( f ^ M _ M 0 )dµ (5.1)

Transcription

1 47 5. LEBESGUE INTEGRAL: GENERAL CASE Although the Lebesgue integral defined in the previous chapter is in many ways much better behaved than the Riemann integral, it shares its restriction to bounded functions and bounded measure spaces only. To remove this restriction one might proceed similarly as for the improper Riemann integral; e.g., by taking limits of A n ( f ^ M _ M 0 )dµ (5.1) as M!, M 0! and n! for some increasing sequence {A n } of measurable sets with µ(a n ) < and S A n = X. Unfortunately, the existence of such a limit is not guaranteed without further conditions on the integrand, and we may not even be able to find a sequence {A n } with the stated properties. The way around this that has generally been adopted is to first define the integral of nonnegative functions on arbitrary measure spaces. This is the unsigned integral below. A key difference between the bounded integral from the previous chapter is that here we abandon integrability as a primary concept and rather put measurability first. Notwithstanding, the bounded integral and approximations of the form (5.1) are key to proving properties such as additivity etc. Once the unsigned integral is defined and its properties established, a more or less formal extension leads to the signed integral that covers also functions that can take both positive and negative values. 5.1 Unsigned integral. Let (X,F, µ) be a measure space with µ(x) not necessarily finite. We now put forward: Definition 5.1 (Lower unsigned integral) Let f : X! [0, ]. Then we set: n o f dµ := sup f dµ : f simple, 0 apple f apple f. (5.2) This is the analogue of what we called the lower Lebesgue integral in the bounded case. The corresponding upper integral will not be defined at all. Observe the following properties (shared, generally, by all lower integrals): Lemma 5.2 Let f,g be functions on X. Then we have: (1) if 0 apple f apple g then f dµ apple gdµ, (5.3) (2) if f,g, 0 then ( f + g)dµ f dµ + gdµ (5.4) (3) if c 2 [0, ) and f 0 then (using the convention 0. = 0) (cf)dµ = c f dµ. (5.5)

2 48 Proof. Let f be a simple function with 0 apple f apple f. For (1) we note that f apple g then implies 0 apple f apple g. For (2) we note that if also 0 apple f 0 apple g is simple, then f + f 0 apple f + g. Finally, for (3) we observe that cf is simple with cf apple f. It is not hard to check that, although the lower integral is superadditive, it is generally not additive unless further conditions on f and g are imposed. It turns out that, as before, it is enough to require measurability. (Since we permit functions to take value +, we mean the notion of measurability for functions taking extended-real values.) What needs to be proved is: Proposition 5.3 (Superadditivity on measurable functions) Let f, g: X! [0, ] be measurable. Then we have: ( f + g)dµ apple f dµ + gdµ (5.6) First we observe that if we reduce the unsigned lower integral to bounded functions with bounded measure of their support, then additivity can be proved directly. This is because we have: supp( f ) := x 2 X : f (x) 6= 0, (5.7) Lemma 5.4 Let f : X! [0, ] be bounded, measurable and assume there is A 2 F such that supp( f ) A and µ(a) <. Then f dµ = f dµ, (5.8) A where the integral on the right is in the sense of Definition 4.8. Proof. As 0 apple f apple f forces f = 0 on A c, we may restrict all definitions of integrals to the measure space (A,F A, µ A ) (see Lemma 4.9) and get exactly the same values. But µ A (A) < and f is bounded, and so the lower unsigned integral coincides with the lower bounded integral. Now ( f A ) 1 (B r {0}) = f 1 (B r {0}) for any set B X shows that the function f A (see again Lemma 4.9) is F A -measurable. Since f A is also bounded, it is integrable with respect to µ A and so the lower bounded integral on measure space (A,F A, µ A ) is equal to the integral R f A µ A. By Lemma 4.9 again, this is the integral on the right of (5.8). The passage from the bounded integral to the unsigned integral will be provided by methods not dissimilar to those mentioned very early in this chapter: Lemma 5.5 (1) We have: Let f : X! [0, ] be measurable. Then f ^ M dµ! M! (2) if {A n } F obey A n " A {f > 0}, then ( f 1 An ) dµ! f dµ, (5.9) f dµ. (5.10) Proof. (1) We have f ^ M apple f and so the limit (which exists by monotonicity) is at most the integral on the right. For the opposite inequality, note that if f is simple with 0 apple f apple f then

