measure_notes.pdf

Measure Theory John K. Hunter Department of Mathematics, University of California at Davis  Abstract. These are some brief notes on measure theory, concentrating on Lebesgue measure on Rn. Some missing topics I would have liked to have in-...

Measure Theory John K. Hunter Department of Mathematics, University of California at Davis  Abstract. These are some brief notes on measure theory, concentrating on Lebesgue measure on Rn. Some missing topics I would have liked to have in- cluded had time permitted are: the change of variable formula for the Lebesgue integral on Rn ; absolutely continuous functions and functions of bounded vari- ation of a single variable and their connection with Lebesgue-Stieltjes measures on R; Radon measures on Rn , and other locally compact Hausdorff topological spaces, and the Riesz representation theorem for bounded linear functionals on spaces of continuous functions; and other examples of measures, including k-dimensional Hausdorff measure in Rn , Wiener measure and Brownian mo- tion, and Haar measure on topological groups. All these topics can be found in the references. c John K. Hunter, 2011  Contents Chapter 1. Measures 1 1.1. Sets 1 1.2. Topological spaces 2 1.3. Extended real numbers 2 1.4. Outer measures 3 1.5. σ-algebras 4 1.6. Measures 5 1.7. Sets of measure zero 6 Chapter 2. Lebesgue Measure on Rn 9 2.1. Lebesgue outer measure 10 2.2. Outer measure of rectangles 12 2.3. Carathéodory measurability 14 2.4. Null sets and completeness 18 2.5. Translational invariance 19 2.6. Borel sets 20 2.7. Borel regularity 22 2.8. Linear transformations 27 2.9. Lebesgue-Stieltjes measures 30 Chapter 3. Measurable functions 33 3.1. Measurability 33 3.2. Real-valued functions 34 3.3. Pointwise convergence 36 3.4. Simple functions 37 3.5. Properties that hold almost everywhere 38 Chapter 4. Integration 39 4.1. Simple functions 39 4.2. Positive functions 40 4.3. Measurable functions 42 4.4. Absolute continuity 45 4.5. Convergence theorems 47 4.6. Complex-valued functions and a.e. convergence 50 4.7. L1 spaces 50 4.8. Riemann integral 52 4.9. Integrals of vector-valued functions 52 Chapter 5. Product Measures 55 5.1. Product σ-algebras 55 iii iv CONTENTS 5.2. Premeasures 56 5.3. Product measures 58 5.4. Measurable functions 60 5.5. Monotone class theorem 61 5.6. Fubini’s theorem 61 5.7. Completion of product measures 61 Chapter 6. Differentiation 63 6.1. A covering lemma 64 6.2. Maximal functions 65 6.3. Weak-L1 spaces 67 6.4. Hardy-Littlewood theorem 67 6.5. Lebesgue differentiation theorem 68 6.6. Signed measures 70 6.7. Hahn and Jordan decompositions 71 6.8. Radon-Nikodym theorem 74 6.9. Complex measures 77 Chapter 7. Lp spaces 79 7.1. Lp spaces 79 7.2. Minkowski and Hölder inequalities 80 7.3. Density 81 7.4. Completeness 81 7.5. Duality 83 Bibliography 89  CHAPTER 1 Measures Measures are a generalization of volume; the fundamental example is Lebesgue measure on Rn , which we discuss in detail in the next Chapter. Moreover, as formalized by Kolmogorov (1933), measure theory provides the foundation of prob- ability. Measures are important not only because of their intrinsic geometrical and probabilistic significance, but because they allow us to define integrals. This connection, in fact, goes in both directions: we can define an integral in terms of a measure; or, in the Daniell-Stone approach, we can start with an integral (a linear functional acting on functions) and use it to define a measure. In probability theory, this corresponds to taking the expectation of random variables as the fundamental concept from which the probability of events is derived. In these notes, we develop the theory of measures first, and then define integrals. This is (arguably) the more concrete and natural approach; it is also (unarguably) the original approach of Lebesgue. We begin, in this Chapter, with some prelimi- nary definitions and terminology related to measures on arbitrary sets. See Folland for further discussion. 1.1. Sets We use standard definitions and notations from set theory and will assume the axiom of choice when needed. The words ‘collection’ and ‘family’ are synonymous with ‘set’ — we use them when talking about sets of sets. We denote the collection of subsets, or power set, of a set X by P(X). The notation 2X is also used. If E ⊂ X and the set X is understood, we denote the complement of E in X by E c = X \ E. De Morgan’s laws state that !c !c ∞ [ \ \ [ c Eα = Eα , Eα = Eαc. α∈I α∈I α∈I α∈I We say that a collection C = {Eα ⊂ X : α ∈ I} of subsets of a set X, indexed by a set I, covers E ⊂ X if [ Eα ⊃ E. α∈I The collection C is disjoint if Eα ∩ Eβ = ∅ for α 6= β. The Cartesian product, or product, of sets X, Y is the collection of all ordered pairs X × Y = {(x, y) : x ∈ X, y ∈ Y }. 1 2 1. MEASURES 1.2. Topological spaces A topological space is a set equipped with a collection of open subsets that satisfies appropriate conditions. Definition 1.1. A topological space (X, T ) is a set X and a collection T ⊂ P(X) of subsets of X, called open sets, such that (a) ∅, X ∈ T ; (b) if {Uα ∈ T : α ∈ I} is an arbitrary collection of open sets, then their union [ Uα ∈ T α∈I is open; (c) if {Ui ∈ T : i = 1, 2,... , N } is a finite collection of open sets, then their intersection N \ Ui ∈ T i=1 is open. The complement of an open set in X is called a closed set, and T is called a topology on X. 1.3. Extended real numbers It is convenient to use the extended real numbers R = {−∞} ∪ R ∪ {∞}. This allows us, for example, to talk about sets with infinite measure or non-negative functions with infinite integral. The extended real numbers are totally ordered in the obvious way: ∞ is the largest element, −∞ is the smallest element, and real numbers are ordered as in R. Algebraic operations on R are defined when they are unambiguous e.g. ∞ + x = ∞ for every x ∈ R except x = −∞, but ∞ − ∞ is undefined. We define a topology on R in a natural way, making R homeomorphic to a compact interval. For example, the function φ : R → [−1, 1] defined by   1 √ if x = ∞ φ(x) = x/ 1 + x2 if −∞ < x < ∞  −1 if x = −∞ is a homeomorphism. A primary reason to use the extended real numbers is that upper and lower bounds always exist. Every subset of R has a supremum (equal to ∞ if the subset contains ∞ or is not bounded from above in R) and infimum (equal to −∞ if the subset contains −∞ or is not bounded from below in R). Every increasing sequence of extended real numbers converges to its supremum, and every decreasing sequence converges to its infimum. Similarly, if {an } is a sequence of extended real-numbers then     lim sup an = inf sup ai , lim inf an = sup inf ai n→∞ n∈N i≥n n→∞ n∈N i≥n both exist as extended real numbers.  1.4. OUTER MEASURES 3 P∞ Every sum i=1 xi with non-negative terms xi ≥ 0 converges in R (to ∞ if xi = ∞ for some i ∈ N or the series diverges in R), where the sum is defined by ∞ ( ) X X xi = sup xi : F ⊂ N is finite. i=1 i∈F As for non-negative sums of real numbers, non-negative sums of extended real numbers are unconditionally convergent (the order of the terms does not matter); we can rearrange sums of non-negative extended real numbers ∞ X ∞ X ∞ X (xi + yi ) = xi + yi ; i=1 i=1 i=1 and double sums may be evaluated as iterated single sums   X∞  X  xij = sup xij : F ⊂ N × N is finite   i,j=1 (i,j)∈F   X∞ X∞ =  xij  i=1 j=1 ∞ ∞ ! X X = xij. j=1 i=1 Our use of extended real numbers is closely tied to the order and monotonicity properties of R. In dealing with complex numbers or elements of a vector space, we will always require that they are strictly finite. 1.4. Outer measures As stated in the following definition, an outer measure is a monotone, countably subadditive, non-negative, extended real-valued function defined on all subsets of a set. Definition 1.2. An outer measure µ∗ on a set X is a function µ∗ : P(X) → [0, ∞] such that: (a) µ∗ (∅) = 0; (b) if E ⊂ F ⊂ X, then µ∗ (E) ≤ µ∗ (F ); (c) if {Ei ⊂ X : i ∈ N} is a countable collection of subsets of X, then ∞ ! ∞ [ X µ∗ Ei ≤ µ∗ (Ei ). i=1 i=1 We obtain a statement about finite unions from a statement about infinite unions by taking all but finitely many sets in the union equal to the empty set. Note that µ∗ is not assumed to be additive even if the collection {Ei } is disjoint. 