Measure Theory and Integration

Transcription

1 CHAPTER 1 Measure Theory and Integration One fundamental result of measure theory which we will encounter inthischapteristhatnoteverysetcanbemeasured. Morespecifically, if we wish to define the measure, or size of subsets of the real line in such a way that the measure of an interval is the length of the interval then, either we find that this measure does not extend all subsets of the real line or some desirable property of the measure must be sacrificed. An example of a desirable property that might need to be sacrificed is this: the (extended) measure of the union of two disjoint subsets might not be the sum of the measures. The Banach-Tarski paradox dramatically illustrates the difficulty of measuring general sets; it is stated in Section???????? without proof. It is with this in mind that our discussion of measure and integration begins, in this Section, by making precise what constitutes a measurable set and a measurable function and then an integrable function. 1. Algebras of Sets Let be a set and P() denote the power set of which is the set of all subsets of. The reader is referred to Chapter 0, Subsection 1 if any notations or set theoretic notions are unfamiliar. Notation: N 0 is the set of nonnegative integers; Q is the set of rational numbers; Z is the set of all integers. Suppose 1 and 2 are two sets and f : 1 2 is a function with domain 1 and range in 2 (but f need not be onto). For any B 2, denote by f 1 (B) = {x 1 : f(x) B}. the inverse image of the set B under f. Of course f need not be one-to-one (injective) and so f 1 need not exist as a function. Some of the elementary properties of the inverse image are worthy of note: If B,B 2 then f 1 (B B ) = f 1 (B) f 1 (B ) f 1 (B B ) = f 1 (B) f 1 (B ) f 1 (B c ) = f 1 (B) c 25

2 26 1. MEASURE THEORY AND INTEGRATION The verification is asked for in the Exercises at the end of the Section. Definition: If A then define the characteristic function χ A on by χ A (x) = 1 if x A and χ A (x) = 0 if x A. Exercise: Determine χ 1 A (B) for an arbitrary B R. There are 4 cases, if A. Exercise: (De Morgan s Laws) If {A α : α I} is a collection of subsets of a set then the operation C of complementation has the properties. C( α I A α ) = α I C(A α ) C( α I A α ) = α I C(A α ) If I = N is countable then we can write n N A n = n N B n where B 1 = A 1, B 2 = A 2 A c 1,...B n = A n A c 1 A c 2... A c n 1. The B n are then disjoint. Definition: Let A n, n N be a countable collection of sets in. The limit supremum of {A n } is lim supa n = n N k n A k n whereas the limit infimum of {A n } is lim inf n A n = n N k n A k. Intuitively limsup n A n is all points that are in infinitely many of the A n whereas liminf n A n is all those points that are in all except possibly finitely many of the A n. If A n is a decreasing sequence of sets, A 1 A 2 A 3... then lim supa n = n N A n = liminfa n n n Definition: A collection F of subsets of is an algebra (field) if a. F. b. If A F then A c F. c. If {A α : α I} F and if I is finite then α I F. Condition c in the definition could equivalently be replaced by either of the following Conditions. c. If A,B F then A B F. c. If A,B F then A B F. because Condition c is equivalent to c; and Condition c in conjunction with Condition b is equivalent to Conditions c and b, as can be seen by observing that C(A B) = CA CB. Definition: AcollectionF ofsubsetsofisaσ-algebra(orσ-field) if F is an algebra and

3 1. ALGEBRAS OF SETS 27 d. If {A α : α I} F and if I is countable, or finite, then α I A α F. Of course condition d implies condition c in the definition of an algebra. Conditions b and d together are equivalent to b and d where d. If {A α : α I} F and if I is countable, or finite, then α I A α F. Example 1.1. (i) The power set P() of a set is a σ- algebra. (ii) {,} is a σ-algebra. (iii) {,A,A c,} is a σ-algebra, if A is any subset of. Example (i) is the largest σ-algebra of subsets of ; Example (ii) is the smallest and Example (iii) is the smallest σ-algebra that contains the set A. Example 1.2. Consider the set of all real intervals of the form [a,b) along with all intervals of the form [a, ) and (,b) where a and b are real numbers. Define F to be the set of all finite unions of disjoint intervals of this form along with the null set and R itself. We claim that F is an algebra. Certainly F. If A F then we mustcheckthata c isinf andthatisobviousifa = ora = Randso we may suppose A = 1 j n [a j,b j ) where a 1 < b 1 < a 2 < b 2 < a 3 <... < b n. Then A c is of the same form: indeed if a 1 > and b n < then A c = (,a 1 ) [b 1,a 2 )... [b n 1,a n ) [b n, ). Therefore A c F; the cases where a 1 = or b 1 = or both are handled similarly. Therefore F is closed under taking complements. It remains to check that F is closed under finite unions and it suffices to show that if A,B F then A B F. If A = or A = R then this is obvious and so we suppose A = 1 j n [a j,b j ) where a 1 < b 1 < a 2 < b 2 < a 3 <... < b n and further it suffices to consider the cases that B = [c,d) where c < d or B = (,d) or B = [c, ) or B = R. If A B = then there is nothing to show but if B meets any subinterval [a j,b j ) of A or shares an endpoint then B [a j,b j ) is again an interval of the same type, that is half open and half closed. (Alternatively one argues that there are four cases: c A or not and d A or not.) Therefore A B is in F. Example 1.3. Sometimes it is more convenient to work with F consisting of all finite disjoint intervals of the form (a, b] (as opposed to [a,b) above) and (,b] and (a, ). Again in this case F is an algebra but not a σ-algebra.

4 28 1. MEASURE THEORY AND INTEGRATION Proposition 1.4. Let {F α : α I} be a non empty collection of σ-algebras (resp. algebras) on a given set indexed by a set I. Then the intersection α I F α is also a σ-algebra (resp. algebra). Proof. We need only check that α I F α satisfies the three properties of σ-algebras The set is certainly in all σ-algebras and hence in α I F α. If A α I F α then A c is in every F α. Finally if (A n ) n N is a sequence of sets in α I F α then n N A n is in every F α and this completes the proof in the case of σ-algebras and the proof for algebras is analogous. Suppose now that S P() is any collection of subsets of a set. Then there is a smallest σ-algebra which contains S which we shall denote this σ(s). To see this observe that the set of all σ-algebras that contains isnonemptybecauseitcontainsp()andsotheintersection of all σ-algebras that contain S is a σ-algebra, by the Proposition and it certainly contains S and so it must be the minimal such σ-algebra. Definition: If F is a σ-algebra then a subset S F with the property that F = σ(s) is said to be a system of generators of F. Consider now the case that = R n, where n 1. Similar considerationsapplyifisanytopologicalspace. OnR n x = (x 1,...,x n ),we define the norm x = ( 1 j n x j 2 ) 1/2 and the corresponding (usual) topology where a base for the neighborhoods of a point x R n is {V r (x) : r > 0} where V r (x) = {y R n : x y < r} is an open ball of radius r centered at x. Then a set U is said to be open in R n, if, for every x U there is r > 0 so that V r (x) U. The set of all open sets is denoted O and we define the Borel subsets of R n to be B = σ(o), that is the smallest σ-algebra that contains O. We shall sometimes write B(R n ) for B when wishing to emphasize that it is the Borel σ-algebra on R n. In general, if is a topological space and O is the set of all open subsets of then the Borel σ-algebra on is σ(o). Remark: If C denotes the set of all closed subsets of R n then σ(c) = B. For recall that a set is closed if and only if its complement is open. Since B contains the open sets and is closed under the operation of taking complements, it follows that B C and hence B σ(c). Conversely the open sets are contained in σ(c): O σ(c). Therefore B = σ(o) σ(c) and so B = σ(c). Therefore O is a system of generators of B by the definition of B and C is a second system of generators. Example: ThesetS ofallintervalsoftheforms = {(,a) : a R} is a system of generators of the Borel σ-algebra B of R: B = σ(s) Indeed, since S consists solely of open sets it is obvious that B σ(s).

