Measure Theory, 2009

Transcription

1 Measure Theory, 2009 Contents 1 Notational issues 2 2 The concept of a measure 4 3 The general notion of integration and measurable function 15 4 Convergence theorems 23 5 Radon-Nikodym and Conditional Expectation 28 6 Standard Borel spaces 33 7 Borel and measurable sets and functions 36 8 Summary of some important terminology 41 9 Review of Banach space theory The Riesz representation theorem Stone-Weierstrass Product Measures Measure disintegration Ergodic theory Examples and the notion of recurrence The ergodic theorem and Hilbert space techniques Mixing properties The ergodic decomposition theorem

2 1 Notational issues As with much of modern mathematics, we will be using the language of set theory and the notation of logic. Let be a set. We say that Y is a subset of if every element of Y is an element of. I will variously denote this by Y, Y, Y, and Y. In the case that I want to indicate that Y is a proper subset of that is to say, Y is a subset of but it is not equal to I will use Y. We write x or x to indicate that x is an element of the set. We write x / to indicate that x is not an element of. For P( ) some property we use {x : P(x)} or {x P(x)} to indicate the set of elements x for which P(x) holds. Thus for x, we have x {x : P(x)} if and only if P(x). There are certain operations on sets. Given A, B we let A B be the intersection and thus x A B if and only if x is an element of A and x is an element of B. We use A B for the union so x A B if and only if x is a member of one of the two. We use A \ B to indicate the elements of A which are not elements of B. In the case that it is understood by context that all the sets we are currently considering are subsets of some fixed set, we use A c to indicate the complement of A in in other words, \ A. A B denotes the symmetric difference of A and B, the set of points which are in one set but not the other. Thus A B = (A \ B) (B \ A). We use A B to indicate the set of all pairs (a, b) with a A, b B. P(), the power set of, indicates the set of all subsets of thus Y P() if and only if Y. A very, very special set is the empty set:. It is the set which has no members. If you like, it is the characteristically zen set. Some other special sets are: N = {1, 2, 3,...}, the set of natural numbers; Z = {..., 2, 1, 0, 1, 2,...}, the set of integers; Q, the set of rational numbers; R the set of real numbers. Given some collection {Y α : α Λ} of sets, we write α Λ or {Yα : α Λ} Y α to indicate the union of the Y α s. Thus x α Λ Y α if and only if there is some α Λ with x Y α. A slight variation is when we have some property P( ) which could apply to the elements of Λ and we write {Yα : α Λ, P(α)} or α Λ,P(α) to indicate the union over all the Y α s for which P(α) holds. The obvious variations hold on this for intersections. Thus we use α Λ or {Yα : α Λ} Y α Y α to indicate the set of x which are members every Y α. Given a infinite list (Y α ) α Λ of sets, we use α Λ to indicate the infinite product which formerly may be thought of as the collection of all functions f : Λ α Λ Y α with f(α) Y α at every α. Given two sets, Y and a function Y α f : Y 2

3 between the sets, the are various set theoretical operations involved with the function f. Given A, f A indicates the function from A to Y which arises from the restriction of f to the smaller domain. Given B Y we use f 1 [B] to indicate the pullback of B along f in other words, {x : f(x) B}. Given A we use f[a] to indicate the image of B the set of y Y for which there exists some x A with f(x) = y. Given A we use χ A to denote the characteristic function or indicator function of A. This is the function from to R which assumes the value 1 at each element of A and the value 0 on each element of A c. A function f : Y is said to be injective or one-to-one if whenever x 1, x 2 with x 1 x 2 we have f(x 1 ) f(x 2 ) different elements of move to different elements of Y. The function is said to be surjective or onto if every element of Y is the image of some point under f in other words, for any y Y we can find some x with f(x) = y. The function f : Y is said to be a bijection or a one-to-one correspondence if it is both an injection and a surjection. In this case of a bijection, we can define f 1 : Y by the requirement that f 1 (y) = x if and only if f(x) = y. A set is said to be countable if it is either finite or it can be placed in a bijection with N, the set of natural numbers. 1 We will not be discussing the cardinality theory of infinite sets in this course, but there are a few basic facts you need to know: Z is countable; Q is countable; the product of finitely many countable sets, such as Q Q or Z Z Q, is again countable; the countable union of countable sets is countable; a subset of a countable set is countable. Not every set is countable however: R is uncountable; P(N) is uncountable. A set is said to have cardinality ℵ 0 if it can be placed in a bijection with N. 1 But be warned: A small minority of authors only use countable to indicate a set which cane be placed in a bijection with N 3

4 2 The concept of a measure Definition For S a set we let P(S) be the set of all subsets of S. Σ P(S) is an algebra if it is closed under complements, unions, and intersections; it is a σ-algebra if it is closed under complements, countable unions, and countable intersections. Here by a countable union we mean one of the form n N A n and by countable intersection one of the form n N A n. Definition A set S equipped with a σ-algebra Σ is said to be a measure space if Σ P(S) is a (non-empty) σ-algebra. A function µ : Σ R 0 { } is a measure if 1. µ( ) = 0, and 2. µ is countably additive given a sequence (A n ) n N of disjoint sets in Σ we have µ( A n ) = n ). n N n Nµ(A Here R 0 refers to the non-negative reals. We require at least at this stage that our measures return non-negative numbers, with the inclusion of positive infinity. In this definition, some sets may have infinite measure. Exercise (a) Show that if Σ is a non-empty σ-algebra on S then S and (the empty set) are both in Σ. (b) Show that if µ : Σ R 0 { } is countably additive and finite (that is to say, µ(a) R all A Σ), then µ( ) = 0. The very first issue which confronts us is the existence of measures. The definition is in the paragraph above bold and confident but without the slightest theoretical justification that there are any non-trivial examples. Simply taking enough classes in calculus one might develop the intuition that Lebesgue measure has the required property of σ-additivity. We will prove a sequence of entirely abstract lemmas which will show that Lebesgue measure is indeed a measure in our sense. The approach in this section will be through the Carathéodory extension theorem. Much later in the notes we will see a different approach to the existence of measures in terms of the Riesz representation theorem for continuous functions on compact spaces. Although the Riesz representation theorem could be used to show the existence of Lebesgue measure, the approach is far more abstract, appealing to various ideas from Banach space theory, and the actual verifications involved in the proof take far longer. The proof in this section may already seem rather abstract and in some sense, it is. Still it is an easier first pass at the notions than the path through Riesz. Definition Let S be a set. A function is an outer measure if λ( ) = 0 and whenever λ : P(S) R 0 { } A n N A n then λ(a) n N λ(a n ). 4

