36-752: Lecture 1. We will use measures to say how large sets are. First, we have to decide which sets we will measure.

Transcription

1 : Lecture How is this course different from your earlier probability courses? There are some problems that simply can t be handled with finite-dimensional sample spaces and random variables that are either discrete or have densities. Example. Try to express the strong law of large numbers without using an infinitedimensional space. Oddly enough, the weak law of large numbers requires only a sequence of finite-dimensional spaces, but the strong law concerns entire infinite sequences. Example. Consider a distribution whose cumulative distribution function (cdf) increases continuously part of the time but has some jumps. Such a distribution is neither discrete nor continuous. How do you define the mean of such a random variable? Is there a way to treat such distributions together with discrete and continuous ones in a unified manner? General Measures. Both of the above examples are accommodated by a generalization of the theories of summation and integration. Indeed, summation becomes a special case of the more general theory of integration. It all begins with a generalization of the concept of size of a set. Example. One way to measure the size of a set is to count its elements. All infinite sets would have the same size (unless you distinguish different infinite cardinals). Example. Special subsets of Euclidean spaces can be measured by length, area, volume, etc. But what about sets with lots of holes in them? For example, how large is the set of irrational numbers between 0 and? We will use measures to say how large sets are. First, we have to decide which sets we will measure. Definition. Let Ω be a set. A collection F of subsets of Ω is called a field if it satisfies Ω F, for each A F, A C F, for all A, A F, A A F. A field F is a σ-field if, in addition, it satisfies for every sequence {A k } k= in F, k= A k F. We will define measures on fields and σ-field s. A set Ω together with a σ-field F is called a measurable space (Ω, F), and the elements of F are called measurable sets.

2 0 0 Example. Let Ω = IR and define U to be the collection of all unions of finitely many disjoint intervals of the form (a, b] or (, b] or (a, ) or (, ), together with. Then U is a field. Example. (Power set) Let Ω be an arbitrary set. The collection of all subsets of Ω is a σ-field. It is denoted Ω and is called the power set of Ω. Example. (Trivial σ-field) Let Ω be an arbitrary set. Let F = {Ω, }. This is the trivial σ-field. Definition. The extended reals is the set of all real numbers together with and. We shall denote this set IR. The positive extended reals, denoted IR + is (0, ], and the nonnegative extended reals, denoted IR +0 is [0, ]. Definition 0. Let (Ω, F) be a measurable space. Let µ : F IR +0 satisfy µ( ) = 0, for every sequence {A k } k= of mutually disjoint elements of F, µ( k= A k) = k= µ(a k). Then µ is called a measure on (Ω, F) and (Ω, F, µ) is a measure space. If F is merely a field, then a µ that satisfies the above two conditions whenever k= A k F is called a measure on the field F. Example. Let Ω be arbitrary with F the trivial σ-field. Define µ( ) = 0 and µ(ω) = c for arbitrary c > 0 (with c = possible). Example. (Counting measure) Let Ω be arbitrary and F = Ω. For each finite subset A of Ω, define µ(a) to be the number of elements of A. Let µ(a) = for all infinite subsets. This is called counting measure on Ω.

3 -: Lecture For every collection C of subsets of Ω, there is a smallest field containing C and a smallest σ-field containing C. These are called the field generated by C and the σ-fieldgenerated by C. Just check that the intersection of an arbitrary collection of fields is a field and the intersection of an arbitrary collection of σ-field s is a σ-field. These collections are nonempty because Ω is always a σ-field that contains every collection of subsets of Ω. The σ-field generated by C is sometimes denoted σ(c) Exercise. Let F, F,... be classes of sets in a common space Ω such that F n F n+ for each n. Show that if each F n is a field, then n=f n is also a field. If each F n is a σ-field, then is n= F n also necessarily a σ-field? Think about the following case: Ω is the set of nonnegative integers and F n σ({{0}, {},..., {n}}). Example. Let C = {A} for some nonempty A that is not itself Ω. Then σ(c) = {, A, A C, Ω}. Example. Let Ω = IR and let C be the collection of all intervals of the form (a, b]. Then the field generated by C is U from Example while σ(c) is larger. Example. (Borel σ-field) Let Ω be a topological space and let C be the collection of open sets. Then σ(c) is called the Borel σ-field. If Ω = IR, the Borel σ-field is the same as σ(c) in Example. The Borel σ-field of subsets of IR k is denoted B k. Exercise. Give some examples of classes of sets C such that σ(c) = B. Exercise. Are there subsets of IR which are not in B? Definition. Let (Ω, F, P) be a measure space. If P(Ω) =, then P is called a probability, (Ω, F, P) is a probability space, and elements of F are called events. Sometimes, if the name of the probability P is understood or is not even mentioned, we will denote P(E) by Pr(E) for events E. Infinite measures pose a few unique problems. Some infinite measures are just like finite ones. Definition 0. Let (Ω, F, µ) be a measure space, and let C F. Suppose that there exists a sequence {A n } n= of elements of C such that µ(a n) < for all n and Ω = n= A n. Then we say that µ is σ-finite on C. If µ is σ-finite on F, we merely say that µ is σ-finite. Example. Let Ω = ZZ with F = Ω and µ being counting measure. This measure is σ-finite. Counting measure on an uncountable space is not σ-finite. Exercise. Prove the claims in Example.

4 Properties of Measures. There are several useful properties of measures that are worth knowing. First, measures are countably subadditive in the sense that () ( ) µ A n n= µ(a n ), n= for arbitrary sequences {A n } n=. The proof of this uses a standard trick for dealing with countable sequences of sets. Let B = A and let B n = A n \ n i= B i for n >. The B n s are disjoint and have the same finite and countable unions as the A n s. The proof of () relies on the additional fact that µ(b n ) µ(a n ) for all n. Next, if µ(a n ) = 0 for all n, it follows that µ ( n= A n) = 0. This gets used a lot in proofs. Similarly, if µ is a probability and µ(a n ) = for all n, then µ ( n= A 0 n) =. Definition. Suppose that some statement about elements of Ω holds for all ω A C where µ(a) = 0. Then we say that the statement holds almost everywhere, denoted a.e. [µ]. If P is a probability, then almost everywhere is often replaced by almost surely, denoted a.s. [P]. Example. Let (Ω, F, P) be a probability space. Let {X n } n= be a sequence of func- a.s. tions from Ω to IR. To say that X n converges to X a.s. [P] (denoted X n X) means that there is a set A with P(A) = 0 and P({ω A C : lim n X n (ω) = X(ω)}) =. Proposition. If µ, µ,... are all measures on (Ω, F) and if {a n } n= of positive numbers, then n= a nµ n is a measure on (Ω, F). is a sequence 0 Exercise. Prove Proposition. Definition. Define the indicator function I A : Ω {0, } for the set A Ω as I A (ω) = if ω A and I A (ω) = 0 if ω A C. Definition. Let (Ω, F, µ) be a measure space. A sequence {A n } n= of elements of F is called monotone increasing if A n A n+ for each n. It is monotone decreasing if A n A n+ for each n. For a general sequence, we define lim sup A n = n lim inf n A n = A n, i= n=i i= n=i A n. 0 If lim sup n A n = lim inf n A n, the common set is called lim n A n. The set lim sup n A n is often called A n infinitely often or A n i.o. because a point ω is in that set if and only if ω is in infinitely many of the A n sets. The set lim inf n A n is often called A n all but finitely often or A n eventually (A n ev.). This set has all those ω such that ω is in all of the A n except possibly finitely many of the A n, i.e., eventually.

