Analysis I. 1 Measure theory. Piotr Haj lasz. 1.1 σ-algebra.

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1 Analysis I Piotr Haj lasz 1 Measure theory 1.1 σ-algebra. Definition. Let be a set. A collection M of subsets of is σ-algebra if M has the following properties 1. M; 2. A M = \ A M; 3. A 1, A 2, A 3,... M = A i M. The pair (, M) is called measurable space and elements of M are called measurable sets. xercise. Let (, M) be a measurable space. Show that 1. M; 2. A, B M = A \ B M; 3. M is closed under finite unions, finite intersections and countable intersections. xamples. 1. 2, the family of all subsets of is a σ-algebra. 2. M = {, } is a σ-algebra. 3. If is a fixed set, then M = {,,, \ } is a σ-algebra. 1

2 Proposition 1 If {M i } i I is a family of σ-algebras, then is a σ-algebra. A simple proof is left to the reader. M = i I M i Let R be a family of subsets of. By σ(r) we will denote the intersection of all σ-algebras that contain R. Note that there is at least one σ-algebra that contains R, namely σ(r) is a σ-algebra that contains R 2. σ(r) is the smallest σ-algebra that contains R in the sense that if M is a σ-algebra that contains R, then σ(r) M. We say that σ(r) is a σ-algebra generated by R. xample. If R = {}, then σ(r) = {,,, \ }. In the sequel we will need the following result. Proposition 2 The family M of subsets of a set is a σ-algebra if and only if it satisfies the following three properties 1. M; 2. If A, B M, then A \ B M. 3. If A 1, A 2, A 3,... M are pairwise disjoint, then A i M. We leave the proof as an exercise cf. the proof of Theorem 3(d). 1.2 Borel sets. Let be a metric space (or more generally a topological space). By B() we will denote the σ-algebra generated by the family of all open sets in. lements of B() are called Borel sets and B() is called σ-algebra of Borel sets. The following properties are obvious. 1. If U 1, U 2, U 3,... are open, then U i is a Borel set; 2. All closed sets are Borel; 3. If F 1, F 2, F 3,... are closed, then F i is a Borel set. 2

3 1.3 Measure. Definition. Let (, M) be a measurable space. A measure (called also positive measure) is a function µ : M [0, ] such that 1. µ( ) = 0; 2. µ is countably additive i.e. if A 1, A 2, A 3,... M are pairwise disjoint, then ( ) µ A i = µ(a i ). The triple (, M, µ) is called measure space. If µ() <, then µ is called finite measure. If µ() = 1, then µ is called probability or probability measure. If we can write = A i, where A i M and µ(a i ) < for all i = 1, 2, 3,..., then we say that µ is σ-finite. xample. If is an arbitrary set, and µ : 2 [0, ] is defined by µ() = m if is finite and has m elements, µ() = if is infinite, then µ is a measure. It is called counting measure. Theorem 3 (lementary properties of measures) Let (, M, µ) be a measure space. Then (a) If the sets A 1, A 2,..., A n M are pairwise disjoint then µ(a 1... A n ) = µ(a 1 ) µ(a n ). (b) If A, B M, A B and µ(b) <, then µ(b \ A) = µ(b) µ(a). (c) If A, B M, A B, then µ(a) µ(b). (d) If A 1, A 2, A 3,... M, then ( ) µ A i µ(a i ). (e) If A 1, A 2, A 3,... M, µ(a i ) = 0, for i = 1, 2, 3,... then µ( A i) = 0. 3

4 (f) If A 1, A 2, A 3,... M, A 1 A 2 A 3... then ( ) µ A i = lim µ(a i ). i (g) If A 1, A 2, A 3,... M, A 1 A 2 A 3... and µ(a 1 ) <, then ( ) µ A i = lim µ(a i ). i Proof. (a) The sets A 1, A 2,..., A n,,,,... are pairwise disjoint. Hence µ(a 1... A n ) = µ(a 1... A n...) = µ(a 1 ) µ(a n ) + µ( ) + µ( ) + µ( ) +... = µ(a 1 ) µ(a n ). (b) Since B = A (B \ A) and the sets A, B \ A are disjoint we have and the claim easily follows. (c) The claim follows from (1). (d) We have A = A 1 A 2 A 3... µ(b) = µ(a) + µ(b \ A) (1) = A }{{} 1 (A 2 \ A 1 ) (A }{{} 3 \ (A 1 A 2 )) (A }{{} 4 \ (A 1 A 2 A 3 ))... }{{} B 1 B 2 B 3 B 4 The sets B i are pairwise disjoint and B i A i. Hence (e) It easily follows from (d). (f) Since A i A i+1, we have µ(a) = µ(b i ) µ(a i ). A = A 1 A 2 A 3... = A }{{} 1 (A 2 \ A 1 ) (A }{{} 3 \ A 2 ) (A }{{} 4 \ A 3 )... }{{} B 1 B 2 B 3 B 4 4

5 The sets B i are pairwise disjoint and hence µ(a) = µ(b i ) = lim i (µ(b 1 ) µ(b i )) = lim i µ(b 1... B i ) = lim i µ(a i ). The limit exist because µ(a i ) µ(a i+1 ) (since A i A i+1 ). (g) It suffice to apply (f) to the sets A 1 \ A i. The proof is complete. xercise. Provide a detailed proof of (g). Where do we use the assumption that µ(a 1 ) <? Show an example that the conclusion of (g) need not be true without the assumption that µ(a 1 ) <. 1.4 Outer measure and Carathéodory construction. It is quite difficult to construct a measure with desired properties, but it is much easier to construct so called outer measure which has less restrictive properties. The Carathéodory theorem shows then how to extract a measure from an outer measure. Definition. Let be a set. A function is called outer measure if 1. µ ( ) = 0; 2. A B = µ (A) µ (B); 3. for all sets A 1, A 2, A 3,... µ : 2 [0, ] ( ) µ A i µ (A i ). We call a set µ -measurable if A µ (A) = µ (A ) + µ (A \ ). (2) Condition (2) is called Carathéodory condition. Since the inequality µ (A) µ (A ) + µ (A \ ) is always satisfied, in order to verify measurability of a set it suffices to show that µ (A) µ (A ) + µ (A \ ). 5

6 Proposition 4 All sets with µ () = 0 are µ measurable. Proof. If µ () = 0, then for an arbitrary set A we have because A. µ (A) µ (A \ ) = µ (A \ ) + µ (A ), }{{} 0 Let M be the class of all µ -measurable sets. Theorem 5 (Carathéodory) M is a σ-algebra and µ : M [0, ] is a measure. Proof. We will split the proof into several steps. Step I: M This is obvious. Step II:, F M = F M. Let A. We have µ (A) = µ (A ) + µ (A \ ) = µ (A ) + µ ((A \ ) F ) + µ ((A \ ) \ F ). Since A = A ( F ) and (A \ ) F = (A ( F )) \, we have µ (A) = µ (A ( F ) ) + µ (A ( F ) \ ) + µ (A \ ( F )) = µ (A ( F )) + µ (A \ ( F )). Step III:, F M = \ F M. It follows from the symmetry of the condition (2) that \ M. Since \ F = \ (( \ ) F ) the claim follows. Step IV: If the sets 1, 2, 3,... M are pairwise disjoint, then for every A µ (A i ) = µ (A i ). If, F M are pairwise disjoint, the for every A µ (A ( F )) = µ (A ( F ) ) + µ (A ( F ) \ ) = µ (A ) + µ (A F ). 6