5 Lemma 2.1 Let S be a set and suppose K P(S) is such that for every A S there is a countable sequence (A n ) n N with: 1. each A n K; 2. A n N A n. Let ρ : K R 0 { } with ρ( ) = 0. Then if we define by letting λ(a) equal the infinum of the set λ : P(S) R 0 { } { ρ(a n ) : A A n ; each A n K}, n N n N then λ is an outer measure. Proof This is largely an unravelling of the definitions. The issue is to check that if λ(b n ) = a n and B n N B n, then λ(b) n N λ(b n). However if we fix ǫ > 0 and if at each n we fix a covering (B n,m ) m N with B n m NB n,m, then n,m N B n,m B and ρ(b n,m ) < a n + ǫ2 m n 1, m ρb n,m < ǫ + ρ(a n ). n,m N n Definition Given an outer measure λ : P(S) R 0 { }, we say that A S is λ-measurable if for any B S we have λ(b) = λ(b A) + λ(b \ A). Here B \A refers to the elements of B not in A. If we adopt the convention that A c is the relative complement of A in S the elements of S not in A then we could as well write this as B A c. Theorem 2.2 (Carathéodory extension theorem, part I) Let λ be an outer measure on S and let Σ be the collection of all λ-measurable sets. Then Σ is a σ-algebra and λ is measure on Σ. Proof The closure of Σ under complements should be clear from the definitions. Before doing closure under countable unions and intersections, let us first do finite intersections. For that purpose, it suffices to do intersections of size two, since any finite intersection can be obtained by repeating the operation of a single intersection finitely many times. Suppose A 1, A 2 Σ and B S. Applying our assumptions on A 1 we obtain λ(b (A 1 A 2 ) c ) = λ((b A c 1 ) (B Ac 2 )) = λ((b A c 1) (B A c 2)) A c 1) + λ((b A c 1) (B A c 2)) A 1 ) = λ(b A c 1) + λ(b A 1 A c 2). Then applying the assumptions on A 1 once more we obtain λ(b) = λ(b A 1 ) + λ(b A c 1), 5

6 and then applying assumptions on A 2, this equals λ(b A 1 A 2 ) + λ(b A 1 A c 2 ) + λ(b Ac 1 ), which after using λ(b (A 1 A 2 ) c ) = λ(b A c 1) + λ(b A 1 A c 2) from above gives λ(b) = λ(b A 1 A 2 ) + λ(b (A 1 A 2 ) c ), as required. Having Σ closed under complements and finite intersections we obtain at once finite unions. Note moreover that our definitions immediately give that λ is finitely additive on Σ, since given A, B Σ disjoint, λ(a B) = λ((a B) A) + λ((a B) A c ), which by disjointness unravels as λ(a) + λ(b). It remains to show closure under countable unions and countable intersections. However, given the previous work, this now reduces to showing closure under countable unions of disjoint sets in Σ. Let (A n ) n N be a sequence of disjoint sets in Σ. Let A = n N A n. Fix B S. Note that since λ is monotone (in the sense, C C λ(c) λ(c ))) we have for any N N λ(b A) λ(b Then any easy induction on N using the disjointness of the sets gives λ(b n N A n) = n N λ(b A n). (For the inductive step: Use that since A N Σ we have λ(b n N A n) = λ((b n N A n) A N ) + λ((b n N A n) A c N ) = λ(b A N) + λ(b n N 1 A n).) On the other hand, the assumption that λ is an outer measure give the inequality in the other way, and hence n N A n ). λ(b A) = n N λ(b A n ). Finally, putting this altogether with the task at hand we have at every N λ(b) = λ(b A n )+λ(b ( A n ) c ) λ(b A n ) c ), n N n N n N A n )+λ(b ( n N A n ) c ) = ( n N λ(b A n ))+λ(b ( n N and thus taking the limit λ(b) ( λ(b A n )) + λ(b ( A n ) c ) = λ(b A) + λ(b A c ); n N n N by λ being an outer measure we get the inequality in the other direction and are done. The argument from the last paragraph also shows that λ is countably additive on Σ, since in the equation λ(b A) = n N λ(b A n ) we could as well have taken B = S. This is good news, but doesn t yet solve the riddle of the existence of non-trivial measures: For all we know at this stage, Σ might typically end up as the two element σ-algebra {, S}. 6

7 Theorem 2.3 (Carathéodory extension theorem, part II) Let S be a set and let Σ 0 P(S) be an algebra and let µ 0 : Σ 0 R 0 { } have µ 0 ( ) = 0 and be σ-additive on its domain. (That is to say, if (A n ) n N is a sequence of disjoint sets in Σ 0 and and if n N A n Σ 0, then µ 0 ( n N A n) = n N µ 0(A n ).) Then µ 0 extends to a measure on the σ-algebra generated by Σ 0. Proof Here the σ-algebra generated by Σ 0 means the smallest σ-algebra containing Σ 0 it can be formally defined as the intersection of all σ-algebras containing Σ 0. First of all we can apply the last theorem to obtain an outer measure λ, with λ(a) being set equal to the infinum of all n N µ 0(A n ) with each A n Σ 0 and A n N A n. The task which confronts us is to show that the σ-algebra indicated in 2.2 extends Σ 0 and that the measure λ extends the function µ 0. These are consequences of the next two claims. Claim: For A Σ 0 and B S λ(b) = λ(b A) + λ(b A c ). Proof of Claim: Clearly λ(b) λ(b A) + λ(b A c ). For the converse direction, suppose (A n ) n N is a sequence of sets in Σ 0 with B A n N. Then at each n by the additivity properties of µ 0. Thus λ(b A) + λ(b A c ) n N µ 0 (A n ) = µ 0 (A n A) + µ 0 (A n A c ) µ 0 (A A n ) + n N µ 0 (A c A n ) = n N µ 0 (A n ). (Claim ) Claim: µ 0 (A) = λ(a) for any A Σ 0. Proof of Claim: Suppose (A n ) n N is a sequence of sets in Σ 0 with A n N A n. We need to show that µ 0 (A) µ 0 (A n ). n N After replacing each A n by A n \ i<n A i we may assume the sets are disjoint. But consider B n = A n A. µ 0 (A) = n N µ 0(B n ) by the σ-additivity assumption on µ 0. Since σ-additivity implies finite additivity and hence that µ 0 is monotone, at each n, µ 0 (B n ) µ 0 (A n ). Thus µ 0 (A) = n N µ 0 (B n ) n Nµ 0 (A n ). (Claim ) Now we are in a position to define Lebesgue measure rigorously. Definition B R N is Borel if it appears in the smallest σ-algebra containing the open sets. Of course one issue is why this is even well defined. Why should there be a unique smallest such algebra? The answer is that we can take the intersection of all σ-algebras containing the open sets and this is easily seen to itself be a σ-algebra. 7