3 f apple ( f ^ M) as soon as M max f. (Simple functions are necessarily bounded.) Hence f dµ apple lim ( f ^ M)dµ, 8f simple,0 apple f apple f. (5.11) M! Taking supremum over such f we get R f dµ on the left. Part (1) is proved. For (2) the limit again exists by monotonicity and it is at most the integral on the right. For the opposite inequality let f = Â k i=1 a k 1 Bi be simple with 0 apple f apple f. Then f1 An is also simple with f1 An apple f1 A. But A {f > 0} {f > 0} and so, in fact, f1 An " f. Hence, B i \ A i " B i whenever a i > 0 and so, by the Monotone Convergence Theorem for sets, k k ( f 1 An )dµ (f1 An )dµ = Â a i µ(a n \ B i )! Â a i µ(b i )= i=1 i=1 f dµ. (5.12) Taking supremum over f, we get that the limit of lower integrals of f 1 An dominates the lower integral of f. Before we use these in the proof of Proposition 5.3, we also generalize the Markov inequality to the unsigned integral: Lemma 5.6 (Markov inequality again) For any measurable f : X! [0, ] and any l > 0, µ f l apple 1 f dµ. (5.13) l Proof. We have l1 { f l} apple f whenever l 0. The claim follows (as before) by integrating these with respect to µ and using Lemma 5.2(1). Proof of Proposition 5.3. Without loss of generality, assume that both lower integrals of f and g are finite. (Otherwise the claim holds by Lemma 5.2(2).) Define B n := { f > 1/n} and C n := {g > 1/n}. (5.14) By Markov s inequality, µ(b n ) < and µ(c n ) < for all n 2 N. Hence also A n := B n \C n obeys µ(a n ) <. Now ( f + g) ^ M apple f ^ M + g ^ M (5.15) for any M 0 and so ( f + g) ^ M dµ apple ( f ^ M + g ^ M)dµ A n A n = ( f ^ M)dµ + (g ^ M)dµ (5.16) A n A n = ( f ^ M)1 An dµ + (g ^ M)1 An dµ apple f dµ + gdµ, where we used the additivity of the bounded integral to get the middle equality and then applied Lemma 5.4 and Lemma 5.2(1) to get the conclusion. However, and so, by Lemma 5.4 and 5.5, A n ( f + g) ^ M dµ! A n "{f + g > 0} (5.17) 49 ( f + g) ^ M dµ! ( f + g)dµ. (5.18) M!

4 50 The desired conclusion follows. In light of this, it now makes sense to put forward: Definition 5.7 (Unsigned integral) For any f : X! [0, ] measurable, we define the unsigned Lebesgue integral of f with respect to µ by f dµ := f dµ. (5.19) For non-measurable f the unsigned integral is left undefined, and so the concept of integrability is now completely tied to measurability. Naturally, once f is measurable, all properties of unsigned lower integral proved above are shared by its unsigned integral. 5.2 Signed integral and convergence theorems. This unsigned integral shares the properties (1) and (3) from Lemma 5.2 and is additive in the integrand thanks to Lemma 5.2(2) and Proposition 5.3. The unsigned integral is actually well behaved even under monotone pointwise limits. Indeed, we have: Theorem 5.8 (Monotone Convergence Theorem) Let f n : X! [0, ] be measurable with f n " f. Then also f is measurable and f n dµ " f dµ. (5.20) Proof. The limit of the integrals of f n exists and, since f n apple f, is at most the integral of f. For the opposite direction we may assume that the limit is actually finite. Let A n := { f n > 1/n}. By Markov s inequality we have µ(a n ) < and so ( f n ^ M)1 AK are bounded functions with support of finite measure. The Bounded Convergence Theorem and Lemma 5.4 then show f n dµ ( f n ^ M)1 AK dµ! ( f ^ M)1 AK dµ (5.21) and so the limit of the integrals is at most the integral on the right. But A K "{f > 0} and so taking K! followed by M! makes the right hand side tend to the integral of f. Two simple but useful consequences are in order: Corollary 5.9 Let { f n } be measurable functions f n : X! [0, ]. Then  f n dµ =  f n dµ. (5.22) Proof. Define g n :=  n k=1 f k and g :=  k2n f k. By finite additivity of the integral, n g n dµ =  k=1 f k dµ (5.23) The left hand side tends to the integral of g by the Monotone Convergence Theorem, while the right-hand side converges to the infinite sum by (3.1).

