Lecture 5 Theorems of Fubini-Tonelli and Radon-Nikodym

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1 Lecture 5: Fubini-onelli and Radon-Nikodym 1 of 13 Course: heory of Probability I erm: Fall 2013 Instructor: Gordan Zitkovic Lecture 5 heorems of Fubini-onelli and Radon-Nikodym Products of measure spaces We have seen that it is possible to define products of arbitrary collections of measurable spaces - one generates the σ-algebra on the product by all finite-dimensional cylinders. he purpose of the present section is to extend that construction to products of measure spaces, i.e., to define products of measures. Let us first consider the case of two measure spaces (,, µ ) and (,, µ ). If the measures are stripped, the product is endowed with the product σ-algebra = σ({a B : A, B }). he family P = {A B : A, B } serves as a good starting point towards the creation of the product measure µ µ. Indeed, if we interpret of the elements in P as rectangles of sorts, it is natural to define (µ µ )(A B) = µ (A)µ (B). he family P is a π-system (why?), but not necessarily an algebra, so we cannot use heorem 2.9 (Caratheodory s extension theorem) to define an extension of µ µ to the whole. It it not hard, however, to enlarge P a little bit, so that the resulting set is an algebra, but that the measure µ µ can still be defined there in a natural way. Indeed, consider the smallest algebra that contains P. It is easy to see that it must contain the family defined by A = { n k=1 A k B k : n N, A k, B k, k = 1,..., n}. Problem 5.1. how that A is, in fact, an algebra and that each element C A can be written in the form C = n k=1 A k B k, for n N, A k, B k, k = 1,..., n, such that A 1 B 1,..., A n B n are pairwise disjoint.

2 Lecture 5: Fubini-onelli and Radon-Nikodym 2 of 13 he problem above allows us to extend the definition of the set function µ µ to the entire A by (µ µ )(C) = n µ (A k )µ (B k ), k=1 where C = n k=1 A k B k for n N, A k, B k, k = 1,..., n is a representation of C with pairwise disjoint A 1 B 1,..., A n B n. At this point, we could attempt to show that the so-defined set function is σ-additive on A and extend it using the Caratheodory extension theorem. his is indeed possible - under the additional assumption of σ-finiteness - but we will establish the existence of product measures as a side-effect in the proof of Fubini s theorem below. Lemma 5.1 (ections of measurable sets are measurable). Let C be an -measurable subset of. For each x the section C x = {y : (x, y) C} is measurable in. Proof. In the spirit of most of the measurability arguments seen so far in these notes, let C denote the family of all C such that C x is -measurable for each x. Clearly, the rectangles A B, A, B are in A because their sections are either equal to or B, for each x. Remember that the set of all rectangles generates. he proof of the theorem will, therefore, be complete once it is established that C is a σ-algebra. his easy exercise is left to the reader. Problem 5.2. how that an analogous result holds for measurable functions, i.e., show that if f : R is a -measurable function, then the function x f (x, y 0 ) is -measurable for each y 0, and the function y f (x 0, y) is -measurable for each y 0. Proposition 5.2 (A simple Cavalieri s principle). Let µ and µ be finite measures. For C, define the functions ϕ C : [0, ) and ψ C : [0, ) by ϕ C (y) = µ (C y ), ψ C (x) = µ (C x ). hen, ϕ C L 0 +( ), ψ C L 0 +() and ϕ C dµ = ψ C dµ. (5.1) Proof. Note that, by Problem 5.2, the function x 1 C (x, y) is - measurable for each y. herefore, 1 C (, y) dµ = µ (C y ) = ϕ C (y), (5.2)