5 1. ALGEBRAS OF SETS 29 It therefore suffices to show that any open set is contained in σ(s). Certainly if a < b then [a,b[=],a[ c ],b[ belongs to σ(s). Thereforeitsufficestoshowthatanyopensetcanbewrittenasaunion ofatmostcountablymanyintervalsoftheform[a,b). Wecandothisas follows: For each n, consider all intervals of the form [k2 n,(k+1)2 n ). where k Z. Given an open set U R, we let A n be the union of all suchintervalsthatareentirelycontainedinu. ThenA n isanincreasing sequence of sets and U = n 0 A n and this expresses U as a countable union of the desired type of intervals. It follows that B = σ(s). Other generating sets for the Borel subsets of R are (1) S 1 = {],a] : a R} (2) S 2 = {]a, [: a R} (3) S 3 = {[a, [: a R} (4) S 4 = {]a,b] : a,b R,a < b} (5) S 5 = {[a,b[: a,b R,a < b} Example: For each a = (a 1,a 2,...,a n ) R n let I(a) = {x R n : x 1 < a 1,x 2 < a 2,...,x n < a n }. Here x R n has components x = (x 1,x 2,...,x n ) is used. Further let I 0 (a,b) = {x R n : a 1 x 1 < b 1,a 2 x 2 < b 2,...,a n x n < b n }. whenever a,b R n and, presumably a 1 < b 1, a 2 < b 2,...a n < b n because I 0 (a,b) is void otherwise. Define S = {I(a) : a R n } and S 0 = {I 0 (a,b) : a,b R n }. Therefore S and S 0 are two collections of subsets of R n and we claim that σ(s) = σ(s 0 ) = B. The rectangular sets I 0 (a,b) can be expressed in terms of unions, intersections and complements of the sets I(a) so that σ(s) σ(s 0 ). Conversely any of the sets I(a) is the limit of an increasing sequence of sets I 0 (b n,a) so that σ(s) = σ(s 0 ). Also σ(s) B since each I(a) is open. Therefore it remains only to show that B σ(s) and for this it suffices to show that an arbitrary open set is in σ(s 0 ). Proceeding as in the previous example, it can be shown that any σ-algebra open set is a countable union of the rectangular sets I 0 (a,b) and so this completes the proof. There will also be occasion to work in the extended real numbers R = [, ], regarded as a two point compactification of the real line. The open sets in this case are all those sets that are subsets of R and are open in R as well as the sets ]a, ] and ],b] for arbitrary real a and b or is a union of the three types of sets. Again we denote the open sets by O and we define the Borel σ-algebra of R to be B(R) = σ(o). We have A B(R) if and only if A {, }, regarded as a subset of R is in B(R), for it is straightforward to check that the set of such sets is closed under complementation and countable union. The generating

6 30 1. MEASURE THEORY AND INTEGRATION setslistedaboveforb(r)arealsogeneratingsetsforb(r)ifoneadjoins the singleton set { } (resp. { }). Monotone Classes We say that a sequence (A p ) p N of subsets of a set is increasing if A 1 A 2 A 3... and decreasing if A 1 A 2 A 3... and monotone if it is either increasing or decreasing. Every monotone sequence has a limit: A = p N A p if (A p ) p N is increasing and A = p N A p in the decreasing case. A set M of subsets of is said to be a monotone class if, for every monotone sequence in M, the limit is also in M. Every σ-algebra is a monotone class. For recall that an increasing sequence of sets can be written as the union of disjoint subsets. The complement of a decreasing sequence forms an increasing sequence and so a σ-algebra is a monotone class. On the other hand if F 0 is both an algebra and a monotone class then it must also be a σ-algebra. If {M α : α I} is a nonempty collection of monotone classes then α I M α is also a monotone class. Since the power set P() is a monotone class it follows that every subset S P() is contained in a smallest monotone class which we denote M(S) Theorem 1.5. (Monotone Class Lemma) If F 0 is an algebra of subsets of a set then σ(f 0 ) = M(F 0 ). Proof: Ithasalreadybeenremarkedthataσ-algebraisamonotone class and so it follows that a M(F 0 ) σ(f 0 ). To prove the converse it suffices to show that M(F 0 ) is an algebra. Introduce the notation M A = {B M(F 0 ) : A B,A B c and A c B are in M(F 0 )} It is not difficult to check that M A is a monotone class. Now if we suppose that A F 0 then M A F 0. By minimality M A = M(F 0 ) if A F 0. Next we observe that, by symmetry that B M A implies A M B. Therefore if A F 0 and B M A then A M B. This shows that, whenever B M(F 0 ) = M A, then we have M B F 0. But then, by minimality again we must have M B = M(F 0 ) even for B M(F 0 ). Therefore for every A,B M(F 0 ) A B; A c B, and A B c are in M(F 0 ). We check now that M(F 0 ) is an algebra. Certainly F 0 M(F 0 ) and if A M(F 0 ) then A c M(F 0 ) for take B =. Finally if A,B M(F 0 ) then A c,b M(F 0 ) so that A c B c M(F 0 ) and we may take the complement to get A B M(F 0 ). This proves M(F 0 ) is an algebra and as a monotone class it is therefore a σ-algebra.