5 Corollary 5.10 Then n is a measure on (X,F ). Let f : X 7! [0, ] be measurable. For each A 2 F define n(a) := f dµ := ( f 1 A )dµ. (5.24) Proof. Obviously, n(/0 )=0. If {A n } F are disjoint, then and so by (3.3) and the previous corollary [ n A n = f A 1 S A n = Â 1 An (5.25) Â 1 An dµ = Â Hence, n is countably additive and so it is a measure. 51 ( f 1 An )dµ = Â n(a n ). (5.26) An interesting questions is when is a measure n on (X,F ) of the form (5.24). We will answer this later in these notes. Another useful consequence of the Monotone Convergence Theorem is: Lemma 5.11 (Fatou s lemma) Let f n : X! [0, ] be measurable. Then lim inf f n dµ apple liminf f n dµ. (5.27) Proof. For each m n we have f m inf k n f k and so inf f m dµ inf m dµ. m n m n (5.28) Since inf m n f m increases to the limes inferior as n!, the claim follows by the Monotone Convergence Theorem. Having discussed the unsigned integral enough, we can move to the signed integral. Given a function f, let be its positive and negative parts. Obviously, f +, f We now put forward: f + := f 1 { f 0} and f = f 1 { f <0} (5.29) are measurable whenever f is and f = f + f. (5.30) Definition 5.12 (Signed integral) For any f : X! R measurable such that the unsigned integrals of f + and f obey min f + dµ, f dµ < (5.31) we define the (signed) Lebesgue integral of f with respect to µ by f dµ := f + dµ f dµ. (5.32) We say that such an f is integrable (with respect to µ).

6 52 Notice that the above does permit the signed integral to be equal to ±. As we will see that the signed integral is considerably weaker than the unsigned one. Notwithstanding, the expected properties still hold, at least with correct provisos: Lemma 5.13 If f : X! R is integrable and c 2 R, then (cf)dµ = c f dµ (5.33) whenever the right-hand side is not of the form 0 times infinity. In particular, c f is integrable whenever f is. Similarly, if f,g: X! R are integrable, then f + g is integrable and ( f + g)dµ = f dµ + gdµ, (5.34) provided the right-hand side is not a difference of two infinities of same sign. Proof. For the first property we note that if c > 0, then (cf) ± = cf ± while for c < 0 we get (cf) ± = cf. The claim then follows from Lemma 5.2(2) and the definition of unsigned integral. For c = 0 we assume that both R f ± dµ < and so the claim again follows by Lemma 5.2(2). That integrability of f implies integrability of cf is a consequence of these. For the second property, observe that if the expression on the right of (5.34) is defined and meaningful, then at least one of f + + g + and f + g has finite signed integral. Since ( f + g) + =(f + + g + )1 { f +g 0} ( f + g )1 { f +g 0} ( f + g) = ( f + + g + )1 { f +g<0} +(f + g )1 { f +g<0}, (5.35) we have ( f + g) + apple f + + g + and ( f + g) apple f + g and so f + g is integrable. Moreover, by the additivity of the unsigned integral we also have ( f + g) + dµ = ( f + + g + )1 { f +g 0} dµ ( f + g )1 { f +g 0} dµ, (5.36) ( f + g) dµ = ( f + g )1 { f +g<0} dµ ( f + + g + )1 { f +g<0} dµ, with both expressions on the right meaningful. Thus ( f + g)dµ := ( f + g) + dµ ( f + g) dµ = ( f + + g + )1 { f +g 0} dµ ( f + g )1 { f +g 0} dµ ( f + g )1 { f +g<0} dµ + ( f + + g + )1 { f +g<0} dµ = ( f + + g + )dµ ( f + g )dµ. (5.37) Breaking this expression into a sum of four integrals, and noting that only one type of infinity can occur among these, we can reassemble this into the right-hand side of (5.34). Thus, also the signed integral is additive provided the resulting expressions make sense. Concerning behavior under limits, here we get:

7 Theorem 5.14 (Dominated Convergence Theorem) Let { f n } and g be measurable functions such that R gdµ < and f n appleg for all n 2 N. Then f n! f a.e. ) f n dµ! f dµ (5.38) and, in particular, f is integrable. Proof. We have g + f n Since g 0 and so, by Fatou s lemma, g + liminf f n)dµ apple liminf 53 (g + f n )dµ (5.39) 0 and R gdµ <, we can separate the integral of g on both sides and get lim inf f n dµ apple liminf f n dµ. (5.40) Similarly we get g f n 0 and, since liminf (g f n )=g limsup f n, Fatou s lemma tells us g lim sup f n )dµ apple liminf (g f n )dµ (5.41) Separating the integral of g again, this can be rewritten as lim sup f n dµ apple limsup f n dµ. (5.42) Combining (5.40) and (5.42), we obtain lim inf n dµ apple liminf f n dµ apple limsup f n dµ apple lim sup f n dµ. (5.43) When f n! f a.e., both limes superior and limes inferior of f n are equal to f a.e. The extreme ends of the inequality thus coincide, the limit of integrals of f n exists and equals the integral of the a.e.-limit. It is instructive to note that without the existence of the dominating function g, the theorem falls. This can be for (one or both of) the following reasons: (1) the functions f n increase to infinity locally, e.g., f n := n1 [0,1/n] which tends to zero Lebesgue a.e. and yet its integral equals one, or (2) (in infinite measure spaces) the functions f n spread over sets of larger and larger measure, e.g., f n := n 1 1 [0,n] which tends to zero pointwise (in fact uniformly) and yet its integral with respect to the Lebesgue measure equals one. As is easy to check, the existence of a dominating function makes these impossible. 5.3 Absolute integrability and L 1 -space. The situation with the signed integral improves if we assume that both the positive and negative part of the function integrates to a finite number. Definition 5.15 A measurable function f : X! R is said to be absolutely integrable if the signed integral of f obeys f dµ <. (5.44)

8 54 We will denote the set of absolutely integrable functions by L 1 (or L 1 (X,F, µ) or L 1 (µ), depending on context). Since f = f + + f, absolute integrability implies integrability and we have the bound f dµ apple f dµ. (5.45) As a consequence of the properties of the signed integral, L 1 is closed under (pointwise) addition and scalar multiplication and is thus a vector space. Note, however, that L 1 is not closed under pointwise limits (even in a.e. sense) as f n 2 L 1 and f n! f does not imply f 2 L 1. (Take, e.g., f n := 1 [0,n].) The notion of convergence that ensures this is given by: Definition 5.16 Let { f n }, f be measurable functions. We say that f n! f in L 1 (µ) if f n f dµ! 0 (5.46) To assess the strength of this convergence, we note: Lemma 5.17 Let { f n } L 1. If f n! f in L 1, then f 2 L 1,f n! f in measure and lim f n dµ = f dµ and lim f n dµ = f dµ (5.47) Proof. Assume without loss of generality that R f n f dµ < for all n. (Otherwise drop those terms from the sequence.) Then we claim that f n dµ <. (5.48) sup Indeed, by the triangle inequality for the absolute value and finite additivity of the unsigned integral, f n dµ apple f 1 dµ + f 1 f dµ + f n f dµ. (5.49) The first two integrals are finite while the last integral converges to zero as n! and so the supremum over n of the right-hand side is finite indeed. Next we pick e > 0 and use Markov s inequality to show µ f n f > e apple 1 f n f dµ! 0. (5.50) e So f n! f in measure. By Lemma 4.21 there is {n k } such that f nk! f a.e. and Fatou s lemma then ensures f dµ apple lim f nk dµ apple sup f n dµ <. (5.51) k! So f 2 L 1 as claimed. But then we use (5.45) to write f n dµ f dµ = ( f n f )dµ apple f n f dµ! 0 (5.52) and so the integral of f n tends to that of f. Replacing f n by f n and f by f and using that a b apple a b, the same holds for the integrals of absolute values.

