x) = 0, there is othig to prove, because this forces κ A k (x) = 0, k = 1,...,. Assume ow f(x) > 0. ice f βκ B, we ow that f(x) (0, β], so there exists a uique k {1,..., }, such that (k 1)β < f(x) kβ, i.e. x A k. We the obviously have g (x) = (k 1)β κ A k (x) = (k 1)β < f(x) kβ = kβ κ A (x) = h (x), k ad we are doe. Fially, let us observe that sice g h h 0, it follows that g ad h are i L 1 +(, A, µ), so g ad h are elemetary itegrable. Notice that h g = β κ A k = β κ A β κ B, so we have I µ elem (h g ) I µ elem ( β κ B ) = βµ(b), so usig Theorem 1.1, we get I µ elem (h ) = I µ elem (g ) + I µ elem (h g ) I µ (g ) + βµ(b). Havig prove the Claim, we immediately see that by the defiitio of the positive itegral, we have J(f) I µ elem (h ) I µ (g ) + βµ(b) ice the iequality J(f) I +(f) µ + βµ(b) J(f) I +(f). µ I +(f) µ + βµ(b). holds for all 1, it will clearly force Our ext goal is to prove a aalogue of Theorem 1.1, for the positive itegral (Theorem 1.2 below). We discuss first a weaker versio. Lemma 1.1. Let (, A, µ) be a measure space. (i) If f L 1 +(, A, µ) ad g L 1 R,elem (, A, µ) are such that g + f 0, the g + f L 1 +(, A, µ), ad I +(g µ + f) = I µ elem (g) + Iµ +(f). (ii) If f L 1 +(, A, µ) ad g L 1 R,elem (, A, µ) are such that g f 0, the g f L 1 +(, A, µ), ad I +(g µ f) = I µ elem (g) Iµ +(f).

11 CHAPTER IV: INTEGRATION THEORY 299 Proof. (i). We start with a weaker versio. Claim: If f L 1 +(, A, µ) ad g L 1 R,elem (, A, µ), are such that g +f 0, the g + f L 1 +(, A, µ), ad I +(g µ + f) I µ elem (g) + Iµ +(f). What we eed to prove is the fact that, for every h A-Elem R (), with 0 h g + f, we have: (a) h L 1 R,elem (, A, µ); (b) I µ elem (h) Iµ elem (g) + Iµ +(f). Cosider the elemetary fuctio h 1 = max{h g, 0}. It is obvious that 0 h 1 f, so by Propositio 1.3, it follows that h 1 L 1 +(, A, µ), ad I +(h µ 1 ) I +(f). µ By Propositio 1.2, this gives h 1 L 1 R,elem (, A, µ), ad (10) I µ elem (h 1) = I µ +(h 1 ) I µ +(f). Usig the obvious iequality g h g h 1, agai by Propositio 1.2, it follows that h g L 1 R,elem (, A, µ), ad (11) I µ elem (h g) Iµ elem (h 1). Of course, by Theorem 1.1, this gives the fact that h = (h g) + g is elemetary µ-itegrable, as well as the equality I µ elem (h) = Iµ elem (h g) + Iµ elem (g). Combiig this with (11) ad (10) immediately gives I µ elem (h) Iµ elem (h 1) + I µ elem (g) Iµ +(f) + I µ elem (g), ad the Claim is prove. Havig prove the above Claim, we ow proceed with the proof of (i). If f L 1 +(, A, µ) ad g L 1 R,elem (, A, µ) are such that g + f 0, the by the Claim, we already kow that g+f L 1 +(, A, µ), ad I µ (g+f) I µ elem (g)+iµ +(f). We apply ow agai the Claim to the fuctios f 1 = g + f ad g 1 = g, to get I µ +(f) = I µ +(g 1 + f 1 ) I µ elem (g 1) + I µ +(f 1 ) = I µ +(g) + I + (g + f), which gives the other iequality I µ elem (g) + Iµ +(f) I +(g µ + f). (ii). tart with f L 1 +(, A, µ) ad g L 1 R,elem (, A, µ), with g f 0. First of all, sice 0 g f g, by Propositio 1.5, it follows that g f L 1 +(, A, µ), ad (12) I µ +(g f) = if { I µ elem (k) : k L1 R,elem(, A, µ), k g f }. ecod, remark that, wheever k L 1 R,elem (, A, µ) is such that g f k, it follows that k + f g, so usig part (i) combied with Propositio 1.3, we see that k + f L 1 +(, A, µ), ad This meas that we have I µ elem (g) = Iµ +(g) I µ +(k + f) = I µ elem (k) + Iµ +(f). I µ elem (k) Iµ elem (g) Iµ +(f), for all k L 1 R,elem (, A, µ), with k g f, ad the by (12), we immediately get I µ +(g f) I µ elem (g) Iµ +(f).