8 Theorem 2.4 There is a σ-additive measure m on the Borel subsets of R with for all a < b in R. m({x R : a < x b}) = b a Proof Let us call a A R a fingernail set if it has the form A = (a, b] = df {x R : a < x b} for some a < b in the extended real number line, { } R {+ }. Let us take Σ 0 to be the collection of all finite unions of finger nail sets. It can be routinely checked that this is an algebra and every non-empty element of Σ 0 can be uniquely written in the form (a 1, b 1 ] (a 2, b 2 ]... (a n, b n ], for some n N and a 1 < b 1 < a 2 <... < b n in the extended real number line. For A = (a 1, b 1 ] (a 2, b 2 ]... (a n, b n ] we define m 0 (A) = (b 1 a 1 ) + (b 2 a 2 ) +...(b n a n ). Claim: If (a, b] n N(a n, b n ] then m 0 ((a, b]) n N m 0((a n, b n ]). Proof of Claim: This amounts to show that if (a, b] n N (a n, b n ] then b n a n b a. n N Suppose instead n N b n a n < b a. Choose ǫ > 0 such that ǫ + n Nb n a n < b a. Let c n = b n + 2 n 1 ǫ. Then (a n, c n ) [a + ǫ/2, b]. n N Applying Heine-Borel 2 we can find some N such that (a n, c n ) [a + ǫ/2, b]. n N 2 The Heine-Borel theorem is one of the most central results of real analysis. It states that a closed bound interval in R is compact: In other words, whenever {U i : i Λ} is a covering by open sets, there will be some finite subcover. Here a covering of a set A is a collection whose union includes A. A subcover is then a subset of the collection whose union again includes A. Thus Heine-Borel states in particular that if d < e are both in R and we have d n < e n at each n N with [d, e] [ n N(d n, e n), then there is some finite N with More details can be found [9]. [d, e] [ (d n, e n). n N 8

9 The next subclaim states that after possibly changing the enumeration of the sequence (a n, c n ) n N we may assume that the intervals are consecutively arranged with overlapping end points. Subclaim: we may assume without loss of generality that at each i < N we have a i+1 c i. Proof of Subclaim: First of all, since the index set is now finite, we may assume that no proper subset Z {1, 2,...N} has [a + ǫ 2, b] n Z (a n, c n ). After possibly reordering the sequence we can assume a 1 a j all j N. Then we have a 1 < a+ ǫ 2 and since 1<n N (a n, c n ) does not include [a + ǫ 2, b] we have c 1 > a + ǫ 2. Since c 1 n N (a n, c n ) (there is another possibility, which is c 1 > b, but then N = 1 and the claim is trivialized) we have some j with a j < c 1 < c j. Without loss of generality j = 2, and then we continue so on. ( Subclaim) Then b i a i c i a i c N a N + a i+1 a i. i N i N i<n This is one of those telescoping sums where the middle terms all cancels out and we are left with c i a i c N a 1, i N which turn must equal at least b a ǫ/2, which contradicts our initial assumption of then b i a i < b a ǫ 2. i N (Claim )Claim: If {(a n, b n ] : n N} is a disjoint sequence of fingernail sets with (a n, b n ] (a, b], n N n Nm 0 ((a n, b n ]) = n N b n a n b a. Proof of Claim: It suffices to show that at each N N we have n N b n a n b a. After reordering we can assume that at any i j N we have a i a j. Then disjointness of the sequence gives a j+1 b j at each j N. Note also that our assumptions imply that each a i a and b i b. Then it all unravels with b n a n b N a N + a n+1 a n n<n n N = b N a 1 b a. (Claim )Claim: m 0 is σ-additive on Σ 0. Proof of Claim: Let A = (a 1, b 1 ] (a 2, b 2 ]...(a N, b N ] be in Σ 0. Suppose {(c n, d n ] : n N} are disjoint fingernail sets with A = n N(c n, d n ]. 9

10 At each i N and n N let B n,i = (c n, d n ] (a i, b i ]. The intersection of two fingernail sets is again a fingernail set, and thus each B n,i can be written in the form B n,i = (c n,i, d n,i ]. The last two claims give that and that d n,i c n,i = d n c n i N b i a i = n N d n,i c n,i. Putting this altogether we have d n c n m 0 ((c n, d n ]) = n N n N = b i a i = m 0 (A). n N,i N d n,i c n,i = i N (Claim )Thus by 2.3 m 0 extends to a measure m which will be defined on the σ-algebra generated by Σ 0, which is the collection of Borel subsets of R. We used the fingernail sets (a, b] because they easily generated a finite algebra. There would be no difference using different kinds of intervals. Exercise Let m be the measure from 2.4. (i) Show that for any x R, m({x}) = 0. (ii) Conclude that m([a, b]) = m((a, b)) = m([a, b)) = b a. (iii) Show that if A is a countable subset of R, then m(a) = 0. A couple of remarks about the proof of 2.4. First of all, we have been rather stingy in our statement. The m from theorem is only defined on the Borel sets, but the proof of 2.2 and 2.3 gives that it is defined on a σ-algebra at least equal to the Borel sets; in fact it is a lot more, though for certain historical and conceptual reasons I am stating 2.4 simply for the Borel sets. Another remark about the proof is that we have only shown the theorem for one dimension, but it certainly makes sense to consider the case for higher dimensional euclidean space. Indeed one can prove that at every N there is a measure m N on the Borel subsets of R N such that for any rectangle of the form A = (a 1, b 1 ] (a 2, b 2 ]...(a N, b N ] we have m N (A) = (b 1 a 1 ) (b 2 a 2 )...(b N a N ). Theorem 2.5 Let N N and Σ P(R N ) the σ-algebra of Borel subsets of N-dimensional Euclidean space. Then there is a measure m N : Σ R such that whenever A = I 1 I 2... I N is a rectangle, each I n an interval of the form (a n, b n ), [a n, b n ), (a n, b n ], or [a n, b n ] we have m N (A) = (b 1 a 1 ) (b 2 a 2 )...(b N a N ). 10

