m(r n ) = Bor(R n ),

Transcription

1 6. The Lebesgue mesure In this section we pply vrious results from the previous sections to very bsic exmple: the Lebesgue mesure on R n. Nottions. We fix n integer n 1. In Section 21 we introduced the semiring of hlf-open boxes in R n : J n = { } { n } [ j, b j ) : 1 < b 1,..., n < b n P(R n ). j=1 For non-empty box A = [ 1, b 1 ) [ n, b n ) J n, we defined its n-dimesnionl volume by n vol n (A) = (b k k ). We lso defined vol n ( ) = 0. By Theorem 4.2, we know tht vol n is finite mesure on J n. Definitions. The mximl outer extension of vol n is clled the n-dimensionl outer Lebesgue mesure, nd is denoted by λ n. The λ n-mesurble sets in R n will be clled n-lebesgue mesurble. The σ- lgebr m λ n (R n ) will be denoted simply by m(r n ). The mesure λ n is m(r n ) simply denoted by λ n, nd is clled the n-dimensionl Lebesgue mesure. Although this nottion my pper to be confusing, it turns out (see Proposition 5.3) tht λ n is indeed the mximl outer extension of λ n. In the cse when n = 1, the subscript will be ommitted. We know (see Section 21) tht S(J n ) = Σ(J n ) = Bor(R n ). Using the fct tht the semiring J n is σ-totl in R n, by the definition of the outer Lebesgue mesure, we hve (1) λ n(a) = inf { vol n (B k ) : (B k ) J n, B k A }, A R n Using Corollry 5.2, we hve the equlity m(r n ) = Bor(R n ), where Bor(R n ) is the completion of Bor(R n ) with respect to the mesure λ n Bor(Rn ). This mens tht subset A R n is Lebesgue mesurble, if nd only if there exists Borel set B nd neglijeble set N such tht A = B N. (The fct N is neglijeble mens tht λ n(n) = 0, nd is equivlent to the existence of Borel set C N with λ n (C) = 0.) Exercise 1. Let A = [ 1, b 1 ) [ n, b n ) be hlf-open box in R n. Assume A (which mens tht 1 < b 1,..., n < b n ). Consider the open box Int(A) nd the closed box A, which re given by Int(A) = ( 1, b 1 ) ( n, b n ) nd A = [ 1, b 1 ] [ n, b n ]. 200

2 6. The Lebesgue mesure 201 Prove the equlities λ n ( Int(A) ) = λn ( A ) = voln (A). Remrks 6.1. If D R n is non-empty open set, then λ n (D) > 0. This is consequence of the bove exercise, combined with the fct tht D contins t lest one non-empty open box. The Lebesgue mesure of countble subset C R n is zero. Using σ-dditivity, it suffices to prove this only in the cse of singletons C = {x}. If we write x in coordintes x = (x 1,..., x n ), nd if we consider hlf-open boxes of the form J ε = [x 1, x 1 + ε) [x n, x n + ε), then the obvious inclusion {x} J ε will force 0 λ n ( {x} ) λn (J ε ) = ε n, so tking the limit s ε 0, we indeed get λ n ( {x} ) = 0. The (outer) Lebesgue mesure is completely determined by its vlues on open sets. More explicitly, one hs the following result. Proposition 6.1. Let n 1 be n integer. For every subset A R n one hs: (2) λ n(a) = inf{λ n (D) : D open subset of R n, with D A}. Proof. Throughout the proof the set A will be fixed. Let us denote, for simplicity, the right hnd side of (2) by ν(a). First of ll, since every open set is Lebesgue mesurble (being Borel), we hve λ n (D) = λ n(d), for ll open sets D, so by the monotonicity of λ n, we get the inequlity λ n(a) ν(a). We now prove the inequlity λ n(a) ν(a). Fix for the moment some ε > 0, nd use (1). to get the existence of sequence (B k ) J n, such tht B k A, nd vol n (B k ) < λ n(a) + ε. For every k 1, we write B k = [ (k) 1, b(k) 1 ) [(k) n, b (k) n ), so tht vol n (B k ) = n j=1 (b(k) 1 (k) j ). Using the obvious continuity of the mp n R t (b (k) 1 (k) j t) R, j=1 we cn find, for ech k 1 some numbers c (k) 1 < (k) 1,..., c(k) n < (k) n, with n (3) (b (k) 1 c (k) j ) < ε n 2 k + (b (k) 1 (k) j ). j=1 Notice tht, if we define the hlf-open boxes E k = [c (k) 1, b(k) j=1 1 ) [c(k) n, b (k) n ),

3 202 CHAPTER III: MEASURE THEORY then for every k 1, we clerly hve B k Int(E k ), nd by Exercise 1, combined with (3), we lso hve the inequlity ( λ n Int(Ek ) ) = vol n (E k ) < ε 2 k + vol n(b k ). Summing up we then get (4) ( λ n Int(Ek ) ) < [ ε 2 k + vol n(b k ) ] = ε + vol n (B k ) < 2ε + λ n(a). Now we observe tht by σ-sub-dditivity we hve ( ) ( λ n Int(E k ) λ n Int(Ek ) ), so if we define the open set D = Int(E k), then using (4) we get (5) λ n (D) < 2ε + λ n(a). It is cler tht we hve the inclusions A B k Int(E k ) = D, so by the definition of ν(a), combined with (5), we finlly get ν(a) λ n (D) < 2ε + λ n(a). Up to this moment ε > 0 ws fixed. Since the inequlity ν(a) < 2ε + λ n(a) holds for ny ε > 0 however, we finlly get the desired inequlity ν(a) λ n(a). The Lebesgue mesure cn lso be recovered from its vlues on compct sets. Proposition 6.2. Let n 1 be n integer. For every Lebesgue mesurble subset A R n one hs: (6) λ n (A) = sup{λ n (K) : K compct subset of R n, with K A}. Proof. Let us denote, for simplicity, the right hnd side of (6) by µ(a). First of ll, by the mononoticity we clerly hve the inequlity λ n (A) µ(a). To prove the inequlity λ n (A) µ(a), we shll first use reduction to the bounded cse. For ech integer k 1, we define the compct box B k = [ k, k] [ k, k]. Notice tht we hve B 1 B 2..., with B k = R n. We then hve B 1 A B 2 A..., with (B k A) = A, so using the Continuity Lemm 4.1, we hve (7) λ n (A) = lim k λ n(b k A) = sup { λ n (B k A) : k 1 }. Fix for the moment some ε > 0, nd use the (7) to find some k 1, such tht λ n (A) λ n (B k A) + ε. Apply Proposition 6.1 to the set B k A, to find n open set D, with D B k A, nd λ n (B k A) λ n (D) ε. On the one hnd, we hve (8) λ n (B k ) = λ n (B k A) + λ n (B k A) λ n (B k A) + λ n (D) ε λ n (B k A) + λ n (B k D) ε.