12 300 LECTURE To prove the other iequality, we use the defiitio of the positive itegral, which gives (13) I µ +(g f) = sup { I µ elem (h) : h L1 R,elem(, A, µ), 0 h g f }. Remark that, wheever h L 1 R,elem (, A, µ) is such that 0 h g f, it follows that 0 h + f g, so usig part (i) combied with Propositio 1.3, we see that h + f L 1 +(, A, µ), ad This meas that we have I µ elem (g) = Iµ +(g) I µ +(h + f) = I µ elem (h) + Iµ +(f). I µ elem (h) Iµ elem (g) Iµ +(f), for all h L 1 R,elem (, A, µ), with 0 h g f, ad the by (13), we immediately get I +(g µ f) I µ elem (g) Iµ +(f). We are ow i positio to prove the followig result (compare with Theorem 1.1). Theorem 1.2. Let (, A, µ) be a measure space. (i) If f 1, f 2 L 1 +(, A, µ), the f 1 + f 2 L 1 +(, A, µ), ad oe has the equality I µ +(f 1 + f 2 ) = I µ +(f 1 ) + I µ +(f 2 ). (ii) If f L 1 +(, A, µ), ad α [0, ), the 1 αf L 1 +(, A, µ), ad oe has the equality I µ +(αf) = αi µ +(f). Proof. (i). Fix f 1, f 2 L 1 +(, A, µ). Claim 1: Wheever h A-Elem R () satisfies 0 h f 1 + f 2, it follows that (a) h L 1 R,elem (, A, µ), (b) I µ elem (h) Iµ +(f 1 ) + I +(f µ 2 ). Fix a elemetary fuctio h A-Elem R (), with 0 h f 1 + f 2, ad let us first show that h is elemetary itegrable. Fix some α h() {0}, ad let us prove that µ ( h 1 ({α}) ) <. If we defie the sets A j = f 1 ( ) j [α/2, ] A, j = 1, 2, the the elemetary fuctios h j = α 2 κ A j satisfy 0 h j f j, j = 1, 2. I particular, it follows that h 1, h 2 L 1 R,elem (, A, µ), which forces µ(a 1) < ad µ(a 2 ) <. Notice however that, for every x h 1 ({α}), we have f 1 (x) + f 2 (x) h(x) = α, which forces either f 1 (x) α 2 or f 2(x) α 2. This argumet shows tha we have the iclusio h 1 ({α}) A 1 A 2, so it follows that we ideed have µ ( h 1 ({α}) ) <. Havig show property (a), let us prove property (b). Defie the sets B = {x : h(x) f 1 (x)} ad D = B. It is obvious that B, D A are pairwise disjoit, ad B D =. Defie the elemetary fuctios h = hκ B, ad h = h h = hκ D. O the oe had, we have f 1 κ B h f 1 κ B + f 2 κ B, which gives 0 h f 1 κ B f 2 κ B. 1 Here we use the covetio that whe α = 0, we take αf = 0.