11 A proof of this is given in [6]. In any case, the existence of such measures in higher dimension will follow from the one dimensional case and the section on product measures and Fubini s theorem later in the notes. Finally, nothing has been said about uniqueness. One might in principle be concerned whether the measure m on the Borel sets in 2.4 has been properly defined or whether there might be many such measures with the indicated properties. Although I did not pause to explicitly draw out this point, the measure indicated is unique for the reason that the measure of 2.3 is unique under modest assumptions. It is straightforward and I will leave it as an exercise. Definition A measure µ on a measure space (, Σ) is σ-finite if can be written as a countable union of sets in Σ on which µ is finite. Exercise Show that any measure m on R is σ-finite satisfying the conclusion of 2.4 is σ-finite. Exercise (i) Let Σ 0 be an algebra, µ 0 : Σ 0 R 0 { } σ-additive on its domain. Suppose S can be written as a countable union of sets in Σ 0 each of which has finite value under Σ 0. Let Σ Σ 0 be the σ-algebra generated by Σ 0 and let µ : Σ R 0 { } be a measure. Then for every A Σ we have µ(a) = inf{ n N µ 0 (A n ) : A n N A n ; each A n Σ 0 }. (Hint: Write S = n N S n, each S n Σ 0, each µ 0 (S n ) <. It suffices to consider the case that A S n for some n. By comparing A with its complement A c and applying additivity, it suffices to show µ(a) inf{ n N µ 0(A n ) : A n N A n; each A n Σ 0 } and µ(a c ) inf{ n N µ 0(A n ) : A c n N A n; each A n Σ 0 }.) (ii) Conclude that there is a unique measure satisfying 2.5. Definition The unique measure described by the above theorem 2.5 is called the Lebesgue measure on R N. Lemma 2.6 Let A R be Borel. Let m be Lebesgue measure on R. Suppose m(a) <. Let ǫ > 0. (i) Then there exists an open set O R such that and (ii) And there exists a closed set C R such that A O m(o) < m(a) + ǫ. C A and m(a) < m(c) + ǫ. Proof (i) This is a consequence of our proof of 2.4. Implicitly we appealed to the existence of an outer measure via 2.3. This means here that for any A in the σ-algebra on which m is defined and for any ǫ > 0 we have some sequence of fingernail sets, (a 1, b 1 ], (a 2, b 2 ],.. with A n N(a n, b n ] and b n a n < m(a) + ǫ. n N 11

12 Choose δ > 0 with n N b n a n < m(a) + ǫ δ. Let c n = b n + δ2 n. We finish with O = n N(a n, c n ). (ii) It suffices to consider the case that A [a, b] for some a < b. Appealing to part (i), let O [a, b] A c be open with m(o) < m([a, b] A c ) + ǫ. Let C = [a, b] \ O. Formally I have just set up the Lebesgue measure on the Borel subsets of R, but the proofs of 2.2 and 2.3 suggest that perhaps it can be sensibly defined on a rather larger σ-algebra. Definition Let m be the outer measure used to define Lebesgue measure in that equals the infinum of m (A) { (b i a i ) : ((a i, b i ]) i N is a sequence of fingernail sets with A i, b i ]}. n N i N(a A subset B R is Lebesgue measurable if for every A R we have m (A) = m (A B) + m (A B c ). From the 2.2 and 2.3 we obtain that the Lebesgue measurable sets in R form a σ-algebra and m extends to a measure on that σ-algebra. In a similar vein: Exercise Show that if B R is Lebesgue measurable then there are Borel A 1, A 2 R with A 1 B A 2 and m(a 2 \ A 1 ) = 0. (Hint: This follows from the proof of 2.6.) One may initially wonder whether the Lebesgue measurable sets are larger than the Borel. The short answer is that not only are there more, there are vastly more. Take a version of the Cantor set with measure zero. For instance, the standard construction where we remove the interval (1/4, 3/4) from [0, 1], then (1/16, 3/16) from [0, 1/4] and (13/16, 15/16) from [3/4, 1], and continute, iteratively removing the middle halves. The final result will be a closed, nowhere dense set. A routine compactness argument shows that it is non-empty and has no isolated points. With a little bit more work we can show it is actually homeomorphic to N {0, 1}, and hence has size 2 ℵ0. Its Lebesgue measure is zero, since the set remaining after n many steps has measure 2 n. Any subset of a Lebesgue measurable set of measure zero is again Lebesgue measurable, thus all its subsets are Lebesgue measurable. Since it has 2 (2ℵ0) many subsets, we obtain 2 (2ℵ0) Lebesgue measurable sets. On the other hand it can be shown (see for instance [5]) that there are only 2 ℵ0 many Borel sets and thus not every Lebesgue measurable set is Borel. Fine. But then of course it is natural to be curious about the other extreme. In fact, there exist subsets of R which are not Lebesgue measurable. Lemma 2.7 There exists a subset V [0, 1] which is not Lebesgue measurable. Proof For each x [0, 1] let Q x be Q + x [0, 1] that is to say, the set of y [0, 1] such that x y Q. Note that Q x = Q z if and only if x z Q. Now let V [0, 1] be a set which intersects each Q x exactly once. Thus for each x [0, 1] there will be exactly one z V with x z Q. I claim V is not Lebesgue measurable. For a contradiction assume it is. Note then that for any q Q the translation V + q = {z + q : z V } has the same Lebesgue as V. This is simply because the entire definition of Lebesgue measure was translation invariant. The first case is that V is null. But then q Q V +q is a countable union of null sets covering [0, 1], with a contradiction to m([0, 1]) = 1. Alternatively, if m(v ) = ǫ > 0, then q Q,0 q 1 V + q is an infinite union of disjoint sets all of measure ǫ, all included in [ 1, 2], with a contradiction to m([ 1, 2]) = 3 <. 12

13 There is something curious about the construction of the set V above. It is created by an appeal to the axiom of choice we choose exactly one point from each set of the form Q x, but the final set V is not itself presented with any easy description. One might then ask if there is an explicit or concrete example of a set which is not Lebesgue measurable, or alternatively whether one can prove the existence of sets which are not Lebesgue measurable without appealing to the axiom of choice. It takes some care to make these questions mathematically precise, but the answer to both ultimately is in the negative see [10]. Bear in mind that there are other kinds of spaces and objects to which we might wish to assign something like a measure. Examples (i) For a finite set and P() the set of all subsets of, we could take the counting measure: µ(a) = A, the size of A. (ii) For {H, T } N, which could be thought of as tossing a coin with outcome either H (heads) or T (tails), we could take the normalized counting measure: µ(a) = A 2 N. (iii) A natural variation on (iii) is to take the infinite product of the coin tossing measure. Let {H, T } N be the collection of all functions f : N {H, T }. For i 1, i 2,..., i N distinct elements of N and S 1, S 2,..., S N {H, T } let µ({f {H, T } : f(i 1 ) = S 1, f(i 2 ) = S 2,..., f(i N ) = S N } = 2 N. N 2.3 shows that µ extends to a measure on the subsets of N {H, T } which are Borel with respect to the product topology. (iv) Another way in which (ii) can be altered is to adjust the measure on {H, T }. We might instead be working with a slightly biased coin, that comes down heads 7 times out of ten. Then for each f : {1, 2,..., N} {H, T } we set µ({f}) = ( 7 10 ) {i:f(i)=h} ( 3 10 ) {i:f(i)=t }. Similarly we could define this lopsided measure on the space of infinite runs. (v) Somewhat more loosely, imagine we are dealing with some experiment, such as tickling a dangerous African land animal with a long feather or shooting gamma rays into a metal alloy, and is the space of all possible outcomes to the experiment. Let f : R be some function which arises from measuring some property of the outcome (e.g. the decibels of the dangerous animal s laugh or the heat of the metal alloy). We could then define a measure on R by letting µ(a) be the probability that f assumes its value in A. We have already fought a considerable battle simply to show that interesting measures, such as Lebesgue measure, indeed exist. From now on we will take this as a given, and consider more the abstract properties of measures. Here there is the key notion of completion of a measure. Definition Let (, Σ) be a measure space and µ : Σ R 0 { } a measure. M is said to be measurable with respect to µ if there are A, B Σ with A M B and µ(b \ A) = 0. Beware: Often authors simply write measurable instead of measurable with respect to µ when context makes the intended measure clear. 13