4 6. The Lebesgue mesure 203 On the other hnd, we hve so using (8) we get the inequlity λ n (B k ) = λ n (B k D) + λ n (B k D), λ n (B k D) + λ n (B k D) λ n (B k A) + λ n (B k D) ε, nd since ll numbers involved in the bove inequlity re finite, we conclude tht λ n (B k D) λ n (B k A) ε λ n (A) 2ε. Obviously the set K = B k D is compct, with K B k A A, so we hve µ(a) λ n (K), hence we get the inequlity µ(a) λ n (A) 2ε. Since this is true for ll ε > 0, the desired inequlity µ(a) λ n (A) follows. Corollry 6.1. For set A R n, the following re equivlent: (i) A is Lebesgue mesurble; (ii) there exists neglijeble set N nd sequence of (K j ) j=1 subsets of R n, such tht A = N K j. j=1 Proof. (i) (ii). Strt by using the boxes B k = [ k, k] [ k, k] of compct which hve the property tht B j = R n, so we get A = (B k A). Fix for the moment k. Apply Proposition 6.2. to find sequence (Cr k ) r=1 of compct subsets of B k A, such tht lim r λ n (Cr k ) = λ n (B k A). Consider the countble fmily (Cr k ) k,r=1 of compct sets, nd enumerte it s sequence (K j) j=1, so tht we hve K j = Cr k. j=1 r=1 If we define, for ech k 1, the sets E k = r=1 Ck r B k A nd N k = (B k A) E k, then, becuse of the inclusion C k r E k B k A, we hve the inequlities (9) 0 λ n (N k ) = λ n (B k A) λ n (E k ) λ n (B k A) λ n (C k r ), r 1. Using the fct tht lim λ n(cr k ) = λ n (B k A) λ n (B k ) <, r the inequlities (9) force λ n (N k ) = 0, k 1. Now if we define the set N = A ( j=1 K j), we hve [ N = (B k A) ( ) ] [ K j = (B k A) ( ) ] E p j=1 [ ] (Bk A) E k = N k, which proves tht λ n (N) = 0. The impliction (ii) (i) is trivil. p=1

5 204 CHAPTER III: MEASURE THEORY Proposition 6.2 does not hold if A R n is non-mesurble. In fct the equlity (6), with λ n replced by λ n, essentilly forces A to be mesurble, s shown by the following. Exercise 2. Let A R n be m rbitrry subset, with λ n(a) <. Prove tht the following re equivlent: (i) A is Lebesgue mesurble; (ii) λ n(a) = sup{λ n (K) : K compct subset of R n, with K A}. Propositions 6.1 nd 6.2 re regulrity properties. The following terminology is useful: Definitions. Suppose A is σ-lgebr on X, nd µ is mesure on A. Suppose we hve sub-collection F A. (i) We sy tht µ is regulr from below, with respect to F, if µ(a) = sup { µ(f ) : F A, F F }. (ii) We sy tht µ is regulr from bove, with respect to F, if µ(a) = inf { µ(f ) : F A, F F }. With this terminology, Proposition 6.1 gives the fct tht the Lebesgue mesure is regulr from bove with respect to open sets, while Proposition 6.2 gives the fct tht the Lebesgue mesure is regulr from below with respect to compct sets. Hint: Exercise 3. For subset A R n, prove tht the following re equivlent: (i) A is Lebesgue mesurble; (ii) There exist sequence of compct sets (K j ) j=1, nd dequence of open sets (D j ) j=1, such tht j=1 K j A j=1 D j, nd the difference ( j=1 D ) ( j j=1 K ) j is neglijeble. For the impliction (i) (ii) nlyze first the cse when λ (A) <. Then write A s countble union of sets of finite outer mesure. In the one-dimensionl cse n = 1, the Lebesgue mesure of open sets cn be computed with the id of the following result. Proposition 6.3. For every open set D R, there exists countble (or finite) pir-wise disjoint collection {J i } i I of open intervls with D = i I J i. Proof. For every point x D, we define x = inf{ < x : (, x) D} nd b x = sup{b > x : (x, b) D}. (The fct tht D is open gurntees the fct tht both sets bove re non-empty.) It is cler tht, for every x D, the open intervl J x = ( x, b x ) is contined in D, so we hve the equlity D = x D J x. The problem t this point is the fct tht the collection {J x } x D is not pir-wise disjoint. Wht we need to find is countble (or finite) subset X D, such tht the sub-collection {J x } x X is pir-wise disjoint, nd we still hve D = x X J x. One wy to do this is bsed on the following Clim: For two points x, y D, the following re equivlent: (i) x J y ; (ii) J x J y ; (ii) J x J y ; (iii) J x = J y.