13 CHAPTER IV: INTEGRATION THEORY 301 By Lemma 1.1.(ii), combied with Propositio 1.4, it follows that h f 1 κ B L 1 +(, A, µ) ad I µ elem (h ) I +(f µ 1 κ B ) = I +(h µ f 1 κ B ) I +(f µ 2 κ B ), so we get (14) I µ elem (h ) I µ +(f 1 κ B ) + I µ +(f 2 κ B ). O the other had, we have which gives h = hκ D f 1 κ D, (15) I µ elem (h ) I µ +(f 1 κ D ) I µ +(f 1 κ D ) + I µ +(f 2 κ D ). ice h = h + h, with h ad h elemetary itegrable, usig Theorem 1.1 combied with Propositio 1.4, by addig the iequalities (14) ad (15) we get I µ elem (h) = Iµ elem (h ) + I µ elem (h ) I µ +(f 1 κ B ) + I µ +(f 2 κ B ) + I µ +(f 1 κ D ) + I µ +(f 2 κ D ) = I µ +(f 1 ) + I µ +(f 2 ), ad the Claim is prove. Claim 1 obviously implies the fact that f 1 + f 2 L 1 +(, A, µ), as well as the iequality I µ +(f 1 + f 2 ) I µ +(f 1 ) + I µ +(f 2 ). To prove the other iequality, we use the followig. Claim 2: For every h A-Elem R (), with 0 h f 1, oe has the iequality I µ elem (h) Iµ +(f 1 + f 2 ) I µ +(f 2 ). Ideed, if h is as above, the h is i L 1 +(, A, µ), hece elemetary itegrable, ad we obviously have 0 h + f 2 f 1 + f 2. The by Lemma 1.1.(i), combied with Propositio 1.3, we get I µ elem (h) + Iµ +(f 2 ) = I µ +(h + f 2 ) I µ +(f 1 + f 2 ), ad the Claim follows. Usig Claim 2, ad the defiitio of the positive itegral, we get I +(f µ 1 ) = sup { I µ elem (h) : h A-Elem } R(), 0 h f 1 I µ +(f 1 + f 2 ) I +(f µ 2 ), which the gives I +(f µ 1 ) + I +(f µ 2 ) I +(f µ 1 + f 2 ). (ii). This part is obvious. Defiitios. Let (, A, µ) be a measure space. Deote the exteded real lie [, ] by R. A measurable fuctio f : R is said to be µ-itegrable, if there exist fuctios f 1, f 2 L 1 +(, A, µ), such that (16) f(x) = f 1 (x) f 2 (x), x [ f1 1 1 ({ }) f2 ({ })]. By Remark 1.3, we kow that the sets f 1 k ({ }), k = 1, 2, have measure zero. The equality (16) gives the the fact f = f 1 f 2, µ-a.e. We defie L 1 R(, A, µ) = { f : R : f µ-itegrable }. We also defie the space of hoest real-valued µ-itegrable fuctios, as L 1 R(, A, µ) = { f L 1 R(, A, µ) : f < f(x) <, x }. Fially, we defie the space of complex-valued µ-itegrable fuctios as L 1 C(, A, µ) = { f : C : Re f, Im f L 1 R(, A, µ) }.