7 This, Step II and the induction argument implies that for every n n n µ (A i ) = µ (A i ). (3) Hence and thus in the limit µ (A µ (A i ) i ) n µ (A i ) µ (A i ). Since the opposite inequality is a property of an outer measure, the claim follows. Step V: If the sets 1, 2, 3,... M are pairwise disjoint, then i M. Step II and the induction argument implies that for every n n i M. Hence (3) gives µ (A) = µ (A ) n i + µ (A \ ) n i ) n µ (A i ) + µ (A \ i, and in the limit ) ) ) µ (A) µ (A i ) + µ (A \ i = µ (A i + µ (A \ i. Since the opposite inequality is a property of an outer measure the claim follows. Step VI: Final step. It follows from Steps I, III, V and Proposition 2 that M is a σ-algebra and then Step IV with A = implies that µ restricted to M is a measure. The proof is complete. We say that a measure µ : M [0, ] is complete if every subset of a set of measure zero is measurable (and hence has measure zero). Therefore it follows from Proposition 4 that the measure described by the Carathéodory theorem is complete. Let (, d) be a metric space. For, F we define dist (, F ) = inf d(x, y), x,y F diam = sup d(x, y). x,y Definition. An outer measure µ defined on subsets of a metric space is called metric outer measure if µ ( F ) = µ () + µ (F ) whenever dist (, F ) > 0. 7

8 Theorem 6 If µ is a metric outer measure, then all Borel sets are µ -measurable i.e. B() M. Proof. Since M is a σ-algebra, it suffices to show that all open sets belong to M. Let G be open. It suffices to show that for every A µ (A) µ (A G) + µ (A \ G) (4) (because the opposite inequality is obvious). otherwise (4) is obviously satisfied. For each positive integer n define We can assume that µ (A) < as G n = {x G : dist (x, \ G) > 1 n }. Then dist (G n, \ G) 1 n > 0. (5) Let Clearly and D n = G n+1 \ G n = {x G : G \ G n = dist (D i, D j ) 1 i j 1 n + 1 < dist (x, \ G) 1 n }. D i (6) i=n > 0 provided i + 2 j. Since the mutual distances between the sets D 1, D 3, D 5,..., D 2n 1 are positive we have µ (A D 1 ) + µ (A D 3 ) +... µ (A D 2n 1 ) = µ (A (D 1 D 3... D 2n 1 )) µ (A) and similarly µ (A D 2 ) + µ (A D 4 ) µ (A D 2n ) µ (A). Hence Now (6) yields and hence µ (A D i ) 2µ (A) <. µ (A (G \ G n )) µ (A D i ) i=n µ (A (G \ G n )) 0 as n. 8

9 Inequality (5) gives µ (A G n ) + µ (A \ G) = µ ((A G n ) (A \ G)) µ (A) and thus µ (A G) + µ (A \ G) µ (A G n ) + µ (A (G \ G n )) + µ (A \ G) µ (A) + µ (A (G \ G n )) and after passing to the limit µ (A G) + µ (A \ G) µ (A). The proof is complete. We will use outer measures to prove the following important result. Theorem 7 Let be a metric space and µ a measure in B(). Suppose that is a union of countably many open sets of finite measure. Then µ() = inf µ(u) = U U open sup µ(c). (7) C C closed Before proving the theorem let s discuss two of its corollaries. The first one shows that in a situation described by the above theorem in order to prove that two measures are equal it suffices to compare them on the class of open sets. Corollary 8 If µ is as in Theorem 7 and ν is another measure on B() that satisfies ν(u) = µ(u) for all open sets U then ν() = µ() for all B(). The second corollary describes an important class of measures satisfying assumptions of theorem 7. Definition. Let be a metric space and µ a measure defined on the σ-algebra of Borel sets. We say that µ is a Radon measure if µ(k) < for all compact sets and µ() = inf µ(u) = U U open sup µ(k) for all B(). (8) K K compact Corollary 9 If is a locally compact and separable metric space 1 and µ is a measure in B() such that µ(k) < for every compact set K, then is a union of countably many open sets of finite measure and µ is a Radon measure. 1 Being locally compact means that ever point has a neighborhood whose closure is compact. IR n is locally compact, but also IR n \ {0} is locally compact. 9

10 Proof. It follows from locally compactness of that is a union of a family of open sets with compact closures. Since is separable, every covering of by open sets has countable subcovering and hence we can write = U i, U i compact. This proves the first part of the corollary because µ(u i ) µ(u i ) <. Now (8) follows from (7) and Theorem 3(f) because ( ) n C = C U i. n=1 } {{ } compact and hence µ(c) = for every closed set C. The proof is complete Proof of Theorem 7. For we set sup µ(k) K C K compact µ () = inf µ(u). U U open It is easy to see that µ is a metric outer measure. Hence µ restricted to the class of Borel sets is a measure. Clearly and µ(u) = µ (U) for all open sets U, (9) µ() µ () for all B(). (10) We can assume that is a union of an increasing sequence of open sets with finite measure. Indeed, if is the union of open sets U i of finite measure, then the sets V n = U 1 U 2... U n satisfy = V n, V n V n+1, µ(v n ) <. n=1 Now inequality (10) implies that for all B() µ(v n \ ) µ (V n \ ) and µ(v n ) µ (V n ), (11) because both V n \ and V n are Borel. Actually we have equalities in both inequalities of (11) because otherwise we would have a sharp inequality µ(v n ) = µ(v n \ ) + µ(v n ) < µ (V n \ ) + µ (V n ) = µ (V n ), 10

11 which contradicts (9). Hence in particular µ(v n ) = µ (V n ). Since µ(v n ) µ() and µ (V n ) µ () as n by Theorem 3(f) we conclude that µ() = µ () and thus µ() = inf µ(u) (12) U U open by the definition of µ. To prove that the measure of a set B() can be approximated by measures of closed subsets of observe that V n \ has finite measure and hence it follows from (12) that there is an open set G n such that V n \ G n, µ(g n \ (V n \ )) < ε 2 n. The set G = n=1 G n is open and C = \ G is closed. Now it suffices to observe that \ C = G n G n \ (V n \ ) n=1 and hence µ( \ C) < ε. The proof is complete. n=1 1.5 Hausdorff measure. Let ω s = π s/2 /Γ(1 + s 2 ), s 0. If s = n is a positive integer, then ω n is volume of the unit ball in IR n. Let be a metric space. For ε > 0 and we define H s ε() = inf ω s 2 s (diam A i ) s where the infimum is taken over all possible coverings A i with diam A i < ε. Since the function ε H s ε() is nonincreasing, the limit exists. H s is called Hausdorff measure. H s () = lim ε 0 H s ε() It is easy to see that if s = 0, H 0 is the counting measure. Theorem 10 H s is a metric outer measure. 11

