Based on the Appendix to B. Hasselblatt and A. Katok, A First Course in Dynamics, Cambridge University press,

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1 NOTE ON ABSTRACT RIEMANN INTEGRAL Based on the Appendix to B. Hasselblatt and A. Katok, A First Course in Dynamics, Cambridge University press, a. Definitions. 1. Metric spaces DEFINITION 1.1. If X is a set then d: X X R is said to be a metric or distance function if (1) d(x, y) = d(y, x) (symmetry), (2) d(x, y) = 0 x = y (positivity), (3) d(x, y) + d(y, z) d(x, z) (triangle inequality). If d is a metric then (X, d) is said to be a metric space. REMARK 1.2. Putting z = x in (3) and using (1) and (2) shows that d(x, y) 0. REMARK 1.3. A subset of a metric space is itself a metric space by using the metric of the space (this is then called the induced metric). The following notions generalize familiar concepts from euclidean space. DEFINITION 1.4. The set B(x, r):={y X d(x, y) < r} is called the (open) r-ball around x. A set O X is said to be open if for every x O there exists r > 0 such that B(x, r) O. (This immediately implies that any union of open sets is open.) The interior of a set S is the union Int S of all open sets contained in it. Equivalently, it is the set of x S such that B(x, r) S for some r > 0. If x X and O is an open set containing x then O is said to be a neighborhood of x. A point x X is called a boundary point of S X if for every neighborhood U of x we have U S and U \ S. The boundary of S is the set A of its boundary points. For A X the set A := {x X B(x, r) A for all r > 0} is called the closure of A. A is said to be closed if A = A. A set A X is said to be dense if A = X and ɛ-dense if X {B(x, ɛ) x A}. A set is said to be nowhere dense if its closure has empty interior (that is, contains no nonempty open set). This is true for finite sets but fails for Q and intervals. A sequence (x n ) n N in X is said to converge to x X if for all ɛ > 0 there exists an N N such that for every n N we have d(x n, x) < ɛ. It is easy to see that a set is closed if and only if its complement is open. (Therefore any intersection of closed sets is closed.) Another way to define a closed set is via accumulation points: DEFINITION 1.5. An accumulation point of a set A is a point x for which every ball B(x, ɛ) intersects A {x}. The set of accumulation points of A is called the derived set of A and denoted by A. A set is closed if A A and the closure A of a set A is A = A A. A set A is said to be perfect if A = A, that is, there are no points missing (all accumulation points are there) nor any extraneous (isolated) ones. Note that x A if and only if there is a sequence of points in A that does not include x but converges to x. 1

2 2 EXAMPLE 1.6. Perfect sets are closed. R is perfect, as are [0, 1], closed balls in R n, S 1 and the middle-third Cantor set (see Section 0.1g). But Z or finite subsets of R n are not (they have no accumulation points) and nor are the rationals Q (they have irrational accumulation points). Finite sets are nowhere dense but this fails for Q and intervals. The ternary Cantor set is nowhere dense, because it is closed and has empty interior (contains no interval). Here is an interesting pertinent fact: PROPOSITION 1.7. All sets that are both perfect and nowhere dense are homeomorphic to the ternary Cantor set. DEFINITION 1.8. A metric space X is said to be connected if it contains no two disjoint nonempty open sets. A totally disconnected space is a space X where for every two points x 1, x 2 X there exist disjoint open sets O 1, O 2 X containing x 1, x 2, respectively, whose union is X. R or any interval of R, as well as R n and open balls in R n, or the circle in R 2 are connected. Examples of totally disconnected spaces are provided by finite subsets of R with at least 2 elements as well as the rationals, and, in fact, any countable subset of R. The ternary Cantor set is an uncountable totally disconnected set. b. Completeness. One important property sets apart the real number system from that of rational numbers. This property is called completeness, and it reflects that fact that the real line has no holes, like the rationals do. There are several equivalent ways of expressing this property precisely, and different versions may be useful in different circumstances. (1) If a nondecreasing sequence of real numbers is bounded above then it is convergent. (2) If a subset of R has an upper bound then it has a smallest upper bound. (3) A Cauchy sequence of real numbers converges. A Cauchy sequence is a sequence (a n ) n N such that for any ɛ > 0 there exists an n N such that a n a m < ɛ for any n, m N. The first two versions of completeness refer to the ordering of the real numbers (by using the notions of upper bound and nondecreasing). The last one does not, and it is used to define completeness of metric spaces. DEFINITION 1.9. A sequence (x i ) i N is said to be a Cauchy sequence if for all ɛ > 0 there exists an N N such that d(x i, x j ) < ɛ whenever i, j N. A metric space X is said to be complete if every Cauchy sequence converges. EXAMPLE For example, R is complete (see also Section 0.1b), whereas an open interval is not, when one uses the usual metric d(x, y) = x y (the endpoints are missing ). If, however, we define a metric on the open interval ( π/2, π/2) by d (x, y) = tan x tan y, then this unusual metric space is indeed complete. The endpoints are no longer perceived as missing because sequences that look like they converge to an endpoint are not Cauchy sequences with respect to this metric since it stretches distances near the endpoints.