7 1. ALGEBRAS OF SETS 31 We shall have occasion to use the Monotone Class Lemma when verifying uniqueness of measures. Exercises: (1) Show that, if B,B 2 then f 1 (B B ) = f 1 (B) f 1 (B ) f 1 (B B ) = f 1 (B) f 1 (B ) f 1 (B c ) = f 1 (B) c (2) Show that, in general, liminf n A n limsup n A n (3) Recall the definition of the limit supremum and limit infimum of a sequence of real numbers a n. lim supa n = inf{sup{a k : k n} : n 1} and n N lim inf a n = sup{inf{a k : k n} : n 1} n N where the value (resp. ) is possible if the sequence is unbounded above (resp. below). (See page two of Ash s book Measure, Integration, and Functional Analysis.) How is limsupχ An related to χ B where B = limsupa n? What happens if we replace limsupχ An by liminfχ An? (4) Suppose that A n = {x R 3 : x (0,0,( 1) n /n) < 1}. Find limsup n A n and liminf n A n. Compare this with Exercise 3, page 3 of Ash s book. (5) Show that F of Example 1.2 is not a σ-algebra. (6) Suppose that E is an arbitrary subset of a set and S is a collection of subsets of. Define E S = {E A : A S}. Show that a. If F 0 is an algebra of subsets of then E F 0 is also an algebra of subsets of E. b. σ(e S) = E σ(s). That is the smallest σ-algebra in P(E) containing E S is the intersection of E with the smallest σ-algebra in P() containing S. c. Conclude further that, if F is a σ-algebra of subsets of, then E F is a σ-algebra of subsets of E. (Reference: Halmos s Measure Theory)

8 32 1. MEASURE THEORY AND INTEGRATION 2. Measurability It is an objective of this Chapter to define the integral of a function. If f is a nonnegative function defined on an interval in the real line then the integral should be able to tell us the area under the graph of f just as the Riemann integral does in calculus. It will tell us much more but even in this limited context we shall discover that f must be restricted: the notion of area under the graph of f will not make sense for every f and we will be forced to restrict the class of functions considered. We introduce in this Section the concept of a measurable function. We shall see that f must be measurable if we are to make sense of the notion of area under the graph. Of course this difficulty arises already in the case of Riemann integration: the Riemann integral is not defined for arbitrary functions f. We shall clarify this remark and discuss the Riemann integral and its relation to the Lebesgue integral, introduced here, at the end of this Chapter. Definition: Let be a set and F be a σ-algebra of subsets of. The we shall refer to (,F) as a measurable space. The sets in F will be referred to as measurable sets. Definition: Suppose ( 1,F 1 ) and ( 2,F 2 ) are two measurable spaces. Then a mapping f : 1 2 is measurable, with respect to F 1 and F 2 if f 1 (B) F 1 whenever B F 2. We shall sometimes say f is measurable function from ( 1,F 1 ) to ( 2,F 2 ) to clarify the choice of σ-algebras F 1 and F 2. Unless otherwise specified a function f : 1 R n is said to be measurable or Borel measurable if f is measurable from ( 1,F 1 ) to (R n,b). An elementary example of a measurable function is given by the characteristic function χ A of a set A. Then χ A is measurable as a mapping from (,F) to (R,B) if and only if A F, that is if and only if A is a measurable set. Further examples of measurable functions will be apparent once some elementary properties have been established. Lemma 2.1. Suppose that ( 1,F 1 ), ( 2,F 2 ) and ( 3,F 3 ) are three measurable spaces and that f : 1 2 and g : 2 3 are two mappings. If f and g are both measurable then the composed function g f is measurable from ( 1,F 1 ) to ( 3,F 3 ). Proof: We must show that, for an arbitrary set C F 3, (g f) 1 (C) F 1. It suffices to show that (2.1) (g f) 1 (C) = f 1 (g 1 (C))

9 2. MEASURABILITY 33 because g 1 (C) F 2 because g is measurable and so f 1 (g 1 (C)) F 1 because f is measurable. The verification of (2.1) is a straightforward exercise in checking the equality of sets. Proposition 2.2. Suppose that f : 1 2 is a function and F 1 and F 2 are σ-algebras on 1 and 2 respectively. Suppose further that S is a system of generators of F 2, which is to say F 2 = σ(s). Then f is measurable from ( 1,F 1 ) to ( 2,F 2 ) if and only if f 1 (B) F 1 for every B S. Proof: It is obvious that, if f is measurable then f 1 (B) F 1 for every B S simply because S F 2. Conversely, suppose f 1 (B) F 1 for every B S. Let T = {B F 2 : f 1 (B) F 1 } so that T S. We shall show that T is a σ-algebra which implies that T = F 2 and so, by the definition of T, f is measurable. This will complete the proof. We check therefore that T is a σ-algebra. Certainly T, because f 1 ( ) = F 1. We check next that if B T then f 1 (B) F 1 and so f 1 (B c ) = f 1 (B) c F 1 because F 1 is closed under taking complements. (Recall f 1 (B c ) = f 1 (B) c by an Exercise.) We check finally that if (B n ) n N is a sequence of sets in T so that f 1 (B n ) T for each n N then f 1 ( n N B n ) = n N f 1 (B n ) F 1 because F 1 is a σ-algebra. This shows that T is closed under countable unions and is therefore a σ-algebra and the proof is complete. Remark: This set T in the proof constitute good sets and the argument that there are many good sets is an instance of what Ash calls the good sets principle in his text page 5. Remark: The above proof uses a special case of the following observation: Iff isamappingf : 1 2 andif(b α ) α I isacollection of subsets of 2 indexed by a set I then f 1 ( α I B α ) = α I f 1 (B α ) f 1 ( α I B α ) = α I f 1 (B α ). On the other hand if (A α ) α I is a collection of subsets of 1 then f( α I A α ) = α I f(a α ) but it may happen that f( α I A α ) α I f(a α ). Corollary 2.3. A continuous function f : R m R n is Borel measurable. Of course the understood σ-algebras here are the Borel σ-algebra B(R m ) and B(R n ).

10 34 1. MEASURE THEORY AND INTEGRATION Proof: Since the B(R n ) is generated by the open sets, it suffices to show that f 1 (U) B(R m ) for an arbitrary open set U. The result will then follow from the Proposition. However f is continuous means that f 1 (U) is open whenever U is and so f 1 (U) B(R m ). The same reasoning applies in a more general setting. Corollary 2.4. If f is a continuous function from one topological space (X,Σ) to another (Y,T ) then f is measurable from (X,B(X)) to (Y,B(Y)) where B(X) is the Borel σ-algebra which is generated by the open sets Σ of X and similarly B(Y) is generated by the open sets T. Proposition 2.5. Suppose that (, F) is a measurable space and f 1, f 2,...f m are m Borel measurable real valued functions, f j : (,F) (R,B), for 1 j m. Suppose that g : (R m,b(r m )) (R, B(R)) is measurable. Then φ(x) = g(f 1 (x),f 2 (x),...,f m (x)) defines a measurable function φ : (,F) (R,B(R)). We shall set aside the proof of the Proposition until later and consider its consequences. Corollary 2.6. If f j : (,F) (R,B), j = 1,2 are Borel measurable functions, then f 1 +f 2 is also Borel measurable. Proof. This is an application of the Proposition with g : R 2 R defined by g(x 1,x 2 ) = x 1 + x 2. Because g is continuous it is Borel measurable. Corollary 2.7. If f j : (,F) (R,B), j = 1,2 are Borel measurable functions, then the product f 1 f 2 is also Borel measurable. In particular, if k is a real constant then kf 2 is Borel measurable. Proof. In this case g(x,y) = xy. The special case follows by defining f 1 (x) = k which makes f 1 measurable because it is continuous. Corollary 2.8. If f : (,F) (R,B), is a Borel measurable function then f is also Borel measurable. Proof. In this case g(x) = x. Corollary 2.9. If f j : (,F) (R,B), j = 1,2 are Borel measurable functions, and if f 2 (x) 0 for all x then the ratio f 1 /f 2 is also Borel measurable.