12 Proof. Clearly H s : 2 [0, ], H s ( ) = 0, A B = H s (A) H s (B). In order to show that H s is an outer measure it remains to show that ( ) H s n n=1 H s ( n ). n=1 We can assume that the right hand side is finite. Hence Hε( s n ) < for every ε > 0. n=1 Fix δ > 0. We can find a covering of each set n such that Hence n=1 H s ε( n ) ω s s s n A ni, H s ε( n ) ω s 2 s diam A ni < ε (diam A ni ) s δ 2 n ( ) (diam A ni ) s δ Hε s n δ, i,n=1 because {A ni } i,n=1 forms a covering of n=1 n. Passing to the limit with δ 0 yields ( ) H s ( n ) Hε( s n ) Hε s n. n=1 n=1 Now passing to the limit with ε 0 yields the result. We are left with the verification of the metric condition. Let dist (, F ) > 0. It suffices to show that for all ε < dist (, F ). Let n=1 n=1 H s ε( F ) = H s ε() + H s ε(f ) (13) F A i, be a covering of F. We can assume that diam A i < ε A i ( F ) (14) 12

13 for all i, as otherwise we could remove A i from the covering. Since diam A i < dist (, F ) it follows from (14) that A i has a nonempty intersection with exactly one set or F. Accordingly, the family {A i } i splits into two disjoint subfamilies {B j } j = all the sets A i such that A i ; Hence ω s 2 s {C j } j = all the sets A i such that A i F. (diam A i ) s = ω s (diam B 2 s j ) s + ω s (diam C 2 s j ) s j j H s ε() + H s ε(f ). Since {A i } i was an arbitrary covering of F such that diam A i < ε, we conclude upon taking the infimum that H s ε( F ) H s ε() + H s ε(f ). Since the opposite inequality is obvious (13) follows. xercise. Show that if H s () <, then H t () = 0 for all t > s; H s () > 0, then H t () = for all 0 < t < s. Definition. The Hausdorff dimension is defined as follows. If H s () > 0 for all s 0, then dim H () =. Otherwise we define dim H () = inf{s 0 : H s () = 0}. It follows from the exercise that there is s [0, ] such that H t () = 0 for t > s and H t () = for 0 < t < s. Hausdorff dimension of equals s. 1.6 Lebesgue measure. For a closed interval we define its volume by and for A IR n we define P = [a 1, b 1 ]... [a n, b n ] IR n (15) P = (b 1 a 1 )... (b n a n ) L n(a) = inf P i 13

14 where the infimum is taken over all coverings A by intervals as in (15). Arguments similar to those in the proof that the Hausdorff measure is an outer metric measure give the following result Theorem 11 L n is a metric outer measure. Definition. L n is called outer Lebesgue measure. L n-measurable sets are called Lebesgue measurable. L n restricted to Lebesgue measurable sets is called Lebesgue measure and is denoted by L n. Corollary 12 All Borel sets are Lebesgue measurable. All sets with L n(a) = 0 are Lebesgue measurable. Then L n (A) = 0. A set has Lebesgue measure zero if and only it it can be covered by a sequence of intervals with whose sum of volumes can be arbitrarily small. Note that this is the same as the sets of measure zero in the setting of Riemann integral. Theorem 13 If P is a closed interval as in (15), then P i L n (P ) = P. In what follows we will often write A to denote the Lebesgue measure of a Lebesgue measurable set A. Proof. In order to prove the theorem we have to show that if P P i, then P P i. We will need the following quite obvious special case of this fact. Lemma 14 If P k P i is a finite covering of a closed interval P by closed intervals P i, then P k P i. Proof. This lemma follows from the theory of Riemann integral. 2 Indeed, volume of the interval equals the integral of the characteristic function. Hence inequality χ P k χ P i implies k k P = χ P χ Pi = P i. IR n IR n 2 This lemma can be proved in a more elementary way without using the Riemann integral, but the proof would be longer. 14

15 The proof is complete. Now we can complete the proof of the theorem. Observe that each interval P i is contained in a slightly bigger open interval Pi ε such that P ε i = P i + ε 2 i, where Pi ε is the closure of Pi ε. Now from the covering P P i ε finite subcovering P (compactness) and hence the lemma yields k j=1 P ε i j we can select a P k j=1 P ε i j P ε i = P i + ε. Since ε > 0 was arbitrary the theorem follows. Proposition 15 If P = (a 1, b 1 )... (a n, b n ) is a bounded interval, then L n ( P ) = 0. L n (P ) = (b 1 a 1 )... (b n a n ). If P is an unbounded interval in IR n and has nonempty interior (i.e. each side has positive length), then L n (P ) =. Proof. The first claim follows from the observation that P can be approximated from inside and from outside by closed intervals that differs arbitrarily small in volume. The second claim follows from the first one L n ( P ) = L n (P ) L n (int P ) = 0, and the lats claim follows from the fact that P contains closed intervals of arbitrarily large measure. IR k IR n, k < n is a subset generated by the first k coordinates. Corollary 16 If k < n, then L n (IR k ) = 0. The next result shows that we can compute Lebesgue measure of an open set by representing it as a union of cubes. We call a cube dyadic if its sidelength equals 2 k for some k Z. 15

16 Theorem 17 An arbitrary open set in IR n is a union of closed dyadic cubes with pairwise disjoint interiors. Hence Lebesgue measure of the open set equals the sum of the measures of these cubes. Proof. Instead of a formal proof we will show a picture which explains how to represent any open set as a union of cubes with sidelength 2 k, k = 0, 1, 2, 3,... The proof is complete Compare the following result with Theorem 7. Theorem 18 For an arbitrary set IR n L n() = inf L n (U), where the infimum is taken over all open sets U such that U. Proof. L n = inf i P i and each interval P i is contained in an open set (interval) of slightly bigger measure. very Borel set B is Lebesgue measurable. very set with L () = 0 is Lebesgue measurable. Hence sets of the form B are Lebesgue measurable. These are all Lebesgue measurable sets. More precisely we have the following characterization. 16