5 Exercise 0. What is the relationship between the definition of the lim sup and lim inf of a sequence of reals {x n } n= and this definition of the lim sup and lim inf of a sequence of sets? 0 Exercise. Define A n to be the set ( /n, ] if n is odd, and to be (, /n] if n is even. What are lim sup n A n and lim inf n A n? It is easy to establish some simple facts about these limiting sets. Proposition. Let {A n } n= be a sequence of sets. lim inf n A n = lim sup n A n, if and only if, for each ω, lim n I An (ω) exists. If the sequence is monotone increasing, then lim n A n = n= A n. If the sequence is monotone decreasing, then lim n A n = n= A n. Exercise. Prove Proposition.

6 -: Lecture Lemma. Let (Ω, F, µ) be a measure space. Let {A n } n= be a monotone sequence of elements of F. Then lim n µ(a n ) = µ (lim n A n ) if either of the following hold: the sequence is increasing, the sequence is decreasing and µ(a k ) < for some k. 0 Proof. Define A = lim n A n. In the first case, write B = A and B n = A n \ A n for n >. Then A n = n k= B k for all n (including n = ). Then µ(a n ) = n k= µ(b k), and ( ) n µ lim A n = µ(a ) = µ(b k ) = lim µ(b k ) = lim µ(a n ). n n n k= In the second case, write B n = A n \ A n+ for all n k. Then, for all n > k, k= A k \ A n = A k \ A = n i=k B i, B i. i=k 0 By the first case, lim µ(a k \ A n ) = µ n ( ) B i = µ(a k \ A ). Because A n A k for all n > k and A A k, it follows that i=k µ(a k \ A n ) = µ(a k ) µ(a n ), µ(a k \ A ) = µ(a k ) µ(a ). It now follows that lim n µ(a n ) = µ(a ). Exercise. Construct a simple counterexample to show that the condition µ(a k ) < is required in the second claim of Lemma. Uniqueness of Measures. There is a popular method for proving uniqueness theorems about measures. The idea is to define a function µ on a convenient class C of sets and then prove that there can be at most one extension of µ to σ(c). Example. Suppose it is given that for any a IR, P((, a]) = a π exp ( u / ) du. Does that uniquely define a probability measure on the class of Borel subsets of the line, B?

7 0 0 0 Definition. A collection A of subsets of Ω is a π-system if, for all A, A A, A A A. A class C is a λ-system if Ω C, for each A C, A C C, for each sequence {A n } n= of disjoint elements of C, n= A n C. Example. The collection of all intervals of the form (, a] is a π-system of subsets of IR. So too is the collection of all intervals of the form (a, b] (together with ). The collection of all sets of the form {(x, y) : x a, y b} is a π-system of subsets of IR. So too is the collection of all rectangles with sides parallel to the coordinate axes. Some simple results about π-systems and λ-systems are the following. Proposition. If Ω is a set and C is both a π-system and a λ-system, then C is a σ-field. Proposition 0. Let Ω be a set and let Λ be a λ-system of subsets. If A Λ and A B Λ then A B C Λ. Exercise. Prove Propositions and 0. Lemma. (π λ theorem) Let Ω be a set and let Π be a π-system and let Λ be a λ-system that contains Π. Then σ(π) Λ. Proof. Define λ(π) to be the smallest λ-system containing Π. For each A Ω, define G A to be the collection of all sets B Ω such that A B λ(π). First, we show that G A is a λ-system for each A λ(π). To see this, note that A Ω λ(π), so Ω G A. If B G A, then A B λ(π), and Proposition 0 says that A B C λ(π), so B C G A. Finally, {B n } n= G A with the B n disjoint implies that A B n λ(π) with A B n disjoint, so their union is in λ(π). But their union is A ( n= B n). So n= B n G A. Next, we show that λ(π) G C for every C λ(π). Let A, B Π, and notice that A B Π, so B G A. Since G A is a λ-system containing Π, it must contain λ(π). It follows that A C λ(π) for all C λ(π). If C λ(π), it then follows that A G C. So, Π G C for all C λ(π). Since G C is a λ-system containing Π, it must contain λ(π). Finally, if A, B λ(π), we just proved that B G A, so A B λ(π) and hence λ(π) is also a π-system. By Proposition, λ(π) is a σ-field containing Π and hence must contain σ(π). Since λ(π) Λ, the proof is complete. The uniqueness theorem is the following. Theorem. Suppose that µ and µ are measures on (Ω, F) and F is the smallest σ-field containing the π-system Π. If µ and µ are both σ-finite on Π and they agree on Π, then they agree on F.

8 0 Proof. First, let C Π be such that µ (C) = µ (C) <, and define G C to be the collection of all B F such that µ (B C) = µ (B C). It is easy to see that G C is a λ-system that contains Π, hence it equals F by Lemma. (For example, if B G C, µ (B C C) = µ (C) µ (B C) = µ (C) µ (B C) = µ (B C C), so B C G C.) Since µ and µ are σ-finite, there exists a sequence {C n } n= Π such that µ (C n ) = µ (C n ) <, and Ω = n= C n. (Since Π is only a π-system, we cannot assume that the C n are disjoint.) For each A F, ( n ) µ j (A) = lim µ j [C i A] for j =,. n i= Since µ j ( n i= [C i A]) can be written as a linear combination of values of µ j at sets of the form A C, where C Π is the intersection of finitely many of C,...,C n, it follows from A G C that µ ( n i= [C i A]) = µ ( n i= [C i A]) for all n, hence µ (A) = µ (A). Exercise. Return to Example. You should now be able to answer the question posed there. Exercise. Suppose that Ω = {a, b, c, d, e} and I tell you the value of P({a, b}) and P({b, c}). For which subset of Ω do I need to define P( ) in order to have a unique extension of P to a σ-field of subsets of Ω?