11 2. MEASURABILITY 35 Proof. In this case we define { x/y if y 0 g(x,y) = 0 if y = 0 so that g is defined on all of R 2 but of course it is not continuous. We shall check however that g is measurable, as a mapping from (R 2,B(R 2 )) to (R,B). It suffices to show that g 1 (,a) B(R 2 ) for any real a by Proposition 2.2. If a < 0 then g 1 (,a) = {(x,y) R 2 : x/y < a} is easily seen to be open in R 2 and hence in B(R 2 ). If a 0 then g 1 (,a) = {(x,y) R 2 : x/y < a} {(x,y) : y = 0} which is theunion ofan open and a closed set and is thereforein B(R 2 ). This proves that g is measurable and so the above Proposition applies which completes the proof. Exercise: Show that if f j : (,F) (R,B), j = 1,2 are Borel measurable functions then max{f 1,f 2 } and min{f 1,f 2 } are also measurable. Remark: max{a,b} = ( a b +a+b)/2 for any reals a and b. It remains to establish the Proposition. Proof of the Proposition: It suffices to show that the mapping, f say, defined by f(x) = (f 1 (x),f 2 (x),...f m (x)) is measurable as a mapping from (,F) to (R m,b(r m ) because the composition of measurable functions is measurable. To verify f is measurable, it suffices to show that, for an arbitrary a = (a 1,a 2,...,a m ) R m, f 1 ({x 1 < a 1,x 2 < a 2,...,x m < a m }) F by Proposition 2.2. However f 1 ({x 1 < a 1,x 2 < a 2,...,x m < a m }) = 1 j m f 1 j (,a j ) and as the intersection of m measurable sets, this set is, itself measurable. Definition A function f : (,F) R is said to be simple if there exist finitely many sets A j F, 1 j m and scalars λ j so that f(x) = λ j χ Aj (x) 1 j m The set of all simple functions is denotes S(,F) or S when the context is clear. Observe that a simple function is measurable because it is the sum of measurable functions. Also a simple function takes on only finitely many values, λ j, 1 j m and possibly 0. Conversely a measurable function that takes on finitely many values is simple. To see this, simply define, for each λ j in the image of f, A j = f 1 ({λ j }). The set S

12 36 1. MEASURE THEORY AND INTEGRATION is closed under addition, scalar multiplication and multiplication and therefore forms a linear algebra over R of functions. (Recall that a linear algebraisavectorspacewithamultiplicationoperation(f,g) fg that is associative and distributes over addition from the left and right and, for any scalar α, α(fg) = (αf)g = f(αg). Reference: Naimark s Normed Algebras 7.) It is further worth noting that the representation f(x) = 1 j m λ jχ Aj (x) in the definition of f S is not unique. Proposition Let f n, n N be a sequence of Borel measurable real valued functions defined on a measurable space (, F). Suppose that lim f n(x) = f(x) exists in R for every x. n N Then the function f : (,F) R, so defined, is measurable. This result says, briefly, that the pointwise limit of measurable functions is measurable. Proof: We will show that (2.2) f 1 ((,a]) = p N n N j n f 1 j ((,a+1/p]) for every a R The set of all sets of the form (,a] generate the Borel σ-algebra B because any open set is generated. Therefore, we will have shown f is measurable by Proposition 2.2. To verify (2.2), let x f 1 ((,a]) so that f(x) a. Then, for any p N there is n so that f j (x) a + 1/p for all j n: x f 1 j ((,a+1/p]) for all j n or x j n f 1 j ((,a+1/p]). But p was arbitrary and so x belongs to the right side of (2.2) and this shows that f 1 ((,a]) is a subset or equal to the right side of (2.2). Conversely suppose that x belongs to the right hand side of (2.2). Then, for every p N, there exists n N so that f j (x) a+1/p for all j n. Taking limits in this last expression as j we see that f(x) a+1/p. Since p is arbitrary f(x) a and this shows the right side of (2.2) is in f 1 ((,a]) which verifies (2.2). This proves that f is measurable.. The Extended Reals R: We introduce the extended real line R = [, ], sometimes referred to as the two point compactification of the real line, and this is just R with two points ± adjoined: R = R { } { }. This will be a convenience when discussing convergence. We introduce the following topology on R. The neighborhoods of (resp. ) are those sets which contain an interval of the form (a, ] for some a R (resp. [,a)) and the neighborhoods of x R are the usual: those sets that contain an interval of the form {y : y x < δ}

13 2. MEASURABILITY 37 for some δ > 0. Of course the topology that R inherits as a subset of R is its usual topology. As a consequence of these definitions we see that R is homeomorphic to the compact interval [ 1,1] with the (usual) topology it inherits as a subset of R. Indeed (2.3) Φ(x) = x x 2 +1 if x R 1 if x = 1 if x = is continuous from R to [ 1,1] with inverse Φ 1 (x) = which is also continuous. x 1 x 2 if 1 < x < 1 if x = 1 if x = 1 Corollary Let f n, n N be a sequence of Borel measurable extended real valued functions defined on a measurable space (, F). Suppose that lim n N f n(x) = f(x) exists in R for every x. Then the function f : (,F) R, so defined, is measurable. Proof. An extended real valued function g is Borel measurable if and only if Φ g is Borel measurable for Φ as in (2.3) because Φ and Φ 1 are continuous and hence measurable. Therefore the previous Proposition 2.11 applied to Φ f n implies implies the present result Measurable functions can be written as the pointwise limit of simple functions. We begin by considering nonnegative bounded functions and later we will extend to all measurable functions. Proposition Suppose f is be a nonnegative, bounded, measurable function defined on a measurable space (, F). Then there is a sequence (f n ) n N of simple functions such that (1) (f n ) n N is increasing. (2) 0 f n f, for every n N (3) (f n ) n N converges to f pointwise and even uniformly. Proof. Let ǫ > 0 be given. We shall construct a simple function g ǫ so that g ǫ f and f(x) g ǫ (x) < ǫ for all x. Since f is bounded we have 0 f M for some positive constant M. We choose a 0 = 0 < a 1 < a 2 <... < a m < a m+1 ] so that a j+1 a j < ǫ, for 0 j m+1 and M < a m+1 < M +ǫ We define A j = {x : a j f(x) < a j+1 } = f 1 ([a j,a j+1 ))