6 6. The Lebesgue mesure 205 To prove the impliction (i) (ii) we observe tht if x J y, then y < x < b y, so we hve ( y, x) D nd (x, b y ) D, which mens tht x y nd b x b y, therefore we hve the inclusion J x = ( x, b x ) ( y, b y ) = J y. The impliction (ii) (iii) is trivil. To prove (iii) (iv), ssume J x J y, nd pick point z J x J y. Using the impliction (i) (ii) we hve the inclusions J z J x nd J z J y. In prticulr we hve x J z, so gin using the inpliction (i) (ii) we get J x J z, which mens tht we hve in fct the equlity J x = J z. Likewise we hve the equlity J y = J z, so (iv) follows. The impliction (iv) (i) is trivil. Going bck to the proof of the Proposition, we now see tht, using the fct tht ny open intervl contins rtionl number, if we put X 0 = D Q, then for ny y D, there exists x X 0, such tht J x = J y. This gives the equlity D = x X 0 J x, this time with the indexing set X 0 countble. Finlly, if we equip the set X 0 with the equivlence reltion x y J x = J y, nd we choose X X 0 to the list of ll equivlence clsses. This mens tht, for every y X 0, there exists unique x X with J x = J y. It is cler now tht we still hve D = x X J x, but now if x, x X re such tht x x, then x x, so we hve J x J x, which by the Clim gives J x J x =. Comments. When we wnt to compute the Lebesgue mesure of n open set D R, we should first try to write D = i I J i with (J i ) i I countble (or finite) pir-wise collection of open intervls. If we succeed, then we would hve λ(d) = i I λ(j i ). For intervls (open or not) the Lebesgue mesure is the sme s the length. There re instnces when we cn mnge only to write given open set D s union D = J k, with the J s not necessrily disjoint. In tht cse we cn only get the estimte λ(d) λ(j k ). Exmple 6.1. Consider the ternry Cntor set K 3 [0, 1], discussed in III.3. We know (see Remrks 3.5) tht one cn find pir-wise sequence (D n ) n=0 of open subsets of (0, 1) such tht K 3 = [0, 1] n=0 D n, nd such tht, for ech n 0, the open set D n is disjoint union of 2 n intervls of length 1/3 n+1. In prticulr, this mens tht λ(d n ) = 2 n /3 n+1, so λ(k 3 ) = λ ( [0, 1] ) ( ) 2 n λ D n = 1 λ(d n ) = 1 = 0. 3n+1 n=0 Wht is interesting here (see Remrks 3.5) is the fct tht crd K 3 = c. Remrk 6.2. An interesting consequence of the bove computtion is the fct tht ll subsets of K 3 re Lebesgue mesurble, i.e. one hs the inclusion P(K 3 ) m(r). This gives the inequlity Since we lso hve m(r) P(R), we get n=0 crdm(r) crd P(K 3 ) = 2 crd K3 = 2 c. crdm(r) crd P(R) = 2 crd R = 2 c, n=0

7 206 CHAPTER III: MEASURE THEORY so using the Cntor-Bernstein Theorem we get the equlity crdm(r) = 2 c. We lso know (see Corollry 2.5) tht crd Bor(R) = c. As consequence of this difference in crdinlities, one gets the fct tht we hve strict inclusion (10) Bor(R) m(r). Lter on we shll construct (more or less) explicitly Lebesgue mesurble set which is not Borel. Exercise 4. The strict inclusion (10) holds lso if R is replced with R n, with n 2. In this cse, insted of using Cntor sets, one cn proceed s follows. Consider the set S = R n 1 {0}. Prove tht λ n (S) = 0. Conclude tht crdm(r n ) = 2 c. One key feture of the Lebesgue (outer) mesure is the trnsltion invrince property, described in the following result. To formulte it we introduce the following nottion. For n integer n 1, point x R n, nd subset A R n, we define the set A + x = { + x : A}. Remrk tht the mp Θ x : R n + x R n is homeomorphism. In prticulr, both Θ x nd Θ 1 x = Θ x re Borel mesurble, which mens tht, for set A R n, one hs the equivlence A Bor(R n ) A + x Bor(R n ). Proposition 6.4. Let n 1 be n integer. For ny set A R n one hs the equlity λ n(a + x) = λ n(a). Proof. Fix A nd x. First remrk tht, for every hlf-open box B J n, its trnsltion B + x is gin hlf-open box, nd we hve the equlity vol n (B + x) = vol n (B). Fix for the moment ε > 0, nd choose sequence (B k ) J n, such tht A B k, nd vol n (B k ) λ n(a) + ε. Then, using the obvious inclusion A + x (B k + x), by the remrk mde t the begining of the proof, combined with the monotonicity of the outer Lebesgue mesure, we hve ( ) λ n(a + x) λ n (B k + x) λ n(b k + x) = = vol n (B k + x) = vol n (B k ) λ n(a) + ε. Since the inequlity λ n(a + x) λ n(a) + ε holds for ll ε > 0, we get λ n(a + x) λ n(a).