14 Lemma 2.8 Let µ be a measure on the measure space (, Σ). Then the measurable sets form a σ-algebra Proof First suppose M is measurable as witnessed by A, B Σ, A M B. Then A c = \A, B c = \B are in Σ and have A c M c B c. Since A c \ B c = B \ A, this witnesses M c measurable. If (M n ) n N is a sequence of measurable sets, as witnessed by (A n ) n N, (B n ) n N, with A n M n B n, then n NA n n N M n n N B n. On the other hand B n \ n n N n NA (B n \ A n ), n N and so n N B n \ n N A n is null, as required to witness n N M n null. Definition Let µ be a measure on the measure space (, Σ). For M measurable, as witnessed by A M B with A, B Σ, µ(b \ A) = 0, we let µ (M) = µ(a)(= µ(b)). We call µ the completion of µ. As with Lebesgue measure, we will commit the minor logical sin of using the same symbol for µ as its extension µ to the measurable sets. A more serious issue is to check the measure is well defined. Lemma 2.9 The completion of µ to the measurable sets is well defined. Proof If A 1 M B 1 and A 2 M B 2 both witness M measurable, then A 1 A 2 M \ A 1 M \ A 2 B 1 \ A 1 B 2 \ A 2. and hence is null. Lemma 2.10 Let µ be a measure on a measure space (, Σ). Let Σ be the σ-algebra of measurable sets. Then the completion defined above, µ : Σ R 0 { }, is a measure. Proof Exercise. 14

15 3 The general notion of integration and measurable function We will give a rigorous foundation to Lebesgue integration as well as integration on general measure spaces. Since the notion of integration is so closely intertwined with linear notions of adding or subtracting, I will first give the definitions of the linear operations for functions. Definition Let be a set and f, g : R. Then we define the functions f + g and f g by f + g : R, and and Similarly, for c R we define and x f(x) + g(x), f g : R, x f(x) g(x), f g : R, x f(x)g(x). cf : R, x cf(x) c + f : R, x c + f(x). Definition Let (, Σ) be a measure space equipped with a measure µ : Σ R. A function f : R is measurable if for any open set U R A function f 1 [U] Σ. h : R is simple if we can partition into finitely many measurable sets A 1, A 2,..., A n with h assuming a constant value a i on each A i. If µ(a i ) < whenever a i 0 we say that h is integrable and let hdµ be the sum of all µ(a i ) a i for a i 0. For B a measurable set, we define hdµ to be the sum of µ(a i B) a i for a i 0. Exercise Show that every simple function is measurable. B 15

16 If, as in [6], we adopt the convention that 0 = 0 then we can simply write this out as hdµ = µ(a i ) a i. i n Definition For (, Σ, µ) as above, f : R measurable and non-negative (f(x) 0 all x ), we let f dµ = sup{ h dµ : h is simple and 0 h f}. S In general given f : R measurable we can uniquely write f = f + f where f +, f are both non-negative and have disjoint supports. Assuming f + dµ, f dµ are both finite we say that f is integrable and let fdµ = f + dµ f dµ. We have implicitly used above that f +, f will be measurable. This is easy to check. You might also want to check that we could have instead used the definition f + = 1 2 ( f + f), and f = 1 2 ( f f). Definition For (, Σ, µ) as above, f : R measurable and B Σ, fdµ = χ B fdµ. Exercise We could alternatively have defined B fdµ to be the supremum of hdµ B for h ranging over simple functions with h f. Show this definition is equivalent to the one above. B Lemma 3.1 Let (, Σ) be a measure space, µ a measure defined on. Let f : R be a simple integrable function. Let c R. Then cfdµ = c fdµ. Proof Exercise. Lemma 3.2 Let (, Σ) be a measure space, µ a measure defined on. Let f, g : R be simple integrable functions with f(x) g(x) at all x. Then fdµ gdµ. Proof Exercise. Lemma 3.3 Let, Σ, µ be as above. Let f 1, f 2 be simple integrable functions. Then (f 1 + f 2 )dµ = f 1 dµ + f 2 dµ. 16

17 Proof Suppose f 1 = i N a iχ Ai, f 2 = i M b iχ Bi. Then at each i N, j M let C i,j = A i B j. f 1 + f 2 = (a i + b j )C i,j. i N,j M Since µ(a i ) = j M µ(c i,j) and µ(b j ) = i N µ(c i,j) we have f 1 dµ = a i µ(c i,j ) i N,j M and Thus which in turn equals (f 1 + f 2 )dµ. f 1 dµ = f 1 dµ + f 2 dµ = i N,j M i N,j M b j µ(c i,j ). (a i + b j )µ(c i,j ), Definition Let S be a set. A partition of S is a collection {S i : i I} such that: 1. each S i S; 2. S = i I S i; 3. for i j we have S i S i =. In other words, a partition is a division of the set into disjoint subsets. Lemma 3.4 Let, Σ, µ be as above. Let f : R be integrable. Let (A i ) i N be a partition of into countably many sets in Σ. Then fdµ = fdµ i N A i. Proof Wlog f 0. First to see that fdµ i N A i fdµ, suppose h i f χ Ai at each i N. Then i N h i f and h i dµ = h i dµ. Conversely, if h f is simple, write it as h = j k a j χ Bj with each a j > 0, which implies each µ(b j ) < by integrability of f. Consider some ǫ > 0. Go to a large enough N with µ( B j ) \ A i ) < ǫ j a. j Then fdµ > A i i N Letting ǫ 0 finishes the proof. j k i>n hdµ µ( B j ) \ j k i>n A i ) j a j > hdµ ǫ. 17