14 38 1. MEASURE THEORY AND INTEGRATION for 0 j m. Then A j is measurable and A j A k = if j k and 0 j m A j =. We define g ǫ (x) = a j χ Aj 0 j m Then g ǫ f < g ǫ +ǫ, in fact, for any x, x A j for some j and so g ǫ (x) = a j and a j f(x) < a j+1. Define f 1 = g 1, f 2 = max{f 1,g 1/2 },...f n = max{f n 1,g 1/n } It follows that f n f < f n + 1/n Also that f n is measurable as the maximum of two simple functions. And (f n ) n N is increasing and so f n has the required properties. Next consider the case that f is not necessarily nonnegative but is still bounded. Proposition Suppose f is a real valued, bounded, measurable function defined on a measurable space (,F). Then there is a sequence (f n ) n N of simple functions such that (1) f n f, for every n N (2) (f n ) n N converges to f pointwise and even uniformly. Remark: The proof is based on the observation that any real valued, measurable function f = f + f where f + = max{f,0} = ( f + f)/2 is nonnegative and measurable and f = min{f,0} = max{ f, 0} is also nonnegative and measurable. Proof of the Proposition 2.14 : Let g n be a sequence of nonnegative simple functions convergent to f + = max{f,0} as guaranteed by the preceding Proposition. Similarly let h n be simple functions convergent to f and define f n = g n h n. Then f n = g n h n g n +h n f + +f = f. Moreover f f n = f + g n (f h n ) f + g n + f h n and since f + g n and f h n go to zero uniformly on, f n converges uniformly to f. Finally we consider the general case when f need not be bounded. Theorem Let f be a nonnegative extended value measurable function, f : (,F) R +. Then there is an increasing sequence f n of simple functions (measurable on (,F)) so that (1) 0 f n f, for every n N (2) (f n ) n N converges to f pointwise. Theorem Let f be a measurable function, f : (,F) R. Then there is a sequence f n of simple functions so that (1) f n f, for every n N (2) (f n ) n N converges to f pointwise.

15 2. MEASURABILITY 39 Remark: It is not true in general that the convergence is uniform in this more general setting. Indeed, if f is unbounded then it cannot be approximated uniformly by bounded functions. For suppose f n be a sequence of functions, uniformly convergent to a function f and suppose each f n is bounded (with bound depending on n). If f f n < 1 then f f n < 1 which says that f is bounded. In the setting of the Theorem above, simple functions are bounded and so we cannot expect uniform convergence. Proof of Theorem 2.15: For each p N, define h p = min{f,p} so that h p is a bounded, nonnegative measurable function and so it is possible to choose a simple function g p 0 so that h p 1 p g p h p by Proposition Then we define f 1 = g 1, f 2 = max{f 1,g 2 },...f p = max{f p 1,g p } so that f p is simple, nonegative and increasing. We also see by induction on p, f p h p f. To check the pointwise convergence, suppose x 0. If f(x 0 ) < then we may choose p 0 N sothatp 0 f(x 0 ). Thenforp p 0, f(x 0 ) = h p (x 0 )sothatf(x 0 ) 1 p = h p (x 0 ) 1 p g p f p (x 0 ) f(x 0 ) and so f(x 0 ) = lim p N f p (x 0 ). On the other hand, if f(x 0 ) = then h p (x 0 ) = p and g p (x 0 ) p (1/p) so that f p (x 0 ) g p (x 0 ) p (1/p). Therefore lim p f p (x 0 ) = = f(x 0 ). Proof of Theorem 2.16 The proof of Proposition 2.14 applies here except the uniform convergence there must be replaced by pointwise convergence. Let g n be a sequence of nonnegative simple functions convergent to f + = max{f,0} as guaranteed by the preceding Theorem. Similarly let h n be simple functions convergent to f and define f n = g n h n. Then f n = g n h n g n + h n f + + f = f. Moreover, if f(x) 0 then h n (x) = 0 and lim f n(x) = limg n (x) = f + (x) = f(x) n N n N Similarly f(x) < 0, then g n (x) = 0 and lim f n(x) = lim h n (x) = f (x) = f(x) n N n N Note that lim n g n (x) = and lim n h n (y) = are both possible, but only if x y. The set of simple functions is an algebra as we have seen. Moreover if φ : R R is Borel measurable then φ f is simple whenever f is. In particular f is simple if f is and max{f,g} = 1 (f +g+ f g ) is 2 simple whenever f and g are. Similarly min{f,g} is simple.

16 40 1. MEASURE THEORY AND INTEGRATION 3. Measures and Premeasures A measure, or more generally a set function is a mapping from a σ-algebra into R = [, ]. Arithmetic on R is defined as follows. Significantly is not defined but +a = if a +a = if a a = if a > 0 Definition: A measure µ on a measurable space (,F) is a mapping of F to R + = [0, ] such that (1) µ( ) = 0 (2) µ( α I A α ) = α I µ(a α) whenever {A α F : α I} is a countable collection of sets in F which are pairwise disjoint which means that A α A β = whenever α β. A triple (,F,µ) where µ is a measure on the measurable space (, F) is a measure space or sometimes measured space to distinguish it from a measurable space. In the special case that µ() < then µ is said to be a finite measure and if µ() = 1 then µ is said to be a probability measure and (, F, µ) as a probability space. Remark: If {x α : α I} is a set of nonnegative numbers then α I x α = sup F α F x α where the supremum is taken over all finite subsets F of I and the sum could be. It is not necessary for this definition that I be countable but of course if {x α > 0 : α I} is uncountable then the sum is necessarily. (reference: General Topology by Bourbaki, Chapter IV, 4.3 and 7.1.) A somewhat more primitive concept than a measure is a premeasure Definition: Let F 0 be an algebra of subsets of a set. A premeasure µ 0 is a mapping of to R + = [0, ] such that (1) µ( ) = 0 (2) µ( α I A α ) = α I µ(a α) whenever {A α F 0 : α I} is a countable collection of sets in F 0 which are pairwise disjoint and provided α I A α is in F 0. Of course a measure is a premeasure and a premeasure is a measure if its domain is a σ-algebra. Property 2 for measures and premeasures is referred to as countable additivity. If a set function µ 0 is countably additive on an algebra F 0 then µ 0 (A) = µ 0 (A)+ for any A F 0 α Nµ( )

17 3. MEASURES AND PREMEASURES 41 Therefore if µ 0 (A) < we must have µ 0 ( ) = 0 that is, property 2 of premeasures implies property 1. Therefore property 1 could equally well be replaced by µ 0 (A) < for some A F. A measure or premeasure must be finitely additive (that is I could be taken to be finite in property 2) because, by property 2 µ 0 ( α I A α ) = µ 0 ([ α I A α ] [ n N ]) = µ 0 (A α )+ µ 0 ( ) = µ 0 (A α ) α I n N α I Anelementaryconsequenceofthisisthat, whenevera B anda,b F then µ 0 (A) µ(b), because µ 0 (B) = µ 0 (A)+µ 0 (B A) µ 0 (A) (Of course, R is ordered so that a if a R.) As a consequence, when we define a measure or premeasure µ 0 to be finite if µ 0 () <, we assure that indeed µ 0 (A) < for every A. Example 3.1. Discrete Measures: Let be any set and F = P() (the power set of ). For every x assign a nonnegative weight 0 p x and define µ(a) = x Ap x. Property 1 of measures is easily checked and property 2 is left as an exercise to the reader. In the special case that p x = 1 for every x then µ is called the counting measure on. Example 3.2. Let be any nonvoid set and F = P() be the power set and let µ( ) = 0 and µ(a) = if A. Then µ is a measure. More interesting examples will be constructed by starting with a distribution function which we now define. Definition: A real valued function F defined on R is a distribution function if (1) F is increasing which means, whenever x < y, F(x) F(y) (2) F is right continuous, which means that lim x a+ F(x) = F(a), for every real a. The archtypal example of a distribution function is F(x) = x but given any distribution function there is a corresponding Lebesgue Stieltjes premeasure which we now introduce. Example 3.3. Lebesgue Stieltjes Premeasures Consider the algebra F 0 of all finite unions of half open, half closed intervals of the form (a,b] or (,b] or (a, ) for any reals a and b. We saw in 1.1 that indeed these sets do form an algebra. If F is a distribution