8 6. The Lebesgue mesure 207 The other inequlity follows from the bove one pplied to the set A + x nd the trnsltion by x. Corollry 6.2. For subset A R n, one hs the equivlence A m(r n ) A + x m(r n ). Proof. Write A = B N, with B Borel, nd N neglijeble. Then we hve A + x = (B + x) (N + x). The set B + x is Borel. By the bove result we hve λ n(n + x) = λ n(n) = 0, i.e. N + x is neglijeble. Therefore A + x is Lebesgue mesurble. As we hve seen, the fct tht there exist Lebesgue mesurble sets tht re not Borel is explined by the difference in crdinlities. Since crdm(r n ) = 2 c = crd P(R n ), it is legitimte to sk whether the inclusion m(r n ) P(R n ) is strict. In other words, do there exist sets tht re not Lebesgue mesurble? The nswer is ffirmtive, s discussed in the following. Exmple 6.2. Equipp R with the equivlence reltion x y x y Q. Denote by R/Q the quotient spce (this is in fct the quotient group of (R, +) with respect to the subgroup Q), nd denote by π : R R/Q the quotient mp. Since for every x R, one cn find some y x, with y [0, 1), it follows tht the mp π [0,1) : [0, 1) R/Q is surjective. Choose then mp φ : R/Q [0, 1), such tht φ π = Id, nd put E = φ(r/q). The set E is complete set of representtives for the equivlence reltion. In other words, E [0, 1) hs the property tht, for every x R, there exists exctly one element y E, with x y. In prticulr, the collection of sets (E + q) q Q is pir-wise disjoint, nd stisfies q Q (E + q) = R. Using σ-sub-dditivity, we get = λ(r) q Q λ (E + q). Since (by Proposition 6.5) we hve λ (E + q) = λ (E), the bove inequlity forces λ (E) > 0. Clim: The set E is not Lebesgue mesurble Assume E is Lebesgue mesurble. If we define the set X = Q [0, 1), then the sets E + q, q X re pir-wirse disjoint. On the one hnd, the mesurbility of E, combined with the Corollry 6.2 would imply the mesurbility of the set S = q X (E + q). On the other hnd, the equlities λ(e + q) = λ(e) > 0 will force λ(s) =. But this is impossible, since we obviously hve S [0, 2), which forces λ(s) 2. Exercise 5. Let E m(r n ). Prove tht the mp is continuous. R n x λ ( E (E + x) ) [0, ] Hint: Anlyze first the cse when E is compct. In this prticulr cse, show tht for every x 0 R n nd every open set D E (E + x 0 ), there exists some neighborhood V of x 0, such tht D E (E + x), x V.

9 208 CHAPTER III: MEASURE THEORY Use then regulrity from bove, combined with the inequlity 1 λ(a) λ(b) λ(a B), for ll A, B m(r n ), with λ(a), λ(b) <. In the generl cse, use regulrity from below. (The cse λ(e) = is trivil.) Exercise 6. Let E m(r n ), be such tht λ n (E) > 0. Prove tht the set is neighborhood of 0. Hint: E E = {x y : x, y E} Assume the contrry, which mens tht there exists sequence (x p) p=1 Rn (E E), with lim p x p = 0. This will force E (E + x p) =, p 1. Use the preceding Exercise to get contrdiction. We re now in position to construct Lebesgue mesurble set which is not Borel. Exmple 6.3. In Section 3 we discussed the compct spce T = {0, 1} ℵ0 nd the mps φ r : T (α n ) α n n=1 (r 1) [0, 1]. rn For ech r 2 the mp φ r : T [0, 1] is continuous so the set K r = φ r (T ) is compct. We hve K 2 = [0, 1], nd K 3 is the ternry Cntor set. We lso know (see Theorem 3.5) tht, for set A T, one hs the equivlence n=1 (11) A Bor(T ) φ r (A) Bor(K r ). Choose now set E [0, 1] which is not Lebesgue mesurble. In prticulr, E is not Borel, so E Bor([0, 1]). Since φ 2 : T [0, 1] is surjective, by (11) it follows tht the set A = φ 1 2 (E) is not in Bor(T ). Agin, by (11) it follows tht the set S = φ 3 (A) is not in Bor(K 3 ). Since Bor(K 3 ) = Bor(R) K3 this gives S Bor(R). Notice however tht since S K 3, it follows tht S is Lebesgue mesurble. Comment. When one wnts to prove tht Lebesgue mesurble set M R hs positive mesure, sufficient condition for this property is tht Int(M) (see Remrk 6.1). It turns out however tht this condition is not lwys necessry, s seen from the following: Exercise 7. Strt with n rbitrry inervl [0, 1], nd list ll rtionl numbers in [0, 1] s sequence Q [0, 1] = {x n } n=1. Fix some ε > 0, nd consider the open set ( D = xn ε 2 n+1, x n + ε ). 2 n+1 n=1 Consider the compct set K = [0, 1] D. (i) Prove tht λ(d) ε. (ii) Prove tht λ(k) 1 ε. (iii) Prove tht Int(K) =. Hint: For (iii) use the fct tht K Q =. 1 This inequlity holds for ny dditive mp defined on ring.

10 6. The Lebesgue mesure 209 Exercise 8*. Prove tht, for every non-empty open set D R, nd ny two positive numbers α, β with α + β < λ(d), there exist compct sets A, B D, with λ(a) > α, λ(b) > β, such tht A B = nd (A B) Q =. Hint: Write D s union of pir-wise disjoint sequence (J n) n=1 of open intervls, so tht λ(d) = n=1 λ(jn). Find then two sequences (αn) n=1 nd (βn) n=1 of positive numbers, such tht n=1 αn > α, n=1 βn > β, nd αn + βn < λ(jn), for ll n 1. This reduces essentilly the problem to the cse when D is n open intervl, for which one cn use the construction outlined in Exercise 7. Exercise 9*. Construct o Borel set A R, such tht, for every open intervl I R one hs λ(i A) > 0 nd λ(i A) > 0. Hints: List ll open intervls with rtionl endpoints s sequence (I n) n=1. Strt (use exercise 8) off by choosing two compct sets A 1, B 1 I 1, with A 1 B 1 =, (A 1 B 1 ) Q =, nd λ(a 1 ), λ(b 1 ) > 0. Use Exercise 5 to construct two sequences (A n) n=1 nd (Bn) n=1 of compct sets, such tht, for ll n 1 we hve: (i) A n B n = ; (ii) (A n B n) Q = ; (iii) λ(a n), λ(b n) > 0; (iv) A n+1 B n+1 I n+1 [ n (A k B k ) ]. Put A = n=1 An nd B = n=1 Bn. Notice tht A B =, λ(a), λ(b) > 0, nd λ(a In), λ(b In) > 0, n 1. In the reminder of this section we discuss some pplictions of the Lebesgue mesure to the theory of Riemnn integrtion. The following techincl result will be very useful. Lemm 6.1. Let f : [, b] R be non-negtive Riemnn integrble function, let A, B [, b] be two disjoint sets, with A B = [, b]. Then one hs the estimtes λ (A) inf z A f(z) Proof. Define the numbers f(t) dt (b ) sup f(x) + λ (B) sup f(y). x A y B α = sup f(x), β = sup f(y), nd γ = inf f(z). x A y B z A Recll first tht, if for ech prtition = ( = x 0 < x 1 < < x n = b) of [, b], we define the lower nd the upper Drboux sums of f with respect to : n L(, f) = (x k x k 1 ) inf f(t), t [x k 1,x k ] U(, f) = n (x k x k 1 ) sup t [x k 1,x k ] f(t), then one hs the equlities (12) f(t) dt = sup { L(, f) : prtition of [, b] } = = inf { U(, f) : prtition of [, b] }. Fix now prtition = ( = x 0 < x 1 < < x n = b) of [, b], nd define the set S = { k {1,..., n} : [x k 1, x k ] A }.