18 Lemma 3.5 Let, Σ, µ be as above. Let f : R be integrable. Then for each ǫ > 0 we can find a set B with µ(b) finite and fdµ < ǫ. \B 1 Proof Wlog, f 0. At each N N let A N = {x : N f(x) < 1 N 1 }. (Note then that f 1 ([1, )) = A 1 ). The measure of each A N is finite since A N fdµ µ(a N ) 1 N. By 3.4, we have some M with SN M (AN) fdµ < ǫ. Lemma 3.6 Let (, Σ) be a measure space. Let µ : Σ R 0 { } be a measure. Let f : R be integrable. Then there is a sequence of simple functions, (f n ) n N such that: 1. for a.e. x, f n (x) f(x); 2. f n (x) < f(x) all x ; 3. at each n N, x, f)n(x) 0 if and only if f(x) 0. As a word on notation, we say that something happens a.e. or µ-a.e if it is true off of some null sets where a null set is some set in Σ whose value under µ is zero. Thus, for a.e. x, f n (x) f(x) means that there is some B Σ with µ(b) = 0 and f n (x) f(x) all x B c. Proof It suffices to consider the case of f(x) 0 all x. First we apply our definition of the integral to find a sequence of simple functions (g n ) n N such that g n (x) f(x) all x and g n dµ fdµ. Replacing g n by f n : R x max{g 1 (x), g 2 (x),...g n (x)} we obtain a fresh sequence which satisfies f n (x) < f(x) all x and f n (x) f n+1 (x) at all x, along with f n dµ fdµ. We only need to see that f n (x) f(x) a.e. At each m N, let B m = {x : n N(f n (x) < f(x) 1 m )}. Claim: Each B m is in Σ. In future I am going to skip proving claims of this nature, but I want to do it once in detail just so it will be clear to you how to do routine calculations. Proof of Claim: Fix m. Let S be the set of pairs of rational numbers, (q 1, q 2 ), such that q 1 < q 2 1 m. For each such pair and n N, let C q1,q 2,n be the set of x such that f n (x) < q 1 and Since f n and f are both measurable, f(x) > q 2. and hence is in Σ. Then we observe that C q1,q 2,n = f 1 n [(, q 1 )] f 1 [(q 2, )] B m = (q 1,q 2) S n N C q1,q 2,n to see B m Σ. (Claim )We will be done if we show each B m is null. But if some B m is non-null, then we can define a new sequence of simple functions, h n = f n + 1 m χ B m. 18

19 These h n s are all still below f, and h n dµ = 1 m µ(b m) + f n dµ, by 3.3. This in turn contradicts f n dµ fdµ. Lemma 3.7 Let (, Σ, µ) and f : R be as in the last lemma. Let (f n ) n N be as in the conclusion that is to say, Then 1. each f n is simple; 2. for a.e. x, f n (x) f(x); 3. f n (x) < f(x) all x ; 4. at each n N, x, f n (x) 0 if and only if f(x) 0. f n dµ fdµ. Remark: Please note, the point of this lemma is not to say that we can have all the indicated properties along with f n f. We already showed that in the proof above. The point is rather that once we have 1-4, then f n f follows automatically. Proof Let the f n s be as indicated. Again we can assume f(x) 0 at all x. Suppose we instead have some simple function h = i N a i χ Bi with all x and at all n h(x) f(x) hdµ > ǫ + f n dµ for some fixed positive ǫ. We can assume that a i 0 at each i N. Then it follows that i N B i has finite measure. We may also assume each b i 0. We want to show that hdµ lim f n dµ Let δ = For each n let D n be the set of x such that ǫ 2(a 1 + a a N + µ(b 1 ) + µ(b 2 ) µ(b N )). f n (x) f(x) < δ. Here n N D n is conull, D n D n+1, and each D n is in Σ. Thus we can find some D k such that µ( B i D k ) < δ. i N Then h(x) (f k (x) + δ)χ Dk + (a 1 + a a N )χ (B1 B 2...B N)\D k. 19

20 Thus by 3.2 and 3.3 we have hdµ < as required. D k (B 1 B 2...B N) D k (B 1 B 2...B N) f k dµ + (f k + δ)dµ + (a 1 + a a N )dµ (B 1 B 2...B N)\D k δdµ + (a 1 + a a N )dµ (B 1 B 2...B N)\D k D k (B 1 B 2...B N) f k dµ + δ µ(b 1 B 2...B N ) + (a 1 + a a N )µ((b 1 B 2...B N ) \ D k ) < f k + ǫ, Lemma 3.8 Let (, Σ) be a measure space, µ a measure defined on. Let f : R be an integrable function. Let c R. Then cfdµ = c fdµ. Proof Exercise. Lemma 3.9 Let, Σ, µ be as above. Let f 1, f 2 be integrable functions. Then (f 1 + f 2 )dµ = f 1 dµ + f 2 dµ. Proof We want to put this into the framework of 3.7 ideally choosing simple functions g 1, g 2..., h 1, h 2,... such that 1. at each x and n, g n (x) f 1 (x), and 2. at each x and n, h n (x) f 2 (x), and 3. g n (x) f 1 (x) for a.e. x, and 4. h n (x) f 2 (x) for a.e. x, and then concluding that g n + h n converge to f 1 + f 2 with g n (x) + h n (x) f 1 (x) + f 2 (x). This will be fine if f 1 and f 2 are either both positive or both negative. The problem is that if they have different sign for instance, if f 1 (x) = 6, f 2 (x) = 5, and we had unluckily chosen g n (x) = 4, h n (x) = 2. Here is the solution to that minor technical issue. Let B 1 = {x : f 1 (x), f 2 (x) 0}, B 2 = {x : f 1 (x), f 2 (x) < 0}, B 3 = {x : f 1 (x) 0, f 2 (x) < 0, f 1 (x) > f 2 (x) }, B 4 = {x : f 1 (x) 0, f 2 (x) < 0, f 1 (x) f 2 (x) }, B 5 = {x : f 2 (x) 0, f 1 (x) < 0, f 1 (x) > f 2 (x) }, B 6 = {x : f 2 (x) 0, f 1 (x) < 0, f 1 (x) f 2 (x) }. It suffices to show that on each B i we have B i f 1 dµ + B i f 2 dµ = B i (f 1 + f 2 )dµ. The sets B 1, B 2 are handled by the argument given above; I will skip the details. It is B i for i 3 which requires more work. All these cases are much the same, and so I will simply do B 3. First choose simple h n s with h n (x) f 2 (x) and h n (x) h n+1 (x) f 2 (x) all x B 3. Now choose g n on B 3 such that g n (x) f 1 (x) and g n (x) g n+1 (x) g 2 (x) and g n (x) h n (x) ; the last point is easily arranged since we can always replace g n with max{g n (x), h n (x) }. Then we have 20

