MEASURE THEORY AND LEBESGUE INTEGRAL 15

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1 MASUR THORY AND LBSGU INTGRAL 15 Proof. Let 2Mbe such that µ() = 0, and f 1 (x) apple f 2 (x) apple and f n (x) =f(x) for x 2 c.settingg n = c f n and g = c f, we observe that g n = f n a.e. and g = f a.e. so that R g n dµ = R f n dµ, for all n and R gdµ = R fdµ, by Cor We may apply the MCT to {g n } and obtain that as we wish to show. gdµ= g n dµ = f n dµ In the MCT the key assumption, besides the non-negativity of the functions, was the monotonicity of the sequence. Such monotonicity guaranteed the existence of the its. In the general case, we have the following. Theorem (Fatou s Lemma.) Let {f n } L +, then inf f n dµ apple inf f n dµ. Proof. As in the proof of Thm. 2.6 (iv), if we set g k (x) =inf japplek f j (x), for k =1, 2,...,wehave that {g k } is an increasing sequence in L + and k g k =inf f n. Moreover, g k apple f k so that g k dµ apple inf f n dµ. Therefore, by the MCT, inf f n as we wished to show. dµ = k!+1 g k dµ = k!+1 k!+1 g k dµ apple inf The following result follows at once from Fatou s Lemma and Cor Corollary Let {f n } L + and suppose f n! f a.e. Then fdµapple inf f n dµ. f n dµ, Proof. Let 2Mbe a set such that f n! f in and µ( c ) = 0. Set g n = f n and g = f. Then g n! g, f n = g n a.e. and f = g a.e. By Fatou s Lemma, gdµapple inf g n dµ =inf f n dµ. Remark We observe that the previousr result cannot be improved to have equality even if we assume that the it of the sequence fn dµ exists. Indeed, consider the measure space (N, P(N),µ), where µ is the counting measure. Let s n be the numerical sequence {s n,k } that is equal to 1 if k = n and equals 0 otherwise. Then the sequence of functions on N {s n } converges to 0 pointwise, but s n dµ = s n,k =1 for all n. Therefore, 0= N N sdµapple s n dµ =1. N

2 16 M. M. PLOSO Remark We also observe that there exists a version of the MCT for decreasing sequences {f n }, but one needs to assume that R R f 1 dµ < +1. Precisely,if{f n } L +, f 1 f 2, and f1 dµ < +1, then f n dµ. Proof. We first observe that if 0 apple g apple f, thenf g 0 and f g) dµ = fdµ gdµ. Next, since {f n } is monotone and non-negative, the it f exists it is non-negative, and f n f for each n. Set g n = f 1 f n. Then {g n } is a sequence of non-negative functions such that g n apple g n+1 for all n and converging to f 1 f. Applying the MCT we obtain f 1 dµ f 1 f) dµ = gdµ= g n dµ = f 1 dµ f n dµ. Therefore, since R f 1 dµ < +1 we can subtractive from both sides to obtain f n dµ, as we wished to prove Integration of complex-valued functions. We now pass to consider generic real, or complex, valued functions. If (X, M,µ) is a measure space, f : X! C is measurable, and u =Ref, v =Imf, wewrite f = u + u + i(v + v ). Then, u ±,v ± : X![0, +1). If f is real-valued, we simply write f = f + f. Definition Given a measure space (X, M,µ), given f : X! C is measurable, we say that f is absolutely integrable, or simply integrable, if f dµ < +1; equivalently, if 0 apple u ± dµ, v ± dµ < +1. For f integrable we set u + dµ u dµ + i v + dµ v dµ. In particular, if f is real-valued, we have f + dµ f dµ. Remark Notice that, if f is integrable, f = u + iv, then0apple u ±,v ± apple f so that 0 apple u ± dµ, v ± dµ apple f dµ < +1.

3 MASUR THORY AND LBSGU INTGRAL 17 Conversely, if 0 apple R u ± dµ, R v ± dµ < +1, then f apple u + v = u + + u + v + + v, so that f dµ apple u + dµ + u dµ + v + dµ + v dµ < +1. Moreover, the set of integrable functions is a complex vector space, as it is easy to check, and for integrable f,g and a, b 2 C we have af + bg) dµ = a fdµ+ b gdµ. Proposition The following properties hold true. (i) If f is integrable, then fdµ apple f dµ. (ii) If f,g are integrable, then gdµ for every 2Mif and only if f g =0 a.e. if and only if f g dµ =0. Proof. (i) If R f = 0 the result is trivial. Next, if f is real fdµ = f + dµ f dµ apple f + dµ + f dµ = f dµ. If f is complex-valued, and R f 6= 0, from Cor. 2.7 that sgn f, f are measurable and we set = R fdµ /( R fdµ). Then fdµ = f dµ so that, in particular R f dµ is non-negative. Therefore, fdµ =Re f dµ = Re( f) dµ apple f dµ = f dµ, since = 1. (ii) If f,g 2 L 1 (µ), from Cor we know that R f g dµ = 0 if and only if f g =0 a.e., that is, if and only if f = g a.e. Now, if R f g dµ = 0, then, for every 2M, fdµ gdµ apple f g dµ apple f g dµ =0, so that gdµ for every 2M. Finally, suppose that the above condition holds, and assume for simplicity that f and g are real-valued. Consider the function f g = m and its decomposition m = m + m.then,by assumption we have that R mdµ = 0 for all 2M. In particular if we take + = x : m(x) apple 0,thenwehave 0= mdµ= + m + dµ = + m + dµ, since m + = 0 on c +.Thus,m + = 0 a.e. and, with a similar argument we obtain also m =0 a.e. Hence, m = 0 a.e., i.e. f = g a.e., as we wished to show.