4 1. MEASURES 1.5. σ-algebras A σ-algebra on a set X is a collection of subsets of a set X that contains ∅ and X, and is closed under complements, finite unions, countable unions, and countable intersections. Definition 1.3. A σ-algebra on a set X is a collection A of subsets of X such that: (a) ∅, X ∈ A; (b) if A ∈ A then Ac ∈ A; (c) if Ai ∈ A for i ∈ N then ∞ [ ∞ \ Ai ∈ A, Ai ∈ A. i=1 i=1 From de Morgan’s laws, a collection of subsets is σ-algebra if it contains ∅ and is closed under the operations of taking complements and countable unions (or, equivalently, countable intersections). Example 1.4. If X is a set, then {∅, X} and P(X) are σ-algebras on X; they are the smallest and largest σ-algebras on X, respectively. Measurable spaces provide the domain of measures, defined below. Definition 1.5. A measurable space (X, A) is a non-empty set X equipped with a σ-algebra A on X. It is useful to compare the definition of a σ-algebra with that of a topology in Definition 1.1. There are two significant differences. First, the complement of a measurable set is measurable, but the complement of an open set is not, in general, open, excluding special cases such as the discrete topology T = P(X). Second, countable intersections and unions of measurable sets are measurable, but only finite intersections of open sets are open while arbitrary (even uncountable) unions of open sets are open. Despite the formal similarities, the properties of measurable and open sets are very different, and they do not combine in a straightforward way. If F is any collection of subsets of a set X, then there is a smallest σ-algebra on X that contains F , denoted by σ(F ). Definition 1.6. If F is any collection of subsets of a set X, then the σ-algebra generated by F is \ σ(F ) = {A ⊂ P(X) : A ⊃ F and A is a σ-algebra}. This intersection is nonempty, since P(X) is a σ-algebra that contains F , and an intersection of σ-algebras is a σ-algebra. An immediate consequence of the definition is the following result, which we will use repeatedly. Proposition 1.7. If F is a collection of subsets of a set X such that F ⊂ A where A is a σ-algebra on X, then σ(F ) ⊂ A. Among the most important σ-algebras are the Borel σ-algebras on topological spaces. Definition 1.8. Let (X, T ) be a topological space. The Borel σ-algebra B(X) = σ(T ) is the σ-algebra generated by the collection T of open sets on X.  1.6. MEASURES 5 1.6. Measures A measure is a countably additive, non-negative, extended real-valued function defined on a σ-algebra. Definition 1.9. A measure µ on a measurable space (X, A) is a function µ : A → [0, ∞] such that (a) µ(∅) = 0; (b) if {Ai ∈ A : i ∈ N} is a countable disjoint collection of sets in A, then ∞ ! ∞ [ X µ Ai = µ(Ai ). i=1 i=1 In comparison with an outer measure, a measure need not be defined on all subsets of a set, but it is countably additive rather than countably subadditive. S∞ A measure µ on a set X is finite if µ(X) < ∞, and σ-finite if X = n=1 An is a countable union of measurable sets An with finite measure, µ(An ) < ∞. A probability measure is a finite measure with µ(X) = 1. A measure space (X, A, µ) consists of a set X, a σ-algebra A on X, and a measure µ defined on A. When A and µ are clear from the context, we will refer to the measure space X. We define subspaces of measure spaces in the natural way. Definition 1.10. If (X, A, µ) is a measure space and E ⊂ X is a measurable subset, then the measure subspace (E, A|E , µ|E ) is defined by restricting µ to E: A|E = {A ∩ E : A ∈ A} , µ|E (A ∩ E) = µ(A ∩ E). As we will see, the construction of nontrivial measures, such as Lebesgue mea- sure, requires considerable effort. Nevertheless, there is at least one useful example of a measure that is simple to define. Example 1.11. Let X be an arbitrary non-empty set. Define ν : P(X) → [0, ∞] by ν(E) = number of elements in E, where ν(∅) = 0 and ν(E) = ∞ if E is not finite. Then ν is a measure, called count- ing measure on X. Every subset of X is measurable with respect to ν. Counting measure is finite if X is finite and σ-finite if X is countable. A useful implication of the countable additivity of a measure is the following monotonicity result. Proposition 1.12. If {Ai : i ∈ N} is an increasing sequence of measurable sets, meaning that Ai+1 ⊃ Ai , then ∞ ! [ (1.1) µ Ai = lim µ(Ai ). i→∞ i=1 If {Ai : i ∈ N} is a decreasing sequence of measurable sets, meaning that Ai+1 ⊂ Ai , and µ(A1 ) < ∞, then ∞ ! \ (1.2) µ Ai = lim µ(Ai ). i→∞ i=1 6 1. MEASURES Proof. If {Ai : i ∈ N} is an increasing sequence of sets and Bi = Ai+1 \ Ai , then {Bi : i ∈ N} is a disjoint sequence with the same union, so by the countable additivity of µ ! ! ∞ [ ∞ [ X∞ µ Ai = µ Bi = µ (Bi ). i=1 i=1 i=1 Sj Moreover, since Aj = i=1 Bi , j X µ(Aj ) = µ (Bi ) , i=1 which implies that ∞ X µ (Bi ) = lim µ(Aj ) j→∞ i=1 and the first result follows. If µ(A1 ) < ∞ and {Ai } is decreasing, then {Bi = A1 \ Ai } is increasing and µ(Bi ) = µ(A1 ) − µ(Ai ). It follows from the previous result that ∞ ! [ µ Bi = lim µ(Bi ) = µ(A1 ) − lim µ(Ai ). i→∞ i→∞ i=1 Since ! ! ∞ [ ∞ \ ∞ [ ∞ \ Bi = A1 \ Ai , µ Bi = µ(A1 ) − µ Ai , i=1 i=1 i=1 i=1 the result follows.  Example 1.13. To illustrate the necessity of the condition µ(A1 ) < ∞ in the second part of the previous proposition, or more generally µ(An ) < ∞ for some n ∈ N, consider counting measure ν : P(N) → [0, ∞] on N. If An = {k ∈ N : k ≥ n}, then ν(An ) = ∞ for every n ∈ N, so ν(An ) → ∞ as n → ∞, but ∞ ∞ ! \ \ An = ∅, ν An = 0. n=1 n=1 1.7. Sets of measure zero A set of measure zero, or a null set, is a measurable set N such that µ(N ) = 0. A property which holds for all x ∈ X \ N where N is a set of measure zero is said to hold almost everywhere, or a.e. for short. If we want to emphasize the measure, we say µ-a.e. In general, a subset of a set of measure zero need not be measurable, but if it is, it must have measure zero. It is frequently convenient to use measure spaces which are complete in the following sense. (This is, of course, a different sense of ‘complete’ than the one used in talking about complete metric spaces.) Definition 1.14. A measure space (X, A, µ) is complete if every subset of a set of measure zero is measurable.  1.7. SETS OF MEASURE ZERO 7 Note that completeness depends on the measure µ, not just the σ-algebra A. Any measure space (X, A, µ) is contained in a uniquely defined completion (X, A, µ), which the smallest complete measure space that contains it and is given explicitly as follows. Theorem 1.15. If (X, A, µ) is a measure space, define (X, A, µ) by A = {A ∪ M : A ∈ A, M ⊂ N where N ∈ A satisfies µ(N ) = 0} with µ(A ∪ M ) = µ(A). Then (X, A, µ) is a complete measure space such that A ⊃ A and µ is the unique extension of µ to A. Proof. The collection A is a σ-algebra. It is closed under complementation because, with the notation used in the definition, (A ∪ M )c = Ac ∩ M c , M c = N c ∪ (N \ M ). Therefore (A ∪ M )c = (Ac ∩ N c ) ∪ (Ac ∩ (N \ M )) ∈ A, since Ac ∩ N c ∈ A and Ac ∩ (N \ M ) ⊂ N. Moreover, A is closed under countable unions because if Ai ∈ A and Mi ⊂ Ni where µ(Ni ) = 0 for each i ∈ N, then ∞ ∞ ! ∞ ! [ [ [ Ai ∪ Mi = Ai ∪ Mi ∈ A, i=1 i=1 i=1 since ! ∞ [ ∞ [ ∞ [ ∞ [ Ai ∈ A, Mi ⊂ Ni , µ Ni = 0. i=1 i=1 i=1 i=1 It is straightforward to check that µ is well-defined and is the unique extension of µ to a measure on A, and that (X, A, µ) is complete.   