11 210 CHAPTER III: MEASURE THEORY It is cler tht so we get (13) (14) inf f(x) α, sup x [x k 1,x k ] inf f(y) β, sup y [x k 1,x k ] Consider now the sets f(x) γ, k S, x [x k 1,x k ] f(y) 0, k {1,..., n} S, y [x k 1,x k ] L(, f) [ (x k x k 1 ) ] α + [ (x k x k 1 ) ] β k S U(, f) [ (x k x k 1 ) ] γ k S k S M = [x k 1, x k ] nd N = k 1, x k ]. k S k S[x Since the intervls involded in both M nd N hve t most singleton overlps, it follows tht we hve the equlities k x k 1 ) = λ(m) nd k S(x (x k x k 1 ) = λ(n), k S so the estimtes (13) nd (14) red (15) (16) L(, f) λ(m) α + λ(n) β U(, f) λ(m) γ Since we clerly hve A M [, b] nd N B, we hve the inequlities so the inequlities (15) nd (16) give λ (A) λ(m) b nd λ(n) λ (B), L(, f) (b ) α + λ (B) β nd U(, f) λ (A) γ. Since is rbitrry, the desired inequlity then follows from (12). One ppliction of the bove result is the following. Proposition 6.5. If f : [, b] R is Riemnn integrble, nd the set is neglijeble, then (17) N = {x [, b] : f(x) 0} f(x) dx = 0. Proof. Since f is bounded, there exists some constnt C > 0, such tht the Riemnn integrble functions C+f nd C f re both non-negtive. Apply Lemm 6.1 to these two functions with A = [, b] N nd B = N. Since f [,b] N = 0, we get (C ± f) [,b] N = C, so we get [C ± f(x)] dx (b ) C,

12 6. The Lebesgue mesure 211 which yields ± f(x) dx = from which (17) immeditely follows. { } b [C ± f(x)] C dx = [C ± f(x)] dx (b ) C 0, In order to mke the exposition bit esier to follow, it will be helpful to introduce the following Convention. Given two functions f 1, f 2 : [, b] R, nd reltion r on R (in our cse r will be either =, or, or ), we write if the set f 1 r f 2,.e. A = { x [, b] : f 1 (x) r f 2 (x) } hs neglijeble complement in [, b], i.e. λ ( [, b] A ) = 0. The brevition.e. stnds for lmost everywhere. For exmple, using this convention, Proposition 6.6 reds: if f : [, b] R is Riemnn integrble, nd f = 0,.e., then f(x) dx = 0. Exercise 10. A. Prove tht =.e is n equivlence reltion, nd.e nd.e re trnsitive reltions on the collection of ll function [, b] R. B. Prove tht f 1 f 2,.e. nd f 1 f 2,.e. imply f 1 = f 2,.e. C. Prove tht these reltions re comptible with the rithmetic opertions, in the exct wy s their honest versions. For exmple, if r is one of =, or, or, nd if f 1 r f 2,.e. nd g 1 r g 2,.e., then (f 1 + g 1 ) r (f 2 + g 2 ),.e. Exercise 11. Let f, g : [, b] R be continuous functions, such tht f g,.e. Prove tht f g. Exercise 12. Let f : [, b] R be non-negtive Riemnn integrble function, with f(x) dx = 0. Prove tht f = 0,.e. Comment. Riemnn integrbility is quite rigid condition. For exmple the chrcteristic function κ Q [,b] of the set of rtionl numbers in [, b] is not Riemnn integrble. By the bove result however, we cn introduce slightly weker notion, which will mke such functions integrble, in weker sense. This will be first improvement of the Riemnn integrtion theory. Eventully (see Chpter IV), more sofisticted theory - the Lebesgue integrl - will emerge. Definition. We sy tht function f : [, b] R is lmost Riemnn integrble, if there exists Riemnn integrble function g : [, b] R, with f = g,.e. Of course, such g is not unique. Notice however tht, if h : [, b] R is nother Riemnn integrble function, with f = h,.e., then g = h,.e., so by Proposition 6.6, we immeditely get the equlity g(x) dx = h(x) dx. This observtion shows tht we cn unmbiguously define f(x) dx = g(x) dx.