9 As we just saw, from an L 1 -converging sequence (or even a sequence converging in measure) one can always choose an a.e.-converging subsequence. However, L 1 -convergence does not imply a.e.-convergence or even just convergence in measure. Indeed, consider a sequence { f n } defined by f 2 n +k := 1 [k2 n,(k+1)2 n ], k = 0,...,2 n 1. (5.53) Then f n! 0 in L 1 and measure and yet limsup f n = 1 [0,1] while liminf f n = 0. The example f n := n1 [0,1/n] shows that convergence in measure does not imply L 1 convergence. Notwithstanding, for bounded functions on finite measure spaces, convergence in measure is equivalent to L 1 convergence. Indeed: Lemma 5.18 Suppose { f n } obey f n applek a.e. for some constant K 2 (0, ) and all n 2 N. If µ(x) <, then f n! f in measure, f n! f in L 1. (5.54) Proof. We already proved (, so let us focus on ). Suppose f n! f in measure. Then µ f K + e apple µ f n f e + µ f n K (5.55) so f n applek a.e. implies also f applek a.e. Therefore, f n f dµ = f n f 1 { fn f >e} dµ + f n f 1 { fn f applee} dµ apple 2Kµ f n f > e + eµ(x). But this tends to zero as n! followed by e # 0 and so f n! f in L 1 as well. 5.4 Uniform integrability. 55 (5.56) A natural question to ask is what conditions on a pointwise a.e. convergent sequence of measurable functions { f n } imply convergence in L 1. For this we first collect the necessary conditions: Proposition 5.19 (Necessary conditions for L 1 -convergence) Suppose that { f n } and f are measurable functions with f n! f in L 1. Then (1) f 2 L 1 and f n dµ! f dµ, (5.57) (2) (3) lim limsup M! lim limsup d#0 n 1 f n 1 { fn M} dµ = 0, (5.58) f n 1 { fn appled} dµ = 0. (5.59) Note that the condition (3) is trivial in finite measure spaces (and, in particular, in probability spaces). The proof will need the following lemma: Lemma 5.20 Let f 2 L 1. Then 8e > 0 9d > 0 8A 2 F : µ(a) < d ) A f dµ < e. (5.60)

10 56 Proof. Suppose the claim fails. Then there is e > 0 such that for each n 1 there is A n 2 F with µ(a n ) < 2 n such that R A n f dµ e. Define B n := S k n A k. Then µ(b n ) apple 2 n+1 and B n # B := limsup A n. By the Downward Monotone Convergence Theorem for sets, µ(b) apple limsup µ(a n )=0. However, the Dominated Convergence Theorem (with f being the dominating function for functions f 1 Bn ) ensures e apple f dµ apple f dµ! f dµ = 0, (5.61) A n B n B a contradiction. So the claim holds after all. As we will see later, the statement of this lemma is related to the properties of absolutely continuous measures. With this lemma in hand, we can now proceed to give: Proof of Proposition (1) is a restatement of part of Lemma For (2) we note that f n apple f n f + f and so f n 1 { fn 2M} apple f n f + f 1 { f M} + f 1 { f fn M}. (5.62) The integral of the first term tends to zero as n! by our assumption of L 1 convergence. The second term is dominated by the integrable function f and tends to zero a.e. (integrability of f implies that f < a.e.). The Dominated Convergence Theorem shows f 1 { f M} dµ! 0 (5.63) M! For the last term in (5.62) we recall that f n! f in L 1 implies that f n! f in measure. This yields lim sup µ f f n M = 0 (5.64) and so lim sup f 1 { f fn M} = 0 (5.65) by Lemma Part (3) is proved analogously. From f n apple f n f + f and f apple f n f + f n we have f n 1 { fn <d} apple f f n + f 1 { f <2d} + f 1 { fn f >d}. (5.66) The integral of the first term again vanishes by assumed L 1 -convergence. The last integral tends to zero in the limit n! by Lemma The middle function tends to zero in the limit d # 0, and so does its integral thanks tot the Dominated Convergence Theorem. As is not hard to check, the limes superior in parts (2) and (3) of Proposition 5.19 can be replaced by supremum. We now come up with a name for families satisfying these sorts of conditions: Definition 5.21 A family { f a : a 2 I} (any cardinality) of measurable functions is said to be uniformly integrable (UI) if the following hold: (1) { f a } L 1 (µ) and sup a2i f a dµ <, (5.67)

11 (2) (3) To see some examples, we note: Lemma 5.22 lim sup M! a2i lim sup d#0 a2i The following families are UI: (1) { f } for f 2 L 1, (2) { f 1,..., f n } L 1, (3) { f a,k : k = 1,...,n} if { f a,k },k= 1,...,n are UI, (4) { f 2 L 1 : f appleg} for any g 2 L 1. Proof. Exercise for the reader. f a 1 { fa M} dµ = 0, (5.68) f a 1 { fa appled} dµ = 0. (5.69) We also note that Lemma 5.20 holds uniformly over UI families (we state this in a slightly weaker form): Lemma 5.23 Then Let { f a : a 2 I} L 1 obey lim M! a2i f a 1 { fa M} dµ = 0. (5.70) 8e > 0 9d > 0 8A 2 F : µ(a) < d ) sup f a dµ < e. (5.71) a2i A Proof. We will give a slightly different proof than given for Lemma Fix e > 0 and suppose that for some A 2 F we have sup a2i RA f a > 2e. By (5.70), there is M 2 (0, ) such that f a 1 { fa M} dµ < e (5.72) sup a2i and then sup f a dµ apple e + sup f a 1 { fa applem}dµ apple e + Mµ(A). (5.73) a2i A n a2i A n Thus, if µ(a) < e/m, the right hand side is at most 2e. The main result of this section is a rather far-reaching generalization of Lemma Theorem 5.24 Let { f n } and f be measurable functions. If { f n } is UI, then f n! f in measure, f n! f in L 1. (5.74) Proof. Thanks to Lemma 5.17, we only need to prove ). Fix d > 0, consider the integral of f n f and split it according to whether f n f > d or not. The first part yields f n f 1 { fn f >d} dµ apple f n + f 1 { fn f >d} dµ (5.75) Splitting the integral and applying Lemma 5.23 for the UI family { f }[{f n : n 2 N} along with f n! f in measure, the right-hand side tends to zero as n!. 57

12 58 For the second part of the integral, we write f n f 1 { fn f appled} dµ = f n f 1 { fn f appled}1 { f >d} dµ + f n f 1 { fn f appled}1 { f appled} dµ (5.76) Since µ( f > d) < by Markov s inequality, the first integral tends to zero as n! by the Bounded Convergence Theorem. The second integral we further bound as f n f 1 { fn f appled}1 { f appled} dµ apple f n 1 { fn apple2d} dµ + f 1 { f appled} dµ. (5.77) The supremum over n of the first term tends to zero as d # 0 thanks to UI of { f n }. By the Dominated Convergence Theorem, the second integral tends to zero as d # 0 as well. Uniform integrability plays very important role in probability, particularly, in martingale convergence theory etc.