CHAPTER 2 Lebesgue Measure on Rn Our goal is to construct a notion of the volume, or Lebesgue measure, of rather general subsets of Rn that reduces to the usual volume of elementary geometrical sets such as cubes or rectangles. If L(Rn ) denotes the collection of Lebesgue measurable sets and µ : L(Rn ) → [0, ∞] denotes Lebesgue measure, then we want L(Rn ) to contain all n-dimensional rect- angles and µ(R) should be the usual volume of a rectangle R. Moreover, we want µ to be countably additive. That is, if {Ai ∈ L(Rn ) : i ∈ N} is a countable collection of disjoint measurable sets, then their union should be measurable and ! [∞ ∞ X µ Ai = µ (Ai ). i=1 i=1 The reason for requiring countable additivity is that finite additivity is too weak a property to allow the justification of any limiting processes, while uncountable additivity is too strong; for example, it would imply that if the measure of a set consisting of a single point is zero, then the measure of every subset of Rn would be zero. It is not possible to define the Lebesgue measure of all subsets of Rn in a geometrically reasonable way. Hausdorff (1914) showed that for any dimension n ≥ 1, there is no countably additive measure defined on all subsets of Rn that is invariant under isometries (translations and rotations) and assigns measure one to the unit cube. He further showed that if n ≥ 3, there is no such finitely additive measure. This result is dramatized by the Banach-Tarski ‘paradox’: Banach and Tarski (1924) showed that if n ≥ 3, one can cut up a ball in Rn into a finite number of pieces and use isometries to reassemble the pieces into a ball of any desired volume e.g. reassemble a pea into the sun. The ‘construction’ of these pieces requires the axiom of choice.1 Banach (1923) also showed that if n = 1 or n = 2 there are finitely additive, isometrically invariant extensions of Lebesgue measure on Rn that are defined on all subsets of Rn , but these extensions are not countably additive. For a detailed discussion of the Banach-Tarski paradox and related issues, see. The moral of these results is that some subsets of Rn are too irregular to define their Lebesgue measure in a way that preserves countable additivity (or even finite additivity in n ≥ 3 dimensions) together with the invariance of the measure under 1Solovay (1970) proved that one has to use the axiom of choice to obtain non-Lebesgue measurable sets. 9 10 2. LEBESGUE MEASURE ON Rn isometries. We will show, however, that such a measure can be defined on a σ- algebra L(Rn ) of Lebesgue measurable sets which is large enough to include all set of ‘practical’ importance in analysis. Moreover, as we will see, it is possible to define an isometrically-invariant, countably sub-additive outer measure on all subsets of Rn. There are many ways to construct Lebesgue measure, all of which lead to the same result. We will follow an approach due to Carathéodory, which generalizes to other measures: We first construct an outer measure on all subsets of Rn by approximating them from the outside by countable unions of rectangles; we then restrict this outer measure to a σ-algebra of measurable subsets on which it is count- ably additive. This approach is somewhat asymmetrical in that we approximate sets (and their complements) from the outside by elementary sets, but we do not approximate them directly from the inside. Jones , Stein and Shakarchi , and Wheeler and Zygmund give detailed introductions to Lebesgue measure on Rn. Cohn gives a similar development to the one here, and Evans and Gariepy discuss more advanced topics. 2.1. Lebesgue outer measure We use rectangles as our elementary sets, defined as follows. Definition 2.1. An n-dimensional, closed rectangle with sides oriented parallel to the coordinate axes, or rectangle for short, is a subset R ⊂ Rn of the form R = [a1 , b1 ] × [a2 , b2 ] × · · · × [an , bn ] where −∞ < ai ≤ bi < ∞ for i = 1,... , n. The volume µ(R) of R is µ(R) = (b1 − a1 )(b2 − a2 )... (bn − an ). If n = 1 or n = 2, the volume of a rectangle is its length or area, respectively. We also consider the empty set to be a rectangle with µ(∅) = 0. We denote the collection of all n-dimensional rectangles by R(Rn ), or R when n is understood, and then R 7→ µ(R) defines a map µ : R(Rn ) → [0, ∞). The use of this particular class of elementary sets is for convenience. We could equally well use open or half-open rectangles, cubes, balls, or other suitable ele- mentary sets; the result would be the same. Definition 2.2. The outer Lebesgue measure µ∗ (E) of a subset E ⊂ Rn , or outer measure for short, is (∞ ) X S∞ ∗ n (2.1) µ (E) = inf µ(Ri ) : E ⊂ i=1 Ri , Ri ∈ R(R ) i=1 where the infimum is taken over all countable collections of rectangles whose union contains E. The map µ∗ : P(Rn ) → [0, ∞], µ∗ : E 7→ µ∗ (E) is called outer Lebesgue measure.  2.1. LEBESGUE OUTER MEASURE 11 P∞ In this definition, a sum i=1 µ(Ri ) and µ∗ (E) may take the value ∞. We do not require that the rectangles Ri are disjoint, so the same volume may contribute to multiple terms in the sum on the right-hand side of (2.1); this does not affect the value of the infimum. Example 2.3. Let E = Q ∩ [0, 1] be the set of rational numbers between 0 and 1. Then E has outer measure zero. To prove this, let {qi : i ∈ N} be an i enumeration of the points in E. S∞Given ǫ > 0, let Ri be an interval of length ǫ/2 which contains qi. Then E ⊂ i=1 µ(Ri ) so ∞ X 0 ≤ µ∗ (E) ≤ µ(Ri ) = ǫ. i=1 Hence µ∗ (E) = 0 since ǫ > 0 is arbitrary. The same argument shows that any countable set has outer measure zero. Note that if we cover E by a finite collection of intervals, then the union of the intervals would have to contain [0, 1] since E is dense in [0, 1] so their lengths sum to at least one. The previous example illustrates why we need to use countably infinite collec- tions of rectangles, not just finite collections, to define the outer measure.2 The ‘countable ǫ-trick’ used in the example appears in various forms throughout measure theory. Next, we prove that µ∗ is an outer measure in the sense of Definition 1.2. Theorem 2.4. Lebesgue outer measure µ∗ has the following properties. (a) µ∗ (∅) = 0; (b) if E ⊂ F , then µ∗ (E) ≤ µ∗ (F ); (c) if {Ei ⊂ Rn : i ∈ N} is a countable collection of subsets of Rn , then ∞ ! ∞ [ X ∗ µ Ei ≤ µ∗ (Ei ). i=1 i=1 Proof. It follows immediately from Definition 2.2 that µ∗ (∅) = 0, since every collection of rectangles covers ∅, and that µ∗ (E) ≤ µ∗ (F ) if E ⊂ F since any cover of F covers E. The main property to prove is the countable subadditivity of µ∗. If µ∗ (Ei ) = ∞ for some i ∈ N, there is nothing to prove, so we may assume that µ∗ (Ei ) is finite for every i ∈ N. If ǫ > 0, there is a countable covering {Rij : j ∈ N} of Ei by rectangles Rij such that ∞ X ∞ [ ǫ µ(Rij ) ≤ µ∗ (Ei ) + , Ei ⊂ Rij. j=1 2i j=1 Then {Rij : i, j ∈ N} is a countable covering of ∞ [ E= Ei i=1 2The use of finitely many intervals leads to the notion of the Jordan content of a set, intro- duced by Peano (1887) and Jordan (1892), which is closely related to the Riemann integral; Borel (1898) and Lebesgue (1902) generalized Jordan’s approach to allow for countably many intervals, leading to Lebesgue measure and the Lebesgue integral. 12 2. LEBESGUE MEASURE ON Rn and therefore ∞ n ǫo X ∗ ∞ X X ∞ µ∗ (E) ≤ µ(Rij ) ≤ µ∗ (Ei ) + = µ (Ei ) + ǫ. i,j=1 i=1 2i i=1 Since ǫ > 0 is arbitrary, it follows that ∞ X ∗ µ (E) ≤ µ∗ (Ei ) i=1 which proves the result.  2.2. Outer measure of rectangles In this section, we prove the geometrically obvious, but not entirely trivial, fact that the outer measure of a rectangle is equal to its volume. The main point is to show that the volumes of a countable collection of rectangles that cover a rectangle R cannot sum to less than the volume of R.3 We begin with some combinatorial facts about finite covers of rectangles. We denote the interior of a rectangle R by R◦ , and we say that rectangles R, S are almost disjoint if R◦ ∩ S ◦ = ∅, meaning that they intersect at most along their boundaries. The proofs of the following results are cumbersome to write out in detail (it’s easier to draw a picture) but we briefly explain the argument. Lemma 2.5. Suppose that R = I1 × I2 × · · · × In is an n-dimensional rectangle where each closed, bounded interval Ii ⊂ R is an almost disjoint union of closed, bounded intervals {Ii,j ⊂ R : j = 1,... , Ni }, Ni [ Ii = Ii,j. j=1 Define the rectangles (2.2) Sj1 j2...jn = I1,j1 × I2,j2 × · · · × In,jn. Then N1 X Nn X µ(R) = ··· µ (Sj1 j2...jn ). j1 =1 jn =1 Proof. Denoting the length of an interval I by |I|, using the fact that Ni X |Ii | = |Ii,j |, j=1 3As a partial justification of the need to prove this fact, note that it would not be true if we allowed uncountable covers, since we could cover any rectangle by an uncountable collection of points all of whose volumes are zero.  2.2. OUTER MEASURE OF RECTANGLES 13 and expanding the resulting product, we get that µ(R) = |I1 ||I2 |... |In |      XN1 N2 X Nn X = |I1,j1 |  |I2,j2 |...  |In,jn | j1 =1 j2 =1 jn =1 X N2 N1 X Nn X = ··· |I1,j1 ||I2,j2 |... |In,jn | j1 =1 j2 =1 jn =1 X N2 N1 X Nn X = ··· µ (Sj1 j2...jn ). j1 =1 j2 =1 jn =1  Proposition 2.6. If a rectangle R is an almost disjoint, finite union of rect- angles {R1 , R2 ,... , RN }, then N X (2.3) µ(R) = µ(Ri ). i=1 If R is covered by rectangles {R1 , R2 ,... , RN }, which need not be disjoint, then N X (2.4) µ(R) ≤ µ(Ri ). i=1 Proof. Suppose that R = [a1 , b1 ] × [a2 , b2 ] × · · · × [an , bn ] is an almost disjoint union of the rectangles {R1 , R2 ,... , RN }. Then by ‘extending the sides’ of the Ri , we may decompose R into an almost disjoint collection of rectangles {Sj1 j2...jn : 1 ≤ ji ≤ Ni for 1 ≤ i ≤ n} that is obtained by taking products of subintervals of partitions of the coordinate intervals [ai , bi ] into unions of almost disjoint, closed subintervals. Explicitly, we partition [ai , bi ] into ai = ci,0 ≤ ci,1 ≤ · · · ≤ ci,Ni = bi , Ii,j = [ci,j−1 , ci,j ]. where the ci,j are obtained by ordering the left and right ith coordinates of all faces of rectangles in the collection {R1 , R2 ,... , RN }, and define rectangles Sj1 j2...jn as in (2.2). Each rectangle Ri in the collection is an almost disjoint union of rectangles Sj1 j2...jn , and their union contains all such products exactly once, so by applying Lemma 2.5 to each Ri and summing the results we see that N X N1 X Nn X µ(Ri ) = ··· µ (Sj1 j2...jn ). i=1 j1 =1 jn =1 Similarly, R is an almost disjoint union of all the rectangles Sj1 j2...jn , so Lemma 2.5 implies that N1 X Nn X µ(R) = ··· µ (Sj1 j2...jn ) , j1 =1 jn =1 14 2. LEBESGUE MEASURE ON Rn and (2.3) follows. If a finite collection of rectangles {R1 , R2 ,... , RN } covers R, then there is a almost disjoint, finite collection of rectangles {S1 , S2 ,... , SM } such that M [ M X N X R= Si , µ(Si ) ≤ µ(Ri ). i=1 i=1 i=1 To obtain the Si , we replace Ri by the rectangle R ∩ Ri , and then decompose these possibly non-disjoint rectangles into an almost disjoint, finite collection of sub-rectangles with the same union; we discard ‘overlaps’ which can only reduce the sum of the volumes. Then, using (2.3), we get M X N X µ(R) = µ(Si ) ≤ µ(Ri ), i=1 i=1 which proves (2.4).  The outer measure of a rectangle is defined in terms of countable covers. We reduce these to finite covers by using the topological properties of Rn. Proposition 2.7. If R is a rectangle in Rn , then µ∗ (R) = µ(R). Proof. Since {R} covers R, we have µ∗ (R) ≤ µ(R), so we only need to prove the reverse inequality. Suppose that {Ri : i ∈ N} is a countably infinite collection of rectangles that covers R. By enlarging Ri slightly we may obtain a rectangle Si whose interior Si◦ contains Ri such that ǫ µ(Si ) ≤ µ(Ri ) + i. 2 Then {Si◦ : i ∈ N} is an open cover of the compact set R, so it contains a finite subcover, which we may label as {S1◦ , S2◦ ,... , SN ◦ }. Then {S1 , S2 ,... , SN } covers R and, using (2.4), we find that N n ǫo X N X X ∞ µ(R) ≤ µ(Si ) ≤ µ(Ri ) + i ≤ µ(Ri ) + ǫ. i=1 i=1 2 i=1 Since ǫ > 0 is arbitrary, we have ∞ X µ(R) ≤ µ(Ri ) i=1 and it follows that µ(R) ≤ µ∗ (R).  2.3. Carathéodory measurability We will obtain Lebesgue measure as the restriction of Lebesgue outer measure to Lebesgue measurable sets. The construction, due to Carathéodory, works for any outer measure, as given in Definition 1.2, so we temporarily consider general outer measures. We will return to Lebesgue measure on Rn at the end of this section. The following is the Carathéodory definition of measurability. Definition 2.8. Let µ∗ be an outer measure on a set X. A subset A ⊂ X is Carathéodory measurable with respect to µ∗ , or measurable for short, if (2.5) µ∗ (E) = µ∗ (E ∩ A) + µ∗ (E ∩ Ac )  2.3. CARATHÉODORY MEASURABILITY 15 for every subset E ⊂ X. We also write E ∩ Ac as E \ A. Thus, a measurable set A splits any set E into disjoint pieces whose outer measures add up to the outer measure of E. Heuristically, this condition means that a set is measurable if it divides other sets in a ‘nice’ way. The regularity of the set E being divided is not important here. Since µ∗ is subadditive, we always have that µ∗ (E) ≤ µ∗ (E ∩ A) + µ∗ (E ∩ Ac ). Thus, in order to prove that A ⊂ X is measurable, it is sufficient to show that µ∗ (E) ≥ µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) for every E ⊂ X, and then we have equality as in (2.5). Definition 2.8 is perhaps not the most intuitive way to define the measurability of sets, but it leads directly to the following key result. Theorem 2.9. The collection of Carathéodory measurable sets with respect to an outer measure µ∗ is a σ-algebra, and the restriction of µ∗ to the measurable sets is a measure. Proof. It follows immediately from (2.5) that ∅ is measurable and the comple- ment of a measurable set is measurable, so to prove that the collection of measurable sets is a σ-algebra, we only need to show that it is closed under countable unions. We will prove at the same time that µ∗ is countably additive on measurable sets; since µ∗ (∅) = 0, this will prove that the restriction of µ∗ to the measurable sets is a measure. First, we prove that the union of measurable sets is measurable. Suppose that A, B are measurable and E ⊂ X. The measurability of A and B implies that µ∗ (E) = µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) (2.6) = µ∗ (E ∩ A ∩ B) + µ∗ (E ∩ A ∩ B c ) + µ∗ (E ∩ Ac ∩ B) + µ∗ (E ∩ Ac ∩ B c ). Since A ∪ B = (A ∩ B) ∪ (A ∩ B c ) ∪ (Ac ∩ B) and µ∗ is subadditive, we have µ∗ (E ∩ (A ∪ B)) ≤ µ∗ (E ∩ A ∩ B) + µ∗ (E ∩ A ∩ B c ) + µ∗ (E ∩ Ac ∩ B). The use of this inequality and the relation Ac ∩ B c = (A ∪ B)c in (2.6) implies that µ∗ (E) ≥ µ∗ (E ∩ (A ∪ B)) + µ∗ (E ∩ (A ∪ B)c ) so A ∪ B is measurable. Moreover, if A is measurable and A ∩ B = ∅, then by taking E = A ∪ B in (2.5), we see that µ∗ (A ∪ B) = µ∗ (A) + µ∗ (B). Thus, the outer measure of the union of disjoint, measurable sets is the sum of their outer measures. The repeated application of this result implies that the finite union of measurable sets is measurable and µ∗ is finitely additive on the collection of measurable sets. Next, we we want to show that the countable union of measurable sets is measurable. It is sufficient to consider disjoint unions. To see this, note that if 16 2. LEBESGUE MEASURE ON Rn {Ai : i ∈ N} is a countably infinite collection of measurable sets, then j [ Bj = Ai , for j ≥ 1 i=1 form an increasing sequence of measurable sets, and Cj = Bj \ Bj−1 for j ≥ 2, C1 = B1 form a disjoint measurable collection of sets. Moreover ∞ [ ∞ [ Ai = Cj. i=1 j=1 Suppose that {Ai : i ∈ N} is a countably infinite, disjoint collection of measur- able sets, and define [j ∞ [ Bj = Ai , B= Ai. i=1 i=1 Let E ⊂ X. Since Aj is measurable and Bj = Aj ∪ Bj−1 is a disjoint union (for j ≥ 2), µ∗ (E ∩ Bj ) = µ∗ (E ∩ Bj ∩ Aj ) + µ∗ (E ∩ Bj ∩ Acj ),. ∗ ∗ = µ (E ∩ Aj ) + µ (E ∩ Bj−1 ). Also µ∗ (E ∩ B1 ) = µ∗ (E ∩ A1 ). It follows by induction that j X µ∗ (E ∩ Bj ) = µ∗ (E ∩ Ai ). i=1 Since Bj is a finite union of measurable sets, it is measurable, so µ∗ (E) = µ∗ (E ∩ Bj ) + µ∗ (E ∩ Bjc ), and since Bjc ⊃ B c , we have µ∗ (E ∩ Bjc ) ≥ µ∗ (E ∩ B c ). It follows that j X ∗ µ (E) ≥ µ∗ (E ∩ Ai ) + µ∗ (E ∩ B c ). i=1 Taking the limit of this inequality as j → ∞ and using the subadditivity of µ∗ , we get ∞ X µ∗ (E) ≥ µ∗ (E ∩ Ai ) + µ∗ (E ∩ B c ) i=1 ∞ ! [ ∗ (2.7) ≥µ E ∩ Ai + µ∗ (E ∩ B c ) i=1 ≥ µ∗ (E ∩ B) + µ∗ (E ∩ B c ) ≥ µ∗ (E).  2.3. CARATHÉODORY MEASURABILITY 17 S∞ Therefore, we must have equality in (2.7), which shows that B = i=1 Ai is mea- surable. Moreover, ! ∞ [ X∞ ∗ µ E ∩ Ai = µ∗ (E ∩ Ai ), i=1 i=1 so taking E = X, we see that µ∗ is countably additive on the σ-algebra of measur- able sets.  