3 1. METRIC SPACES 3 REMARK This is an example of the pullback of a metric. If (Y, d) is a metric space and h: X Y an injective map then d (x, y) := d(h(x), h(y)) defines a metric on X. Here we took X = ( π/2, π/2), Y = R, and h = tan. LEMMA A closed subset Y of a complete metric space X is itself a complete metric space. PROOF. A Cauchy sequence in Y is a Cauchy sequence in X and hence converges to some x X. Then x Y because Y is closed. An important example is the space if continuous functions (Definition 1.16). THEOREM The space C([0, 1], R n ) := {f : [0, 1] R n f is continuous} is a complete metric space with the metric induced by the norm f := max x [0,1] f(x). PROOF. Suppose (f n ) n N is a Cauchy sequence in C([0, 1], R n ). Then it is easy to see that (f n (x)) n N is a Cauchy sequence in R n for all x [0, 1]. Therefore f(x) := lim n f n (x) is well-defined by completeness of R n. To prove f n f uniformly fix any ɛ > 0 and find N N such that f k f l < ɛ/2 whenever k, l N. Now fix k N. For any x [0, 1] there is an N x such that l N x f l (x) f(x) < ɛ/2. Taking l N gives f k (x) f(x) f k (x) f l (x) + f l (x) f(x) < ɛ. This proves the claim because k was chosen independently of x. Likewise one proves completeness of the space of bounded sequences. THEOREM The space l of bounded sequences (x n ) n N0 with the sup-norm (x n ) n N0 := sup n N0 x n is complete. PROOF. The proof is the same, except that the domain is N rather than [0, 1]. (Boundedness is assumed to make the norm well-defined, for continuous functions on [0, 1] it is automatic.) Given any metric space X which fails to be complete there is a standard procedure to extend it to a complete metric space. This process is used, for example, to obtain the real numbers from the rationals. This process, as well as the complete space obtained by it, is referred to as completion of X. THEOREM 1.15 (Completion). Suppose X is a metric space. Consider the set of Cauchy sequences of X and define an equivalence relation by a b: lim n d(a n, b n ) = 0. Denote by X the set of equivalence classes with the metric d([a], [b]):=lim n d(a n, b n ). Then X is a complete metric space. PROOF. We need to show that this is well-defined and a metric. To see that it is welldefined take a, a [a] and b, b [b] and note that lim d(a n, b n ) lim d(a n, a n n n ) + lim n d(a n, b n ) + lim n d(b n, b n) = lim n d(a n, b n ) lim n d(a n, a n) + lim n d(a n, b n ) + lim n d(b n, b n ) = lim n d(a n, b n ), so lim n d(a n, b n ) = lim n d(a n, b n ) as required.

4 4 Symmetry and positivity are obvious. To verify the triangle inequality take [a], [b], [c] X with representatives (a n ) n N, (b n ) n N, and (c n ) n N. Then d(a, b) + d(b, c) = lim n d(a n, b n ) + lim n d(b n, c n ) = lim n d(a n, b n ) + d(b n, c n ) lim n d(a n, c n ) = d(a, c). We outline the proof of completeness. Given a Cauchy sequence in X choose representatives (a n,m ) m N for the terms. The representatives are Cauchy sequences, so δ > 0 M n,δ N k, l M n,δ d(a n,k, a n,l ) < δ. They represent a Cauchy sequence, so ɛ > 0 N ɛ k, l N ɛ d(a k,, a l, ) = lim m d(a k,m, a l,m ) N. The required limit is then represented by the limiting sequence (a n ) n N defined by taking k n N 1/n, m n M kn,1/n, a n := a kn,m n. It is straightforward to check that d(a n, a n+l ) < 3/n for all n, l N, hence this is a Cauchy sequence. Now one checks that it represents the limit. c. Continuity. DEFINITION Let (X, d), (Y, d ) be metric spaces. A map f : X Y is said to be an isometry if d (f(x), f(y)) = d(x, y) for all x, y X. It is said to be continuous at x X if for every ɛ > 0 there exists a δ > 0 such that f(b(x, δ)) B(f(x), ɛ), or equivalently, if d(x, y) < δ implies d (f(x), f(y)) < ɛ. f is said to be continuous if f is continuous at x for every x X. An equivalent characterization is that the preimage of each open set is open. f is said to be uniformly continuous if the choice of δ does not depend on x, that is, for all ɛ > 0 there is a δ > 0 such that for all x, y X with d(x, y) < δ we have d (f(x), f(y)) < ɛ. f is said to be an open map if it maps open sets to open sets. A continuous bijection (one-to-one and onto map) with continuous inverse is said to be a homeomorphism. A map f : X Y is said to be Lipschitz continuous (or Lipschitz) with Lipschitz constant C, or C-Lipschitz, if d (f(x), f(y)) Cd(x, y). A map is said to be a contraction (or, more specifically, a λ-contraction) if it is Lipschitz continuous with Lipschitz constant λ < 1. Continuity does not imply that the image of an open set is open. For example, the map x 2 sends ( 1, 1) or R to sets that are not open. There are various ways in which two metrics can be similar, or equivalent. The easiest way to describe these is to view the process of changing metrics as taking the identity map on X as a map between two different metric spaces. DEFINITION We say that two metrics are isometric if the identity establishes an isometry between them. Two metrics are said to be uniformly equivalent (sometimes just equivalent) if the identity and its inverse are Lipschitz maps between the two metric spaces. Finally, two metrics are said to be homeomorphic (sometimes also equivalent) if the identity is a homeomorphism between them.