18 42 1. MEASURE THEORY AND INTEGRATION function then we define µ 0 ((a,b]) = F(b) F(a) and extend µ 0 by finite additivity to F 0. Later we shall see that µ 0 is indeed a premeasure, that is we will check the countable additivity property 2. It will further be shown, in 1.4 that µ 0 has an extension defined on the Borel subsets of the real line which is a measure. This extension is a Lebesgue Stieltjes measure and its construction and properties is an objective of this Chapter. Some properties of measures and premeasures will now be recorded. Proposition 3.4. If µ and ν are measures on a measurable space (,F) and a 0 and b 0 the aµ + bν is also a measure. Similarly if µ 0 and ν 0 are premeasures on an algebra of sets F 0 then aµ 0 +bν 0 is also a measure. Proof: The proof is left as an exercise. Theorem 3.5. Suppose that µ 0 is a premeasure on an algebra of subsets F 0 of a set. Then (1) If B A and A,B F then µ 0 (B) µ 0 (A). (2) If {A α : α I} is a countable collection of sets in F 0 and if α I A α F 0 then µ 0 ( α I A α ) α I µ 0 (A α ) (3) If A 1 A 2 A 3... is an increasing sequence of sets in F 0 and if n N A α F 0 then lim µ 0(A n ) = µ 0 ( n N A n ) n N (4) If B 1 B 2 B 3... is a decreasing sequence of sets in F 0 such that n N B n is also in F 0 and if µ 0 (B k ) < for some k N then lim µ 0(B n ) = µ 0 ( n N B n ) n N Proof. Part 1 was discussed early in this section. For Part 2 we recall from 1.1 that α I A α can be written as the union of disjoints sets B n F. Indeed if α : N I is an enumeration of I and if with B n = A α(n) A c α(n 1)... Ac α(1). µ 0 ( α I A α ) = µ 0 ( n N B n ) = n N µ 0 (B n ) n Nµ 0 (A α(n) ) where the last inequality follows because B n A α(n). Of course the order of summation is irrelevant since the terms are nonnegative.

19 3. MEASURES AND PREMEASURES 43 Consider Part 3. We observe that, if µ 0 (A k ) = for some k, then µ 0 ( n N A n ) µ 0 (A k ) = and so there is nothing to check and so we cansupposethatµ 0 (A k )isfiniteforeveryk. Wehave, bytheadditivity property for measures, µ 0 ( n N A n ) = µ 0 (A 1 )+µ 0 (A 2 A 1 )+...+µ 0 (A p A p 1 )+... = lim p N µ 0 (A 1 )+µ 0 (A 2 A 1 )+...+µ 0 (A p A p 1 ) = lim p N µ 0 (A p ) We shall suppose that µ 0 (B 1 ) < ; the general case is very similar. Let B = n N B n. Then B 1 B n is an increasing sequence of sets whose union is B 1 B and so by Part 3 µ 0 (B 1 ) µ 0 (B) = µ 0 (B 1 B) = µ 0 ( n N (B 1 B n ) This implies Part 4 because µ 0 (B 1 ) <. = lim n N µ 0 (B 1 B n ) = µ 0 (B 1 ) lim n N µ 0 (B n ) Remark: In Part 4 of the preceding Theorem, the hypothesis µ 0 (B k ) < for some k cannot be dispensed with. For example if µ is the counting measure on the rational numbers Q then µ( 1/n,1/n) = for every n but µ( n N ( 1/n,1/n)) = µ({0}) = 1. The Theorem is of course true for measures in place of premeasures in which case F 0 would be replaced by a σ-algebra F say and so the α I A α and n N B n are automatically in F and so the statement simplifies slightly in this case. There is a partial converse to parts 3 and 4 of the Theorem that is helpful for checking the properties of premeasures. Proposition 3.6. Suppose that F 0 is an algebra of subsets of a set and µ 0 : F 0 R + is a mapping so that (1) µ 0 ( ) = 0 (2) µ( α I A α ) = α I µ(a α) whenever {A α F 0 : α I} is a finite collection of sets in F 0 which are pairwise disjoint. Suppose in addition that either one of the following two conditions is valid. a. Whenever A 1 A 2 A 3... is an increasing sequence of sets in F 0 such that the limit A = α I A α is also in F 0 then lim n N µ 0 (A n ) = µ 0 (A) b. Whenever B 1 B 2 B 3... is a decreasing sequence of sets in F 0 such that the limit n N B n = then lim n N µ 0 (B n ) = 0

20 44 1. MEASURE THEORY AND INTEGRATION Then, in either case, µ 0 is a premeasure. Proof. Let C 1, C 2, C 3...be a sequence of pairwise disjoint sets in F 0 and such that C = p N C p is also in F 0. Define A n = 1 p n C p. Assume Condition a. It implies lim n N µ 0 (A n ) = µ 0 (C) because A n ր C and C F 0. By finite additivity, µ 0 (A n ) = 1 p n µ 0(C p ) so that µ 0 (C p ) = lim µ 0 (C p ) = µ 0 (C) n N 1 p n p N which verifies µ 0 is countably additive and hence a premeasure under Condition a. Assume Condition b. We define B n = C A n so that B n ց and therefore Condition b implies lim n µ 0 (B n ) = 0 We have µ 0 (C) = µ 0 (A n )+µ 0 (B n ) = µ 0 (C p )+µ 0 (B n ) 1 p n Now if we take the limit in n N and we have µ 0 (C) = n µ 0(C p ) which says µ 0 is countably additive. We recall from Example 3.3 that each distribution function F definesasetfunctionµ 0 sothatµ 0 ((a,b]) = F(b) F(a)whichisafinitely additive set function on the set F 0 of all finite disjoint unions of intervalsoftheform(a,b]and(,b])and(a, ). Ofcourseµ 0 ((,b]) = lim n N = µ 0 (( n,b]) = lim n N F(b) F( n), with the value of possible. In general, if A F 0 then µ 0 (A) = lim n N µ 0 (A ( n,n]). We are now ready to show that µ 0 is a premeasure. Lemma 3.7. Let F be any distribution function and let µ 0 be the corresponding finitely additive set function defined on the algebra F 0 as in Example 3.3 (so that, in particular µ 0 ((a,b]) = F(b) F(a)). Then µ 0 is a premeasure. Proof: Consider first the finite case where F(x) is constant outside ( N,N] for some N so that µ 0 is finite. We shall use Part b of the preceding proposition and so we assume that B p F 0 forms a decreasing sequence of sets and B p ց. Given ǫ > 0, we will show that µ 0 B p ) < ǫ for all suffiiciently large p and this will imply the countable additivity of µ 0. We claim that there is another sequence C p F 0 with compact closures C p B p and such that µ 0 (B p C p ) < ǫ/2 p This is possible because for each subinterval (a,b] of B p we have µ 0 ((a,b]) = F(b) F(a) = lim a a+f(b) F(a ) = lim a a+µ((a b]) by the right continuity of F. We may also suppose that C p ( N,N] for all p. Since p N C p p N B p = it follows that, for some n, 1 p n C p =