Returning to Lebesgue measure on Rn , the preceding theorem shows that we get a measure on Rn by restricting Lebesgue outer measure to its Carathéodory- measurable sets, which are the Lebesgue measurable subsets of Rn. Definition 2.10. A subset A ⊂ Rn is Lebesgue measurable if µ∗ (E) = µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) for every subset E ⊂ Rn. If L(Rn ) denotes the σ-algebra of Lebesgue measurable sets, the restriction of Lebesgue outer measure µ∗ to the Lebesgue measurable sets µ : L(Rn ) → [0, ∞], µ = µ∗ |L(Rn ) is called Lebesgue measure. From Proposition 2.7, this notation is consistent with our previous use of µ to denote the volume of a rectangle. If E ⊂ Rn is any measurable subset of Rn , then we define Lebesgue measure on E by restricting Lebesgue measure on Rn to E, as in Definition 1.10, and denote the corresponding σ-algebra of Lebesgue measurable subsets of E by L(E). Next, we prove that all rectangles are measurable; this implies that L(Rn ) is a ‘large’ collection of subsets of Rn. Not all subsets of Rn are Lebesgue measurable, however; e.g. see Example 2.17 below. Proposition 2.11. Every rectangle is Lebesgue measurable. Proof. Let R be an n-dimensional rectangle and E ⊂ Rn. Given ǫ > 0, there is a cover {Ri : i ∈ N} of E by rectangles Ri such that X∞ µ∗ (E) + ǫ ≥ µ(Ri ). i=1 We can decompose Ri into an almost disjoint, finite union of rectangles {R̃i , Si,1 ,... , Si,N } such that N [ Ri = R̃i + Si,j , R̃i = Ri ∩ R ⊂ R, Si,j ⊂ Rc. j=1 From (2.3), N X µ(Ri ) = µ(R̃i ) + µ(Si,j ). j=1 Using this result in the previous sum, relabeling the Si,j as Si , and rearranging the resulting sum, we get that ∞ X ∞ X µ∗ (E) + ǫ ≥ µ(R̃i ) + µ(Si ). i=1 i=1 18 2. LEBESGUE MEASURE ON Rn Since the rectangles {R̃i : i ∈ N} cover E ∩ R and the rectangles {Si : i ∈ N} cover E ∩ Rc , we have ∞ X ∞ X µ∗ (E ∩ R) ≤ µ(R̃i ), µ∗ (E ∩ Rc ) ≤ µ(Si ). i=1 i=1 Hence, µ∗ (E) + ǫ ≥ µ∗ (E ∩ R) + µ∗ (E ∩ Rc ). Since ǫ > 0 is arbitrary, it follows that µ∗ (E) ≥ µ∗ (E ∩ R) + µ∗ (E ∩ Rc ), which proves the result.  ◦ An open rectangle R is a union of an increasing sequence of closed rectangles whose volumes approach µ(R); for example (a1 , b1 ) × (a2 , b2 ) × · · · × (an , bn ) ∞ [ 1 1 1 1 1 1 = [a1 + , b1 − ] × [a2 + , b2 − ] × · · · × [an + , bn − ]. k k k k k k k=1 Thus, R◦ is measurable and, from Proposition 1.12, µ(R◦ ) = µ(R). Moreover if ∂R = R \ R◦ denotes the boundary of R, then µ(∂R) = µ(R) − µ(R◦ ) = 0. 2.4. Null sets and completeness Sets of measure zero play a particularly important role in measure theory and integration. First, we show that all sets with outer Lebesgue measure zero are Lebesgue measurable. Proposition 2.12. If N ⊂ Rn and µ∗ (N ) = 0, then N is Lebesgue measurable, and the measure space (Rn , L(Rn ), µ) is complete. Proof. If N ⊂ Rn has outer Lebesgue measure zero and E ⊂ Rn , then 0 ≤ µ∗ (E ∩ N ) ≤ µ∗ (N ) = 0, so µ∗ (E ∩ N ) = 0. Therefore, since E ⊃ E ∩ N c , µ∗ (E) ≥ µ∗ (E ∩ N c ) = µ∗ (E ∩ N ) + µ∗ (E ∩ N c ), which shows that N is measurable. If N is a measurable set with µ(N ) = 0 and M ⊂ N , then µ∗ (M ) = 0, since µ∗ (M ) ≤ µ(N ). Therefore M is measurable and (Rn , L(Rn ), µ) is complete.  In view of the importance of sets of measure zero, we formulate their definition explicitly. Definition 2.13. A subset N ⊂ Rn has Lebesgue measure zero if for every ǫ > 0 there exists a countable collection of rectangles {Ri : i ∈ N} such that ∞ [ ∞ X N⊂ Ri , µ(Ri ) < ǫ. i=1 i=1  2.5. TRANSLATIONAL INVARIANCE 19 The argument in Example 2.3 shows that every countable set has Lebesgue measure zero, but sets of measure zero may be uncountable; in fact the fine structure of sets of measure zero is, in general, very intricate. Example 2.14. The standard Cantor set, obtained by removing ‘middle thirds’ from [0, 1], is an uncountable set of zero one-dimensional Lebesgue measure. Example 2.15. The x-axis in R2  A = (x, 0) ∈ R2 : x ∈ R has zero two-dimensional Lebesgue measure. More generally, any linear subspace of Rn with dimension strictly less than n has zero n-dimensional Lebesgue measure. 2.5. Translational invariance An important geometric property of Lebesgue measure is its translational in- variance. If A ⊂ Rn and h ∈ Rn , let A + h = {x + h : x ∈ A} denote the translation of A by h. Proposition 2.16. If A ⊂ Rn and h ∈ Rn , then µ∗ (A + h) = µ∗ (A), and A + h is measurable if and only if A is measurable. Proof. The invariance of outer measure µ∗ result is an immediate consequence of the definition, since {Ri + h : i ∈ N} is a cover of A + h if and only if {Ri : i ∈ N} is a cover of A, and µ(R + h) = µ(R) for every rectangle R. Moreover, the Carathéodory definition of measurability is invariant under translations since (E + h) ∩ (A + h) = (E ∩ A) + h.  The space Rn is a locally compact topological (abelian) group with respect to translation, which is a continuous operation. More generally, there exists a (left or right) translation-invariant measure, called Haar measure, on any locally compact topological group; this measure is unique up to a scalar factor. The following is the standard example of a non-Lebesgue measurable set, due to Vitali (1905). Example 2.17. Define an equivalence relation ∼ on R by x ∼ y if x − y ∈ Q. This relation has uncountably many equivalence classes, each of which contains a countably infinite number of points and is dense in R. Let E ⊂ [0, 1] be a set that contains exactly one element from each equivalence class, so that R is the disjoint union of the countable collection of rational translates of E. Then we claim that E is not Lebesgue measurable. To show this, suppose for contradiction that E is measurable. Let {qi : i ∈ N} be an enumeration of the rational numbers in the interval [−1, 1] and let Ei = E +qi denote the translation of E by qi. Then the sets Ei are disjoint and ∞ [ [0, 1] ⊂ Ei ⊂ [−1, 2]. i=1 20 2. LEBESGUE MEASURE ON Rn The translational invariance of Lebesgue measure implies that each Ei is measurable with µ(Ei ) = µ(E), and the countable additivity of Lebesgue measure implies that ∞ X 1≤ µ(Ei ) ≤ 3. i=1 P∞ But this is impossible, since i=1 µ(Ei ) is either 0 or ∞, depending on whether if µ(E) = 0 or µ(E) > 0. The above example is geometrically simpler on the circle T = R/Z. When reduced modulo one, the sets {Ei : i ∈ N} partition T into a countable union of disjoint sets which are translations of each other. If the sets were measurable, their measures would be equal so they must sum to 0 or ∞, but the measure of T is one. 2.6. Borel sets The relationship between measure and topology is not a simple one. In this section, we show that all open and closed sets in Rn , and therefore all Borel sets (i.e. sets that belong to the σ-algebra generated by the open sets), are Lebesgue measurable. Let T (Rn ) ⊂ P(Rn ) denote the standard metric topology on Rn consisting of all open sets. That is, G ⊂ Rn belongs to T (Rn ) if for every x ∈ G there exists r > 0 such that Br (x) ⊂ G, where Br (x) = {y ∈ Rn : |x − y| < r} is the open ball of radius r centered at x ∈ Rn and | · | denotes the Euclidean norm. Definition 2.18. The Borel σ-algebra B(Rn ) on Rn is the σ-algebra generated by the open sets, B(Rn ) = σ (T (Rn )). A set that belongs to the Borel σ-algebra is called a Borel set. Since σ-algebras are closed under complementation, the Borel σ-algebra is also generated by the closed sets in Rn. Moreover, since Rn is σ-compact (i.e. it is a countable union of compact sets) its Borel σ-algebra is generated by the compact sets. Remark 2.19. This definition is not constructive, since we start with the power set of Rn and narrow it down until we obtain the smallest σ-algebra that contains the open sets. It is surprisingly complicated to obtain B(Rn ) by starting from the open or closed sets and taking successive complements, countable unions, and countable intersections. These operations give sequences of collections of sets in Rn (2.8) G ⊂ Gδ ⊂ Gδσ ⊂ Gδσδ ⊂... , F ⊂ Fσ ⊂ Fσδ ⊂ Fδσδ ⊂... , where G denotes the open sets, F the closed sets, σ the operation of countable unions, and δ the operation of countable intersections. These collections contain each other; for example, Fσ ⊃ G and Gδ ⊃ F. This process, however, has to be repeated up to the first uncountable ordinal before we obtain B(Rn ). This is because if, for example, {Ai : i ∈ N} is a countable family of sets such that A1 ∈ Gδ \ G, A2 ∈ Gδσ \ Gδ , A3 ∈ Gδσδ \ Gδσ ,... S T∞ and so on, then there is no guarantee that ∞ i=1 Ai or i=1 Ai belongs to any of the previously constructed families. In general, one only knows that they belong to the ω + 1 iterates Gδσδ...σ or Gδσδ...