5 1. METRIC SPACES 5 d. Compactness. An important class of metric spaces is that of compact ones: DEFINITION A metric space (X, d) is said to be compact if any open cover of X has a finite subcover, that is, whenever {O i i I} is a collection of open sets of X indexed by I such that X i I O i then there is a finite subcollection {O i1, O i2,..., O in } such that X n l=1 O i l. PROPOSITION Compact sets are closed and bounded. PROOF. Suppose X is a metric space and C X is compact. If x / C then the sets O n := {y X d(x, y) > 1/n} form an open cover of X {x}, hence of C. There is a finite subcover O of {O n } n N. Let n 0 := max{n N O n O}. Then d(x, y) > 1/n 0 for all y C, so x / C. This proves C C. C is bounded because the open cover {B(x, r) r > 0} has a finite subcover. The Heine Borel Theorem tells us that in euclidean space a set is compact if and only if it is closed and bounded. In some important metric spaces closed bounded sets may fail to be compact, however, and this definition of compactness describes the property that is useful in a general metric space. Indeed, this definition uses the metric only to the extent that it involves open sets. If a metric is given compactness is equivalent to being both complete and totally bounded: DEFINITION A metric space is said to be totally bounded if for any r > 0 there is a finite set C such that the r-balls with center in C cover the space. PROPOSITION Compact sets are totally bounded. PROOF. If C is compact and r > 0 then {B(x, r) x C} has a finite subcover. Some facts that make compactness useful are PROPOSITION If (X, d) and (Y, d ) are metric spaces, X is compact, and f : X Y is a continuous map then f is uniformly continuous and f(x) Y is compact, hence closed and bounded. If Y = R this shows that f attains its minimum and maximum. Among the most used facts about compact spaces is this last observation that a continuous real-valued function on a compact set attains its minimum and maximum. PROOF. For every ɛ > 0 there is a δ = δ(x, ɛ) > 0 such that d (f(x), f(y)) < ɛ/2 whenever d(x, y) < δ. The balls B(x, δ(x, ɛ)/2) cover X, so by compactness of X there is a finite subcover by balls B(x i, δ(x i, ɛ)/2). Let δ 0 = (1/2) min{δ(x i, ɛ)}. If x, y X with d(x, y) < δ 0 then d(x, x i ) < δ 0 < δ(x i, ɛ) for some x i and by the triangle inequality d(y, x i ) d(x, x i ) + d(x, y) < δ 0 + δ 0 δ(x i, ɛ). These two facts imply d (f(x), f(y)) d (f(x), f(x i )) + d (f(y), f(x i )) < ɛ/2 + ɛ/2 = ɛ. To see that f(x) Y is compact consider any open cover f(x) i I O i of f(x). Then the sets f 1 (O i ) = {x f(x) O i } cover X and hence there is a finite subcover X n l=1 f 1 (O il ). But then f(x) n l=1 O i l. A further useful property of compact sets is the following: PROPOSITION Suppose {C i i I} is a collection of compact sets in a metric space X such that n l=1 C i for any finite subcollection {C il 1 l n}. Then i I C i.