14 302 LECTURE The ext result collects the basic properties of L 1 R. states that it is a almost vector space. Theorem 1.3. Let (, A, µ) be a measure space. Amog other thigs, it (i) For a measurable fuctio f : R, the followig are equivalet: (a) f L 1 R(, A, µ); (b) f L 1 +(, A, µ). (ii) If f, g L 1 R(, A, µ), ad if h : R is a measurable fuctio, such that h(x) = f(x) + g(x), x [ f 1 ({, }) g 1 ({, }) ], the h L 1 R(, A, µ). (iii) If f L 1 R(, A, µ), ad α R, ad if g : R is a measurable fuctio, such that the g L 1 R(, A, µ). (iv) Oe has the iclusio g(x) = αf(x), x f 1 ({, }), L 1 R,elem(, A, µ) L 1 +(, A, µ) L 1 R(, A, µ). Proof. (i). Cosider the fuctios measurable fuctios f ± : [0, ] defied as f + = max{f, 0} ad f = max{ f, 0}. To prove the impliactio (a) (b), assume f L 1 R(, A, µ), which meas 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 iequalities (17) (18) f + f 1, µ-a.e., f f 2, µ-a.e.. Ideed, if we put N = f1 1 1 ({ }) f2 ({ }), the µ(n) = 0, ad if we start with some x N, we either have f 1 (x) f 2 (x) 0, i 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), i which case we get I 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) ad f (x) f 2 (x), x N, so we ideed get (17) ad (18). Usig these iequalities, ad Propositio 1.3, it follows that f ± L 1 +(, A, µ), so by Theorem 1.2, it follows that f + + f = f also belogs to L 1 +(, A, µ).

15 CHAPTER IV: INTEGRATION THEORY 303 To prove the implicatio (b) (a), start by assumig that f L 1 +(, A, µ). The, sice we obviously have the iequalities 0 f ± f, agai by Propositio 1.3, it follows that f ± L 1 +(, A, µ). ice we obviously have f(x) = f + (x) f (x), x f 1 ({, }), it follows that f ideed belogs to f ± L 1 R(, A, µ). (ii). Assume f, g, ad h are as i (ii). By (i), both fuctios f ad g are i L 1 +(, A, µ). By Theorem 1.2, it follows that the fuctio k = f + g also belogs to L 1 +(, A, µ). Notice that we have the equality so the hypothesis o h reads which the 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, sice µ ( k 1 ({ }) ) = 0, this gives h k, µ-a.e., ad usig (i) it follows that h ideed belogs to L 1 R(, A, µ). (iii). Assume f, α, ad g are as i (iii). Exactly as above, we have g = α f, µ-a.e., ad the by Theorem 1.2 it follows that g L 1 +(, A, µ). (iv). The iclusio L 1 +(, A, µ) L 1 R(, A, µ) is trivial. To prove the iclusio L 1 R,elem (, A, µ) L1 R(, A, µ), we use parts (ii) ad (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 ow obvious, because ay such fuctio belogs to L 1 +(, A, µ) L 1 R(, A, µ). Corollary 1.1. Let (, A, µ) be a measure space, ad let K be oe of the fields R or C. (i) For a K-valued measurable fuctio f : K, the followig are equivalet: (a) f L 1 K (, A, µ); (b) f L 1 +(, A, µ). (ii) Whe equipped with the poitwise additio ad scalar multiplicatio, the space L 1 K (, A, µ) becomes a K-vector space. Proof. (i). The case K = R is immediate from Theorem 1.3 I the case whe K = C, we use the obvious iequalities (19) max { Re f, Im f } f Re f + Im f. If f L 1 C (, A, µ), the both Re f ad Im f belog to L1 R (, A, µ), so by Theorem 1.3, both Re f ad Im f belog to L 1 +(, A, µ). By Theorem 1.2, the fuctio g = Re f + Im f belogs to L 1 +(, A, µ), ad the usig the secod iequality i (19), it follows that f belogs to L 1 +(, A, µ). Coversely, if f belogs to L 1 +(, A, µ), the usig the first iequality i (19), it follows that both Re f ad Im f belog to L 1 +(, A, µ), so by Theorem 1.3, both Re f ad Im f belog to L 1 R (, A, µ), i.e. f belogs to L1 C (, A, µ). (ii). This part is pretty clear. If f, g L 1 K (, A, µ), the by (i) both f ad g belog to L 1 +(, A, µ), ad by Theorem 1.2, the fuctio f + g will