δ , respectively, where ω is the ordinal number  2.6. BOREL SETS 21 of N. A similar argument shows that in order to obtain a family which is closed under countable intersections or unions, one has to continue this process until one has constructed an uncountable number of families. To show that open sets are measurable, we will represent them as countable unions of rectangles. Every open set in R is a countable disjoint union of open intervals (one-dimensional open rectangles). When n ≥ 2, it is not true that every open set in Rn is a countable disjoint union of open rectangles, but we have the following substitute. Proposition 2.20. Every open set in Rn is a countable union of almost disjoint rectangles. Proof. Let G ⊂ Rn be open. We construct a family of cubes (rectangles of equal sides) as follows. First, we bisect Rn into almost disjoint cubes {Qi : i ∈ N} of side one with integer coordinates. If Qi ⊂ G, we include Qi in the family, and if Qi is disjoint from G, we exclude it. Otherwise, we bisect the sides of Qi to obtain 2n almost disjoint cubes of side one-half and repeat the procedure. Iterating this process arbitrarily many times, we obtain a countable family of almost disjoint cubes. The union of the cubes in this family is contained in G, since we only include cubes that are contained in G. Conversely, if x ∈ G, then since G is open some suf- ficiently small cube in the bisection procedure that contains x is entirely contained in G, and the largest such cube is included in the family. Hence the union of the family contains G, and is therefore equal to G.  In fact, the proof shows that every open set is an almost disjoint union of dyadic cubes. Proposition 2.21. The Borel algebra B(Rn ) is generated by the collection of rectangles R(Rn ). Every Borel set is Lebesgue measurable. Proof. Since R is a subset of the closed sets, we have σ(R) ⊂ B. Conversely, by the previous proposition, σ(R) ⊃ T , so σ(R) ⊃ σ(T ) = B, and therefore B = σ(R). From Proposition 2.11, we have R ⊂ L. Since L is a σ-algebra, it follows that σ(R) ⊂ L, so B ⊂ L.  Note that if ∞ [ G= Ri i=1 is a decomposition of an open set G into an almost disjoint union of closed rectan- gles, then [∞ G⊃ Ri◦ i=1 is a disjoint union, and therefore X∞ ∞ X µ(Ri◦ ) ≤ µ(G) ≤ µ(Ri ). i=1 i=1 Since µ(Ri◦ ) = µ(Ri ), it follows that ∞ X µ(G) = µ(Ri ) i=1 22 2. LEBESGUE MEASURE ON Rn for any such decomposition and that the sum is independent of the way in which G is decomposed into almost disjoint rectangles. The Borel σ-algebra B is not complete and is strictly smaller than the Lebesgue σ-algebra L. In fact, one can show that the cardinality of B is equal to the cardinal- ity c of the real numbers, whereas the cardinality of L is equal to 2c. For example, the Cantor set is a set of measure zero with the same cardinality as R and every subset of the Cantor set is Lebesgue measurable. We can obtain examples of sets that are Lebesgue measurable but not Borel measurable by considering subsets of sets of measure zero. In the following example of such a set in R, we use some properties of measurable functions which will be proved later. Example 2.22. Let f : [0, 1] → [0, 1] denote the standard Cantor function and define g : [0, 1] → [0, 1] by g(y) = inf {x ∈ [0, 1] : f (x) = y}. Then g is an increasing, one-to-one function that maps [0, 1] onto the Cantor set C. Since g is increasing it is Borel measurable, and the inverse image of a Borel set under g is Borel. Let E ⊂ [0, 1] be a non-Lebesgue measurable set. Then F = g(E) ⊂ C is Lebesgue measurable, since it is a subset of a set of measure zero, but F is not Borel measurable, since if it was E = g −1 (F ) would be Borel. Other examples of Lebesgue measurable sets that are not Borel sets arise from the theory of product measures in Rn for n ≥ 2. For example, let N = E ×{0} ⊂ R2 where E ⊂ R is a non-Lebesgue measurable set in R. Then N is a subset of the x-axis, which has two-dimensional Lebesgue measure zero, so N belongs to L(R2 ) since Lebesgue measure is complete. One can show, however, that if a set belongs to B(R2 ) then every section with fixed x or y coordinate, belongs to B(R); thus, N cannot belong to B(R2 ) since the y = 0 section E is not Borel. As we show below, L(Rn ) is the completion of B(Rn ) with respect to Lebesgue measure, meaning that we get all Lebesgue measurable sets by adjoining all subsets of Borel sets of measure zero to the Borel σ-algebra and taking unions of such sets. 2.7. Borel regularity Regularity properties of measures refer to the possibility of approximating in measure one class of sets (for example, nonmeasurable sets) by another class of sets (for example, measurable sets). Lebesgue measure is Borel regular in the sense that Lebesgue measurable sets can be approximated in measure from the outside by open sets and from the inside by closed sets, and they can be approximated by Borel sets up to sets of measure zero. Moreover, there is a simple criterion for Lebesgue measurability in terms of open and closed sets. The following theorem expresses a fundamental approximation property of Lebesgue measurable sets by open and compact sets. Equations (2.9) and (2.10) are called outer and inner regularity, respectively. Theorem 2.23. If A ⊂ Rn , then (2.9) µ∗ (A) = inf {µ(G) : A ⊂ G, G open} , and if A is Lebesgue measurable, then (2.10) µ(A) = sup {µ(K) : K ⊂ A, K compact}.  2.7. BOREL REGULARITY 23 Proof. First, we prove (2.9). The result is immediate if µ∗ (A) = ∞, so we suppose that µ∗ (A) is finite. If A ⊂ G, then µ∗ (A) ≤ µ(G), so µ∗ (A) ≤ inf {µ(G) : A ⊂ G, G open} , and we just need to prove the reverse inequality, (2.11) µ∗ (A) ≥ inf {µ(G) : A ⊂ G, G open}. Let ǫ > 0. There is a cover {Ri : i ∈ N} of A by rectangles Ri such that ∞ X ǫ µ(Ri ) ≤ µ∗ (A) +. i=1 2 Let Si be an rectangle whose interior Si◦ contains Ri such that ǫ µ(Si ) ≤ µ(Ri ) + i+1. 2 Then the collection of open rectangles {Si◦ : i ∈ N} covers A and ∞ [ G= Si◦ i=1 is an open set that contains A. Moreover, since {Si : i ∈ N} covers G, ∞ X ∞ X ǫ µ(G) ≤ µ(Si ) ≤ µ(Ri ) + , i=1 i=1 2 and therefore (2.12) µ(G) ≤ µ∗ (A) + ǫ. It follows that inf {µ(G) : A ⊂ G, G open} ≤ µ∗ (A) + ǫ, which proves (2.11) since ǫ > 0 is arbitrary. Next, we prove (2.10). If K ⊂ A, then µ(K) ≤ µ(A), so sup {µ(K) : K ⊂ A, K compact} ≤ µ(A). Therefore, we just need to prove the reverse inequality, (2.13) µ(A) ≤ sup {µ(K) : K ⊂ A, K compact}. To do this, we apply the previous result to Ac and use the measurability of A. First, suppose that A is a bounded measurable set, in which case µ(A) < ∞. Let F ⊂ Rn be a compact set that contains A. By the preceding result, for any ǫ > 0, there is an open set G ⊃ F \ A such that µ(G) ≤ µ(F \ A) + ǫ. Then K = F \ G is a compact set such that K ⊂ A. Moreover, F ⊂ K ∪ G and F = A ∪ (F \ A), so µ(F ) ≤ µ(K) + µ(G), µ(F ) = µ(A) + µ(F \ A). It follows that µ(A) = µ(F ) − µ(F \ A) ≤ µ(F ) − µ(G) + ǫ ≤ µ(K) + ǫ, 24 2. LEBESGUE MEASURE ON Rn which implies (2.13) and proves the result for bounded, measurable sets. Now suppose that A is an unbounded measurable set, and define (2.14) Ak = {x ∈ A : |x| ≤ k}. Then {Ak : k ∈ N} is an increasing sequence of bounded measurable sets whose union is A, so (2.15) µ(Ak ) ↑ µ(A) as k → ∞. If µ(A) = ∞, then µ(Ak ) → ∞ as k → ∞. By the previous result, we can find a compact set Kk ⊂ Ak ⊂ A such that µ(Kk ) + 1 ≥ µ(Ak ) so that µ(Kk ) → ∞. Therefore sup {µ(K) : K ⊂ A, K compact} = ∞, which proves the result in this case. Finally, suppose that A is unbounded and µ(A) < ∞. From (2.15), for any ǫ > 0 we can choose k ∈ N such that ǫ µ(A) ≤ µ(Ak ) +. 2 Moreover, since Ak is bounded, there is a compact set K ⊂ Ak such that ǫ µ(Ak ) ≤ µ(K) +. 2 Therefore, for every ǫ > 0 there is a compact set K ⊂ A such that µ(A) ≤ µ(K) + ǫ, which gives (2.13), and completes the proof.  It follows that we may determine the Lebesgue measure of a measurable set in terms of the Lebesgue measure of open or compact sets by approximating the set from the outside by open sets or from the inside by compact sets. The outer approximation in (2.9) does not require that A is measurable. Thus, for any set A ⊂ Rn , given ǫ > 0, we can find an open set G ⊃ A such that µ(G) − µ∗ (A) < ǫ. If A is measurable, we can strengthen this condition to get that µ∗ (G \ A) < ǫ; in fact, this gives a necessary and sufficient condition for measurability. Theorem 2.24. A subset A ⊂ Rn is Lebesgue measurable if and only if for every ǫ > 0 there is an open set G ⊃ A such that (2.16) µ∗ (G \ A) < ǫ. Proof. First we assume that A is measurable and show that it satisfies the condition given in the theorem. Suppose that µ(A) < ∞ and let ǫ > 0. From (2.12) there is an open set G ⊃ A such that µ(G) < µ∗ (A) + ǫ. Then, since A is measurable, µ∗ (G \ A) = µ∗ (G) − µ∗ (G ∩ A) = µ(G) − µ∗ (A) < ǫ, which proves the result when A has finite measure.  