17 Definition. Let be a metric space. By a G δ set we mean a set of the form A = G i, where the sets G i are open and by F σ we mean a set of the form B = F i, where the sets F i are closed. Clearly all G δ and F σ sets are Borel. Theorem 19 Let A IR n. Then the following conditions are equivalent 1. A is Lebesgue measurable; 2. For every ε > 0 there is an open set G such that A G and L n(g \ A) < ε; 3. There is a G δ set H such that A H and L n(h \ A) = 0; 4. For every ε > 0 there is a closed set F such that F A and L n(a \ F ) < ε; 5. There is a F σ set M such that M A and L n(a \ M) = 0; 6. For every ε > 0 there is an open set G and a closed set F such that F A G and L n (G \ F ) < ε. Proof. (1) (2) very measurable set can be represented as a union of sets with finite measure A = A i, L n (A i ) <. It follows from Theorem 18 that for every i there is an open set G i such that A i G i and L n (G i \ A i ) < ε/2 i. Hence A G = G i, L n (G \ A) < ε. (2) (3) We define H = U i, where U i are open sets such that A U i, L n(u i \A) < 1/i. (3) (1) This implication is obvious. (1) (4) (5) this equivalence follows from the equivalence of the conditions (1), (2) and (3) applied to the set IR n \ A. (1) (6) If A is Lebesgue measurable then the existence of the sets F and G follows from the conditions (2) and (4). (6) (1) Take closed and open sets F i, G i such that F i A G i, L n (G i \ F i ) < 1/i. Then the set H = G i is G δ and L n(h \ A) = 0. The Lebesgue measure has an important property of being invariant under translations i.e. if A is a translation of a Lebesgue measurable set B, then A is Lebesgue measurable and L n (A) = L n (B). This property follows immediately from the definition of the Lebesgue measure. We will show not that this property implies that there is a set which is not Lebesgue measurable. Theorem 20 (Vitali) There is a set IR n which is not Lebesgue measurable. 3 3 The reader will easily check that the same proof applies to any translation invariant measure µ such that the measure of the unit cube if positive and finite; see also Theorem

18 Proof. We will prove the theorem for n = 1 only, but the same argument works for an arbitrary n. The idea of the proof is to find a countable family of pairwise disjoint and isometric (by translation)sets whose union contains (0, 1) and is contained in ( 1, 2). If the sets in the family are isometric to a set V, then V cannot be Lebesgue measurable, because measurability of V would imply that a number between 1 and 3 (measure of ) be equal to infinite sum of equal numbers (equal to measure of V ) which is impossible. It should not be surprising that the construction of the set V has to involve axiom of choice. For x, y (0, 1) we write x y if x y is a rational number. Clearly is an equivalence relation and hence (0, 1) is the union of a family F of pairwise disjoint sets [x] = {y (0, 1) : x y}. It follows from the axiom of choice that there is a set V (0, 1) which contains exactly one element from each set in the family F. Let = V a, where V a = V + a = {x + a : x V }. It is easy to see that a Q ( 1,1) V a V b = if a, b Q, a b; (0, 1) ( 1, 2). Suppose that V is measurable. Then each V a is measurable, L 1 (V ) = L 1 (V a ) and is measurable. If L 1 (V ) > 0, then L 1 () = which is impossible and if L 1 (V ) = 0, then L 1 () = 0 which is also impossible. That proves that the set V cannot be measurable. The Lebesgue measure is translation invariant and L n ([0, 1] n ) = 1. It turns out that the above two properties uniquely determine Lebesgue measure. This proves that the Lebesgue measure is in a sense the only natural way of measuring length, area, volume of general sets in one, two, three,... dimensional uclidean spaces. Theorem 21 If µ is a measure on B(IR n ) such that µ(a + ) = µ() for all a IR n, B(IR n ) and µ([0, 1] n ) = 1, then µ() = L n () on B(IR n ). Proof. According to Corollary 8 it suffices to prove that µ(u) = L n (U) for all open sets U. Consider an open cube Q k = (0, 2 k ) n, where k is a positive integer. There are 2 kn pairwise disjoint open cubes contained in the unit cube [0, 1] n, each being a translation of Q k. Since the measure is invariant under translations we conclude that µ(q k ) 2 kn. 18

19 It is easy to see that we can cover the boundary Q by translations of Q k in a way that the sum of µ-measures of the cubes covering Q goes to zero as k. This proves that µ( Q) = 0. Now [0, 1] n can be covered by 2 kn cubes each being a translation of the closed cube Q k. Since the measure is invariant under translations we conclude that and hence µ(q k ) 2 kn µ(q k ) = µ(q k ) = 2 kn = L n (Q k ). (16) As we know (Theorem 17) an arbitrary open set U is a union of cubes with pairwise disjoint interiors and sidelength 2 k (k depends on the cube). Cubes are not disjoint, but meet along the boundary which has µ-measure zero. Hence the µ(u) equals the sum of the µ-measures of the cubes which according to (16) equals L n (U). The proof is complete. Theorem 22 H n = L n on B(IR n ). Sketch of the proof. Observe that the Hausdorff measure H n in IR n has the property that H n ( + a) = H n () for all B(IR n ). Therefore in order to prove that H n = L n on B(IR n ) it suffices to prove that H n ([0, 1] n ) = 1 This innocently looking result is however very difficult and we will not prove it here. xercise. Using Theorem 22 prove that H n = L n on the class of all subsets of IR n. If s < n, then H s is the s-dimensional measure in IR n. Not that s is not required to be an integer. If k < n is an integer and M IR n is a k-dimensional manifold, then one can prove that H k coincides on M with the natural k-dimensional Lebesgue measure on M. 4 xamples. If C [0, 1] is the standard ternary Cantor set, then one can prove that dim H (C) = log 2/ log 3 = Moreover H s (C) = 1 for s = log 2/ log 3. The Van Koch curve is defined as the limiting curve for the following construction 4 We have already seen how in the course of Advanced Calculus to measure subsets of M in the setting of the Riemann integral. This construction can easily be generalized to the Lebesgue measure on M. We will discuss it with more details later. 19

20 The Van Koch curve, denoted by K is homeomorphic to the circle S 1, but dim H K = log 4/ log 3 = 1, Proposition 23 A set IR n has Lebesgue measure zero if and only if for every ε > there is a family of balls B(x i, r i )} such that B(x i, r i ), ri n < ε. Proof. ach bass B(x i, r i ) is contained in a cube with sidelength 2r i and hence can be covered by a sequence of cubes whose sum of volumes is (2r i) n < 2 n ε. If a set IR n has Lebesgue measure zero, then for every ε > 0 there is an open set U such that U and L n (U) < ε. The set U can be represented as union of cubes with pairwise disjoint interiors (Theorem 17). each cube Q of sidelength l is contained in a ball of radius r = nl and hence r n = n n/2 Q. Therefore can be covered by a sequence of balls {B(x i, r i )} such that rn i < n n/2 ε. Definition. We say that a function between metric spaces f : (, d) (Y, ρ) is Lipschitz continuous if there is a constant L > 0 such that ρ(f(x), f(y)) Ld(x, y) for all x, y. In this case we say that f is L-Lipschitz continuous and call L Lipschitz constant if f. Proposition 24 If f : IR n IR n is Lipschitz continuous and IR n has Lebesgue measure zero, then f() has Lebesgue measure zero too. Proof. It follows from Proposition 23 and the fact that the L-Lipschitz image of a ball of radius r is contained in a ball of radius Lr. xercise. Let f : IR n IR n be a homeomorphism. Prove that f(a) is Borel if and only if A is Borel. Although the theory of Lebesgue measure seems to work very well something very important is missing: we haven t proved so far that if Q is a unit cube whose sides are not parallel to coordinate axes, then its measure equals 1. Indeed we considered only intervals with sides parallel to coordinate directions. Fortunately this important fact is contained in a result that we will prove now. 20