21 1. at each x and n, g n (x) f 1 (x), and 2. at each x and n, h n (x) f 2 (x), and 3. g n (x) f 1 (x) for a.e. x, and 4. h n (x) f 2 (x) for a.e. x, and 5. g n (x) + h n (x) f 1 (x) + f 2 (x). Apply 3.7 to f 1 and the g n s we get B i g n dµ B i f 1 dµ; to f 2 and the h n s, B i h n dµ B i f 2 dµ; finally, B i (g n + h n )dµ B i (f 1 + f 2 )dµ and for simple functions we already know that B i (g n + h n )dµ = B i g n dµ + B i h n dµ. Lemma 3.10 Let C O R with C closed and O open. Show that there is a continuous function f : R [0, 1] with f(x) = 1 at every point on C and f(x) = 0 at every point outside O. Proof Assume C is non empty and O R, or else the task is somewhat trivialized. For any set A R let d(x, A) = inf{ x a : a A} this is a continuous function in x. Then let f(x) = min{1, d(x, Oc ) d(x, C) }. Lemma 3.11 Let h be a simple function on R. Suppose h is integrable. Let ǫ > 0. Then there is a continuous function f : R R such that if we let g(x) = h(x) f(x) then R gdm < ǫ. (Technically, when saying h is simple we should specify the corresponding σ-algebra on R. The convention is to default to the Borel sets thus, we intend that there be a partition on R into finitely many Borel sets and h is constant on each element of the partition.) Proof It suffices to prove this in the case when h is the characteristic function of a single Borel set. But then this follows from 2.6 and Frequently we will want to integrate compositions of functions, just as in the last exercise. Here there is some very specific notation used in this context to guide us. Notation Given some expression involving various variables, x, y, z... and various functions, say G(x, y, z,...), the expression G(x, y,...)dµ(x) indicates that we are integrating the function x G(x, y, z,...) against the measure µ (and keeping y, z,... as fixed but possibly unknown quantities). 21

22 Thus in the exercise just above, for g(x) = f(x) h(x), instead of writing gdm < ǫ we could just as easily written R R f(x) h(x) dµ(x). The definition of integration can be extended to other settings. Definition Let (, Σ, µ) be a measure space equipped with a measure µ; we say that a function from to C is measurable if the pullback of any open set in C is measurable. For f : C measurable, we write f = Ref + iimf, where Ref : R and Imf : R are the real and imaginary parts. We say that f is integrable if both these functions are integrable in our earlier sense and let fdµ = Refdµ + i Imf dµ. In fact, it does not stop there. Given a suitable linear space B we can define integrals for suitably bounded functions f : B. In general terms, if the space B allows us to form finite sums and averages, then it makes sense to define integrals on B-valued functions. 22

23 4 Convergence theorems The order of presentation is following [6], but I am going to present the proofs without making any reference to L 1 (µ) or the theory of Banach spaces. Eventually we will have to engage with these concepts, but not just yet. Lemma 4.1 Let (, Σ) be a measure space. Let µ : Σ R 0 { } be a measure. Let (f n ) n N be a sequence of functions, each f n : R integrable. Suppose each f n 0 and n= n=1 f n dµ <. Then for µ-a.e. x, is well defined, and moreover f(x) = f n (x) n=1 fdµ = f n dµ. n=1 Proof Let B be the set of x such that the partial sums N n=1 f n(x) are unbounded. It is routine to show that this set is in Σ. I claim it is null. Claim: µ(b) = 0. Proof of Claim: Suppose instead that µ(b) > 0. Then at each c > 0 and N N let B c,n = {x : N n=1 f n(x) > c}. B c,n B c,n+1 and B c,n = B. N N Thus there exists an N with µ(b c,n ) > 1 2 µ(b). Then we obtain N f n dµ = B c,n n=1 B c,n n=1 N f n dµ > c 1 2 µ(b). Since c > 0 was arbitrary, we have contradicted n= n=1 fn dµ <. (Claim )Now define f on \ B as f(x) = f n (x) (and set f 0 on B though in terms of calculating the integral, the value of f on a null set is irrelevant). Claim: fdµ n=1 f ndµ. Proof of Claim: At each N N we have fdµ n N f n dµ = n N f n dµ. Claim: fdµ n=1 f ndµ. Let h f be simple. Wlog h 0. Write h as a i χ Ai, i L (Claim ) 23

24 where each a i > 0. Let C N = {x B : M N M ǫ f n (x) f(x) < 2 µ(a i ) }. n=1 Again the C N s are increasing and their union is conull. Fix ǫ > 0. Subclaim: There is some N with \C N hdµ < ǫ 2. Proof of Subclaim: Let D n = C n \ ( i<n D i). Then hdµ = n N D n hdµ by 3.4. Since the integral is finite, we can find some N with hdµ < ǫ D n 2. Letting ǫ 0 we obtain n>n hdµ = hdµ < ǫ S 2 + i L Ai n=1 C N ( S i L Ai) hdµ < ǫ N 2 C + ǫ N ( S ( f n i L Ai) 2 i L µ(a i) )dµ ǫ + = ǫ + N n=1 C N ( S i L Ai) f n dµ ǫ + ǫ + hdµ n=1 n=1 f n dµ. f n dµ. N n=1 C N ( S i L Ai) n=1 f n dµ (Proof of Subclam ) Then N f n dµ Since the integral of f is defined as the supremum of the integral of the simple functions h f, this completes the proof of the claim. (Claim ) Theorem 4.2 Let (, Σ) be a measure space. Let µ : Σ R 0 { } be a measure. Let (f n ) n N be a sequence of functions, each f n : R integrable. Suppose Then for µ-a.e. x, n= n=1 f(x) = f n dµ <. f n (x) n=1 24

25 is well defined, and moreover fdµ = n=1 f n dµ. Proof We have completed the special case of each f n 0 in 4.1. Thus if let g(x) = f n (x) n=1 then we obtain that g is defined on a conull set, is integrable, and has gdµ = n N fn dµ. We then let f(x) = n N f n(x) and have that f is well defined on all the points at which g is well defined. f will be integrable, because its absolute value is bounded by the integrable function g. Fix ǫ > 0. Appealing to 3.5 we find some measurable D with µ(d) finite and gdµ < ǫ 5. At each N let D N be the set {x D : \D f n (x) < n=n ǫ 5µ(D). The D N s are increasing and their union is conull in D. Thus we may find some N with D\D N gdµ < ǫ 5. Then at all M N we have n=1 fdµ M n=1 M f f n dµ + f dµ + D N D\D N D N ǫ 5µ(D) dµ + gdµ + D\D N f n dµ = fdµ D\D N M f n dµ n=1 M M f n dµ + f dµ + n=1 \D \D n=1 gdµ + gdµ + gdµ < ǫ. D\D N \D \D f n dµ Theorem 4.3 (Monotone convergence theorem) Let (, Σ) be a measure space. Let µ : Σ R 0 { } be a measure. Let (f n ) n N be sequence of functions, each f n : R integrable. Assume they are monotone, in the sense that either f n f n+1 all n or f n f n+1 all n. Suppose f n dµ is bounded. Then there exists an integrable f with f n (x) f(x) for µ-a.e. x and f n f dµ 0. 25