6 6 PROOF. We prove the contrapositive: Suppose {C i i I} is a collection of compact sets with i I C i =. Let O i = C 1 C i for i I. Then i I C i = implies that i I O i = C 1, that is, the O i form an open cover of the compact set C 1. Thus there is a finite subcover n l=1 O i l = C 1. This means that n l=1 C i l =. Some other easy but useful observations about compact sets are the following: PROPOSITION (1) A closed subset of a compact set is compact. (2) The intersection of compact sets is compact. (3) A continuous bijection between compact spaces is a homeomorphism. (4) A sequence in a compact set has a convergent subsequence. PROOF. (1) Suppose C X is a closed subset of a compact space and i I O i is an open cover of C. If O = X C then X = O C O i I O i is an open cover of X and hence has a finite subcover O n l=1 O i l. Since O C = we get a finite subcover n l=1 O i l of C. (2) The intersection of compact sets is an intersection of closed subsets and hence a closed subset of any of these compact sets. Therefore is is compact by (1). (3) We need to show that the image of an open set is open. Using bijectivity note that the complement of the image of an open set O is the image of the complement O c of O. O c is a closed subset of a compact space, hence compact, and thus its image is compact, hence closed. Its complement, the image of O, is then open, as required. (4) Given a sequence (a n ) n N let A n := {a i i n} for n N. Then the closures A n satisfy the hypotheses of Proposition 1.23 and there exists an a 0 n N A n. This means that for every k N there exists an n k > n k 1 such that a nk B(a 0, 1/k), i.e., a nk a 0. An interesting example of a metric space is given by the Hausdorff metric: DEFINITION If (X, d) is a compact metric space and K(X) denotes the collection of closed subsets of X then the Hausdorff metric d H on K(X) is defined by d H (A, B) := sup d(a, B) + sup d(b, A), a A b B where d(x, Y ) := inf y Y d(x, y) for Y X. Notice that d H is symmetric by construction and is zero if and only if the two sets coincide (here we use that these sets are closed, hence compact, so the sup are actually max ). Checking the triangle inequality requires a little extra work. To show that d H (A, B) d H (A, C)+d H (C, B) note that d(a, b) d(a, c)+d(c, b) for a A, b B, c C, so taking the infimum over b we get d(a, B) d(a, c) + d(c, B) for a A, c C. Therefore d(a, B) d(a, C) + sup c C d(c, B) and sup a A d(a, B) sup a A d(a, C) + sup c C d(c, B). Likewise one gets sup b B d(b, A) sup b B d(b, C) + sup c C d(c, A). Adding the last two inequalities gives the triangle inequality. DEFINITION A metric space (X, d) is said to be locally compact if for every x and every neighborhood O of x there is a compact set K in O which contains x. It is said to be separable if it contains a countable dense subset (such as the rationals in R).

7 1. METRIC SPACES 7 e. Norms define metrics in R n. There is a particular class of metrics in the euclidean space R n that are invariant under translations. DEFINITION A function N on R n (or any linear space) is said to be a norm if (1) N(λx) = λ N(x) for λ R (homogeneity), (2) N(x) 0 and N(x) = 0 x = 0 (positivity), (3) N(x + y) N(x) + N(y) (convexity). Any norm determines a metric by setting the distance function d(x, y) = N(x y). For the metric thus defined positivity follows from the positivity of the norm, symmetry follows from homogeneity for λ = 1, and triangle inequality from convexity. For such a metric the translations T v : x x+v are isometries by definition. Furthermore, the central symmetry x x is an isometry, and any homothety x λx multiplies all distances by λ (we call the last property homogeneity of the metric). EXAMPLE The maximum distance on R n is given by (1.1) d(x, y) = max 1 i n x i y i. Of course, the standard Euclidean metric is of that kind (it is also invariant under rotations, which we do not require), as is the maximum metric (1.1). PROPOSITION All metrics in R n determined by norms are uniformly equivalent. PROOF. First, since the property of uniform equivalence is transitive it is sufficient to show that any metric determined by a norm is uniformly equivalent to the standard Euclidean metric. Second, since translations are isometries it is suffient to consider distances from the origin, that is, we can work with the norms directly. Third, by homogeneity it is sufficient to consider norms of vectors whose Euclidean norm is equal to one, that is, the points on the unit sphere. But then the other norm is a convex, hence continuous, function with respect to Euclidean distance, so by compactness of the sphere it is bounded from above. It also achieves its minimum on the unit sphere which is nonnegative. The minimum cannot be zero because this would imply existence of a nonzero vector with zero norm. Thus the ratio of the standard and other norm is bounded between two positive constants. f. Product spaces. The construction of the torus as a product of circles illustrates the usefulness of considering products of metric spaces in general. To define the product of two metric spaces (X, d X ) and (Y, d Y ) we need to define a metric on the cartesian product X Y, such as d X Y ((x 1, y 1 ), (x 2, y 2 )) := (d X (x 1, x 2 )) 2 + (d Y (y 1, y 2 )) 2. That this defines a metric is checked in the same way as checking that the euclidean norm on R 2 defines a metric. There are other choices of equivalent metrics on the product. Two evident ones are and d X Y ((x 1, y 1 ), (x 2, y 2 )) := d X (x 1, x 2 ) + d Y (y 1, y 2 ) d X Y ((x 1, y 1 ), (x 2, y 2 )) := max(d X (x 1, x 2 ), d Y (y 1, y 2 )).