3 Lecture 5: Fubini-onelli and Radon-Nikodym 3 of 13 and the function ϕ C is well-defined. Let C denote the family of all sets in such that (5.1) holds. First, observe that C contains all rectangles A B, A, B, i.e., it contains a π-system which generates. herefore, by the π-λ heorem (heorem 2.12), it will be enough to show that C is a λ-system. We leave the details to the reader 1 Proposition 5.3 (imple Cavalieri holds for σ-finite measures). he conclusion of Proposition 5.2 continues to hold if we assume that µ and µ are only σ-finite. 1 Use representation (5.2) and the monotone convergence theorem. Where is the finiteness of the measures used? Proof (*). hanks to σ-finiteness, there exists pairwise disjoint sequences {A n } and {B n } in and, respectively, such that n A n =, m B m = and µ (A n ) < and µ (B m ) <, for all m, n N. For m, n N, define the set-functions µ n and µm on and respectively by µ n (A) = µ (A n A), µ m (B) = µ (B m B). It is easy to check that all µ n and µm, m, n N are finite measures on and, respectively. Moreover, µ (A) = n=1 µn (A), µ (B) = m=1 µm (B). In particular, if we set ϕn C (y) = µn (C y) and ψc m(x) = µ m (C x), for all x and y, we have ϕ C (y) = µ (C y ) = ψ C (x) = µ (C x ) = n=1 µ n (C y) = n=1 ϕc n (y), and µ m (C x) = ψc m(x), m=1 m=1 for all x, y. We can apply the conclusion of Proposition 5.2 to all pairs (,, µ n ) and (,, µ m ), m, n N, of finite measure spaces to conclude that all elements of the sums above are measurable functions and that so are ϕ C and ψ C. imilarly, the sequences of non-negative functions i=1 n ϕn C (y) and i=1 m ψi C (x) are non-decreasing and converge to ϕ C and ψ C. herefore, by the monotone convergence theorem, ϕ C dµ = lim n n i=1 ϕ i C dµ, and ψ C dµ = lim n n i=1 ψ i C dµ. On the other hand, we have ϕc n dµm = ψc m dµn, by Proposition 5.2. umming over all n N we have ϕ C dµ m = ψ m C dµn = ψ m C dµ,

4 Lecture 5: Fubini-onelli and Radon-Nikodym 4 of 13 where the last equality follows from the fact (see Problem 5.3 below) that f dµ n = f dµ, for all f L 0 +. Another summation - this time over m N - completes the proof 2. Problem 5.3. Let {A n } be a measurable partition of, and let the measure µ n be defined by µ n (A) = µ(a A n ) for all A. how that for f L 0 +, we have f dµ = f dµ n. 2 he argument of the proof above uncovers the fact that integration is a bilinear operation, i.e., that the mapping ( f, µ) f dµ, is linear in both arguments. Proposition 5.4 (Finite products of measure spaces). Let ( i, i, µ i ), i = 1,..., n be σ-finite measure spaces. here exists a unique measure 3 - denoted 3 he measure µ 1 µ n is called by µ the product measure, and the measure space ( 1 n, 1 1 µ n - on the product space ( 1 n, 1 n ) with the property that n, µ 1 µ n ) the product (measure (µ 1 µ n )(A 1 A n ) = µ 1 (A 1 )... µ n (A n ), space) of measure spaces ( 1, 1, µ 1 ),..., ( n, n, µ n ). for all A i i, i = 1,..., n. uch a measure is necessarily σ-finite. Proof. o simplify the notation, we assume that n = 2 - the general case is very similar. For C 1 2, we define (µ 1 µ 2 )(C) = ϕ C dµ 2, 2 where ϕ C (y) = µ 1 (C y ) and C y = {x 1 : (x, y) C}. It follows from Proposition 5.3 that µ 1 µ 2 is well-defined as a map from 1 2 to [0, ]. Also, it is clear that (µ 1 µ 2 )(A B) = µ 1 (A)µ 2 (B), for all A 1, B 2. It remains to show that µ 1 µ 2 is a measure. We start with a pairwise disjoint sequence {C n } in 1 2. For y 2, the sequence {(C n ) y } is also pairwise disjoint, and so, with C = n C n, we have ( ) ϕ C (y) = µ 1 (C y ) = µ 2 (C n ) y = ϕ Cn (y), y 2. herefore, by the monotone convergence theorem (see Problem 3.13 for details) we have (µ 1 µ 2 )(C) = ϕ C dµ 2 = ϕ Cn dµ = (µ 1 µ 2 )(C n ). 2 2 Finally, let {A n }, {B n } be sequences in 1 and 2 (respectively) such that µ 1 (A n ) < and µ 2 (B n ) < for all n N and n A n = 1, n B n = 2. Define {C n } as an enumeration of the countable family {A i B j : i, j N} in 1 2. hen (µ 1 µ 2 )(C n ) < and all n N and n C n = 1 2. herefore, µ 1 µ 2 is σ-finite.