26 Proof Assume each f n f n+1, since the other case is symmetrical. Note then that in fact the integrals f n dµ converge, since they form a bounded monotone sequence. After possibly replacing each f n by f n f 1 we may assume the functions are all positive. Let g 1 = f 1 and for n > 1 let g n = f n f n 1. Then f N = n N g n and Now it follows from 4.2 that f is integrable and Since f f n at each n we have f N dµ = g n = g n dµ. n N n N f n dµ = N n=1 g n dµ = N g n dµ f. n=1 f f n dµ = f f n dµ = fdµ f n dµ, as required. Theorem 4.4 (Dominated convergence theorem) Let (, Σ) be a measure space. Let µ : Σ R 0 { } be a measure. Let (f n ) n N be sequence of functions, each f n : R integrable. Suppose g : R is an integrable function with f n (x) < g(x) all x, n N. Suppose f : R is a function to which the f n s converge pointwise that is to say, f n (x) f(x) all x. Then f is integrable and f f n dµ 0. Proof Fix ǫ > 0. Apply 3.5 to find some C with µ(c) < and \C gdµ < ǫ/6. At each N we can let C N be the set of x C for which n N( f(x) f n (x) < ǫ 3µ(C) ). The N s form an increasing set whose union is C, and thus we can find some large enough N with C\C N gdµ < ǫ 6. Then f f N dµ < < \C f f N dµ + f f N dµ + C N 2gdµ + C N ǫ 3µ(C) dµ + < 2 6ǫ + µ(c N) 3µ(C N ) + 2 6ǫ = ǫ. C\C N 2gdµ C\C N f f N dµ 26

27 Theorem 4.5 (Fatou s lemma) Let (, Σ) be a measure space. Let µ : Σ R 0 { } be a measure. Let (f n ) n N be sequence of functions, each f n : R integrable, each f n 0. Suppose that liminf n f n dµ <. Then for a.e. x liminff n (x) exists, and liminf n f n (x)dµ liminf n f n dµ. Proof Let Since g n (x) can be expressed as g n (x) = inf{f m (x) : m n}. lim m min{f n+i (x) : i m} and min{f n+i (x) : i m} min{f n+i (x) : i m + 1} 0 we can apply monotone convergence at 4.3 to get that each g n is integrable with g n dµ = lim min{f n+i : i m}. g n g n+1 and each g n dµ is bounded by liminf n fn dµ, so we can apply monotone convergence once more to get that liminf n f n dµ = lim n g n dµ = lim n g n dµ. At each n and k n we have g n f k, and hence g n dµ inf k n f k dµ, and thus = lim n liminf n dµ = lim n g n dµ g n dµ lim n inf k n f k dµ = liminf n f n dµ. 27

28 5 Radon-Nikodym and Conditional Expectation One of the most important theorems in measure theory is Radon-Nikodym. It can be proved without a large amount of background and we may as well do so now. Definition Let be a set and Σ P() a σ-algebra. µ : Σ R is said to be a signed measure if (a) µ( ) = 0; (b) if (A n ) n N is a sequence of disjoint sets in Σ, then µ( n N A n ) = µ(a n ). Note here we are assuming finiteness of the measure and in (b) above we are demanding convergence of the series. Here in fact (a) is redundant following from (b) and µ only taking finite values. Lemma 5.1 If Σ is a σ-algebra on and µ : Σ R is a signed measure, then whenever (B n ) n N is a sequence of sets in Σ µ( B n ) = lim N µ( B n ). n N Proof First some cosmetic rearrangement. Let C n = i n B i. So at every n we have C n C n+1, but the sequence has the same infinite intersection. Now consider the difference sets and define D n = C n \C n+1 ; the D n s are now disjoint and if we let B = i N B i represent the infinite intersection we have the equalities n=1 n N C n = B D n D n+1 D n+2... at every n. Thus µ(c n ) = µ(b ) + m n µ(d m ). This in particular implies m=1 µ(d m) is convergent and µ(d m ) 0 m n as n, which is all we need to ensure µ(c n ) µ(b ). Theorem 5.2 (Hahn Decomposition Theorem) Let Σ be a σ-algebra on and µ : Σ R a signed measure. Then there exists A Σ such that for all B Σ, B A µ(b) 0 and for all C Σ, C \ A µ(c) 0. Proof Let δ be the supremum of the set {µ(a) : A Σ}. Let (B n ) n N be a sequence of sets in Σ with µ(b n ) δ. 28

29 Then at each n let A n be the algebra of sets generated by (B i ) i n. Let s pause for a moment and observe some of the properties of these A n algbras. First, each is finite, since it is generated by finitely many sets. Moreover A n will have atoms of the form B i B i : i S i n,i/ S each of these atoms contains no smaller no empty set in A n and every element of A n is the finite union of such atoms. Finally note that B n is an element of A n. At each n, let C n be element of A n with maximum value under µ. Since B n A n we have µ(b n ) µ(c n ) and µ(c n ) δ and C n consists of the finite union of atoms in A n with positive measure. Claim: At each n, µ( i n C i) µ(c n ). Proof of Claim: Since at any j n, C j \ n i<j C i equals the union of finitely many atoms with positive measure. A = n N C i. i n (Claim ) Now let Then by the last lemma µ(a) = lim n µ( i n C i) = δ. Now note that is immediately implies δ is finite, since µ is finite and we attained the value δ with µ(a). Since A has attained this maximum value δ we must have every B A in Σ with µ(b) 0 (for otherwise µ(a \ B) would be greater than µ(a)). Similarly for any C Σ disjoint to A we must have µ(c) 0. Theorem 5.3 (Hahn-Jordan Decomposition) Let Σ be a σ-algebra on a set. Let µ : Σ R be a signed measure. Then we can find to measures µ +, µ : Σ R 0 with (i) µ = µ + µ ; (ii) µ +, µ have disjoint support. Proof The statement of the theorem should be quickly clarified. (i) states that for any set C Σ we have µ(c) = µ + (C) µ (C). (ii) states that we can some A Σ with µ + (B) = 0 when B \A and µ (B) = 0 when B A. With this clarification in mind, the proof of the theorem is an immediate consequence of 5.2: Choose A for µ as there, and then let µ + (B) = µ(a B) and µ (B) = µ(b \ A). Definition Given two measures µ, ν : Σ R { }, we say that µ is absolutely continuous with respect to ν, written µ << ν, if whenever B Σ has ν(b) = 0 then µ(b) = 0. We have dealt with the concept of measurable functions in different contexts already. So there is no possibility of confusion, let us fix a convention for the entirely general context of σ-algebras. Definition Let Σ be a σ-algebra on a set. f : R is measurable with respect to Σ if f 1 [U] is in Σ for any open U R. Exercise Show that f : R is measurable with respect to Σ if for any q Q we have f 1 [(, q)] Σ. Exercise Show that if f : R is measurable with respect to Σ then for any Borel B R we have f 1 [B] Σ. 29