21 Theorem 25 If L : IR n IR n is a nondegenerate linear transformation with the matrix A (det A 0), then L() IR n is a Borel (Lebesgue measurable) set if and only if IR n is Borel (Lebesgue measurable) set. Moreover L n (L()) = det A L n (). (17) Proof. Since L is a homeomorphism it preserves the class of Borel sets (see xercise). Since both L and L 1 are Lipschitz continuous L preserves the class of sets of measure zero (Proposition 24). Therefore L preserves the class of Lebesgue measurable sets (because each Lebesgue measurable set is the union of a Borel set and a set of measure zero). It follows from Proposition 24 that the formula (17) is satisfied for sets of measure zero. Therefore it remains to prove (24) for Borel sets. Define a new measure µ() = L n (L()). Clearly µ is a measure on B(IR n ) that is translation invariant. Let µ([0, 1] n ) = a > 0. Hence the measure a 1 µ satisfies all the assumptions of Theorem 21 and therefore a 1 µ() = L n (), so we have L n (L()) = al n () for all B(IR n ). It remains to prove that a = det A. Clearly a = a(a) : GL (n) (0, ) is a function defined on the class of invertible matrices. We have a(a 1 A 2 ) = a(a 1 )a(a 2 ). (18) Indeed, if A 1, A 2 are matrices representing linear transformations L 1 and L 2, then Also a(a 1 A 2 )L n () = L n (L 1 (L 2 ()) = a(l 1 )L n (L 2 ()) = a(l 1 )a(l 2 )L n (). a(s n I) = s n, s > 0. (19) Here I is the identity matrix. This property is quite obvious and we leave a formal proof to the reader. Now it remains to prove the following fact. Lemma 26 Let a : GL (n) (0, ) be a function with properties (18) and (19). Then a(a) = det A. Proof. Let A i (s) be the diagonal matrix with s on the ith place and s on all other places on the diagonal. Since A i (s) 2 = s 2 I we have a(a i (s)) 2 = a(a i (s) 2 ) = s 2n and hance a(a i (s)) = s n = det A i (s). For k l let B kl (s) = [a ij ] be the matrix such that a kl = s, a ii = 1, i = 1, 2,... and all other entries equal zero. Multiplication by the matrix B kl (s) from the right (left) is equivalent to adding kth column (lth row) multiplied by s to lth column (kth row). It is well known from linear algebra (and easy to prove) that applying such operations to any nonsingular matrix A it can be transformed to a matrix of the form ti or 21

22 A n (t). Since multiplication by B kl (s) does not change determinant, t = det A 1/n. It remains to prove that a(b kl (s)) = 1. Since B kl ( s) = A k (1)B kl (s)a k (1) we have that a(b kl ( s)) = a(b kl (s)). On the other hand B kl (s)b kl ( s) = I and hence a(b kl (s)) 2 = 1, so a(b kl (s)) = 1. The proof of the lemma and hence the proof of the theorem is complete. 2 Integration 2.1 Measurable functions. Definition. Let (, M) be a measurable space and Y a metric space. We say that a function f : Y is measurable if f 1 (U) M for every open set U Y. If M is a measurable set, then we say that f : Y is measurable if f 1 (U) M for every open U Y. Clearly, f : Y is measurable if it can be extended to a measurable function f : Y, but there is also another point of view. M = {A : A M} = {B : B M} is a σ-algebra and measurability of f : Y is equivalent to measurability of f with respect to M. Theorem 27 Let (, M) be a measurable space, f : Y a measurable function and g : Y Z a continuous function. Then the function g f : Z is measurable. Proof. For every open set U Z we have (g f) 1 (U) = f 1 (g 1 (U)) M. }{{} open in Y The proof is complete. Theorem 28 Let (, M) be a measurable space and Y a metric space. If the functions u, v : IR are measurable and Φ : IR 2 Y is continuous, then the function is measurable. h(x) = Φ(u(x), v(x)) : Y 22

23 Proof. Put f(x) = (u(x), v(x)), f : IR 2. According to Theorem 27 it suffices to prove that f is measurable. Let R = (a, b) (c, d) be an open rectangle. Then f 1 (R) = f 1 ((a, b) (c, d)) = u 1 ((a, b)) u 1 ((c, d)) M. very open set U IR 2 is a countable union of open rectangles U = R i and hence f 1 (U) = f 1 (R i ) M. The proof is complete. This theorem has many applications. (a) If f = u + iv : C, where u, v : IR are measurable, then f is measurable. Indeed, we take Φ(u, v) = u + iv. (b) If f = u + iv : C is measurable, then the functions u, v, f : IR are measurable. Indeed, we take Φ 1 (u + iv) = u, Φ 2 (u + iv) = v, Φ 3 (u + iv) = u + iv. (c) If f, g : C are measurable, then the functions f +g, fg : C are measurable. Indeed, we take Φ 1 : C C C, Φ 1 (z 1, z 2 ) = z 1 + z 2 and Φ 2 : C C C, Φ 2 (z 1, z 2 ) = z 1 z 2. (d) A set is measurable if and only if the characteristic function if { 1 if x, χ (x) = 0 if x, is measurable. Indeed, this follows from the fact that for any open set U IR if 0 U, 1 U, (χ ) 1 if 0 U, 1 U, (U) = \ if 0 U, 1 U, if 0 U, 1 U, 23

24 (e) If f : C is a measurable function, then there is a measurable function α : C such that α = 1 and f = α f. The set = {x : f(x) = 0} is measurable because = f 1 ( B(0, 1 i ) }{{} ). disc of radius 1/i Hence the function h(x) = f(x)+χ (x) is measurable and h(x) 0 for all x. Therefore h is a measurable function as a function h : C \ {0}. Since ϕ : C \ {0} C, ϕ(z) = z/ z is continuous we conclude that α(x) = ϕ(f(x) + χ (x)) is measurable. If x, then α(x) = 1 and if x, then α(x) = f(x)/ f(x). Therefore f(x) = α(x) f(x) for all x. Definition. Let, Y be two metric spaces. We say that f : Y is a Borel mapping if it is measurable with respect to the σ-algebra of Borel sets in i.e. f 1 (U) B() for every open set U Y. Clearly every continuous mapping is Borel. Theorem 29 Let (, M) be a measurable space and Y a metric space. Let also f : Y be a measurable mapping. (a) If Y is a Borel set, then f 1 () M. (b) If Z is another metric space and g : Y Z is a Borel mapping, then g f : Z is measurable. Proof. (a) Let R = {A Y f 1 (A) M}. Clearly R contains all open sets. If we prove that R is a σ-algebra, we will have that B(Y ) R which is the claim. Y R Indeed, f 1 (Y ) = M. If A R, then Y \ A R. Indeed, f 1 (Y \ A) = \ f 1 (A) M. 24