5 Lecture 5: Fubini-onelli and Radon-Nikodym 5 of 13 Now that we know that product measures exist, we can state and prove the important theorem which, when applied to integrable functions bears the name of Fubini, and when applied to non-negative functions, of onelli. We state it for both cases simultaneously (i.e., on L 0 1 ) in the case of a product of two measure spaces. An analogous theorem for finite products can be readily derived from it. When the variable or the underlying measure space of integration needs to be specified, we write f (x) µ(dx) for the Lebesgue integral f dµ. heorem 5.5 (Fubini, onelli). Let (,, µ ) and (,, µ ) be two σ- finite measure spaces. For f L 0 1 ( ) we have ( ) f d(µ µ ) = f (x, y) µ (dy) µ (dx) ( ) (5.3) = f (x, y) µ (dx) µ (dy) Proof. All the hard work has already been done. We simply need to crank the tandard Machine. Let H denote the family of all functions in L 0 + ( ) with the property that (5.3) holds. Proposition 5.3 implies that H contains the indicators of all elements of. Linearity of all components of (5.3) implies that H contains all simple functions in L 0 +, and the approximation theorem 3.10 implies that the whole L0 + is in H. Finally, the extension to L 0 1 follows by additivity. ince f is always in L 0 1, we have the following corollary Corollary 5.6. For f L 0 ( ), we have ( ) f L 0 1 ( ) if and only if f (x, y) µ (dy) µ (dx) <. Example 5.7. he assumption of σ-finiteness cannot be left out of the statement of heorem 5.5. Indeed, let (,, µ) = ([0, 1], B([0, 1]), λ) and (,, ν) = ([0, 1], 2 [0,1], γ), where γ is the counting measure on 2 [0,1], so that (,, ν) fails the σ-finite property. Define f L 0 ( ) (why it is product-measurable?) by 1, x = y, f (x, y) = 0, x = y. hen and so f (x, y) µ(dx) = λ({y}) = 0, f (x, y) µ(dx) ν(dy) = 0 γ(dy) = 0. [0,1]

6 Lecture 5: Fubini-onelli and Radon-Nikodym 6 of 13 On the other hand, and so f (x, y) ν(dy) = γ({x}) = 1, f (x, y) ν(dy) µ(dx) = [0,1] 1 λ(dx) = 1. Example 5.8. he integrability of either f + or f for f L 0 ( ) is (essentially) necessary for validity of Fubini s theorem, even if all iterated integrals exist. Here is what can go wrong. Let (,, µ) = (,, ν) = (N, 2 N, γ), where γ is the counting measure. Define the function f : N N R by 1, m = n, f (n, m) = 1, m = n + 1, 0, otherwise hen f (n, m) γ(dm) = f (n, m) = ( 1) = 0, m N and so On the other hand, i.e., herefore, f (n, m) γ(dm) γ(dn) = 0. f (n, m) γ(dn) = f (n, m) = f (n, m) γ(dn) = 1 {m=1}. f (n, m) γ(dn) γ(dm) = 1, m = 1, 0, m > 1, 1 {m=1} γ(dm) = 1. If you think that using the counting measure is cheating, convince yourself that it is not hard to transfer this example to the setup where (,, µ) = (,, ν) = ([0, 1], B([0, 1]), λ). he existence of the product measure gives us an easy access to the Lebesgue measure on higher-dimensional Euclidean spaces. Just as λ on R measures the length of sets, the Lebesgue measure on R 2 will measure area, the one on R 3 volume, etc. Its properties are collected in the following problem:

7 Lecture 5: Fubini-onelli and Radon-Nikodym 7 of 13 Problem 5.4. For n N, show the following statements: 1. here exists a unique measure λ (note the notation overload) on B(R n ) with the property that ( ) λ [a 1, b 1 ) [a n, b n ) = (b 1 a 1 )... (b n a n ), for all a 1 < b 1,..., a n < b n in R. 2. he measure λ on R n is invariant with respect to all isometries 4 of 4 An isometry of R n is a map f : R n R n. R n with the property that d(x, y) = d( f (x), f (y)) for all x, y R n. It can be Note: Feel free to use the following two facts without proof: shown that the isometries of R 3 are precisely translations, rotations, reflections (a) A function f : R n R n is an isometry if and only if there exists and compositions thereof. x 0 R n and an orthogonal linear transformation O : R n R n such that f (x) = x 0 + Ox. (b) Let O be an orthogonal linear transformation. hen R 1 and OR 1 Note: probably the least painful way to have the same Lebesgue measure, where R prove this fact is by using the change-ofvariable formula for the Riemann inte- 1 denotes the unit rectangle [0, 1) [0, 1). gral. he Radon-Nikodym heorem We start the discussion of the Radon-Nikodym theorem with a simple observation: Problem 5.5 (Integral as a measure). For a function f L 0 ([0, ]), we define the set-function ν : [0, ] by 1. how that ν is a measure. ν(a) = A f dµ. (5.4) 2. how that µ(a) = 0 implies ν(a) = 0, for all A. 3. how that the following two properties are equivalent µ(a) = 0 if and only if ν(a) = 0, A, and f > 0, a.e. Definition 5.9 (Absolute continuity, etc.). Let µ, ν be measures on the measurable space (, ). We say that 1. ν is absolutely continuous with respect to µ - denoted by ν µ - if ν(a) = 0, whenever µ(a) = 0, A. 2. µ and ν are equivalent if ν µ and µ ν, i.e., if µ(a) = 0 ν(a) = 0, for all A,