16 304 LECTURE also belog to L 1 +(, A, µ). ice f + g f + g, it follows that f + g itself belogs to L 1 +(, A, µ), so usig (i) agai, it follows that f + g ideed belogs to L 1 K (, A, µ). If f L1 K (, A, µ) ad α K, the f belogs to L1 +(, A, µ), so αf = α f agai belogs to L 1 +(, A, µ), which by (i) gives the fact that αf belogs to L 1 K (, A, µ). Remark 1.5. Let (, A, µ) be a measure space. The oe has the equalities (20) (21) 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 (). Ideed, by Theorem 1.3 that we have the iclusio L 1 +(, A, µ) { f L 1 R(, A, µ) : f() [0, ] }. The iclusio i the other directio follows agai from Theorem 1.3, sice ay fuctio that belogs to the right had side of (20) satisfies f = f. The iclusio L 1 K,elem(, A, µ) L 1 K(, A, µ) A-Elem K () is agai cotaied i Theorem 1.3. To prove the iclusio i the other directio, it suffices to cosider the case K = R. tart with h L 1 R (, A, µ) A-Elem R(), which gives h L 1 +(, A, µ). The fuctio h is obviously i A-Elem R (), so we get h L 1 R,elem (, A, µ). ice L1 R,elem (, A, µ) is a vector space, it will also cotai the fuctio h. The fact that h itself belogs to L 1 R,elem (, A, µ) the follows from Propositio 1.1, combied with the obvious iequalities h h h. The followig result deals with the costructio of the itegral. Theorem 1.4. Let (, A, µ) be a measure space. There exists a uique map I µ R(, A, µ) R, with the followig properties: (i) Wheever 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) Wheever f, g L 1 R(, A, µ) ad 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 existece. tart with some f L 1 R(, A, µ), ad defie the fuctios f ± : [0, ] by f + = max{f, 0} ad f = max{ f, 0} so that f = f + f, ad f +, f L 1 +(, A, µ). We the defie I µ R(f) = I µ +(f + ) I µ +(f ). It is obvious that I µ R satisfies coditio (iii). The key fact that we eed is cotaied i the followig.

17 CHAPTER IV: INTEGRATION THEORY 305 Claim: Wheever f L 1 R(, A, µ), ad f 1, f 2 L 1 +(, A, µ) are such that f(x) = f + (x) f (x), x. [ 1 ({ }) f 1 2 ({ })], it follows that we have the equality I µ R(f) = I µ +(f 1 ) I µ +(f 2 ). Ideed, sice we have f = f + f, it follows immediately that we have the equality which gives f 2 (x) + f + (x) = f 1 (x) + f (x), x. [ f1 1 1 ({ }) f2 ({ })], By Theorem 1.2, this immediately gives which the gives f 2 + f + = f 1 + f, µ-a.e. 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). Havig prove the above Claim, let us show ow that I µ R has properties (i) ad (ii). Assume f, g ad h are as i (i). Notice that if we defie h 1 = f + + g + ad h 2 = f + g, the we clearly have 0 h 1 f + g ad 0 h 2 f + g, so h 1 ad h 2 both belg to L 1 +(, A, µ). By Theorem 1.2, we the have (22) I µ +(h 1 ) = I µ +(f + ) + I µ +(g + ) ad I µ +(h 2 ) = I µ +(f ) + I µ +(g ). Notice also that, because of the equalities h 1 1 ({ }) = f 1 ({ } g 1 ({ }) ad h 1 2 ({ }) = 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, combied with (22), 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 uiqueess is also obvious. If we start with a map J : L 1 R(, A, µ) R with properties (i)-(iii), the 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 secod equality we use coditio (iii), combied with the fact that both f + ad f belog to L 1 +(, A, µ).) Corollary 1.2. Let (, A, µ) be a measure space, ad let K be either R or C. There exists a uique liear 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. I this case, we have the iclusio L 1 R(, A, µ) L 1 R(, A, µ), so we ca defie I µ R as the restrictio of Iµ R to L 1 R (, A, µ). The uiqueess is agai clear, because of the equalities I µ R (f) = Iµ R (f + ) I µ R (f ) = I µ +(f + ) I µ +(f ).