8 8 Showing that these metrics are pairwise uniformly equivalent is done in the same way as showing that the euclidean norm, the norm (x, y) 1 := x + y, and the maximum norm (x, y) := max( x, y ) define pairwise equivalent metrics (Proposition 1.29). Indeed, this follows from it. For products of finitely many spaces (X i, d Xi ) (i = 1,..., n) one can define several uniformly equivalent metrics on the product as follows: Fix a norm on R n and for any two points (x 1, x 2,..., x n ) and (x 1, x 2,..., x n) define their distance to be the norm of the vector in R n whose entries are d Xi (x i, x i ). That the resulting metrics are uniformly equivalent follows from uniform equivalence of any two norms on R n (Proposition 1.29). We also encounter products of infinitely many metric spaces (or, usually, a product of infinitely many copies of the same metric space). In an infinite cartesian product of a set X every element is specified by its components, that is, if the copies of the set X are indexed by a label i that ranges over an index set I, then an individual element of the product set is specified by assigning to each value of i an element of X, the ith coordinate. This leads to the formal definition of the infinite product i I X =: XI as the set of all functions from I to X. Unlike in the case of finite products we have to choose our product metric carefully. Not only do we have to keep in mind questions of convergence, but different choices are much more likely to give metrics that are not equivalent, even up to homeomorphism. To define a product metric assume I is countable. In case I = N and if the metric on X is bounded, that is, d(x, y) 1, say, for all x, y X, we can define several homeomorphic metrics by setting (1.2) d λ (x, y) := i=1 d(x i, y i ) λ i. This converges for any λ > 1 by comparison with the corresponding geometric series. If I = Z we make the same definition with summation over Z (this is the reason for writing i in (1.2). A useful and deep fact is that with any of these product metrics the space X I is compact whenever X is compact. As a particular case we can perform this construction with X = [0, 1], the unit interval. The product thus obtained is called the Hilbert cube. This is a new way to think of the collection of all sequences whose entries are in the unit interval. g. Sequence spaces. Generalizing from the standard middle-third Cantor set we now define a more general class of metric spaces of which there are many important examples. DEFINITION A Cantor set is a metric space homeomorphic to the middle-third Cantor set. A natural and important example is the space Ω R 2 of sequences ω = (ω i) i=0 whose entries are either 0 or 1. This set is the product {0, 1} N0 of countably many copies of the set {0, 1} of two elements, so it is natural to endow it with a product metric. Up to multiplication by a constant there is only one metric on {0, 1}, which we define by setting d(0, 1) = 1. Referring to (1.2) we can endow Ω R 2 with the product metric d(ω, ω d(ω i, ω i ) := ) 3 i+1. i=0

9 1. METRIC SPACES 9 PROPOSITION The space Ω R 2 = {0, 1} N0 equipped with the product metric d(ω, ω ) := i=0 d(ω i, ω i )3 (i+1) is a Cantor set. To prove this we need a homeomorphism between the ternary Cantor set C and Ω R 2 : LEMMA The one-to-one correspondence between the ternary Cantor set C and Ω R 2 defined by mapping each point x = 0.α 1 α 2 α 3 = α i i=1 3 i C (α i 1) to the sequence f(x) := {α i /2} i=0 is a homeomorphism. i=0 PROOF. If x = 0.α 0 α 1 α 2 = i=0 β i 3 i+1 (β i 1) in C then d(x, y) = x y = i=0 α i 3 i+1 β i 3 i+1 = i=0 i=0 α i 3 i+1 (α i 1) and y = 0.β 0 β 1 β 2 = α i β i 3 i+1 i=0 α i β i 3 i+1 = 2d(f(x), f(y)). Now let α = f(x), β = f(y). Then d(f 1 (α), f 1 (β)) = d(x, y) 2d(α, β), so f 1 is Lipschitz continuous with Lipschitz constant 2. If ω, ω Ω R 2 are two sequences with d(ω, ω ) 3 n then ω i ω i for some i n because otherwise d(ω, ω ) 3 i 1 = 3 n = 3 n 1 /2 < 3 n. 3 i=n+1 Consequently f 1 (ω) and f 1 (ω ) differ in the ith digit for some i n. This implies d(f 1 (ω), f 1 (ω)) 3 (n+1) because the two points are in different pieces of C n+1. Taking x = f 1 (ω), x = f 1 (ω ) we get d(x, x ) < 3 (n+1) d(f(x), f(y)) < 3 n. This shows that f is Lipschitz continuous as well. We have shown in particular that Ω R 2 is compact and totally disconnected. Let us note in addition that every sequence in Ω R 2 can be approximated arbitrarily well by different sequences in Ω R 2 by changing only very remote entries. Thus every point of Ω R 2 is an accumulation point and Ω R 2 is a perfect set. PROPOSITION Cantor sets are compact, totally disconnected, and perfect. It is not hard to see that the space Ω 2 = {0, 1} Z with a product metric is in turn homeomorphic to Ω R 2, and therefore also is a Cantor set. To that end let { 2n if n 0 α: Z N 0, n 1 2n if n < 0 and f : Ω R 2 Ω 2, ω ω a = (... ω 3 ω 1 ω 0 ω 2 ω 4... ). Endowing Ω 2 and Ω R 2 with any two of the product metrics (1.2) makes f a homeomorphism because two sequences α, α are close if and only if they agree on a large stretch of initial entries. Then the resulting sequences ω = f(α) and ω = f(α ) agree on a long stretch of entries around the 0th entry and hence are also close. Thus f is a continuous bijection between compact spaces and