9 : Lecture Measures Based on Increasing Functions. Let F be a cdf (nondecreasing, rightcontinuous, limits equal 0 and at and respectively). Let U be the field in Example. Define µ : U [0, ] by µ(a) = n k= F(b k) F(a k ) when A = n k= (a k, b k ] and {(a k, b k ]} are disjoint. This set-function is well-defined and finitely additive. To see that it is welldefined, look at an alternative representation as µ(a) = m j= F(d j) F(c j ). Consider the partition of A into the refinement of the two partitions given. The sum over the refinement is the same as both of the two sums we started with. Is µ countably additive as probabilities are supposed to be? That is, if A = i= A i where the A i s are disjoint, each A i is a union of finitely many disjoint intervals, and A itself is the union of finitely many disjoint intervals (a k, b k ] for k =,...,n, does µ(a) = i= µ(a i)? First, take the collection of intervals that go into all of the A i s and split them, if necessary, so that each is a subset of at most one of the (a k, b k ] intervals. Then apply the following result to each (a k, b k ]. Lemma. Let (a, b] = k= (c k, d k ] with the (c k, d k ] s disjoint. Then F(b) F(a) = k= F(d k) F(c k ). Proof. Since (a, b] n k= (c k, d k ] for all n, it follows that F(b) F(a) n k= F(d k) F(c k ), hence F(b) F(a) k= F(d k) F(c k ). We need to prove the opposite inequality. Suppose first that both a and b are finite. Let ǫ > 0. For each k, there is e k > d k such that F(d k ) F(e k ) F(d k ) + ǫ k. Also, there is f > a such that F(a) F(f) ǫ. Now, the interval [f, b] is compact and [f, b] k= (c k, e k ). So there are finitely many (c k, e k ) s (suppose they are the first n) such that [f, b] n k= (c k, e k ). Now, n n F(b) F(a) F(b) F(f) + ǫ ǫ + F(e k ) F(c k ) ǫ + F(d k ) F(c k ). k= It follows that F(b) F(a) ǫ + k= F(d k) F(c k ). Since this is true for all ǫ > 0, it is true for ǫ = 0. If = a < b <, let g > be such that F(g) < ǫ. The above argument shows that F(b) F(g) F(d k g) F(c k g) F(d k ) F(c k ). k= Since lim g F(g) = 0, it follows that F(b) k= F(d k) F(c k ). Similar arguments work when a < b = and = a < b =. In Lemma you can replace F by an arbitrary nondecreasing right-continuous function with only a bit more effort. (See the supplement following at the end of this lecture.) The function µ defined in terms of a nondecreasing right-continuous function is a measure on the field U. There is an extension theorem that gives conditions under which a measure on a field can be extended to a measure on the generated σ-field. Furthermore, the extension is unique. k= k=

10 0 0 Example. (Lebesgue measure) Start with the function F(x) = x, form the measure µ on the field U and extend it to the Borel σ-field. The result is called Lebesgue measure, and it extends the concept of length from intervals to more general sets. Example. Every distribution function for a random variable has a corresponding probability measure on the real line. Another concept that is occasionally useful is that of a complete measure space. Definition. A measure space (Ω, F, µ) is complete if, for every A F such that µ(a) = 0 and every B A, B F. Theorem 0. (Caratheodory extension) Let µ be a σ-finite measure on the field C of subsets of Ω. There exists a σ-field A that contains C and a unique extension µ of µ to a measure on (Ω, A). Furthermore (Ω, A, µ ) is a complete measure space. Exercise. In this exercise, we prove Theorem 0. First, for each B Ω, define () µ (B) = inf µ(a i ), i= 0 where the inf is taken over all {A i } i= such that B i= A i and A i C for all i. Since C is a field, we can assume that the A i s are mutually disjoint without changing the value of µ (B). Let A = {B Ω : µ (C) = µ (C B) + µ (C B C ), for all C Ω }. Now take the following steps:. Show that µ extends µ, i.e. that µ (A) = µ(a) for each A C.. Show that µ is monotone and subadditive.. Show that C A.. Show that A is a field.. Show that µ is finitely additive on A.. Show that A is a σ-field.. Show that µ is countably additive on A.. Show that µ is the unique extension of µ to a measure of (Ω, A).. Show that (Ω, A, µ ) is a complete measure space.

11 0 Supplement: Measures from Increasing Functions Lemma deals only with functions F that are cdf s. Suppose that F is an unbounded nondecreasing function that is continuous from the right. If < a < b <, then the proof of Lemma still applies. Suppose that (, b] = k= (c k, d k ] with b < nd all (c k, d k ] disjoint. Suppose that lim x F(x) =. We want to show that k= F(d k) F(c k ) =. If one c k =, the proof is immediate, so assume that all c k >. Then there must be a subsequence {k j } j= such that lim j c kj =. For each j, let {(c j,n, d j,n]} n= be the subsequence of intervals that cover (c kj, b]. For each j, the proof of Lemma applies to show that () F(b) F(c kj ) = F(d j,n ) F(c j,n ). n= As j, the left side of () goes to while the right side eventually includes every interval in the original collection. A similar proof works for an interval of the form (a, ) when lim x F(x) =. A combination of the two works for (, ).