13 212 CHAPTER III: MEASURE THEORY Exmple 6.4. Consider the function f = κ Q [,b]. Since Q [, b] is neglijeble, we hve f = 0,.e. So f is lmost Riemnn integrble (lthought it is not Riemnn integrble), nd we hve f(x) dx = 0. We now focus our ttention to (honest) Riemnn integrbility, with n eye on the role plyed by continuity. For function f : [, b] R we define the set D f = { x [, b] : f not continuous t x }. It is well-known tht continuous functions re Riemnn integrble. There re discontinuous functions which re still Riemnn integrble, for instnce we know tht (18) D f finite = f Riemnn integrble. Nottions. Let f : [, b] R be bounded function. Suppose = ( = x 0 < x 1 < < x n = b) is prtition. For ech k {1,..., n} we consider the numbers M k = nd we define the functions sup f(t) nd m k = t [x k 1,x k ] inf f(t), t [x k 1,x k ] f = m 1 κ [x0,x 1] + m 2 κ (x1,x 2] + + m n κ (xn 1,x n], f = M 1 κ [x0,x 1] + M 2 κ (x1,x 2] + + M n κ (xn 1,x n]. Clerly the functions f nd f hve only finitely mny points of discontinuity, so they re Riemnn integrble. With these nottions we hve the following Proposition 6.6. For bounded function f : [, b] R, the following re equivlent: (i) f is Riemnn integrble; (ii) inf { [f (x) f (x)] dx : prtition of [, b] } = 0; (iii) there exists sequence ( p ) p=1 of prtitions of [, b], with , [ nd lim p f p (x) f p (x) ] dx = 0. Proof. From the definition of Riemnn integrbility, we know tht (i) is equivlent to ny of the following two conditions (ii ) inf { U(, f) L(, f) : prtition of [, b] } = 0; (iii ) there exists sequence ( p ) p=1 of prtitions of [, b], with , [ nd lim p U( p, f) L( p, f) ] = 0. Then the Proposition follows immeditely from the fct tht, for every prtition one hs the equlities f (x) dx = L(, f) nd f (x) dx = U(, f). The following result gives complete description of the reltionship between Riemnn integrbility nd continuity. Theorem 6.1 (Lebesgue s criterion for Riemnn integrbility). Let f : [, b] R be bounded function. The following re equivlent:

14 6. The Lebesgue mesure 213 (i) f is Riemnn integrble; (ii) the discontinuity set D f is neglijeble. Proof. (i) (ii). Assume f is Riemnn integrble. Using Proposition 6.7, there exists sequence ( p ) p=1 of prtitions of [, b], such tht nd Notice tht lim p [ f p (x) f p (x) ] dx = 0. (19) f 1 f 2 f 3 f f 3 f 2 f 1. Define the Riemnn integrble functions h p = f p f p, p N. We then clerly hve (α) h p h p+1 0, p N; (β) lim p h p(x) dx = 0. Using (α) we cn define the function h : [, b] R by h(x) = lim p h p(x), x [, b]. Clim 1: The set N = {x [, b] : h(x) 0} is neglijeble. First of ll, the functions h p re ll Lebesgue mesurble. Secondly, since h is point-wise limit of sequence of Lebesgue mesurble functions, it follows (see Theorem 3.2) tht h itself is Lebesgue mesurble. In prticulr N is Lebesgue mesurble. For every integer j 1, define N j = { x [, b] : h(x) > 1 j }, so tht the sets N j, j 1 re gin Lebesgue mesurble, nd N = j=1 N j. In order to prove tht N is neglijeble, it then suffices to prove tht λ(n j ) = 0, for ll j 1. Fix for the moment j 1. Since h p h 0, it follows tht so by Lemm 6.1 we get the inequlity inf h p (x) 1 x N j j, p 1, λ(n j ) j h p (x) dx, p 1, so by (β) we indeed get λ(n j ) = 0. Define the set S = p=1 p. Clim 2: If y [, b] (N S), then f is continuous t y. Fix y [, b] (N S). In order to prove tht f is continuous t y, we must find, for every ε > 0, some open intervl J ε y, such tht (20) f(z) f(y) < ε, z J ε [, b]. Since y N, we hve lim p h p (y) = 0. 0 h p (y) < ε. Write the prtition p s p = ( = x 0 < x 1 < < x n = b). Fix ε nd choose p 1, such tht

15 214 CHAPTER III: MEASURE THEORY Using the fct tht y p, if we define k = min { j {1,..., n} : y < x j }, we hve y (x k 1, x k ). In prticulr, we get f p (y) = sup f(t) nd f p (y) = inf f(s), t [x k 1,x k ] s [x k 1,x k ] so the inequlity 0 h p (y) < ε gives [ sup f(t) ] [ inf f(s)] < ε, t [x k 1,x k ] s [x k 1,x k ] so if we choose J ε = (x k 1, x k ), we clerly hve (20). Now we re done, becuse using the fct tht S is countble, it follows tht S is neglijeble, so N S is lso neglijeble. Since by Clim 2, we hve D f N S, it follows tht D f itself is neglijeble. (ii) (i). Assume now the discontinuity set D f is neglijeble, nd let us prove tht f is Riemnn integrble. Fix sequence ( p ) p=1 of prtitions of [, b], with , nd 2 lim p p = 0. As before, we define the set S = p=1 p. Clim 3: For ny point y [, b] (D f S), one hs the equlities lim f p (y) = lim f p (y) = f(y). p p Fix for the moment ε > 0. Since f is continuous t y, there exists some δ ε > 0, such tht (21) f(z) f(y) < ε, z (y δ ε, y + δ ε ) [, b]. Choose now q 1, such tht q < δ ε. Write q = ( = x 0 < x 1 < < x n = b). Using the fct tht y q, we cn find k {1,..., n} such tht y (x k 1, x k ). Since x k x k 1 < δ ε, we hve the inclusion [x k 1, x k ] (y δ ε, y + δ ε ), so by (21) we immeditely get f(y) f q (y) = f(y) f q (y) = sup f(z) f(y) + ε; z [x k 1,x k ] inf f(z) f(y) ε. z [x k 1,x k ] Since the sequence ( f p (y) ) p=1 is non-incresing, nd the sequence ( f p (y) ) p=1 is non-decresing, the bove inequlities give f p (y) f(y) ε nd f p (y) f(y) ε, for ll p q, nd the Clim follows. Going bck to the proof of the Theorem, we will now prove tht f stsifies condition (iii) in Proposition 6.6. Fix ε > 0. Since D f S is lso neglijeble, using regulrity from bove with respect to open sets, we cn find n open set E R such tht E D f S, nd λ(e) < ε. Define the compct set A = [, b] E, nd put B = [, b] E. We clerly hve (22) λ(b) λ(e) < ε. Define the sequence (h p ) p=1 by h p = f p f p. Since A p =, it follows tht h p A is continuous, for ech p 1. Since A (D f S) =, by Clim 3, we know 2 Recll tht, for prtition = ( = x0 < < x n = b), the number is defined s = mx { x k x k 1 : 1 k n }.