4 18 M. M. PLOSO From now on we will identify functions that di er only on a set of measure 0. Indeed, we have the following definition. Definition Given a measure space (X, M,µ), we consider the equivalent relation on the space of integrable saying that f g if f = g a.e. We define the space L 1 (µ) as the space of equivalent classes of integrable function modulo the relation and we define the norm kfk L 1 (µ) = f dµ. Remark We need to check that the above definition gives in fact a norm. It is clear that kfk L 1 (µ) 0 and that f =0inL 1 (that is, f = 0 a.e.) implies kfk L 1 (µ) = 0. Conversely, if 0=kfk L 1 (µ) = R f dµ, then Cor implies that f = 0 a.e., ie. f = 0 a.e. and f =0inL 1. The symmetry and triangular inequality of the norm follow easily. In Thm we will show that L 1 is a complete normed space, that is, a Banach space. Therefore, we have a norm on L 1 (µ), hence a metric, given by the expression d(f,g) =kf gk L 1 (µ). Then, given {f n } L 1 (µ) we say that f n! f in L 1 (µ) (or, simply in L 1 if the measure µ is understood) if kf n fk L 1! 0 as n! +1. Theorem (Dominated Convergence Theorem.) Let (X, M, µ) be a measure space and {f n } L 1 (µ). Suppose that f n! f pointwise a.e. and that there exists a non-negative g 2 L 1 (µ) such that g f n for all n. Then, f n! f in L 1 (µ) and f n dµ. Proof. Since f n appleg for all n, sincef n! f pointwise a.e., it follows that f is measurable and, by the comparison theorem, that f appleg. Therefore, f 2 L 1 (µ). Observing that f n f apple2g, we apply Fatou s Lemma to the sequence {2g f n f } and obtain 2gdµapple inf = 2gdµ 2g f n f dµ = sup f n f dµ. 2gdµ+inf f n f dµ Since R 2gdµ<+1, we can subtract it from both sides of the inequality and obtain that sup f n f dµ apple 0. R This implis that fn f dµ = 0, that is, f n! f in L 1 (µ). Finally, f n dµ fdµ apple f n f dµ! 0 and the last conclusion follows.

5 MASUR THORY AND LBSGU INTGRAL The space L 1 (µ). Our next goal is to show that L 1 (µ) is a Banach space, that is it is complete in its norm. We recall that a normed space (X, k k X )iscomplete if it is complete as a metric space w.r.t the metric d(f,g) =kf gk X. We also recall that a series P n f n of elements in X is said to be absolutely convergent in X if the numerical series P n kf nk X is convergent. We have the following result. Theorem Anormedspace(X, k k X ) is a Banach space if and only if every absolutely convergent series P n f n is convergent in X. Proof. Suppose first X is a Banach space and that P n f n is an absolutely convergent series. Let {s N } be the sequence of partial sums, that is, s N = P +1 f n. Then, given ">0, there exists n " such that for n " <M<N, NX NX d(s N,s M )=ks N s M k X = kf n k X <". n=m+1 f n X apple n=m+1 Hence, {s N } is a Cauchy sequence in X, that converges, since X is complete. Conversely, suppose every absolutely convergent series is convergent in X. Let {g n } be a Cauchy sequence in X. Hence, for every k =1, 2,..., there exists an integer N k such that if m, n N k, d(g n,g m )=kg n g m k X < 2 k. Notice that we may assume that N k+1 >N k for all k. Then,thesubsequence{g Nk } is such that kg Nk+1 g Nk k X < 2 k.then, P +1 kg N k+1 g Nk k X converges, so that by assumption, P +1 g N k+1 g Nk converges to an element g 2 X. But, the series P +1 g N k+1 g Nk is telescopic and the partial sum MX g Nk+1 g Nk = g NM+1 g N1, so that g Nk g N1! g as k! +1. Hence, the original sequence must converge too, and the conclusion follows. Theorem Let {f n } be a sequence of functions in L 1 (µ) such that P +1 kfk L 1 (µ) < +1. Then, there exists f 2 L 1 (µ) such P +1 f n converges to f in L 1 (µ) and f n dµ. As a consequence, L 1 (µ) is a Banach space. Finally, the simple functions are dense in L 1 (µ). Proof. By Cor (ii) we know that X f n dµ = X n n f n dµ apple kf n k L 1 (µ) < +1. Then, setting g = P +1 f n, wehaveg 2 L 1 (µ), and hence it is finite a.e. This implies that P +1 f n converges but on a set of measure 0, call f such it. We apply the DCT to the sequence of partial sums s n = P n f k. Clearly, s n! f pointwise, and also nx nx s n = apple f k = g. f k