12 0 0 -: Lecture Measurable Functions. Measurable functions are the types of functions that we can integrate with respect to measures in much the same way that continuous functions can be integrated dx. Recall that the Riemann integral of a continuous function f over a bounded interval is defined as a limit of sums of lengths of subintervals times values of f on the subintervals. The measure of a set generalizes the length while elements of the σ- field generalize the intervals. Recall that a real-valued function is continuous if and only if the inverse image of every open set is open. This generalizes to the inverse image of every measurable set being measurable. Definition. Let (Ω, F) and (S, A) be measurable spaces. Let f : Ω S be a function that satisfies f (A) F for each A A. Then we say that f is F/A-measurable. If the σ-field s are to be understood from context, we simply say that f is measurable. Example. Let F = Ω. Then every function from Ω to a set S is measurable no matter what A is. Example. Let A = {, S}. Then every function from a set Ω to S is measurable, no matter what F is. Proving that a function is measurable is facilitated by noticing that inverse image commutes with union, complement, and intersection. That is, f (A C ) = [f (A)] C for all A, and for arbitrary collections of sets {A α } α ℵ, ( ) f A α = f (A α ), α ℵ α ℵ ( ) f A α = f (A α ). α ℵ α ℵ 0 Exercise. Is the inverse image of a σ-field is a σ-field? That is, if f : Ω S and if A is a σ-field of subsets of S, then f (A) is a σ-field of subsets of Ω. Proposition. If f is a continuous function from one topological space to another (each with Borel σ-field s) then f is measurable. The proof of this makes use of Lemma 0. Definition. Let f : Ω S, where (S, A) is a measurable space. The σ-field f (A) is called the σ-field generated by f. The σ-field f (A) is sometimes denoted σ(f). It is easy to see that f (A) is the smallest σ-field C such that f is C/A-measurable. We can now prove the following helpful result.

13 0 0 0 Lemma 0. Let (Ω, F) and (S, A) be measurable spaces and let f : Ω S. Suppose that C is a collection of sets that generates A. Then f is measurable if f (C) F. Exercise. Prove Proposition 0. Definition. If (Ω, F, P) is a probability space and X : Ω IR is measurable, then X is called a random variable. In general, if X : Ω S, where (S, A) is a measurable space, we call X a random quantity. Exercise. Prove the following. Let S = IR in Lemma 0. Let D be a dense subset of IR, and let C be the collection of all intervals of the form (, a), for a D. To prove that a real-valued function is measurable, one need only show that {ω : f(ω) < a} F for all a D. Similarly, we can replace < a by > a or a or a. Exercise. Show that a monotone increasing function is measurable. Example. Suppose that f : Ω IR takes values in the extended reals. Then f ({, }) = [f ((, ))] C. Also f ({ }) = {ω : f(ω) > n}, n= and similarly for. In order to check whether f is measurable, we need to see that the inverse images of all semi-infinite intervals are measurable sets. If we include the infinite endpoint in these intervals, then we don t need to check anything else. If we don t include the infinite endpoint, and if both infinite values are possible, then we need to check that at least one of { } or { } has measurable inverse image. Definition. A measurable function that takes at most finitely many values is called a simple function. Example. Let (Ω, F) be a measurable space and let A,...,A n be disjoint elements of F, and let a,..., a n be real numbers. Then f = n i= a ii Ai defines a simple function since f ((, a)) is a union of at most finitely many measurable sets. Definition. Let f be a simple function whose distinct values are a,..., a n, and let A i = {ω : f(ω) = a i }. Then f = n i= a ii Ai is called the canonical representation of f. Lemma. Let f be a nonnegative measurable extended real-valued function from Ω. Then there exists a sequence {f n } n= of nonnegative (finite) simple functions such that f n f for all n and lim n f n (ω) = f(ω) for all ω. Proof. For each n, define A n,k = f ((k/n, (k + )/n]) for k =,...,n and A n,0 = f ([0, /n] (n, )). Let A = f ({ }). Define f n (ω) = n n k=0 ki A n,k (ω)+na. The proof is easy to complete now. Lemma says that each nonnegative measurable function f can be approximated arbitrarily closely from below by simple functions. It is easy to see that if f is bounded the approximation is uniform once n is greater than the bound. Many theorems about real-valued functions are easier to prove for nonnegative measurable functions. This leads to the common device of splitting a measurable function f as follows.

14 0 0 Definition 0. Let f be a real-valued function. The positive part f + of f is defined as f + (ω) = max{f(ω), 0}. The negative part f of f is f (ω) = min{f(ω), 0}. Notice that both the positive and negative parts of a function are nonnegative. It follows easily that f = f + f. It is easy to prove that the positive and negative parts of a measurable function are measurable. Here are some simple properties of measurable functions. Theorem. Let (Ω, F), (S, A), and (T, B) be measurable spaces.. If f is an extended real-valued measurable function and a is a constant, then af is measurable.. If f : Ω S and g : S T are measurable, then g(f) : Ω T is measurable.. If f and g are measurable real-valued functions, then f + g and fg are measurable. Proof. For a = 0, part is trivial. Assume a 0. Because { {ω : f(ω) < c/a} if a > 0, {ω : af(ω) < c} = {ω : f(ω) > c/a} if a < 0, we see that af is measurable. For part, just notice that [g(f)] (B) = f (g (B)). For part, let h : IR IR be defined by h(x, y) = x + y. This function is continuous, hence measurable. Then f + g = h(f, g). We now show that (f, g) : Ω IR is measurable, where (f, g)(ω) = (f(ω), g(ω)). To see that (f, g) is measurable, look at inverse images of sets that generate B, namely sets of the form (, a] (, b], and apply Lemma 0. We see that (f, g) ((, a] (, b]) = f ((, a]) g ((, b]), which is measurable. Hence, (f, g) is measurable and h(f, g) is measurable by part. Similarly f g is measurable. You can also prove that f/g is measurable when the ratio is defined to be an arbitrary constant when g = 0. Similarly, part can be extended to extended real-valued functions so long as care is taken to handle cases of and 0.

15 0 0 -: Lecture Theorem. Let f n : Ω IR be measurable for all n. Then the following are measurable:. lim sup n f n,. lim inf n f n,. {ω : lim n f n exists}. { limn f. f = n where the limit exists, 0 elsewhere. Exercise. Prove Lemma. Random Variables and Induced Measures. Example. Let Ω = (0, ) with the Borel σ-field, and let µ be Lebesgue measure, a probability. Let Z 0 (ω) = ω. For n, define X n (ω) = Z n (ω) and Z n (ω) = Z n (ω) X n (ω). All X n s and Z n s are random variables. Each X n takes only two values, 0 and. It is easy to see that µ({ω : X n (ω) = }) = /. It is also easy to see that µ({ω : Z n (ω) c}) = c for 0 c. Each measurable function from a measure space to another measurable space induces a measure on its range space. Lemma. Let (Ω, F, µ) be a measure space and let (S, A) be a measurable space. Let f : Ω S be a measurable function. Then f induces a measure on (S, A) defined by ν(a) = µ(f (A)) for each A A. Proof. Clearly, ν 0 and ν( ) = 0. Let {A n } n= be a sequence of disjoint elements of A. Then ( ) ( [ ]) ν A n = µ f A n n= n= ( ) = µ f [A n ] = = n= µ(f [A n ]) n= ν(a n ). n= The measure ν in Lemma is called the measure induced on (S, A) from µ by f. This measure is only interesting in special cases. First, if µ is a probability then so is ν.