16 6. The Lebesgue mesure 215 tht lim p h p (y) = 0, y A. Since (h p ) p=1 is monotone, by Dini s Theorem (see??) it follows tht [ lim mx h p(y) ] = 0. p y A In prticulr, there exists p ε 1, such tht (23) h pε (y) ε, y A. Let M = sup f(x) nd m = inf f(x). x [,b] x [,b] Using Lemm 6.1 for h pε nd the sets A nd B, combined with (22), we hve h pε (x) dx (b ) sup h pε (y) + λ (B) sup h pε (z) y A z B ε(b ) + λ (B)(M m) ε(b + M m). Since h pε h p 0, for ll p p ε, we get the inequlities 0 h p (x) dx ε(b + M m), p p ε. The bove rgument proves tht lim p h p(x) dx = 0, i.e. lim p [f p (x) f p (x)] dx = 0. By Proposition 6.6, it follows tht f is Riemnn integrble. Exercise 13. Prove tht Riemnn integrble function f : [, b] R is Lebesgue mesurble. Hint: Use sequence of prtitions ( p) p=1, with , nd lim p p = 0. Use the rguments given in the proof of the impliction (ii) (i), to find neglijeble set N [, b], such tht lim f p p (x) = f(x), x [, b] N. The sequence (f p ) p=1 is non-decresing, so it hs point-wise limit, sy g, which is Lebesgue mesurble. Use the fct tht to show tht f itself is Lebesgue mesurble. f(x) = g(x) x [, b] N, Exercise 14. Let K [0, 1] be compct set with K Q =, nd λ(k) > 0 (see Exercise 7 for the existence of such sets). Prove tht the chrcteristic function κ K : [0, 1] R is not Riemnn integrble. In fct, f cnnot be lmost Riemnn integrble either. Hint: Exmine the discontinuity set D f, nd prove tht K D f. Exercise 15. Let f n : [, b] R, n 1 be sequence of Riemnn integrble functions. Consider the product spce P = n=1 Rn f n, equipped with the product topology (the sets Rn f n, n 1, re equipped with the topology induced from R), nd the function F : [, b] P, defined by F (x) = ( f n (x) ). Prove tht, for n=1 every bounded continuous function g : P R, the composition g F : [, b] R is Riemnn integrble. In other words, the result of bounded continuous opertion, involving sequence of Riemnn integrble functions, is gin Riemnn integrble function.

17 216 CHAPTER III: MEASURE THEORY Hint: D fn, n 1. Exmine the reltionship between the discountinuity set D g F nd the dsicontinuity sets Exercise 16. Let M be n rbitrry subset of [, b], nd let f : [, b] R be Riemnn integrble function, such tht f κ M. Prove the inequlity f(x) dx λ (M). Hint: Consider the function g : [, b] R defined by g(x) = mx{f(x), 1}. Then f g κ M, nd g is still Riemnn integrble. Apply Lemm 6.1 (the first inequlity) to the function 1 g. Exercise 17*. Let f : [, b] R be bounded function. Prove tht the following re equivlent: Hints: (i) f is Riemnn integrble; (ii) for every ε > 0, there exist continuous functions g, h : [, b] R with g f h, nd [g(x) h(x)] dx < ε; (iii) for every ε > 0, there exist Riemnn integrble functions g, h : [, b] R with g f h, nd [g(x) h(x)] dx < ε. For the impliction (i) (ii) nlyze first the prticulr cse when f = κ J, with J sub-intervl of [, b]. Then nlyze the functions of the type f nd f. For the impliction (iii) (i), nlyze the reltionship mong lower/upper Drboux sums of f, g nd h. Comment. The sttement of Theorem 6.1 shows tht, pprt from trivil cses, the problem of checking tht function f : [, b] R is Riemnn integrble, is rther difficult one. The min difficulty rises from the fct tht, if N [, b] is neglijeble set, nd f [,b] N is continuous, then f need not be continuous t ll points in [, b] N. For instnce, if we consider the chrcteristic function f = κ Q [,b] of the rtionls in [, b], nd N = Q [, b], then clerly N is neglijeble, f [,b] N is continuous (becuse it is constnt zero), but D f = [, b]. As erlier suggested, in the hope tht such n nomly cn be eliminted, it is resonble to consider the slightly weker notion of lmost Riemnn integrbilty. In the reminder of this section, we tke closer look t this notion, nd we will eventully show (see Theorem 6.2) tht this indeed removes the bove nomly. We begin with n lmost version of Exercise 17. Lemm 6.2. For function f : [, b] R, the following re equivlent: (i) f is lmost Riemnn integrble; (ii) for every ε > 0, there exist continuous functions g, h : [, b] R with g f h.e., nd [g(x) h(x)] dx < ε; (iii) for every ε > 0, there exist Riemnn integrble functions g, h : [, b] R with g f h.e., nd [g(x) h(x)] dx < ε. Proof. The impliction (i) (iii) is trivil. The impliction (iii) (ii) follows from Exercise 17. We now prove (ii) (i). Assume f hs property (ii). For ech integer n 1, choose continuous functions g n, h n : [, b] R, such tht g n f h n,.e., nd