8 Lecture 5: Fubini-onelli and Radon-Nikodym 8 of µ and ν are (mutually) singular - denoted by µ ν - if there exists D such that µ(d) = 0 and ν(d c ) = 0. Problem 5.6. Let µ and ν be measures with ν finite and ν µ. how that for each ε > 0 there exists δ > 0 such that for each A, we have µ(a) δ ν(a) ε. how that the assumption that ν is finite is necessary. Problem 5.5 states that the prescription (5.4) defines a measure on which is absolutely continuous with respect to µ. What is surprising is that the converse also holds under the assumption of σ-finiteness: all absolutely continuous measures on are of that form. hat statement (and more) is the topic of this section. ince there is more than one measure in circulation, we use the convention that a.e. always uses the notion of the null set as defined by the measure µ. heorem 5.10 (he Lebesgue decomposition). Let (, ) be a measurable space and let µ and ν be two σ-finite measures on. hen there exists a unique decomposition ν = ν a + ν s, where 1. ν a µ, 2. ν s µ. Furthermore, there exists an a.e.-unique function h L 0 + such that ν a (A) = h dµ. A he proof is based on the following particular case of the Riesz representation theorem Proposition Let (,, µ) be a measure space, and let L : L 2 (µ) R be a linear map with the property that L f C f L 2 for all f L 2 for some constant C 0. hen, there exists an a.e.-unique element g L 2 such that L f = f g dµ, for all f L 2. Proof. If L = 0, the statement clearly holds with g = 0, so we assume Lh = 0 for some h L 2. he set Nul L = { f L 2 : L f = 0} is a nonempty and closed (why?) subspace of L 2 with h Nul L. herefore, there exists a sequence { f n } in Nul L such that a n = f n h M = inf f h > 0. f Nul L

9 Lecture 5: Fubini-onelli and Radon-Nikodym 9 of 13 It is easy to check that f n f m 2 = f n + f m 2h f n h f m h 2 4M 2 + 2a 2 n + 2a 2 m 0 as m, n, and, hence, that { f n } is a Cauchy sequence in L 2. By completeness (Proposition 4.13) there exists ˆf L 2 such that f n ˆf in L 2. It follows immediately that ĝ = M, where ĝ = ˆf h. herefore, ĝ 2 ĝ + ε f 2, for all ε > 0, f Nul L, and so ε f f ĝ 0, for all ε > 0, f Nul L, which, in turn, immediately implies that 5 5 Compare this argument to Problem 4.9 f ĝ dµ = 0, for all f Nul L. In words, ĝ is orthogonal to Nul L. tarting from the identity L f = Lαh, where α = L f /Lh, we deduce that f αh Nul L and, so, ( f αh)ĝ dµ = 0, i.e., f ĝ dµ = L f Lh hĝ dµ. herefore, noting that h cannot be orthogonal to ĝ, we get L f = f g dµ where g = Lh hĝ dµ ĝ. Proof of heorem Uniqueness: uppose that ν 1 a + ν 1 s = ν = ν 2 a + ν 2 s are two decompositions satisfying (1) and (2) in the statement. Let D 1 and D 2 be as in the definition of mutual singularity applied to the pairs µ, ν 1 s and µ, ν 2 s, respectively. et D = D 1 D 2, and note that µ(d) = 0 and ν 1 s (D c ) = ν 2 s (D c ) = 0. For any A, we have µ(a D) = 0 and so, thanks to absolute continuity, ν 1 a (A D) = ν 2 a (A D) = 0 so that ν 1 s (A D) = ν 2 s (A D) = ν(a D). By singularity, we have ν 1 s (A D c ) = ν 2 s (A D c ) = 0, and, consequently, ν 1 a (A D c ) = ν 2 a (A D c ) = ν(a D c ). Finally, ν 1 a (A) = ν 1 a (A D) + ν 1 a (A D c ) = ν 2 a (A D) + ν 2 a (A D c ) = ν 2 (A), and, similarly, ν 1 s = ν 2 s.