2.7. BOREL REGULARITY 25 If µ(A) = ∞, define Ak ⊂ A as in (2.14), and let ǫ > 0. Since Ak is measurable with finite measure, the argument above shows that for each k ∈ N, there is an open set Gk ⊃ Ak such that ǫ µ(Gk \ Ak ) < k. 2 S Then G = ∞ k=1 Gk is an open set that contains A, and ∞ ! ∞ ∞ [ X X ∗ ∗ µ (G \ A) = µ Gk \ A ≤ µ∗ (Gk \ A) ≤ µ∗ (Gk \ Ak ) < ǫ. k=1 k=1 k=1 Conversely, suppose that A ⊂ Rn satisfies the condition in the theorem. Let ǫ > 0, and choose an open set G ⊃ A such that µ∗ (G \ A) < ǫ. If E ⊂ Rn , we have E ∩ Ac = (E ∩ Gc ) ∪ (E ∩ (G \ A)). Hence, by the subadditivity and monotonicity of µ∗ and the measurability of G, µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) ≤ µ∗ (E ∩ A) + µ∗ (E ∩ Gc ) + µ∗ (E ∩ (G \ A)) ≤ µ∗ (E ∩ G) + µ∗ (E ∩ Gc ) + µ∗ (G \ A) < µ∗ (E) + ǫ. Since ǫ > 0 is arbitrary, it follows that µ∗ (E) ≥ µ∗ (E ∩ A) + µ∗ (E ∩ Ac ) which proves that A is measurable.  This theorem states that a set is Lebesgue measurable if and only if it can be approximated from the outside by an open set in such a way that the difference has arbitrarily small outer Lebesgue measure. This condition can be adopted as the definition of Lebesgue measurable sets, rather than the Carathéodory definition which we have used c.f. [5, 8, 11]. The following theorem gives another characterization of Lebesgue measurable sets, as ones that can be ‘squeezed’ between open and closed sets. Theorem 2.25. A subset A ⊂ Rn is Lebesgue measurable if and only if for every ǫ > 0 there is an open set G and a closed set F such that G ⊃ A ⊃ F and (2.17) µ(G \ F ) < ǫ. If µ(A) < ∞, then F may be chosen to be compact. Proof. If A satisfies the condition in the theorem, then it follows from the monotonicity of µ∗ that µ∗ (G \ A) ≤ µ(G \ F ) < ǫ, so A is measurable by Theo- rem 2.24. Conversely, if A is measurable then Ac is measurable, and by Theorem 2.24 given ǫ > 0, there are open sets G ⊃ A and H ⊃ Ac such that ǫ ǫ µ∗ (G \ A) < , µ∗ (H \ Ac ) <. 2 2 Then, defining the closed set F = H c , we have G ⊃ A ⊃ F and µ(G \ F ) ≤ µ∗ (G \ A) + µ∗ (A \ F ) = µ∗ (G \ A) + µ∗ (H \ Ac ) < ǫ. Finally, suppose that µ(A) < ∞ and let ǫ > 0. From Theorem 2.23, since A is measurable, there is a compact set K ⊂ A such that µ(A) < µ(K) + ǫ/2 and ǫ µ(A \ K) = µ(A) − µ(K) <. 2 26 2. LEBESGUE MEASURE ON Rn As before, from Theorem 2.24 there is an open set G ⊃ A such that µ(G) < µ(A) + ǫ/2. It follows that G ⊃ A ⊃ K and µ(G \ K) = µ(G \ A) + µ(A \ K) < ǫ, which shows that we may take F = K compact when A has finite measure.  From the previous results, we can approximate measurable sets by open or closed sets, up to sets of arbitrarily small but, in general, nonzero measure. By taking countable intersections of open sets or countable unions of closed sets, we can approximate measurable sets by Borel sets, up to sets of measure zero Definition 2.26. The collection of sets in Rn that are countable intersections of open sets is denoted by Gδ (Rn ), and the collection of sets in Rn that are countable unions of closed sets is denoted by Fσ (Rn ). Gδ and Fσ sets are Borel. Thus, it follows from the next result that every Lebesgue measurable set can be approximated up to a set of measure zero by a Borel set. This is the Borel regularity of Lebesgue measure. Theorem 2.27. Suppose that A ⊂ Rn is Lebesgue measurable. Then there exist sets G ∈ Gδ (Rn ) and F ∈ Fσ (Rn ) such that G ⊃ A ⊃ F, µ(G \ A) = µ(A \ F ) = 0. Proof. For each k ∈ N, choose an open set Gk and a closed set Fk such that Gk ⊃ A ⊃ Fk and 1 µ(Gk \ Fk ) ≤ k Then \∞ ∞ [ G= Gk , F = Fk k=1 k=1 are Gδ and Fσ sets with the required properties.  In particular, since any measurable set can be approximated up to a set of measure zero by a Gδ or an Fσ , the complexity of the transfinite construction of general Borel sets illustrated in (2.8) is ‘hidden’ inside sets of Lebesgue measure zero. As a corollary of this result, we get that the Lebesgue σ-algebra is the comple- tion of the Borel σ-algebra with respect to Lebesgue measure. Theorem 2.28. The Lebesgue σ-algebra L(Rn ) is the completion of the Borel σ-algebra B(Rn ). Proof. Lebesgue measure is complete from Proposition 2.12. By the previous theorem, if A ⊂ Rn is Lebesgue measurable, then there is a Fσ set F ⊂ A such that M = A \ F has Lebesgue measure zero. It follows by the approximation theorem that there is a Borel set N ∈ Gδ with µ(N ) = 0 and M ⊂ N. Thus, A = F ∪ M where F ∈ B and M ⊂ N ∈ B with µ(N ) = 0, which proves that L(Rn ) is the completion of B(Rn ) as given in Theorem 1.15.   2.8. LINEAR TRANSFORMATIONS 27 2.8. Linear transformations The definition of Lebesgue measure is not rotationally invariant, since we used rectangles whose sides are parallel to the coordinate axes. In this section, we show that the resulting measure does not, in fact, depend upon the direction of the coordinate axes and is invariant under orthogonal transformations. We also show that Lebesgue measure transforms under a linear map by a factor equal to the absolute value of the determinant of the map. As before, we use µ∗ to denote Lebesgue outer measure defined using rectangles whose sides are parallel to the coordinate axes; a set is Lebesgue measurable if it satisfies the Carathéodory criterion (2.8) with respect to this outer measure. If T : Rn → Rn is a linear map and E ⊂ Rn , we denote the image of E under T by T E = {T x ∈ Rn : x ∈ E}. First, we consider the Lebesgue measure of rectangles whose sides are not paral- lel to the coordinate axes. We use a tilde to denote such rectangles by R̃; we denote closed rectangles whose sides are parallel to the coordinate axes by R as before. We refer to R̃ and R as oblique and parallel rectangles, respectively. We denote the volume of a rectangle R̃ by v(R̃), i.e. the product of the lengths of its sides, to avoid confusion with its Lebesgue measure µ(R̃). We know that µ(R) = v(R) for parallel rectangles, and that R̃ is measurable since it is closed, but we have not yet shown that µ(R̃) = v(R̃) for oblique rectangles. More explicitly, we regard Rn as a Euclidean space equipped with the standard inner product, n X (x, y) = xi yi , x = (x1 , x2 ,... , xn ), y = (y1 , y2 ,... , yn ). i=1 If {e1 , e2 ,... , en } is the standard orthonormal basis of Rn , e1 = (1, 0,... , 0), e2 = (0, 1,... , 0),... en = (0, 0,... , 1), and {ẽ1 , ẽ2 ,... , ẽn } is another orthonormal basis, then we use R to denote rectangles whose sides are parallel to {ei } and R̃ to denote rectangles whose sides are parallel to {ẽi }. The linear map Q : Rn → Rn defined by Qei = ẽi is orthogonal, meaning that QT = Q−1 and (Qx, Qy) = (x, y) for all x, y ∈ Rn. Since Q preserves lengths and angles, it maps a rectangle R to a rectangle R̃ = QR such that v(R̃) = v(R). We will use the following lemma. Lemma 2.29. If an oblique rectangle R̃ contains a finite almost disjoint collec- tion of parallel rectangles {R1 , R2 ,... , RN } then N X v(Ri ) ≤ v(R̃). i=1 This result is geometrically obvious, but a formal proof seems to require a fuller discussion of the volume function on elementary geometrical sets, which is included in the theory of valuations in convex geometry. We omit the details. 28 2. LEBESGUE MEASURE ON Rn Proposition 2.30. If R̃ is an oblique rectangle, then given any ǫ > 0 there is a collection of parallel rectangles {Ri : i ∈ N} that covers R̃ and satisfies ∞ X v(Ri ) ≤ v(R̃) + ǫ. i=1 Proof. Let S̃ be an oblique rectangle that contains R̃ in its interior such that v(S̃) ≤ v(R̃) + ǫ. Then, from Proposition 2.20, we may decompose the interior of S into an almost disjoint union of parallel rectangles ∞ [ S̃ ◦ = Ri. i=1 It follows from the previous lemma that for every N ∈ N N X v(Ri ) ≤ v(S̃), i=1 which implies that ∞ X v(Ri ) ≤ v(S̃) ≤ v(R̃) + ǫ. i=1 Moreover, the collection {Ri } covers R̃ since its union is S̃ ◦ , which contains R̃.  