18 306 LECTURE I the case K = C, we defie I µ C (f) = Iµ R (Re f) + iiµ R (Im f). The liearity is obvious. The uiqueess is also clear, because the restrictio of I µ C to L 1 R (, A, µ) must agree with Iµ R. Defiitio. Let (, A, µ) be a measure space, ad let K be oe of the symbols R, R, or C. For ay f L 1 K (, A, µ), the umber Iµ K (f) (which is real, if K = R or R, ad is complex if K = C) will be deoted by f dµ, ad is called the µ-itegral of f. This otatio is uambiguous, because if f L 1 R (, A, µ), the we have Iµ R(f) = I µ C (f) = Iµ R (f). Remark 1.6. If (, A, µ) is a measure space, the for every A A, with µ(a) <, usig the above Corollary, we get κ A dµ = I +(κ µ A ) = µ(a). By liearity, if K = R, C, oe has the the equality h dµ = I µ elem (h), h L1 K,elem(, A, µ). To make the expositio a bit easier, it will adopt the followig. Covetio. If (, A, µ) is a measure space, ad if f : [0, ] is a measurable fuctio, which does ot belog to L 1 +(, A, µ), the we defie f dµ =. Remarks 1.7. Let (, A, µ) be a measure space. A. Usig the above covetio, whe h A-Elem R () is a fuctio with h() [0, ), the coditio h dµ = is equivalet to the existece of some α h() {0}, with µ ( h 1 ({α}) ) =. B. Usig the above covetio, for every measurable fuctio f : [0, ], oe has the equality { } f dµ = sup h dµ : h A-Elem R (), 0 h f. C. If f, g : [0, ] are measurable, the oe has the equalities (f + g) dµ = f dµ + g dµ, (αf) dµ = α f dµ, α [0, ), eve i the case whe some term is ifiite. (We use the covetio + t =, t [0, ], as well as α =, α (0, ), ad 0 = 0.) D. If f, g : [0, ] are measurable, ad f g, µ-a.e., the (usig B) oe has the iequality f dµ g dµ, eve if oe side (or both) is ifiite.

19 CHAPTER IV: INTEGRATION THEORY 307 E. Let K be oe of the symbols R, R, or C, ad let f : K be a measurable fuctio. The the fuctio f : [0, ] is measurable. Usig the above covetio, the coditio that f belogs to L 1 K (, A, µ) is equivalet to the iequality f dµ <. I the remaider of this sectio we discuss several properties of itegratio that are aaloguous to those of the positive/elemetary itegratio. We begi with a useful estimate Propositio 1.6. Let (, A, µ) be a measure space, ad let K be oe of the symbols R, R, or C. For every fuctio f L 1 K (, A, µ), oe has the iequality f dµ f dµ. Proof. Let us first examie the case whe K = R, R. I this case we defie f + = max{f, 0} ad f = max{ f, 0}, so we have f = f + f, as well as f = f + + f. Usig the iequalities 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. I other words, we have ± f dµ f dµ, ad the desired iequality immediately follows. Let us cosider ow the case K = C. Cosider the umber λ = f dµ, ad let us choose some complex umber α C, with α = 1, ad αλ = λ. (If λ 0, we take α = λ 1 λ ; otherwise we take α = 1.) Cosider the measurable fuctio g = αf. Notice ow that ( ) ( ) Re g dµ + i Im g dµ = g dµ = α f dµ = αλ = λ 0, so i particular we get λ = Re g dµ. If we apply the real case, we the get (23) λ Re g dµ. Notice ow that, we have the iequality Re g g = f, which gives I +( µ ) Re g I µ ( f ) = f dµ, so the iequality (23) immediately gives f dµ = λ f dµ. Corollary 1.3. Let (, A, µ) be a measure space, ad let K be oe of the symbols R, R, or C. If a measurable fuctio f : K satisfies f = 0, µ-a.e, the f L 1 K (, A, µ), ad f dµ = 0.