16 0 0 0 Definition. Let (Ω, F, P) be a probability space and let (S, A) be a measurable space. Let X : Ω S be a random quantity. Then the measure induced on (S, A) from P by X is called the distribution of X. We typically denote the distribution of X by µ X. In this case, µ X is a measure on the space (S, A). Example. Consider the random variables in Example. The distribution of each X n is the Bernoulli distribution with parameter /. The distribution of each Z n is the uniform distribution on the interval (0, ). These were each computed in Example. If µ is infinite and f is not one-to-one, then the induced measure may be of no interest at all. Exercise. Either prove or create a counterexample to the following conjecture: If µ is a σ-finite on some measurable space (Ω, F), then for any measurable function f from Ω to S, the induced measure is also σ-finite. Example. (Jacobians) If Ω = S = IR k and f is one-to-one with a differentiable inverse, then ν is the measure you get from the usual change-of-variables formula using Jacobians. We have just seen how to construct the distribution from a random variable. Oddly enough, the opposite construction is also available. First notice that every probability ν on (IR, B ) has a distribution function F defined by F(x) = ν((, x]). Now, we can construct a probability space (Ω, F, P) and a random variable X : Ω IR such that ν = P(X ). Indeed, just let Ω = IR, F = B, P = ν, and X(ω) = ω. Integration. Let (Ω, F, µ) be a measure space. The definition of integral is done in three stages. We start with simple functions. Definition 0. Let f : Ω IR +0 be a simple function with canonical representation f(ω) = n i= a ii Ai (ω) The integral of f with respect to µ is defined to be n i= a iµ(a i ). The integral is denoted variously as fdµ, f(ω)µ(dω), or f(ω)dµ(ω). The values ± are allowed for an integral. We use the following convention whenever necessary in defining an integral: ± 0 = 0. This applies to both the case when the function is 0 on a set of infinite measure and when the function is infinite on a set of 0 measure. Proposition. If f g and both are nonegative and simple, then fdµ gdµ. Definition. We say that f is integrable with respect to µ if fdµ is finite. Example. A real-valued simple function is always integrable with respect to a finite measure. Notation: When X is a random quantity and B is a set in the space where X takes its values, we use the following two symbols interchangeably: X (B) and X B. Both of these stand for {ω : X(ω) B}. Finally, for all B, µ X (B) = Pr(X B) = P(X (B)).

17 -: Lecture The second step in the definition of integral is to consider nonnegative measurable functions. Definition. For nonnegative measurable f, define the integral of f with respect to µ by fdµ = sup gdµ. nonnegative finite simple g f That is, if f is nonnegative and measurable, fdµ is the least upper bound (possibly infinite) of the integrals of nonnegative finite simple functions g f. Proposition helps to show that Definition 0 is a special case of Definition, so the two definitions do not conflict when they both apply. Finally, for arbitrary measurable f, we first split f into its positive and negative parts, f = f + f. Definition. Let f be measurable. If either f + or f is integrable with respect to µ, we define the integral of f with respect to µ to be f + dµ f dµ, otherwise the integral does not exist. It is easy to see that Definition is a special case of Definition, so the two definitions do not conflict when they both apply. The reason for splitting things up this way is to avoid ever having to deal with. One unfortunate consequence of this three-part definition is that many theorems about integrals must be proven in three steps. One fortunate consequence is that, for most of these theorems, at least some of the three steps are relatively straightforward. Definition. If A F, we define A fdµ by I A fdµ. Proposition. If f g and both integrals are defined, then fdµ gdµ. Example. Let µ be counting measure on a set Ω. (This measure is not σ-finite unless Ω is countable.) If A Ω, then µ(a) = #(A), the number of elements in A. If f is a nonnegative simple function, f = n i= a ii Ai, then n fdµ = a i #(A i ) = f(ω). All ω i= It is not difficult to see that the equality of the first and last terms above continues to hold for all nonnegative functions, and hence for all integrable functions. Before we study integration in detail, we should note that integration with respect to Lebesgue measure is the same as the Riemann integral in many cases. Theorem. Let f be a continuous function on a closed bounded interval [a, b]. Let µ be Lebesgue measure. Then the Riemann integral b f(x)dx equals fdµ. a [a,b]

18 Exercise 0. Prove Theorem. Example. A case in which the Riemann integral differs from the Lebesgue integral is that of improper Riemann integrals. These are defined as limits of Riemann integrals that are each defined in the usual way. For example, integrals of unbounded functions and integrals over unbounded regions cannot be defined in the usual way because the Riemann sums would always be or undefined. Consider the function f(x) = sin(x)/x over the interval [, ). It is not difficult to see that neither f + nor f is integrable with respect to Lebesgue measure. Hence, the integral that we have defined here does not exist. However, T the improper Riemann integral is defined as lim T f(x)dx, if the limit exists. In this case, the limit exists Some simple properties of integrals include the following: For c a constant, cfdµ = c fdµ if the latter exists. If f 0, then fdµ 0. If f is extended real-valued, then fdµ < only if µ(f ({± })) = 0. if f = g a.e. [µ] and if either fdµ or gdµ exists, then so does the other, and they are equal. Similarly, if one of the integrals doesn t exist, then neither does the other. Definition. If P is a probability and X is a random variable, then XdP is called the mean of X, expected value of X, or expectation of X and denoted E(X). If E(X) = µ is finite, then the variance of X is Var(X) = E[(X µ) ]. The mean and variance of a random variable have an interesting relation to the tail of the distribution. Proposition. (Markov inequality) Let X be a nonnegative random variable. Then Pr(X c) E(X)/c. There is also a famous corollary. Corollary. (Tchebychev inequality) Let X have finite mean µ. Then Pr( X µ c) Var(X)/c. Exercise. Show that there is some random variable X for which Pr( X µ c) = Var(X)/c. Thus, without additional assumptions, Tchebychev s inequality cannot be improved. We would like to be able to prove that (f + g)dµ = fdµ + gdµ whenever at least two of them are finite. We could prove this for nonnegative simple functions now, but not in general. Proposition. Let f and g be nonnegative simple functions defined on a measure space (Ω, F, µ). Then (f + g)dµ = fdµ + gdµ. The proof for general functions requires some limit theorems first.