Conversely, by reversing the roles of the axes, we see that if R is a parallel rectangle and ǫ > 0, then there is a cover of R by oblique rectangles {R̃i : i ∈ N} such that X∞ (2.18) v(R̃i ) ≤ v(R) + ǫ. i=1 Theorem 2.31. If E ⊂ R and Q : Rn → Rn is an orthogonal transformation, n then µ∗ (QE) = µ∗ (E), and E is Lebesgue measurable if an only if QE is Lebesgue measurable. Proof. Let Ẽ = QE. Given ǫ > 0 there is a cover of Ẽ by parallel rectangles {Ri : i ∈ N} such that X ∞ ǫ v(Ri ) ≤ µ∗ (Ẽ) +. i=1 2 From (2.18), for each i ∈ N we can choose a cover {R̃i,j : j ∈ N} of Ri by oblique rectangles such that X∞ ǫ v(R̃i,j ) ≤ v(Ri ) + i+1. i=1 2 Then {R̃i,j : i, j ∈ N} is a countable cover of Ẽ by oblique rectangles, and ∞ X ∞ X ǫ v(R̃i,j ) ≤ v(Ri ) + ≤ µ∗ (Ẽ) + ǫ. i,j=1 i=1 2  2.8. LINEAR TRANSFORMATIONS 29 If Ri,j = QT R̃i,j , then {Ri,j : j ∈ N} is a cover of E by parallel rectangles, so ∞ X µ∗ (E) ≤ v(Ri,j ). i,j=1 Moreover, since Q is orthogonal, we have v(Ri,j ) = v(R̃i,j ). It follows that ∞ X ∞ X µ∗ (E) ≤ v(Ri,j ) = v(R̃i,j ) ≤ µ∗ (Ẽ) + ǫ, i,j=1 i,j=1 and since ǫ > 0 is arbitrary, we conclude that µ∗ (E) ≤ µ∗ (Ẽ). By applying the same argument to the inverse mapping E = QT Ẽ, we get the reverse inequality, and it follows that µ∗ (E) = µ∗ (Ẽ). Since µ∗ is invariant under Q, the Carathéodory criterion for measurability is invariant, and E is measurable if and only if QE is measurable.  It follows from Theorem 2.31 that Lebesgue measure is invariant under rotations and reflections.4 Since it is also invariant under translations, Lebesgue measure is invariant under all isometries of Rn. Next, we consider the effect of dilations on Lebesgue measure. Arbitrary linear maps may then be analyzed by decomposing them into rotations and dilations. Proposition 2.32. Suppose that Λ : Rn → Rn is the linear transformation (2.19) Λ : (x1 , x2 ,... , xn ) 7→ (λ1 x1 , λ2 x2 ,... , λn xn ) where the λi > 0 are positive constants. Then µ∗ (ΛE) = (det Λ)µ∗ (E), and E is Lebesgue measurable if and only if ΛE is Lebesgue measurable. Proof. The diagonal map Λ does not change the orientation of a rectan- gle, so it maps a cover of E by parallel rectangles to a cover of ΛE by paral- lel rectangles, and conversely. Moreover, Λ multiplies the volume of a rectangle by det Λ = λ1... λn , so it immediate from the definition of outer measure that µ∗ (ΛE) = (det Λ)µ∗ (E), and E satisfies the Carathéodory criterion for measura- bility if and only if ΛE does.  Theorem 2.33. Suppose that T : Rn → Rn is a linear transformation and E ⊂ Rn. Then µ∗ (T E) = |det T | µ∗ (E), and T E is Lebesgue measurable if E is measurable Proof. If T is singular, then its range is a lower-dimensional subspace of Rn , which has Lebesgue measure zero, and its determinant is zero, so the result holds.5 We therefore assume that T is nonsingular. 4Unlike differential volume forms, Lebesgue measure does not depend on the orientation of Rn ; such measures are sometimes referred to as densities in differential geometry. 5In this case T E, is always Lebesgue measurable, with measure zero, even if E is not measurable. 30 2. LEBESGUE MEASURE ON Rn In that case, according to the polar decomposition, the map T may be written as a composition T = QU √ of a positive definite, symmetric map U = T T T and an orthogonal map Q. Any positive-definite, symmetric map U may be diagonalized by an orthogonal map O to get U = OT ΛO where Λ : Rn → Rn has the form (2.19). From Theorem 2.31, orthogonal mappings leave the Lebesgue measure of a set invariant, so from Proposition 2.32 µ∗ (T E) = µ∗ (ΛE) = (det Λ)µ∗ (E). Since | det Q| = 1 for any orthogonal map Q, we have det Λ = | det T |, and it follows that µ∗ (T E) = |det T | µ∗ (E). Finally, it is straightforward to see that T E is measurable if E is measurable.  2.9. Lebesgue-Stieltjes measures We briefly consider a generalization of one-dimensional Lebesgue measure, called Lebesgue-Stieltjes measures on R. These measures are obtained from an increasing, right-continuous function F : R → R, and assign to a half-open interval (a, b] the measure µF ((a, b]) = F (b) − F (a). The use of half-open intervals is significant here because a Lebesgue-Stieltjes mea- sure may assign nonzero measure to a single point. Thus, unlike Lebesgue measure, we need not have µF ([a, b]) = µF ((a, b]). Half-open intervals are also convenient because the complement of a half-open interval is a finite union of (possibly infi- nite) half-open intervals of the same type. Thus, the collection of finite unions of half-open intervals forms an algebra. The right-continuity of F is consistent with the use of intervals that are half- open at the left, since ∞ \ (a, a + 1/i] = ∅, i=1 so, from (1.2), if F is to define a measure we need lim µF ((a, a + 1/i]) = 0 i→∞ or lim [F (a + 1/i) − F (a)] = lim F (x) − F (a) = 0. i→∞ x→a+ Conversely, as we state in the next theorem, any such function F defines a Borel measure on R. Theorem 2.34. Suppose that F : R → R is an increasing, right-continuous function. Then there is a unique Borel measure µF : B(R) → [0, ∞] such that µF ((a, b]) = F (b) − F (a) for every a < b.  2.9. LEBESGUE-STIELTJES MEASURES 31 The construction of µF is similar to the construction of Lebesgue measure on Rn. We define an outer measure µ∗F : P(R) → [0, ∞] by (∞ ) X S∞ ∗ µF (E) = inf [F (bi ) − F (ai )] : E ⊂ i=1 (ai , bi ] , i=1 and restrict µ∗Fto its Carathéodory measurable sets, which include the Borel sets. See e.g. Section 1.5 of Folland for a detailed proof. The following examples illustrate the three basic types of Lebesgue-Stieltjes measures. Example 2.35. If F (x) = x, then µF is Lebesgue measure on R with µF ((a, b]) = b − a. Example 2.36. If  1 if x ≥ 0, F (x) = 0 if x < 0, then µF is the δ-measure supported at 0,  1 if 0 ∈ A, µF (A) = 0 if 0 ∈ / A. Example 2.37. If F : R → R is the Cantor function, then µF assigns measure one to the Cantor set, which has Lebesgue measure zero, and measure zero to its complement. Despite the fact that µF is supported on a set of Lebesgue measure zero, the µF -measure of any countable set is zero.  CHAPTER 3 Measurable functions Measurable functions in measure theory are analogous to continuous functions in topology. A continuous function pulls back open sets to open sets, while a measurable function pulls back measurable sets to measurable sets. 3.1. Measurability Most of the theory of measurable functions and integration does not depend on the specific features of the measure space on which the functions are defined, so we consider general spaces, although one should keep in mind the case of functions defined on R or Rn equipped with Lebesgue measure. Definition 3.1. Let (X, A) and (Y, B) be measurable spaces. A function f : X → Y is measurable if f −1 (B) ∈ A for every B ∈ B. Note that the measurability of a function depends only on the σ-algebras; it is not necessary that any measures are defined. In order to show that a function is measurable, it is sufficient to check the measurability of the inverse images of sets that generate the σ-algebra on the target space. Proposition 3.2. Suppose that (X, A) and (Y, B) are measurable spaces and B = σ(G) is generated by a family G ⊂ P(Y ). Then f : X → Y is measurable if and only if f −1 (G) ∈ A for every G ∈ G. Proof. Set operations are natural under pull-backs, meaning that f −1 (Y \ B) = X \ f −1 (B) and ! ! ∞ [ ∞ [ ∞ \ ∞ \ −1 −1 −1 f Bi = f (Bi ) , f Bi = f −1 (Bi ). i=1 i=1 i=1 i=1 It follows that  M = B ⊂ Y : f −1 (B) ∈ A is a σ-algebra on Y. By assumption, M ⊃ G and therefore M ⊃ σ(G) = B, which implies that f is measurable.  It is worth noting the indirect nature of the proof of containment of σ-algebras in the previous proposition; this is required because we typically cannot use an explicit representation of sets in a σ-algebra. For example, the proof does not characterize M, which may be strictly larger than B. If the target space Y is a topological space, then we always equip it with the Borel σ-algebra B(Y ) generated by the open sets (unless stated explicitly otherwise). 33 34 3. MEASURABLE FUNCTIONS In that case, it follows from Proposition 3.2 that f : X → Y is measurable if and only if f −1 (G) ∈ A is a measurable subset of X for every set G that is open in Y. In particular, every continuous function between topological spaces that are equipped with their Borel σ-algebras is measurable. The class of measurable function is, however, typically much larger than the class of continuous functions, since we only require that the inverse image of an open set is Borel; it need not be open. 3.2. Real-valued functions We specialize to the case of real-valued functions f :X →R or extended real-valued functions f : X → R. We will consider one case or the other as convenient, and comment on any differ- ences. A positive extended real-valued function is a function