21 3. MEASURES AND PREMEASURES 45 by a compactness argument. Consequently µ 0 (B n ) = µ 0 (B n 1 p n C p ) = µ 0 ( 1 p n B n C p ) µ 0 ( 1 p n B p C p ) 1 p n ǫ/2 p < ǫ Since (B p ) p N is a decreasing sequence and ǫ > 0 was arbitrary this says µ 0 (B p ) ց 0 as p and this verifies µ 0 is countably additive, and so a premeasure, in the case F(x) is constant outside ( N,N] for some N. Consider now the general case. For each N N let F N (x) = F(x) if x N and F N (x) = F( N) if x N and F N (x) = F(N) if x N. Then F N is a distribution function and the corresponding finitely additive set function µ N is in fact a premeasure by the first part of the proof and µ N (A) = µ 0 (A) if A F 0 and A ( N,N]. Moreover, for any A F 0, lim N N µ N (A) = µ(a) by the discussion preceding the statement of this result. Let (A p ) p N be a sequence of pairwise disjoint sets such that A = p N A p is in F 0. Then, simply because µ 0 is finitely additive we have, for any n N µ 0 (A) µ 0 ( 1 p n A p ) = 1 p nµ 0 (A p ) so that µ 0 (A) µ 0 (A p ) p N Conversely, by the special case µ 0 (A) = lim N N µ N (A) = lim N N p N µ N (A p ) lim 0 (A p ) = N N p Nµ µ 0 (A p ) p N because µ N (B) µ 0 (B) for any B F 0. Combining these two inequalities we have µ 0 (A) = p N µ 0(A p ) and this verifies the µ 0 is a premeasure and completes the proof. Thus every distribution function corresponds to a premeasure (and indeed, we shall see, to a measure). The following result provides a partial converse. Definition: We shall say that µ is a Lebesgue-Stieltjes measure if µ is a measure on the Borel subsets B of R and µ(k) < if K is a compact subset of R Proposition 3.8. Suppose that µ is a Lebesgue-Stieltjes measure and C is any real constant. Define F(x) = µ((0,x])+c if x 0 and F(x) = C µ((x,0]) if x < 0. (Here (0,0] = by convention.) Then F is a distribution function and µ((a,b]) = F(b) F(a) whenever a < b.

22 46 1. MEASURE THEORY AND INTEGRATION Therefore every Lebesgue-Stieltjes measure corresponds to a distribution function which is unique up to an additive constant. Proof: Since µ(a) µ(b) whenever A B are sets in B, µ((0,x]) is increasing x 0 and µ((x,0]) is decreasing for x 0 and so F is increasing on R. Suppose a 0. Then lim F(x) = C +µ( x>a(0,x]) = C +µ((0,a]) = F(a) x a+ so that F is right continuous on [0, ). Next suppose a < 0. Then lim F(x) = C µ( x>a(x,0]) = C µ((a,0]) = F(a) x a+ so that in fact F is right continuous everwhere and is therefore a distribution function. Next check that F(b) F(a) = µ((a,b]) whenever a < b. If 0 a then thisis obvious and similarly if b < 0. Supposetherefore that a < 0 andb 0. ThenF(b) F(a) = C+µ((0,b]) (C µ((a,0])) = µ((a,b]). The uniqueness, up to additive constants, of F with the property F(b) F(a) = µ((a,b]) follows because if G is another distribution function and if G(0) = F(0) then G(b) G(0) = µ((a,b]) = F(b) F(0) implies G(b) = F(b) for b 0. Similarly G(0) G(a) = µ((a,0]) = F(0) F(a) so that G(a) = F(a) for a < 0. Thus G = F. In general G = F F(0)+G(0) so that the choice of G is completely determined by the choice of G(0).

23 4. EXTENSIONS AND MEASURES: Extensions and Measures: In this Section we show that certain premeasures extend to measures and that a measure may be completed. As an application we are able to construct Lebesgue Stieltjes measures. Recall the definition: Definition: A real valued function F defined on R is a distribution function if (1) F is increasing which means, whenever x < y, F(x) F(y) (2) F is right continuous, which means that lim x a+ F(x) = F(a), for every real a. We begin by considering a premeasure µ 0 defined on an algebra F 0. Suppose that (A p ) p N is an increasing sequence in F 0 and A p ր A so that A 1 A 2 A 3... and A = p N A n but A may not be itself in F 0. We would like to extend µ 0 to µ 0 as µ 0 (A) = lim p N µ 0 (A p ). The limit exists in R + but we should verify that it does not depend on the particular sequence and that is the object of the Lemma below. Lemma 4.1. Suppose that µ 0 is a premeasure defined on an algebra F 0 and (A p ) p N and (B p ) p N are two increasing sequences of sets in F 0 with limits, A = p N A p and B = p N B p. If A B then lim p N µ(a p ) lim p N µ(b p ) Proof: Because µ 0 is a premeasure we have for any q N µ 0 (A q ) = lim p N µ 0 (A q B p ) lim p N µ 0 (B p ) Taking the limit in q gives the result. Define therefore G to consist of all A such that there exists an increasing sequence (A p ) p N F 0, so that A p ր A and define also a set function µ 0 on G by µ 0 (A) = lim p N µ 0 (A p ). TheprecedingLemmaassuresthatµ 0 iswelldefinedbecauseitdoesnot depend on the particular sequence (A p ) p N F 0 which approximates A. We see further that µ 0 extends µ 0 because any A F 0 can be written as the limit of a constant sequence. A n = A. Lemma 4.2. Suppose that µ 0 is a finite premeasure on an algebra F 0 and µ 0 is its extension to G as described above. Then G is closed under finite intersections and countable unions and a. If G 1, G 2 are in G then µ 0 (G 1 G 2 )+µ 0 (G 1 G 2 ) = µ 0 (G 1 )+µ 0 (G 2 )