6 20 M. M. PLOSO Hence, the DCT applies to give that s n! f in L 1 R (µ) and sn dµ = R fdµ, that is nx nx f k dµ = f k dµ = f n dµ. In particular, every absolutely convergent series also converges in L 1 (µ), hence L 1 (µ) is complete. Finally, let f 2 L 1 (µ). Then, f is finite a.e. and the sequence of simple functions {' n } in Thm. 2.9 (2), is such that ' n! f pointwise, and ' n apple f. Therefore, we can apply the DCT to obtain that ' n! f in L 1 (µ). We conclude this part by comparing L 1 (µ) withl 1 (µ). If f 2 L 1 (µ) and g is given by by Prop. 2.12, then g = f µ-a.e. and g 2 L 1 (µ). Conversely, if g 2 L 1 (µ), then g is also M-measurable and R g dµ = R g dµ. Then g 2 L 1 (µ). Therefore, we have shown that the spaces L 1 (µ) and L 1 (µ) can be identified and we will do in what follows. 3. The Lebesgue measure in R Having established the definition and main properties of abstract integrals, we now go back to the construction of measures and in particular of the Lebesgue measure and of other Borel measures in R, and eventually in R n Outer measures. Given a set X, an outer measure on X is a set function µ defined on the set of parts of X, P(X ) such that (i) µ (;) = 0; (ii) if A B X,thenµ (A) apple µ (B); (iii) µ [ +1 A j apple P +1 µ (A j ), for all {A j } X. The reason for this definition is that outer measures arise naturally when one has a family of elementary sets and a notion of measure for such sets. Proposition 3.1. Let be a collection of sets in P(X ) such that ;, X 2, and let :! [0, +1] such that (;) =0be given. Define Then µ is an outer measure. n X +1 µ (A) =inf ( j ): A +1 [ o j, j 2. The key example to keep in mind is when is the collection of (all, or a subclass of) intervals I in R, and (I) is the length of I, which could be +1 if I is a ray. Proof. First of all we observe that the definition makes sense, since for any A X there exists a covering of A with sets in, e.g. {X, ;, ;,...}, where;, X2. Also, it is clear that µ (;) = 0, since ; can be covered by {;, ;,...}. Property (ii) in the definition of an outer measure is satisfied since, if A B, and { j } is a covering of B with sets in, { j } is also a covering of A. Thus, the infimum in the definition of µ (A) is taken with respect a larger collection of converings than in the case of µ (B); hence (ii) holds.

7 MASUR THORY AND LBSGU INTGRAL 21 Finally, let {A j } P(X ). Fix ">0. Then, by the properties of the infimum of real numbers, for each j, thereexistsacovering{ j,k } of sets in of A j such that ( j,k ) <µ (A j )+"2 j. Therefore, { j,k } for j, k =1, 2,... is a covering of [ +1 A j by sets in such that µ +1 [ n X A j =inf ( n ): n 2, apple = n µ (A j )+"2 j µ (A j )+". Since ">0 was arbitrary, (iii) follows. +1 [ A j +1 [ n o apple j, ( k,j ) The key step to pass from an outer measure to an actual measure is based on the following definition. Definition 3.2. Let µ be an outer maesure on a set X. We say that a set X is µ - maesurable if µ (A) =µ (A \ )+µ (A \ c ), for all A X. Observe that we always have the inequality µ (A) apple µ (A \ )+µ (A \ c ), by subadditivity. Then, in order to prove the µ -maesurability of a set, we only need to prove the reverse inequality µ (A) µ (A \ )+µ (A \ c ), for all sets A X. This inequality is trivial if µ (A) =+1. Therefore, X is µ -maesurable if and only if µ (A) µ (A \ )+µ (A \ c ) for all A with µ (A) < +1. (10) The key step to pass from an outer measure to an actual measure is the following result, known as Carathéodory s theorem. Theorem 3.3. (Carathéodory s Theorem.) Let X be a set and µ an outer measure on X. Let M be the collection of µ -maesurable sets, and µ the restriction of µ to M. Then M is a -algebra, and µ is a complete measure on M. Proof. We first observe that 2Mimplies that also c 2Msince the condition (10) is symmetric in and c. In order to show that M is a -algebra, it su ces to show that it is closed under countable unions of disjoint sets see the comment regarding formula (1). To this end, we first show that M is closed under finite unions. Let,F 2Mand let A X.Then