10 Lecture 5: Fubini-onelli and Radon-Nikodym 10 of 13 o establish the uniqueness of the function f with the property that ν a (A) = A f dµ for all A, we assume that there are two such functions, f 1 and f 2, say. Define the sequence {B n } by B n = { f 1 f 2 } C n, where {C n } is a pairwise-disjoint sequence in with the property that ν(c n ) <, for all n N and n C n =. hen, with g n = f 1 1 Bn f 2 1 Bn L 1 + we have g n dµ = f 1 dµ B n f 2 dµ = ν a (B n ) ν a (B n ) = 0. B n By Problem 3.9, we have g n = 0, a.e., i.e., f 1 = f 2, a.e., on B n, for all n N, and so f 1 = f 2, a.e., on { f 1 f 2 }. A similar argument can be used to show that f 1 = f 2, a.e., on { f 1 < f 2 }, as well. Existence: Assume, first, that µ and ν are probability measures and set ϕ = µ + ν. We define L f = f dν for f L 2 (ϕ), and note that, L f f dν f dϕ f L 2 (ϕ), Proposition 5.11 can be applied to yield the existence of g L 2 (ϕ) such that L f = f g dϕ, for all f L 2 (ϕ). ince L f 0, for f 0, ϕ-a.e., we conclude (why?) that g 0, ϕ-a.e. By approximation (via the monotone convergence theorem) we then have F dν = Fg dν + Fgdµ, for all F L 0 +(ϕ). (5.5) If we plug F = 1 {g 1+ε} in (5.5), we obtain ν(g 1 + ε) (1 + ε)ϕ(g 1 + ε), and conclude that g 1, ϕ-a.e. It also follows, by taking F = 1 D, where D = {g = 1}, that µ(d) = 0. he function h = g 1 g 1 {g<1} is nonnegative, measurable and finitely valued ϕ-a.e., so we can use F = f 1 g 1 D c, for f L0 + (ϕ) in (5.5) to obtain f 1 D c dν = f h dµ. It follows immediately that ν(a) = 1 A h dµ + ν(a D), which provides the required decomposition ν a (A) = 1 A h dµ, ν s (A) = ν(a D).

11 Lecture 5: Fubini-onelli and Radon-Nikodym 11 of 13 Corollary 5.12 (Radon-Nikodym). Let µ and ν be σ-finite measures on (, ) with ν µ. hen there exists f L 0 + such that ν(a) = A f dµ, for all A. (5.6) For any other g L 0 + with the same property, we have f = g, a.e. Any function f for which (5.6) holds is called the Radon-Nikodym derivative of ν with respect to µ and is denoted by f = dµ dν, a.e. 6 6 he Radon-Nikodym derivative f = dν dµ is defined only up to a.e.-equivalence, and there is no canonical way of picking Problem 5.7. Let µ, ν and ρ be σ-finite measures on (, ). how that a representative defined for all x. For that reason, we usually say that a 1. If ν µ and ρ µ, then ν + ρ µ and function f L 0 + is a version of the 2. If ν µ and f L 0 +, then dν dµ + dρ dµ f dν = 3. If ν µ and ρ ν, then ρ µ and d(ν + ρ) =. dµ g dµ where g = f dν dµ. dν dρ dµ dν = dρ dµ. Note: Make sure to pay attention to the fact that different measure give rise to different families of null sets, and, hence, to different notions of almost everywhere. 4. If µ ν, then dµ dν = ( ) dν 1. dµ Problem 5.8. Let µ 1, µ 2, ν 1, ν 2 be σ-finite measures with µ 1 and ν 1, as well as µ 2 and ν 2, defined on the same measurable space. If ν 1 µ 1 and ν 2 µ 2, show that ν 1 ν 2 µ 1 µ 2. Radon-Nikodym derivative of ν with respect to µ if (5.6) holds. Moreover, to stress the fact that we are talking about a whole class of functions instead of just one, we usually write dν dµ L0 + and not dν dµ L0 +. We often neglect this fact notationally, and write statements such as If f L 0 + and f = dµ dν then.... What we really mean is that the statement holds regardless of the particular representative f of the Radon-Nikodym derivative we choose. Also, when we write dµ dν = dρ dµ, we mean that they are equal as elements of L 0 +, i.e., that there exists f L0 +, which is both a version of dν dµ and a version of dρ dµ. Example Just like in the statement of Fubini s theorem, the assumption of σ-finiteness cannot be omitted. Indeed, take (, ) = ([0, 1], B([0, 1])) and consider the Lebesgue measure λ and the counting measure γ on (, ). Clearly, λ γ, but there is no f L 0 + such that λ(a) = A f dγ. Indeed, suppose that such f exists and set D n = {x : f (x) > 1/n}, for n N, so that D n { f > 0} = { f = 0}. hen 1 1 λ(d n ) = f dγ n dγ = n 1 #D n, D n D n