19 -: Lecture One of the famous limit theorems is the following. Theorem. (Fatou s lemma) Let {f n } n= be a sequence of nonnegative measurable functions. Then lim inf f n dµ lim inf f n dµ. n n 0 The proof of Theorem is given in a separate document. Here is an outline of the proof. Let f = lim inf n f n, and let φ be an arbitrary nonnegative simple function such that φ f. We need to show that φdµ lim inf n fn dµ. The set {ω : φ(ω) > 0} can be written as the union of the sets A n = {ω : f k (ω) > ( ǫ)φ(ω), for all k n}. For each n, f n dµ ( ǫ) A n φdµ. The liminf of the right sides can be shown to equal ( ǫ) φdµ. Since lim inf n fn dµ ( ǫ) φdµ for all ǫ > 0, we have what we need. The first of the two most useful limit theorems is the following. Theorem. (Monotone convergence theorem) Let {f n } n= be a sequence of measurable nonnegative functions, and let f be a measurable function such that f n f a.e. [µ] and lim n f n = f(x) a.e. [µ]. Then, lim n f n dµ = fdµ. 0 Proof. Since f n f for all n, f n dµ fdµ for all n. Hence lim inf n f n dµ lim sup n By Fatou s lemma, fdµ lim inf n fn dµ. f n dµ fdµ. Exercise. Why is it called the monotone convergence theorem? We are now in a position to prove that the integral of the sum is the sum of the integrals. Theorem 00. If fdµ and gdµ are defined and they are not both infinite and of opposite signs, then [f + g]dµ = fdµ + gdµ. Proof. If f, g 0, then by Lemma, there exist sequences of nonnegative simple functions {f n } n= and {g n } n= such that f n f and g n g. Then (f n + g n ) (f + g) and [fn +g n ]dµ = f n dµ+ g n dµ by Proposition. The result now follows from the monotone

20 0 0 0 convergence theorem. For integrable f and g, note that (f+g) + +f +g = (f+g) +f + + g +. What we just proved for nonnegative functions implies that (f + g) + dµ + f dµ + g dµ = [(f + g) + + f + g ]dµ = [(f + g) + f + + g + ]dµ = (f + g) dµ + f + dµ + g + dµ. Rearranging the terms in the first and last expressions gives the desired result. If both f and g have infinite integral of the same sign, then it follows easily that f + g has infinite integral of the same sign. Finally, if only one of f and g has infinite integral, it also follows easily that f + g has infinite integral of the same sign. For proving theorems about integrals, there is a common sequence of steps that is often called the standard machinery or standard machine. It is illustrated in the next result, the measure-theoretic version of the change-of-variables formula. Lemma 0. Let (Ω, F, µ) be a measure space and let (S, A) be a measurable space. Let f : Ω S be a measurable function. Let ν be the measure induced on (S, A) by f from µ. (See Definition.) Let g : S IR be A/B measurable. Then (0) gdν = g(f)dµ, if either integral exists. Proof. First, assume that g = I A for some A A. Then (0) becomes ν(a) = µ(f (A)), which is the definition of ν. Next, if g is a nonnegative simple function, then (0) holds by linearity of integrals. If g is a nonnegative function, then use the monotone convergence theorem and a sequence of nonnegative simple functions converging to g from below to see that (0) holds. Finally, for general g, (0) holds if either g + or g is integrable. Exercise 0. Suppose that f n is integrable for each n and sup n fn dµ <. Show that, if f n f, then f is integrable and f n dµ fdµ. Exercise 0. Show that if f and g are integrable, then fdµ gdµ f g dµ. 0 Exercise 0. Assume the sequence of functions f n is defined on a measure space (Ω, F, µ) such that µ(ω) <. Further, suppose that the f n are uniformly bounded and that f n f uniformly. Show that f n dµ fdµ.

21 Fatou s Lemma Theorem. (Fatou s lemma) Let {f n } n= be a sequence of nonnegative measurable functions. Then lim inf f n dµ lim inf f n dµ. n n Proof. Let f(ω) = lim inf n f n (ω). Because fdµ = sup finite simple φ f φdµ, 0 we need only prove that, for every finite simple φ f, φdµ lim inf f n dµ. n Let φ f be finite and simple, and let ǫ > 0. For each n, define A n = {ω A : f k (ω) ( ǫ)φ(ω), for all k n}. Since ( ǫ)φ(ω) f(ω) for all ω with strict inequality wherever either side is positive, n= A n = Ω and A n A n+ for all n. Let B n = A A C n. (0) f n dµ f n dµ ( ǫ) A n φdµ. A n Let the canonical representation of φ be m i= c ii Ci. Then, for all n. A n φdµ = m c i µ(c i A n ). i= Because the A n s form an increasing sequence whose union is Ω, lim n µ(c i A n ) = µ(c i ) for all i. Taking the lim inf n of both sides of (0) yields lim inf n f n dµ ( ǫ) m i= c i µ(c i ) = ( ǫ) φdµ. 0 Since this is true for every ǫ > 0, lim inf n f n dµ φdµ.

22 -: Lecture Lemma 0 has a widely-used corollary. Corollary 0. (Law of the unconscious statistician) If X : Ω S is a random quantity with distribution µ X and if f : S IR is measurable, then E[f(X)] = fdµx. Another useful application of monotone convergence is the following. Theorem 0. Let (Ω, F, µ) be a measure space, and let f : Ω IR +0 be measurable. Then ν(a) = fdµ is a measure on (Ω, F). A Exercise 0. Prove Theorem 0. 0 If µ is σ-finite and if f is finite a.e. [µ], then ν in Theorem 0 is σ-finite. What goes wrong with the conclusion to Theorem 0 if f is integrable but not necessarily nonnegative? If f can take negative values then ν(a) = fdµ might be negative. Let A A = {ω : f(ω) < 0}. Suppose that µ(a) > 0. Write A = n= A n, where A n = {ω : f(ω) < /n}. If µ(a) > 0, then there exists n such that µ(a n ) > 0. (This argument is used often in proving probability results.) Then ν(a) = I A ( f)dµ I An ( f)dµ n µ(a n) > Here is another application of the standard machinery. Theorem 0. Assume the same conditions as Theorem 0. Integrals with respect to ν can be computed as gdν = gfdµ, if either exists. Proof. We prove the result in four stages. First, assume that g is a indicator I A of some set A F. Then the definition of ν says that gdν = ν(a) = I A fdµ. Second, assume that g is a nonnegative simple function. The result holds for g by linearity of integrals. Third, assume that g is nonnegative. Approximate g from below by nonnegative simple functions {g n } n=. Then g n dν = g n fdµ for each n and the monotone convergence theorem says that the left side converges to gdν and the right side converges to gfdµ. Finally, if g is measurable, write g = g + g (the positive and negative parts). Then g + dν = g + fdµ and g dν = g fdµ. We see that gdν exists if and only if gfdµ exists, and if either exists they are equal. The standard machinery corresponds to the three stages in defining integrals. The first stage is split into indicators and nonnegative simple functions to make four steps in the standard machinery. Definition. The function f in Theorem 0 is called the density of ν with respect to µ.