10 10 therefore a homeomorphism by Proposition (It is as easy to see directly that f 1 is continuous also.) h. General properties of Cantor sets. It is useful to note a general fact about Cantor sets: PROPOSITION Every perfect totally disconnected compact metric space is a Cantor set. We have seen that sequence spaces are perfect and compact; it is easy to see in general that they are totally disconnected: If α β are sequences then α i β i for some index i. The set of sequences ω with ω i = α i is open, and likewise the set of sequences with ω i = β i. But these sets are disjoint and their union is the entire space. COROLLARY Every nonempty, perfect, bounded, nowhere dense set on the line is a Cantor set. PROOF. A perfect bounded set on the line is compact by the Heine Borel Theorem (A closed bounded subset of R n is compact). Being perfect, it also contains more than one point. If it is not totally disconnected then it has a connected component with more than one point and hence contains a nontrivial interval, contrary to being nowhere dense. i. Dyadic integers. Define the following metric d 2 on the group Z of all integers: d(n, n) = 0 and d 2 (m, n) = m n 2 for n M, where n 2 = 2 k if n = 2 k l with an odd number l. The completion of Z with respect to that metric is called the group of dyadic integers and is usually denoted by Z 2. It is a compact topological group. 1 Exercises EXERCISE 1.1. Prove that open intervals are open sets and closed intervals are closed sets. EXERCISE 1.2. What is the interior of an interval [a, b]? (Give a full proof.) EXERCISE 1.3. Prove that the construction in Remark 1.11 defines a metric. EXERCISE 1.4. Show that any homogeneous metric in R n that is invariant under translations is determined by a unique norm. 2. Riemann integration in metric spaces a. The Riemann integral. The basic noion is that of Riemann integration with respect to length, area or volume in space, its subsets and related spaces such as spheres, cylinders and tori. An important question is: What functions are Riemann integrable? In the standard definition through upper and lower sums boundedness is obviously necessary. Similarly, the function must be compactly supported, that is, should vanish outside of a compact set. This is no restriction if the ambient space itself is compact, such as a closed interval, a rectangle, a sphere, or a torus. Under these assumptions every continuous function is integrable. Some of the most important fuctions that appear in connection with 1 For additional properties of metric spaces see Section 2 of the Overview.

11 2. RIEMANN INTEGRATION IN METRIC SPACES 11 integration are discontinuous, though. First of all, there are characteristic funtions of intervals, rectangles and other nice sets where the integral is equal to the length, area or volume depending on the dimension. It turns out there is a powerful necessary and sufficient condition for integrability. THEOREM 2.1 (Lebesgue). A function defined in a bounded domain of euclidean space, or on a sphere, a torus or a similar compact differentiable manifold is Riemann integrable if and only if it is bounded and the set of its discontinuity points is a null set. PROOF IDEA. The main issue is to connect the countably many sets in the definition of null set with the finitely many rectangles in upper and lower sums. The method is to note that the set of points near which f varies by at least ɛ can be covered by countably many rectangles whose volumes sum to arbitrarily little, and then compactness leads to a finite subcover that can be made part of a legitimate partition. 2 Applying this criterion to the characteristic function of a compact set A (whose discontinuity points are exactly the boundary points of A) we immediately obtain COROLLARY 2.2. For a compact set A the length, area or volume is defined if and only if the boundary of A is a null set. This of course immediately extends to sets with compact closure. b. Weighted integration. A natural extension of Riemann integration is weighted integration with respect to a nonnegative density ρ which can be reduced to the standard case my simply multiplying the integrand by ρ. Naturally, in order for this procedure to work the function ρ must be Riemann integrable itself. This is also sufficient due to the following fact. PROPOSITION 2.3. The sum, product and uniform limit of Riemann integrable functions is Riemann integrable. Moreover, the integral behaves naturally with respect to linear combinations and limits, that is, the integral of a linear combination (or limit) is the corresponding linear combination (or limit) of the integrals. The notion of integral can be extended beyond the above setting. Most texts on calculus and elementary real analysis deal with situations (which often appear in real world) when either the domain of the function is not compact (as for f(x) = 1 1+x on the real 2 line), or the function is unbounded (as for f(x) = log x on the interval (0, 1]), or both. In such cases natural approximation procedures often give a notion of integral called improper integral. c. The Riemann Stieltjes integral on the line. Bernoulli measures on the interval [0, 1] defined as the independent distribution of digits of a number in base 2 with given probabilities p and q = 1 p of 0 and 1 correspondingly, constitute a situation where integration is naturally defined but cannot be reduced to any of the situations described above. Integration with respect any nonsymmetric Bernoulli measure cannot be reduced to Riemann integration with respect to a density since a null set may have positive Bernoulli measure (this is referred to as singularity of the measure). Still, integration with respect to See Jerrold E. Marsden, Michael J. Hoffman: Elementary Classical Analysis, W. H. Freeman, New York,