20 308 LECTURE Proof. Cosider the measurable fuctio f : [0, ], which satisfies f = 0, µ-a.e. By Propositio 1.3, it follows that f L 1 +(, A, µ), hece f L 1 K (, A, µ), ad 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 fiite measure space, the every bouded measurable fuctio f : K belogs to L 1 K (, A, µ), ad satisfies f dµ µ() sup f(x). x Proof. If we put β = sup x f(x), the we clearly have f βκ, which shows that f L 1 +(, A, µ), ad also gives f dµ βκ dµ = µ() β. The everythig follows from Propositio 1.6. Commet. The itroductio of the space L 1 R(, A, µ), of exteded real-valued µ-itegrable fuctios, is useful mostly for techical reasos. I effect, everythig ca be reduced to the case whe oly hoest real-valued fuctios are ivolved. The followig result clarifies this matter. Lemma 1.2. Let (, A, µ) be a measure space, ad let f : R be a measurable fuctio. The followig ar equivalet (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 equivalet coditios, the ay fuctio g, satisfyig (ii), also has the property f dµ = g dµ. Proof. Cosider the set F = {x : < f(x) < }, which belogs to A. We obviously have the equality F = f 1 ({ }). (i) (ii). Assume f L 1 R(, A, µ), which meas that f L 1 +(, A, µ). I particular, we get µ( F ) = 0. Defie the measurable fuctio g = fκ F. O the oe had, it is clear, by costructio, that we have < g(x) <, x. O the other had, it is clear that g F = f F, so usig µ( F ) = 0, we get the fact that f = g, µ-a.e. Fially, the iequality 0 g f, combied with Propositio 1.3, gives g L 1 +(, A, µ), so g ideed belogs to L 1 R (, A, µ). (ii) (i). uppose there exists g L 1 R (, A, µ), with f = g, µ-a.e., ad let us prove that (a) f L 1 R(, A, µ); (b) f dµ = g dµ. The first assertio is clear, because by Propositio 1.3, the equality f = g, µ-a.e., combied with g L 1 +(, A, µ), forces f L 1 +(, A, µ), i.e. f L 1 R(, A, µ). To prove (b), we cosider the differece h = f g, which is a measurable fuctio h : R, ad satisfies h = 0, µ-a.e. By Corollary 1.3, we kow that h L 1 R(, A, µ), ad h dµ = 0. By Theorem 1.3, we get f dµ = g dµ + h dµ = g dµ. The followig result is a aalogue of Propositio 1.1 (see also Propositio 1.3). x

21 CHAPTER IV: INTEGRATION THEORY 309 Propositio 1.7. Let (, A, µ), ad let f 1, f 2 L 1 R(, A, µ). uppose f : R is a measurable fuctio, such that f 1 f f 2, µ-a.e. The f L 1 R(, A, µ), ad oe has the iequality f 1 dµ f dµ f 2 dµ. Proof. First of all, sice f 1 ad f 2 belog to L 1 R(, A, µ), it follows that f 1 ad f 2, hece also f 1 + f 2, belog tp L 1 +(, A, µ). ecod, sice we have f 2 f 2 f 1 + f 2 ad f 1 f 1 f 1 f 2 (everyhwere!), the iequalities f 1 f f 2, µ-a.e., give f 1 f 2 f f 1 + f 2, µ-a.e., which reads f f 1 + f 2, µ-a.e. ice f 1 + f 2 L 1 +(, A, µ), by Propositio 1.3., we get f L 1 +(, A, µ), so f ideed belogs to L 1 R(, A, µ). To prove the iequality for itegrals, we use Lemma 1.2, to fid fuctios 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., ad f = g, µ-a.e. Lemma 1.2 also gives the equalities f 1 dµ = g 1 dµ, f 2 dµ = g 2 dµ, ad f dµ = g dµ, so what we eed to prove are the iequalities (24) g 1 dµ g dµ g 2 dµ. Of course, we have g 1 g g 2, µ-a.e. To prove the first iequality i (24), we cosider the fuctio h = g g 1 L 1 R (, A, µ), ad we prove that h dµ 0. But this is quite clear, because we have h 0, µ-a.e., which meas that h = h, µ-a.e., so by Lemma 1.2, we get h dµ = h dµ = I +( h ) µ 0. The secod iequality i (24) is prove the exact same way. The ext result is a aalogue of Propositio 1.4. Propositio 1.8. Let (, A, µ) be a measure space, ad let K be oe of the symbols R, R, or C. uppose (A k ) A is a pairwise disjoit fiite sequece, with A 1 A =. For a measurable fuctio f : K, the followig are equivalet. (i) f L 1 K (, A, µ); (ii) fκ Ak L 1 K (, A, µ), k = 1,...,. Moreover, if f satisfies these equivalet coditios, oe has (25) f dµ = fκ Ak dµ. Proof. It is fairly obvious that fκ Ak = f κ Ak. The the equivalece (i) (ii) follows from Propositio 1.4 applied to the fuctio f : [0, ]. I the cases whe K = R, C, the equality (25) follows immediately from liearity, ad the obvious equality f = fκ A k. I the case whe K = R, we take g L 1 R (, A, µ), such that f = g, µ-a.e. The we obviously have fκ A k = gκ Ak,

22 310 LECTURE µ-a.e., for all k = 1,...,, ad the equality (25) follows from the correspodig equality that holds for g. Remark 1.8. The equality (25) also holds for arbitrary measurable fuctios f : [0, ], if we use the covetio that preceded Remarks 1.7. This is a immediate cosequece of Propositio 1.4, because the left had side is ifiite, if a oly if oe of the terms i the right had side is ifiite. The followig is a obvious extesio of Remark 1.4. Remark 1.9. Let K be oe of the symbols R, R, or C, let (, A, µ) be a measure space. For a set A, ad a measurable fuctio f : K, oe has the equivalece fκ L 1 K(, A, µ) f L 1 K(, A, µ ). If this is the case, oe has the equality (26) fκ dµ = f dµ. The above equality also holds for arbitrary measurable fuctios f : [0, ], agai usig the covetio that preceded Remarks 1.7. Notatio. The above remark states that, whever the quatities i (26) are defied, they are equal. (This oly requires the fact that f is measurable, ad either f LK( 1, A, µ ), or f() [0, ].) I this case, the equal quatities i (26) will be simply deoted by f dµ. Exercise 1. Let I be some o-empty set. Cosider the σ-algebra P(I), of all subsets of I, equipped with the coutig measure { Card A if A is fiite µ(a) = if A is ifiite 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, the L 1 K(I, P(I), µ) = l 1 K(I), the Baach space discussed i II.2 ad II.3. Exercise 2. There is a istace whe the etire theory developped here is essetially vacuous. Let be a o-empty set, ad let A be a σ-algebra o. For a measure µ o A, prove that the followig are equivalet (i) L 1 +(, A, µ) = { f : [0, ] : f measurable, ad f = 0, µ-a.e. } ; (ii) for every A A, oe has µ(a) {0, }. A measure space (, A, µ), with property (ii), is said to be degeerate. Exercise 3. Let (, A, µ) be a measure space, ad let f : [0, ] be a measurable fuctio, with f dµ = 0. Prove that f = 0, µ-a.e. Hit: Defie the measurable sets A = {x : f(x) 1 }, ad aalyze the relatioship betwee f ad κ A.