23 0 Example. (Probability density functions) Consider a continuous random variable X having a density f. That is, Pr(X a) = a f(x)dx. Then the distribution of X, defined by µ X (B) = Pr(X B) for B B, satisfies µ X (B) = fdλ, where λ is Lebesgue measure. That is, the probability density functions of the usual continuous distributions that you learned about in earlier courses are also densities with respect to Lebesgue measure in the sense defined above. Example. (Probability mass functions) Consider a typical discrete random variable X with mass function f, i.e., f(x) = Pr(X = x) for all x. There are at most countably many x such that f(x) > 0. Let µ X be the distribution of X. For each set B, we know that µ X (B) = Pr(X B) = f(x). x B The rightmost term in this equation is fdµ, where µ is counting measure on the range space of X. So, f is the density of µ X with respect to µ. The other major limit theorem is the following. B Theorem. (Dominated convergence theorem) Let {f n } n= be a sequence of measurable functions, and let f and g be measurable functions such that f n f a.e. [µ], 0 f n g a.e. [µ], and gdµ <. Then, lim n f n dµ = fdµ. Proof. We have g f n g a.e. [µ], hence g + f n 0, a.e. [µ], g f n 0, a.e. [µ], lim n] = g + f n a.e. [µ], lim n] = g f n a.e. [µ]. It follows from Fatou s lemma and Theorem 00 that [g + f]dµ lim inf [g + f n ]dµ n = gdµ + lim inf f n dµ, n fdµ lim inf f n dµ. n

24 Similarly, it follows that [g f]dµ lim inf [g f n ]dµ n = gdµ lim sup f n dµ, n fdµ lim sup f n dµ. n 0 Together, these imply the conclusion of the theorem. Example. Let µ be a finite measure. Then limits and integrals can be interchanged whenever the functions in the sequence are uniformly bounded. An alternate version of the dominated convergence theorem is the following. Proposition. Let {f n } n=, {g n } n= be sequences of measurable functions such that f n g n, a.e. [µ]. Let f and g be measurable functions such that lim n f n = f and lim n g n = g, a.e. [µ]. Suppose that lim n gn dµ = gdµ <. Then, lim n fn dµ = fdµ. The proof is the same as the proof of Theorem, except that g n replaces g in the first three lines and wherever g appears with f n and a limit is being taken. For finite measure spaces (i.e. (Ω, F, µ) with µ(ω) < ), the minimal condition that guarantees convergence of integrals is uniform integrability. Definition. A sequence of integrable functions {f n } n= is uniformly integrable (with respect to µ) if lim c sup n {ω: f f n(ω) >c} n dµ = 0. Theorem. Let µ be a finite measure. Let {f n } n= be a sequence of integrable func- tions such that lim n f n = f a.e. [µ]. Suppose that {f n } 0 n= is uniformly integrable. Then lim n fn dµ = fdµ. If the f n s in Theorem are nonnegative and integrable and f n f, then lim n fn dµ = fdµ implies that {fn } n= are uniformly integrable. We will not use this result, however.

25 -: Lecture 0 Here are some more useful properties of integrals. Theorem. Let (Ω, F, µ) be a measure space. Let f and g be measurable extended real-valued functions.. If f is nonnegative and µ({ω : f(ω) > 0}) > 0, then fdµ > 0.. If f and g are integrable and if fdµ = gdµ for all A F, then f = g a.e. [µ]. A A. If µ is σ-finite and if fdµ = gdµ for all A F, then f = g a.e. [µ]. A A. Let Π be a π-system that generates F. Suppose that Ω is a finite or countable union of elements of Π. If f and g are integrable and if fdµ = gdµ for all A Π, then A A f = g a.e. [µ]. 0 Proof.. Let A c = {ω : f(ω) > c} for each c 0. Because µ(a 0 ) > 0 and A 0 = n= A /n, it follows from Lemma that there exists n such that µ(a /n ) > 0. Since f fi A/n, we have fdµ A /n fdµ. But (/n)i A/n is a simple function that is fi A/n and dµ = µ(a (/n)ia/n /n) > 0. It follows that fdµ > 0.. This will appear on a homework assignment.. First, assume that f and g are real-valued. Let {A n } n= be disjoint elements of F such that µ(a n ) < and n= A n = Ω. Let B m = {ω : f(ω) < m, g(ω) < m} for each integer m. For each pair (n, m), fi An B m and gi An B m satisfy the conditions of the 0 previous part, so fi An B m = gi An B m a.e. [µ]. Let C = {ω : f(ω) g(ω)}. Since C = [C B m A n ], n= m= and each µ(c B m A n ) = 0, it follows that µ(c) = 0. Next, suppose that f and/or g is extended real-valued. Let E = {f = } {g = }, the set where one function is but the other is not. If µ(e) > 0, then there is a subset A of E such that 0 < µ(a) < and one of the functions is bounded above on A while the other is infinite. This contradicts fdµ = gdµ. A similar result holds A A for. 0. Define ν + (A) = A f+ dµ, ν + (A) = A g+ dµ, ν (A) = f dµ, and ν A (A) = A g dµ. These are all finite measures according to Theorem 0. The additional condition implies that they are all σ-finite on Π. The equality of the integrals implies that ν + + ν = ν + ν+ for all sets in Π. Theorem implies that ν+ + ν = ν + ν+ for all sets in F. Hence, the condition of part hold and the result is proven.