12 12 a Bernoulli measure can be defined following the familiar procudure of upper and lower Riemann sums associated to a partition of [0, 1] into small intervals and taking the joint limit when the length of the longest partition element congerves to zero. A general construction of this kind is called the Riemann Stieltjes integral. It is defined for functions on an interval I R. It depends on a distribution function F on I which is monotone, bounded from above and below and continuous from the left, that is, for an increasing sequence x n I one has lim n F (x n ) = F (lim n x n ). We first consider the case of continuous F ; this is not a serious restriction in dynamical considerations. Define the measure of an interval [a, b] I as F (b) F (a). Using this measure instead of length define upper and lower Riemann sums for a function with respect to a finite partition of I. In this case there is no distinction between closed, open or halfopen intervals. In general, that is, if there are discontinuity points, one considers first intervals whose endpoints are continuity points of F and defines the measure the same way. To avoid ambiguities in the definiton of Riemann sums one has to restrict to partitions where the division points are continuity points of F. This is because the discontinuity points are not null sets: The measure of any such point x is given by the jump of the distribution funstion at x. The proof that any continuous function is integrable (that is, upper and lower Riemann sums have the same limit for any sequence of partitions for which the maximal length of elements goes to zero) is effectively identical to the proof for the standard Riemann integral. The distribution function construction provides the most general treatment of integration in one dimension; it can in fact be extended to the case of the whole line or half-line. EXERCISE 2.1. Generalize the construction of Riemann Stieltjes integral to the case of function of n variables by considering the distribution function on a cube. First consider the case of continuous distribution function with proper monotonicity conditions and then discuss allowable discontinuities. Consider the case n = 2 in detail. d. Integrals as positive functionals. In the rest of this section we describe a general framework for integration in metric spaces, which in particular addresses all these situations. In fact, what we describe is going from an integral defined for continuous function to the measure for nice sets and back. DEFINITION 2.4. Let X be a compact metric space. A Riesz integral is a nonzero linear functional I on the space C(X) of continuous real-valued functions on X that is continuous in the uniform topology and nonnegative, that is, I(f) 0 if f 0. The weighted Riemann integral I(f) = ρf dx as well as any Riemann Stieltjes integral satisfy these conditions. A Riesz integral is defined on the characteristic function χ A only if χ A is continuous, that is, if A is simultaneously open and closed. While this is the case with cylinder sets in sequence spaces, it is impossible for connected spaces. We extend the definition of the Riesz integral to some characteristic functions and many other functions by approximation. DEFINITION 2.5. For a function f : X R define the upper integral as I + (f) := inf{i(g) g C(X), g f},