24 48 1. MEASURE THEORY AND INTEGRATION b. If (G p ) p N is an increasing sequence in G then G = p N G n is in G and lim p N µ 0 (G n ) = µ 0 (G) c. If G p is a sequence in G then µ 0 ( p N G p ) p Nµ 0 (G p ) Proof: Suppose that A p, B p are increasing sequences of sets in F 0 and A p ր G 1 and B p ր G 2 so that G 1,G 2 G. It follows that A p B p ր G 1 G 2 and A p B p ր G 1 G 2 so that G is closed under finite unions and intersections. Moreover since µ 0 is a premeasure we can take the limit in p N in the relation µ 0 (A p B p )+µ 0 (A p B p ) = µ 0 (A p )+µ 0 (B p ) to verify Part a. To verify Part b, suppose that G p is an increasing sequence in G, G p ր G. For each p, suppose A pq is an increasing sequence in F 0 convergent to G p = q N A pq. A 11 A A 1q... ր G 1 A 21 A A 2q... ր G A p1 A p2... A pq... ր G p.. Define B q = 1 p q A pq. Then B q is an increasing sequence in F 0 and it is convergent to G so that G G and so µ 0 (G) = lim q N µ 0 (B q ). Certainly µ 0 (G) µ 0 (G p ) for any p by the preceding Lemma. On the other hand µ 0 (G p ) µ 0 (B p ) so that lim p N µ 0 (G p ) lim p N µ 0 (B p ) = µ 0 (G). This establishes Part b. To verify Part c, it suffices to check that, for any n N µ 0 ( 1 p n G p ) µ 0 (G p ). 1 p n because we may take the limit in n N and applying Part b above. In the case n = 2 we have µ 0 (G 1 G 2 ) µ 0 (G 1 )+µ 0 (G 2 ) by Part a. The case of general n follows by induction. The proof is complete.. Thus extension µ 0 is a countably additive set function and G F 0 is closed under countable unions, but A G does not imply A c G. Also unions of sets in G may not be expressible as unions of disjoint sets in G. Definition: A mapping µ on the power set P() of a set to R + is said to be an outer measure if (1) µ ( ) = 0..

25 4. EXTENSIONS AND MEASURES: 49 (2) If A B then µ (A) µ (B). (3) If A α I is a countable collection of sets in then µ ( α I A α ) α I µ (A α ) Frequently an outer measure will not be a measure because it is not additive but it is possible that a restriction to a σ-algebra of subsets of may indeed be a measure. We now suppose that µ 0 is a finite premeasure on an algebra F 0 and that µ 0 is the extension of µ 0 to the G of the preceding Lemma. Later in this Section we weaken the assumption of finite to σ-finite to be defined below. Define (4.1) µ (A) = inf{µ 0 (G) : G A,G G} for any A. Alternatively, in terms of µ 0 itself, we define { } λ(a) = inf µ 0 (A p ) : A p F 0, p N A p A p N We claim µ (A) = λ(a). For suppose that A p, p N is a sequence in F 0 and p N A p A, Define G = p N A p so that G G and µ 0 (G) = lim n µ 0 ( 1 p n A p ) p N µ 0(A p ). Consequently µ (A) λ(a). Conversely suppose G G so that there is a sequence A p F 0 so that A p ր G and we can form a sequence B p F 0 which is pairwise disjoint and p N B p = G so that µ 0 (G) = p N µ 0(B p ). This shows that µ (A) λ(a) and complete the verification that λ = µ. Now let us check that µ is indeed an outer measure and record some of its properties. Lemma 4.3. Suppose that µ 0 is a finite premeasure on an algebra F 0 of subsets of a set and that µ is defined as indicated above; (see (4.1)). Then µ is an outer measure and agrees with µ 0 on F 0 and µ 0 on G. Moreover (1) µ (A B)+µ (A B) µ (A)+µ (B) for any sets A,B. In particular µ (A)+µ (A c ) µ 0 (). (2) If A p, p N is an increasing sequence of sets in then lim p N µ (A p ) = µ ( p N A p ). Proof: It is clear that µ (G) = µ 0 (G) if G G, by the definition (4.1)ofµ andbecauseµ 0 (G) µ 0 (G )ifg G andg G. Similarly one sees that, if A B then µ (A) µ (B), by the definition (4.1) of µ.

26 50 1. MEASURE THEORY AND INTEGRATION Therefore, to complete the check that µ is an outer measure, we suppose that A p, p N is a sequence of subsets. of. We choose G p G, G p A p so that µ 0 (G p ) µ (A p )+2 p ǫ for each p N. Then µ ( p N A p ) µ 0 ( p N G p ) p N µ 0 (G p ) p N µ (A p )+ǫ/2 p where we have applied Part c of the preceding Lemma. Therefore µ is an outer measure. Check next Part 1. We suppose that A,B and ǫ > 0 is given. Choose G 1,G 2 G A G 1 and B G 2 and µ 0 (G 1 ) < µ (A)+ǫ and µ 0 (G 2 ) < µ (B)+ǫ. Then µ (A)+µ (B) > µ 0 (G 1 )+µ 0 (G 2 ) 2ǫ = µ 0 (G 1 G 2 )+µ 0 (G 1 G 2 ) 2ǫ by Part a of the preceding Lemma. Since G 1 G 2 A B and G 1 G 2 A B and G 1 G 2 G and G 1 G 2 G, we have µ (A)+µ (B) > µ 0 (G 1 )+µ 0 (G 2 ) 2ǫ µ (A B)+µ (A B) 2ǫ and, since ǫ > 0 was arbitrary this implies Part 1. It remains to check Part 2. Certainly µ (A n ) µ ( p N A p ) for any n N so that lim p N µ (A p ) µ ( p N A p ). To check the converse, suppose that ǫ > 0 is arbitrary, and that, for each p N, G p G is chosen so that G p A p and µ 0 (G p ) µ (A p ) + 2 p ǫ. We shall show that there is a increasing sequence H p G so that H p A p µ 0 (H p ) µ (A p ) + 1 q p 2 q ǫ. The existence of such a sequence would imply that µ ( p N A p ) µ 0 ( p N H p ) = limµ 0 (H p ) limµ (A p )+ 2 q ǫ. p N p N 1 q p so that µ ( p N A p ) lim p N µ (A p ) + ǫ and since ǫ is arbitrary this would complete the proof. To construct the sequence H p we proceed by induction on n supposing H 1 H 2... H n have been constructed in G and H p A p and µ 0 (H p ) µ (A p )+ 1 q p 2 q ǫ for 1 p n. Define H n+1 = H n G n+1 so that H n+1 G and µ 0 (H n+1 ) = µ 0 (H n G n+1 ) = µ 0 (H n )+µ 0 (G n+1 ) µ 0 (H n G n+1 ) µ (A n )+ 2 p ǫ+µ (A n+1 )+ǫ/2 n+1 µ (A n ) 1 p n since A n H n G n+1. Thus µ 0 (H n+1 ) µ (A n+1 ) + 1 p n+1 2 p ǫ and that completes the construction and therefore the proof. Thus every finite premeasure µ 0 extends to an outer measure µ. One of the less desirable features of µ is that there may exist sets A