26 The condition about unions in part of the above theorem holds for the π-systems in Example. Corollary 0. If µ is σ-finite and ν is related to µ as in Theorem 0, then the density of ν with respect to µ is unique, a.e. [µ]. There is an interesting characterization of σ-finite measures in terms of integrals. Theorem. Let (Ω, F, µ) be a measure space. Then µ is σ-finite if and only if there exists a strictly positive integrable function. Exercise. Prove Theorem. 0 Absolute Continuity. There is a special relationship between measures on the same space that is very useful in Probability theory. Definition. Let ν and µ be measures on the space (Ω, F). We say that ν µ (read ν is absolutely continuous with respect to µ) if for every A F, µ(a) = 0 implies ν(a) = 0. That is, ν µ if and only if every measure 0 set under µ is also a measure 0 set under ν. Example. Let (Ω, F, µ) be a measure space. Let f be a nonnegative function, and define ν(a) = fdµ. Then ν is a measure and ν µ. If f < a.e. [µ] and if µ is σ-finite, A then ν is σ-finite as well. 0 Example. Let µ and µ be measures on the same space. Let µ = µ + µ. Then µ i µ for i =,. Absolute continuity has a connection with continuity of functions. Proposition. Let ν and µ be measures on the space (Ω, F). Suppose that, for every ǫ > 0, there exists δ such that for every A F, µ(a) < δ implies ν(a) < ǫ. Then ν µ. A concept related to absolute continuity is singularity. Definition. Two measures µ and ν on the same space (Ω, F) are (mutually) singular (denoted µ ν) if there exist disjoint sets S µ and S ν such that µ(sµ C ) = ν(sc ν ) = 0. Example. Let f and g be nonnegative functions such that fg = 0 a.e. [µ]. Define 0 ν (A) = A fdµ and ν (A) = A gdµ. Then ν ν. The main theoretical result on absolute continuity is the Radon-Nikodym theorem which says that, in the σ-finite case, all absolute continuity is of the type in Example. Theorem. (Radon-Nikodym) Let µ and ν be σ-finite measures on the space (Ω, F). Then ν µ if and only if there exists a nonnegative measurable f such that ν(a) = fdµ for all A F. The function f is unique a.e. [µ]. A One proof of this result is given in a separate course document. Another proof is given later after we introduce conditional expectation. Definition 0. The function f in Theorem is called a Radon-Nikodym derivative of ν with respect to µ. It is denoted dν/dµ. Each such function is called a version of dν/dµ.

27 : Lecture The uniqueness of Radon-Nikodym derivatives is only a.e. [µ]. If f = dν/dµ, then every measurable function that equals f a.e. [µ] could also be called dν/dµ. All of these functions are called versions of the Radon-Nikodym derivative. Definition. If µ ν and ν µ, we say that µ and ν are equivalent. If µ and ν are equivalent, then dµ dν =. dν dµ If ν µ η, then the chain rule for R-N derivatives says dν dη = dν dµ dµ dη. Absolute continuity plays an important role in statistical inference. Parametric families are collections of probability measures that are all absolutely continuous with respect to a single measure. Theorem. Let (Ω, F, µ) be a σ-finite measure space. Let {µ θ : θ Θ} be a collection of measures on (Ω, F) such that µ θ µ for all θ Θ. Then there exists a sequence of nonnegative numbers {c n } n= and a sequence of elements {θ n} n= of Θ such that n= c n and µ θ n= c nµ θn for all θ Θ. We will not prove this theorem here. (See Theorem A. in Schervish.) Random Vectors. In Definition we defined random variables and random quantities. A special case of the latter and generalization of the former is a random vector. Definition. Let (Ω, F, P) be a probability space. Let X : Ω IR k be a measurable function. Then X is called a random vector. There arises, in this definition, the question of what σ-field of subsets of IR k should be used. When left unstated, we always assume that the σ-field of subsets of a multidimensional real space is the Borel σ-field, namely the smallest σ-field containing the open sets. However, because IR k is also a product set of k sets, each of which already has a natural σ-field associated with it, we might try to use a σ-field that corresponds to that product in some way. Product Spaces. The set IR k has a topology in its own right, but it also happens to be a product set. Each of the factors in the product comes with its own σ-field. There is a way of constructing σ-field s of subsets of product sets directly without appealing to any additional structure that they might have.

28 0 0 0 Definition. Let (Ω, F ) and (Ω, F ) be measurable spaces. Let F F be the smallest σ-field of subsets of Ω Ω containing all sets of the form A A where A i F i for i =,. Then F F is the product σ-field. Lemma. Let (Ω, F ) and (Ω, F ) be measurable spaces. Suppose that C i is a π- system that generates F i for i =,. Let C = {C C : C C, C C }. Then σ(c) = F F, and C is a π-system. Proof. Because σ(c) is a σ-field, it contains all sets of the form C A where A F. For the same reason, it must contain all sets of the form A A for A i F i (i =, ). Because (C C ) (D D ) = (C D ) (C D ), we see that C is a π-system. Example. Let Ω i = IR for i =,, and let F and F both be B. Let C i be the collection of all intervals centered at rational numbers with rational lengths. Then C i generates F i for i =, and the product topology is the smallest topology containing C as defined in Lemma. It follows that F F is the smallest σ-field containing the product topology. We call this σ-field B. Example. This time, let Ω = IR and Ω = IR. The product set is IR and the product σ-field is called B. It is also the smallest σ-field containing all open sets in IR. The same idea extends to each finite-dimensional Euclidean space, with Borel σ-field s B k, for k =,,.... Lemma. Let (Ω i, F i ) and (S i, A i ) be measurable spaces for i =,. Let f i : Ω i S i be a function for i =,. Define g(ω, ω ) = (f (ω ), f (ω )), which is a function from Ω Ω to S S. Then f i is F i /A i -measurable for i =, if and only if g is F F /A A - measurable. Proof. For the only if direction, assume that each f i is measurable. It suffices to show that for each product set A A (with A i A i for i =, ) g (A A ) F F. But, it is easy to see that g (A A ) = f (A ) f (A ) F F. For the if direction, suppose that g is measurable. Then for every A A, g (A S ) F F. But g (A S ) = f (A ) Ω. The fact that f (A ) F will now follow from the first claim in Proposition 0. (Sorry for the forward reference.) So f is measurable. Similarly, f is measurable. Proposition. Let (Ω, F), (S, A ), and (S, A ) be measurable spaces. Let X i : Ω S i for i =,. Define X = (X, X ) a function from Ω to S S. Then X i is F/A i measurable for i =, if and only if X is F/A A measurable. Lemma and Proposition extend to higher-dimensional products as well. The product σ-field is also the smallest σ-field such that the coordinate projection functions are measurable. The coordinate projection functions for a product set S S are the functions f i : S S S i (for i =, ) defined by f i (s, s ) = s i (for i =, ).