12 Lecture 5: Fubini-onelli and Radon-Nikodym 12 of 13 and so #D n n. Consequently, the set { f > 0} = n D n is countable. his leads to a contradiction since the Lebesgue measure does not charge countable sets, and so 1 = λ([0, 1]) = f dγ = f dγ = λ({ f > 0}) = 0. Additional Problems { f >0} H = {(x, r) [0, ) : f (x) r} be the region under the graph of f. how that f dµ = (µ λ)(h). Problem 5.10 (A layered representation). Let ν be a measure on B([0, )) such that N(u) = ν([0, u)) <, for all u R. Let (,, µ) be a σ-finite measure space. For f L 0 + (), show that 1. N f dµ = [0, ) µ({ f > u}) ν(du). 2. for p > 0, we have f p dµ = p [0, ) up 1 µ({ f > u}) λ(du), where λ is the Lebesgue measure. Problem 5.9 (Area under the graph of a function). For f L 0 +, let Note: he equality in this problem is consistent with our intuition that the value of the integral f dµ corresponds to the area under the graph of f. Problem 5.11 (A useful integral). 1. how that sin x 0 x dx =. 2. For a > 0, let f : R 2 R be given by e xy sin(x), 0 x a, 0 y, f (x, y) = 0, otherwise. Hint: Find a function laying below sin x x which is easier to integrate. how that f L 1 (R 2, B(R 2 ), λ), where λ denotes the Lebesgue measure on R Establish the equality a sin x x dx = π 2 cos(a) e ay dy sin(a) y Conclude that for a > 0, a sin(x) 0 x dx π 2 2 a, so that a sin(x) lim a x dx = π 2. 0 ye ay 1+y 2 dy. Problem 5.12 (he Cantor measure). Let ({ 1, 1} N, B({ 1, 1} N ), µ C ) be the coin-toss space. Define the mapping f : { 1, 1} N [0, 1] by f (s) = (1 + s n )3 n, for s = (s 1, s 2,... ). Let δ be the push-forward of µ C by the map f. It is called the Cantor measure.

13 Lecture 5: Fubini-onelli and Radon-Nikodym 13 of Let d be the metric on { 1, 1} N (as given by the equation (1.1)). how that for α = log 3 (2) and s 1, s 2 { 1, 1} N, we have d(s 1, s 2 ) α f (s 2 ) f (s 1 ) 3d(s 1, s 2 ) α. 2. how that δ is atom-free, i.e., that δ({x}) = 0, for all x [0, 1], 3. For a measure µ on the σ-algebra of Borel sets of a topological space X, the support of µ is collection of all x X with the property that µ(o) > 0 for each open set O with x O. Describe the support of δ. Hint: Guess what it is and prove that your guess is correct. Use the result in (1). 4. Prove that δ λ. Note: he Cantor measure is an example of a singular measure. It has no atoms, but is still singular with respect to the Lebesgue measure. Problem 5.13 (Joint measurability). 1. Give an example of a function f : [0, 1] [0, 1] [0, 1] such that Hint: You can use the fact that there exists a subset of [0, 1] which is not Borel measurable. x f (x, y) and y f (x, y) are B([0, 1])-measurable functions for each y [0, 1] and x [0, 1], respectively, but that f is not B([0, 1] [0, 1])-measurable. 2. Let (, ) be a measurable space. A function f : R R is called a Caratheodory function if x f (x, y) is -measurable for each y R, and y f (x, y) is continuous for each x R. how that Caratheodory functions are B(R)-measurable. Hint: Express a Caratheodory function as limit of a sequence of the form f n = k Z g n,k (x)h n,k (r), n N.