25 If A 1, A 2,... R, then A i R. Indeed, f 1 ( A i) = f 1 (A i ) R. The above three properties imply that R is a σ-algebra. (b) If U Z is open, then g 1 (U) Y is Borel and hence (g f) 1 (U) = f 1 (g 1 (U)) M by (a). Definition. Let (, M) be a measurable space. Let IR = IR {, }. We say that a function f : IR is measurable if f 1 (U) is measurable for every open set U IR and if the sets f 1 (+ ), f 1 ( ) are measurable. Similarly for M we define measurable functions f : IR. xercise. Let (, M) be a measurable space. Prove that a function f : IR is measurable if and only if f 1 ((a, ]) M for all a IR. and Let {a n } be a sequence in IR. For k = 1, 2,... put We call β upper limit of {a n } and write b k = sup{a k, a k+1, a k+2,...} (20) β = inf{b 1, b 2,...}. (21) β = lim sup a n n Clearly {b k } is a decreasing sequence, so β = lim k b k. It is easy to see that there is a subsequence {a ni } of {a n } such that lim i a ni = β. Moreover β is the largest possible value for limits of all subsequences of {a n }. The lower limit is defined by lim inf n a n = lim sup( a n ). (22) n quivalently the lower limit can be defined by interchanging sup and inf in (20) and (21). The lower limit equals to the lowest possible value for limits of subsequences of {a n }. For a sequence of functions f n : IR we define new functions sup n f n, lim sup n f n : IR by ( ( sup n lim sup n ) f n (x) = sup f n (x) for every x, n ) f n (x) = lim sup f n (x) for every x. n 25

26 Similarly we can define functions inf n f n and lim inf n f n. If f(x) = lim n f n (x) for every x, then we say that f is a pointwise limit of {f n }. Theorem 30 Suppose that f n : IR is a sequence of measurable functions. Then the functions sup n f n, inf n f n, lim sup n f n, lim inf n f n are measurable. Proof. Since ( ) 1 sup f n ((a, ]) = n n=1 f 1 n ((a, ]) M, measurability of sup n f n follows from the exercise. By a similar argument we prove that the function inf n f n is measurable. The two facts imply measurability of lim sup f n because lim sup f n = inf {sup f i }. n k 1 Now measurability of lim inf n f n is an obvious consequence of the equality lim inf n i k f n = lim sup( f n ). n The proof is complete. Corollary 31 The limit of every pointwise convergent sequence of functions f n : C (or f n : IR) is measurable. Corollary 32 If f, g : IR are measurable functions, then the functions max{f, g}, min{f, g} are measurable. In particular the fuctions are measurable. f + = max{f, 0}, f = min{f, 0} The functions f + and f are called positive part and negative part of f. Note that f = f + f, f = f + + f. Corollary 31 has the following generalization to the case of metric space valued mappings. Theorem 33 Let f n : Y be a sequence of measurable mappings with values in a separable metric space. If f n (x) f(x) for every x, then f : Y is measurable. 26

27 Proof. Let {y i } be a countable dense set. Then the functions d i : Y IR, d i (y) = d(y i, y) are continuous. Hence d i f n : IR are measurable. Since d i f n d i f, the functions d i f : IR are measurable too by Corollary 31. In particular the sets {x : d(y i, f(x)) d(y i, F )} = (d i f) 1 ([d(y i, F ), )) are measurable. To prove measurability of f it suffices to prove measurability of f 1 (F ) for every closed set F and it follows from the equality f 1 (F ) = {x : d(y i, f(x)) d(y i, F )}. (23) To prove this equality observe that if x f 1 (F ), then f(x) F and hence d(y i, f(x)) d(y i, F ) for every i, so x belongs to the right hand side of (23). On the other hand if x f 1 (F ), then f(x) F and hence there is i such that d(y i, f(x)) < d(y i, F ) (because {y i } is a dense subset of Y and so we can find y i arbitrarily close to f(x)), therefore x cannot belong to the right hand side of (23). Definition. Let (, M) be a measurable space. A measurable function s : C is called simple function if it has only a finite number of values. Simple functions can be represented in the form n s = α i χ Ai, where α i α j for i j and A i are disjoint measurable sets. Theorem 34 If f : [0, ] is measurable, then there is a sequence of simple functions s n on such that 1. 0 s 1 s 2... f; 2. lim n s n (x) = f(x) for every x. If in addition f is bounded, then s n converges to f uniformly. Proof. It is easy to see that the sequence has all properties that we need. { n if f(x) n, s n (x) = k k if f(x) < k+1 n, 2 k 2 n 2 n 27

28 2.2 Integral In the integration theory we will assume that 0 = 0. Definition. If s : [0, ) is a simple function of the form s = n α i χ Ai, where α i α j for i j and A i are pairwise disjoint measurable sets, then for any M we define n s dµ = α i µ(a i ). According to our assumption 0 = 0, so 0 µ(a i ) = 0, even if µ(a i ) =. If f : [0, ] is measurable, and M, we define the Lebesgue integral of f over by f dµ = sup s dµ, where the supremum is over all simple functions s such that 0 s f. (a) If 0 f g, then f dµ g dµ. (b) If A B and f 0, then f dµ f dµ. A B (c) If f 0 and c is a constant 0 c <, then f dµ = c f dµ. (d) If f(x) = 0 for all x, then f dµ = 0, even if µ() =. (e) If µ() = 0, then f dµ = 0, even if f(x) = for all x. (f) If f 0, then f dµ = χ f dµ. In the proof of the next two theorems we will need the following lemma. Lemma 35 Let s and t be nonnegative measurable simple functions on. Then the function ϕ() = s dµ, M defines a measure in M. Moreover (s + t) dµ = s dµ + t dµ. (24) 28