18 6. The Lebesgue mesure 217 [g n(x) h n (x)] dx 1/n. Define the functions G n, H n : [, b] R, n 1, by It is cler tht G n (x) = min { g 1 (x),..., g n (x) }, H n (x) = mx { g 1 (x),..., g n (x) }. (α) G m f H n,.e., m, n 1; (β) G 1 G 2... nd H 1 H 2... ; (γ) [G n(x) H n (x)] dx [g n(x) h n (x)] dx 1/n, n 1. Notice tht, since the G m s nd the H n s re continuous, by Exercise??, we lso hve (α ) G m H n (everywhere!), m, n 1. Use (β) to define the functions G, H : [, b] R, by G(x) = lim n G n(x) nd H(x) = lim n H n(x), x [, b], so by (α ) we clerly hve G n G H H n, n 1. Using then (γ), by Exercise 17 it follows tht both G nd H re Riemnn integrble. Moreover, we hve G H 0 nd 0 [G(x) H(x)] dx [G n (x) H n (x)] dx 1/n, n 1, Which forces [G(x) H(x)] dx = 0, so by Exercise??, we get G = H,.e. By (α) it follows tht f = G,.e., so f in indeed lmost Riemnn integrble. We re now in position to prove the lmost version of Theorem 6.1. Theorem 6.2. Let f : [, b] R be bounded function. The following re equivlent: (i) f is lmost Riemnn integrble; (ii) there exists neglijeble set N [, b] such tht f [,b] N is continuous. Proof. (i) (ii). Assume f is lmost Riemnn integrble, so there exists Riemnn integrble function g : [, b] R, such tht f = g,.e. By Theorem 6.1, the discontinuity set D g is neglijeble. Tke M = {x [, b] : f(x) g(x)}. Since f = g,.e., the set M is neglijeble, nd so is the set N = M D g. On the one hnd, since D g N, the restriction g [,b] N, is continuous. On the other hnd, since M N, we hve f [,b] N = g [,b] N, so (ii) follows. (ii) (i). We re going to imitte the proof of Theorem 6.1, with some minor modifictions. Fix N [, b] neglijeble, such tht f [,b] N is continuous. Fix lso sequence ( p ) p=1 of prtitions, with , nd lim p p = 0. Put S = p=1 p. Since S is countble, the set N S is still neglijeble. We put T = [, b] (N S), nd we define the nlogues of the functions f p nd f p s follows. Write ech prtition s p = ( = x p 0 < xp 1 < < xp n p = b), nd define, for ech k {1,..., n p }, the numbers M p k = sup { f(t) : t [x p k 1, xp k ] T } nd m p k = inf { f(t) : t [x p k 1, xp k ] T }.

19 218 CHAPTER III: MEASURE THEORY We then define, for ech p 1, the functions g p = m p 1 κ [x p 0,xp 1 ] + mp 2 κ (x p 1,xp 2 ] + + mp n κ (x p n 1,xp n], g p = M p 1 κ [x p 0,xp 1 ] + M p 2 κ (x p 1,xp 2 ] + + M p n κ (x p n 1,xp n]. Note tht we hve the inequlities g p (x) f(x) g p (x), x T, which give (24) g p f g p,.e., p 1. It is obvious tht g p nd g p, p 1, re ll Riemnn integrble. We re now going to estimte the integrls [gp (x) g p (x)] dx. Put h p = g p g p, p 1. First we observe tht, since f T is continuous, nd T p =, p 1, we clerly hve the equlities lim p g p (x) = lim p g p (x) = f(x), x T, which give (25) lim p h p(x) = 0, x T. Fix some ε > 0, nd use regulrity from bove, to find n open set D with D N S nd λ(d) < ε. Tke the compct set A = [, b] D. Note tht f A is continuous, since A [, b] N. Note lso tht, since A [, b] S, the functions g p A nd g A p re lso continuous, nd so will be h A p, for every p 1. Since ( g p (x) ) p=1 is non-incresing, nd ( g p (x) ) is non-decresing, for ll x, it follows tht the p=1 sequence (h p ) p=1 is monotone, so by Dini s Theorem, (25) gives [ lim mx h p(x) ] = 0. p x A In prticulr, there exists some p ε 1, such tht (26) h p (x) ε, p p ε, x A. Put B = [, b] A, nd tke M = sup x [,b] f(x) nd m = inf x [,b] f(x). Using the inclusion B D, we get λ (B) λ(d) ε, so by Lemm 6.1, (the functions h p, p 1, re clerly non-negtive), combined with (26), we get h p (x) dx (b ) sup h p (x) + λ (B) sup h p (x) x A x B (b )ε + λ (B)(M m) ε(b + M m), p p ε. This estimte then proves tht lim p h p(x) dx = 0, i.e. lim p [g p (x) g p (x)] dx = 0. Combining this with (24), nd pplying Lemm 6.2, yields the fct tht f is lmost Riemnn integrble. Comment. The hypothesis tht f is bounded cn be replced with slightly weker one, which ssumes tht f is lmost bounded, mening tht there exists neglijeble set U [, b], such tht f [,b] U is bounded. Exercise 18. Let f n : [, b] R, n 1, be lmost Riemnn integrble functions, such tht (i) f n f n+1 0,.e., n 1; (ii) lim n f n (x) = 0, for lmost ll x [, b], i.e. there exists neglijeble set N [, b], such tht lim n f n (x) = 0, x [, b] N.

20 6. The Lebesgue mesure 219 Prove tht lim f n (x) dx = 0. n