13 2. RIEMANN INTEGRATION IN METRIC SPACES 13 and similarly the lower integral is I (f) := sup{i(g) g C(X), g f}. The function f is integrable if I + (f) = I (f). In this case this common value is denoted by I(f).. Obviously a linear combination of integrable functions is integrable. It is slightly less trivial but still not hard to see that the product of two integrable functions is integrable and the uniform limit of integrable functions is integrable. The following proposition shows the abundance of integrable functions among characteristic functions. PROPOSITION 2.6. Given x X for all but at most countably many values of r the characteristic functions of both closed and open r balls around x are integrable. PROOF. The upper integral of the characteristic function χ B(x,r) of the open r- ball around x is monotone nondecreasing in r and hence continuous for all but at most countably many values of r. For any such value of r the characteristic functions of both open and close r-balls are integrable. This immediately follows from the fact that for any ɛ > 0 there exists continuous functions φ + ɛ and φ ɛ such that χ B(x,r ɛ) φ ɛ χ B(x,r) φ + ɛ χ B(x,r+ɛ). DEFINITION 2.7. A set whose characteristic function is integrable is said to be (Riemann) measurable with respect to I. I(χ A ) is called the measure of A. PROPOSITION 2.8. Any finite union or finite intersection of measurable sets is measurable. There is an obvious weakness in this definition which is not fully apparent in the case of standard Riemann integration. Namely some fairly nice sets may not be measurable. The simplest example is the δ-measure: δ x0 (f) = f(x 0 ) for a given point x 0 X. This is obviously a Riesz integral and obviously it is concentrated at the point x 0, since it vanishes at any function which vanishes at that point. However the one-point set {x 0 } is not measurable according to our definition. One way out of this problem is to extend the notions of integrability and measurability. This leads to the theory of Lebesgue integration. Let us outline another resolution of this difficulty based on the idea that certain parts of the space are insignificant form the point of view of the given integral and hence can be disregarded, DEFINITION 2.9. The point x X is significant with respect to a Riesz integral I if for any open set U x there exists a continuous function f which vanishes outside of U and such that I(f) > 0. The set of all significant points is called the support of I and is denoted by supp I. It follows for the definition that the support of any Riesz integral is closed. Since every continuous function on a closed subset of a metric space can be extended to a continuous function on the whole space one can in fact define I on the space C(supp I).

14 14 e. Partitions and Riemann sums. Now we show how a Riesz integral can be reconstructed using a procedure similar to the construction of the standard Riemann integral through partitions into rectangles and Riemann sums. DEFINITION Given a Riesz integral I a measurable partition of X is a decomposition of X into the union of finitely many measurable sets. The size of a measurable partition is the supremum of distances betwen points in the same element PROPOSITION For any ɛ there is a measurable partition of size less than ɛ. PROOF. By Proposition 2.6 for every point there exists a ball around that point of radius less than ɛ/3 whose characteristic function is measurable. Take a finite cover B 1,..., B n by such balls and let C k = B k k 1 i=1 B i, k = 1,..., n. The sets C 1,..., C n form a partition with the desired property. Given a bounded function f on X and a measurable partition ξ = (C 1,..., C n ) of X we define the upper and lower Riemann sums as n U(f, ξ) := measure(c i ) sup{f(x) x C i } and correspondingly. L(f, ξ) := i=1 n measure(c i ) inf{f(x) x C i }, i=1 THEOREM If f is integrable and ξ m is a sequence of partitions whose size goes to zero then lim U(f, ξ m) = lim L(f, ξ m) = I(f). m m PROOF. First consider a continuous function f and let ɛ > 0. Since f is uniformly continuous there exists N N such that for any m > N and for any element C of the partition ξ m, sup{f(x) x C} inf{f(x) x C} < ɛ. Since measure is additive this implies that U(f, ξ m ) L(f, ξ m ) < ɛ. For an arbitrary Riemann integrable function f there are continuous functions f + and f such that f f f + and (f + ) (f ) < ɛ. Applying the previous argument to these functions gives the statement. f. The general Riemann integral. Finally we show that a Riesz integral arises from a measure defined on a sufficient collection of sets through the construction of Riemann sums. In analogy with the classical Riemann integral construction we call such sets rectangles. We assume that a collection of rectangles is fixed together with a measure defined for each rectangle. We consider a compact metric space X. Sufficiency: The whole space is a rectangle. Refinement: Given any ɛ > 0 a rectangle can be partitioned into finitely many rectangles, each of which fits into an ɛ-ball. Intersection: The intersection of two rectangles is a rectangle. Additivity: Whenever a rectangle R is partitioned into rectangles R i (i = 1,..., k) then measure(r) = k i=1 measure(r i).

15 2. RIEMANN INTEGRATION IN METRIC SPACES 15 Suppose f is a bounded real-valued function defined on a bounded rectangle A. For any partition R = {R 1,..., R k } (that is, A = k i=1 R i) into rectangles define the upper sum to be k U(f, R) := measure(r i ) sup{f(x) x R i } and the lower sum L(f, R) := i=1 k measure(r i ) inf{f(x) x R i }. i=1 LEMMA If R and R are partitions then L(f, R ) U(f, R). PROOF. If R = R this is obvious. To reduce to this case use the common refinement R := {R R R R, R R }. This is a partition by rectangles and it is not hard to check that L(f, R ) L(f, R) U(f, R) U(f, R) using additivity. This lemma implies that f := inf U(f, R) and A R are well-defined and finite, and A f A f. A f := sup L(f, R) R DEFINITION A function f defined on a rectangle A is said to be Riemann integrable over A if A f = f. In this case A A f := f is called the Riemann integral of A f over A. Using the refinement property above an argument very similar to the proof of Theorem 2.12 shows that continuous functions are Riemann integrable and that the Riemann integral thus defined on C(X) is a Riesz integral as in Definition 2.4.

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