29 Proof. Since ϕ( ) = 0, it remains to prove that ϕ is countably additive. If 1, 2,... are disjoint measurable sets, then ( ) ( n ) n ϕ k = α i µ k A i = µ( k A i ) k=1 = k=1 n α i µ( k A i ) = k=1 k=1 This completes the proof of the first part. Let now α i k=1 ϕ( k ). s = k α i χ Ai, t = m β j χ Bj j=1 be a representation of the simple functions as in the definition, then for ij = A i B j we have s + t dµ = (α i + β j )µ( ij ) = ij s dµ + ij t dµ. ij Since is a union of disjoint sets ij we conclude (24) from the first part of the lemma. The following theorem is one of the most important results of the theory of integration. Theorem 36 (Lebesgue monotone convergence theorem) Let {f n } be a sequence of measurable functions on such that: 1. 0 f 1 f 2... for every x ; 2. lim n f n (x) = f(x) for every x. Then f is measurable and lim f n dµ = f dµ n Proof. Measurability of f follows from Corollary 31. The sequence f n dµ is increasing and hence it has a limit α = lim n f n dµ. Since f n f, we have f n dµ f dµ and hence α f dµ. Let 0 s f be a simple function. Fix a constant 0 < c < 1 and define n = {x : f n (x) cs(x)}, n = 1, 2, 3,... It is easy to see that and n=1 n = (because f n (x) f(x) for every x.) Hence f n dµ f n dµ c n s dµ. n (25) 29

30 Since ϕ() = s dµ is a measure and we conclude (Theorem 3(f)) that ( ) s dµ = ϕ( n ) ϕ n = ϕ() = s dµ. n Thus after passing to the limit in (25) we obtain α c s dµ and hence f dµ α s dµ, n=1 because 0 < c < 1 was arbitrary. Taking supremum over all simple functions 0 s f yields f dµ α f dµ, i.e. f dµ = α and the theorem follows. Theorem 37 If f n : [0, ] is measurable, for n = 1, 2, 3,... and then f is measurable and f(x) = f n (x) for all x, n=1 f dµ = f n dµ. Proof. The function f is a limit of measurable functions, so it is measurable. Let {s i } and {t i } be increasing sequences of simple nonnegative function such that s i f 1, t i f 2 (Theorem 34). Then s i + t i f 1 + f 2 and hence (24) combined with the monotone convergence theorem gives f 1 + f 2 dµ = f 1 dµ + f 2 dµ. Applying an induction argument we obtain (f 1 + f f N ) dµ = N n=1 f n dµ. Since g N = f f N is an increasing sequence of measurable functions and g N f the theorem follows from the monotone convergence theorem. Theorem 38 (Fatou s lemma) If f n : [0, ] is measurable, for n = 1, 2, 3,..., then ( ) lim inf f n dµ lim inf f n dµ. n n 30

31 xercise. Construct a sequence of measurable functions f n : IR [0, ] such that the inequality in Fatou s lemma is sharp. Proof of the theorem. Let g n = inf{f n, f n+1, f n+2,...}. Then g n f n and hence g n dµ f n dµ. (26) Since 0 g 1 g 2..., all the functions g n are measurable (Theorem 30) and g n lim inf k f n, inequality (26) combined with the monotone convergence theorem yields the result. Theorem 39 Let f : [0, ] be measurable. Then ϕ() = f dµ for M defines a measure in M. Moreover g dϕ = for every measurable function g : [0, ]. gf dµ (27) Proof. Clearly ϕ( ) = 0. To prove that ϕ is a measure it remains to prove countable additivity. Let 1, 2,... M be disjoint sets and = n=1 n. Then obviously χ f = χ n f n=1 and ϕ() = = χ f dµ = χ n f dµ n=1 χ n f dµ = n=1 n=1 ϕ( n ), by Theorem 37. To prove (27) observe that it is true if g = χ, M. Hence it is also true for simple functions and the general case follows from the monotone convergence theorem combined with Theorem 34. Definition. Let (, M, µ) be a measure space. The space L 1 (µ) consists of all measurable functions f : C for which f 1 = f dµ <. lements of L1 (µ) are called Lebesgue integrable functions or summable functions. 31

32 If f = u + iv, then the functions u +, u, v +, v are measurable and Lebesgue integrable as bounded by f. For a set M we define ( ) f dµ = u + dµ u dµ + i v + dµ v dµ. If f : IR is measurable and each of the integrals f + dµ, f dµ is finite, then we also define f dµ = f + dµ f dµ. Theorem 40 Suppose that f, g L 1 (µ). If α, β C, then αf + βg L 1 (µ) and (αf + βg) dµ = α f dµ + β g dµ. We leave the proof as a simple exercise. This theorem says that L 1 (µ) is a linear space and f f dµ is a linear functional in L 1 (µ). Similarly we define L 1 (µ, IR n ) as the class of all measurable mappings f = (f 1, f 2,..., f n ) : IR n such that f dµ < and then we set ( ) f dµ = f 1 dµ,..., f n dµ. Theorem 41 If f L 1 (µ, IR n ), then f dµ f dµ. Proof. Let α = (α 1,..., α n ) IR n be a unit vector such that f dµ = α, f dµ. Then n f dµ = α, f dµ = α i f i dµ by Schwarz inequality. = n α i f i dµ = α, f dµ f dµ Theorem 42 (Lebesgue s dominated convergence theorem) Suppose {f n } is a sequence of complex valued functions on such that the limit f(x) = lim n f n (x) exists for every x. If there is a function g L 1 (µ) such that f n (x) g(x) for every x and all n = 1, 2,..., 32

33 then f L 1 (µ), and lim f f n dµ = 0 n lim f n dµ = f dµ. n Proof. Clearly f is measurable and f g. Hence f L 1 (µ). It is easy to see that 2g f f n 0 and hence Fatou s lemma yields 2g dµ lim inf (2g f f n ) dµ n ( ) = 2g + lim inf f f n dµ n = 2g lim sup f f n dµ. n The integral 2g dµ is finite and hence lim sup f n f dµ 0. n This immediately implies Now so The proof is complete. f n dµ lim f n f dµ = 0. n lim n f dµ f n dµ = f n f dµ 0, f dµ. 2.3 Almost everywhere Definition. If there is a set N M of measure zero, µ(n) = 0 such that a property P (x) is satisfied for all x \ N, then we say that the property P (x) is satisfied almost everywhere (a.e.) For example f n f a.e. means that there is a set N M, µ(n) = 0 such that f n (x) f(x) for all x \ N. 33

34 If f = g a.e., then we write f g. It is easy to see that is an equivalence relation (f g, g h f h follows from the fact that the union of two sets of measure zero is a set of measure zero). Note that if f g, then f dµ = g dµ for every M. Therefore from the point of view of the integration theory, functions which are equal a.e. cannot be distinguished. Moreover if f dµ = g dµ for all M, then f = g a.e. This follows from part (b) of Theorem 43 below applied to f g. Thus two functions f and g cannot be distinguished in the integration theory if and only if they are equal a.e. Therefore it is natural to identify such functions. Theorem 43 (a) Suppose f : [0, ] is measurable, M and f dµ = 0, then f = 0 a.e. in. (b) Suppose f L 1 (µ) and f dµ = 0 for every M. Then f = 0 a.e. in. Proof. (a) Suppose A n = {x : f(x) 1/n}. Then 1 n µ(a n) f dµ f dµ = 0. A n Hence µ(a n ) = 0 and therefore the set {x : f(x) > 0} = n=1 A n has measure zero. (b) Let = u + iv. Define = {x : u(x) 0}. Then 0 = real part of f dµ = u dµ = u + dµ, and hence u + = 0 a.e. by (a). Similarly u, v +, v = 0 a.e. Accordingly, f = 0 a.e. Recall that a measure µ is complete if every subset of a set of measure zero is measurable (and hence has measure zero). All the measures obtained through the Carathéodory construction are complete (Proposition 4), e.g. the Hausforff measure and the Lebesgue measure are complete. However, we can assume that any measure is complete, because we can add all subsets of sets of measure zero to the σ-algebra. More precisely one can prove. Theorem 44 Let (, M, µ) be a measure space, let M be the collection of all for which there exist sets A, B M such that A B and µ(b \ A) = 0 and define µ() = µ(a) in this situation. Then M is a σ-algebra and µ is a complete measure on M. 34

35 For a proof, see Rudin page 28. Therefore, whenever needed, we can assume that the measure is complete. Remark. If µ is a measure on B(IR n ) and there is an uncountable set B(IR n ) such that µ() = 0, then µ cannot be complete, because one can prove that the cardinality of all Borel sets in B(IR n ) is the same as the cardinality of real numbers and the cardinality of all subsets of is the same as the cardinality of all subsets of real numbers which is bigger than the cardinality of real numbers. Hence in order to make the measure µ complete we have to enlarge B(IR n ) to a bigger σ-algebra by including also sets which are not Borel. This is the case of the Lebesgue measure. The σ-algebra of Lebesgue measurable sets is larger, than the σ-algebra of Borel sets. Definition. Suppose now that µ is a complete measure. If a function with values in a metric space Y is defined only at almost all points of, i.e. f : \ N Y, µ(n) = 0, then we say that f is measurable if there is y 0 Y such that f(x) = { f(x) if x \ N, y 0 if x N, is measurable. In such a situation we say that f : Y is a measurable function defined a.e. Note that it follows from the assumption that µ is complete that f can be defined in N in an arbitrary way and still the resulting function f is measurable and f f. In particular functions in L 1 (µ) need to be defined a.e. only. L 1 (µ) is a linear space. Since we identify functions that are equal a.e. we identify L 1 (µ) with the quotient space L 1 (µ)/ (check that L 1 (µ)/ is a linear space). Therefore f L 1 (µ) satisfies f = 0 in L 1 (µ) if and only if f = 0 a.e., if and only if f 1 = f dµ = 0 (see Theorem 43(a)). In all the previously discussed convergence theorems (Theorem 36, Theorem 37, Theorem 38, Theorem 42) we can replace the requirement of everywhere convergence by the a.e. convergence. For example the Lebesgue dominated convergence theorem can be formulated as follows. Theorem 45 (Lebesgue dominated convergence theorem) Suppose that {f n } is a sequence of complex valued functino on defined a.e. If the limit f(x) = lim n f n (x) exists a.e. and if there is a function g L 1 (µ) such that f n (x) g(x) a.e., for all n = 1, 2, 3,... then f L 1 (µ), lim f f n dµ = 0 n 35

36 and lim f n dµ = f dµ. n Not surprisingly, this theorem is an easy consequence of the previous version of the Lebesgue dominated convergence theorem. We leave details to the reader. Recall that if Y is a metric space, then a mapping f : IR n Y is Lebesgue measurable if f 1 (U) is a Lebesgue measurable set for every open U Y. Since the σ-algebra of Lebesgue measurable sets is larger, than that of Borel sets, the mapping f need not be Borel. I turns out, however, that f equals a.e. to a Borel mapping, at least when Y is separable. Theorem 46 If Y is a separable metric space and f : IR n Y is Lebesgue measurable, then there is a Borel mapping g : IR n Y such that f = g a.e. Proof. Since Y is separable, there is a countable family of open sets {U i } in Y such that every open set in Y is a union of open sets from the family. Indeed, it suffices to take the family of balls with rational length radii and centers in a countable dense set in Y. For i = 1, 2,... let M i IR n be a F σ set such that M i f 1 (U i ), L n (f 1 (U i ) \ M i ) = 0, see Theorem 19. Clearly the set = (f 1 (U i ) \ M i ) has Lebesgue measure zero and hence there is a G δ set H such that H, L n (H) = 0 (Theorem 19). Fix y 0 Y and define { f(x) if x H, g(x) = y 0 if x H. Clearly f = g a.e. and it remains to prove that the mapping g is Borel. It is easy to see that 5 if y 0 U i, then g 1 (U i ) = M i \ H and if y 0 U i, then g 1 (U i ) = M i H. In each case g 1 (U i ) is a Borel set. Since every open set U Y can be represented as U = j=1 U i j, the set g 1 (U) = g 1 (U ij ) is Borel. 5 A good picture is useful here. j=1 36

37 2.4 Theorems of Lusin and gorov Theorem 47 (Lusin) Let be a metric space and µ a measure in B() such that is a union of countably many open sets of finite measure. If f : Y is a Borel mapping with values in a separable metric space, then for every ε > 0 there is a closed set F such that µ( \ F ) < ε and f F is continuous. Proof. Let {U i } be a countable family of open sets such that every open set in Y is a union of open sets from the family. It follows from Theorem 7 that for every i there is an open set V i such that 6 Let f 1 (U i ) V i, µ(v i \ f 1 (U i )) < ε2 i 1. = V i \ f 1 (U i ). Clearly µ() < ε/2. We will prove that the mapping g = f \ is continuous. Observe that 7 g 1 (U i ) = V i ( \ ). (28) Indeed, g 1 (U i ) V i ( \ ), but on the other hand so V i ( \ ) V i ( \ (V i \ f 1 (U i ))) = f 1 (U i ), V i ( \ ) f 1 (U i ) ( \ ) = g 1 (U i ). In order to prove that g is continuous it suffices to prove that for every open set U Y there is an open set V such that 8 g 1 (U) = V ( \ ). Let U = j=1 U i j. Then (28) gives g 1 (U) = ( j=1 V i j ) ( \ ) which proves continuity of g in \. The set need not be closed, but it follows from Theorem 7 that there is an open set G such that G and µ(g \ ) < ε/2. Now f F is continuous, where F = \ G, F is closed and µ( \ F ) = µ(g) = µ() + µ(g \ ) < ε. As a corollary we will prove the following characterization of Lebesgue measurable functions also known as the Lusin theorem. Let IR n be a Lebesgue measurable set. Recall that f : IR is Lebesgue measurable if f 1 (U) is Lebesgue measurable for every open U IR (f 1 (U) is a subset of ). 6 Actually to prove this fact we need first to split f 1 (U i ) into sets of finite measure and then apply Theorem 7 to each such set. 7 It is easy to see if you make an appropriate picture, but a formal proof is enclosed below. 8 Because the class of open sets in the metric space \ is exactly the class of sets of the form V ( \ ), where V is open. 37