 Welcome to lecture number 26 on measure and integration. We have been studying the properties of the product measure spaces. We defined what is the product of two measure spaces and then we started looking at the problem of how to compute the measure of an element E in the product sigma algebra. So, we will continue that study in the today's lecture. So, today's lecture, the main aim is going to be computing the product measure. .. So, let us as we call that given sigma finite measure space is x a mu and y b nu, we have defined the product measure space namely x cross y and the sigma algebra on x cross y is a times b, which is the sigma algebra generated by all measurable rectangles that is sets of the type E cross f where E belongs to a and f belongs to b and mu cross nu is extension of the measure, which is defined on the rectangles by the property that mu cross nu of a rectangle E cross f is mu E times nu of f. Then via outer measures, we extend that to the sigma algebra a cross b. So, the problem we wanted to analyze was that given a set E in the product sigma algebra a times b, how do we compute the product measure mu cross nu of E? Because at present, we only know that mu cross nu is defined on the sigma algebra a times b via the extension theory. So, to do that, we had said that we should try to look at what are called the sections of the set E. So, let us recall what is called the section. So, for a point element x in E, E lower x. So, this is a notation used for the set of all points y this is not R this should be in y. So, all the points y in y say that x comma y belongs to E and similarly, we can define the section with respect to a point y in y. So, the main questions that we had formulated in the previous lecture were can we say that this section is a element in the sigma algebra b on this is a subset of y. So, is it an element in the sigma algebra b? If yes, then we can define nu of E x, which depends on x. So, we get a function x going to nu of E x. So, the question is, is that a measurable function as a function of x? If it is measurable, we can define its integral with respect to mu and then ask whether that is equal to the product measure eta of E. So, and similarly, can we interchange the role of x and y? So, these are the questions we want to analyze. If that turns out to be true, then this will give us the way of computing the product measure of a set E by looking at the sections, taking the new measure and then adding up or integrating and similarly, the other way round. So, let us start analyzing these problems one by one. So, for the sake of once again clarity, let us note down E for any element E in x cross y, E is a subset of x cross y, x n element in the set x and y n element y, the section E lower x is defined as all points y in y such that x comma y belongs to E and similarly, the set E upper y. So, whenever the point is coming from x, we will write it on the bottom E lower script x and whenever it is coming from the set y, we will write as E superscript y on the right side as all points x in x such that x comma y belongs to E. So, these we were called as the sections. So, E lower x is called the x section of E at the point x and it is a subset of y and similarly, the y section of E or the section of E at y is a subset of x. So, that is E upper y. So, the questions we will like to analyze are the following and or we want to claim that the following holds. So, let us look at the some general properties of these sections. The first is if E and F are subsets such that E is a subset of F, then the section of E at x is a subset of the section of F at x and the section of E at y is a subset of the section of F at y. So, let us prove these properties before going further. So, let us verify namely, so we are given E is a subset of x and F is a subset of y. So, for x belonging to x, let us look at what is E x and we are also given that E is a subset of F. So, for any point x in x, let us look at the section E lower x. So, that is by definition all points y in y such that x comma y belongs to E. Now, but so if x comma y belongs to E and E is a subset of F. So, that means from here, so what we are saying is note for y belonging to y, we have x comma y belongs to E that is the property here and E is a subset of F. So, x comma y belongs to F. So, that implies that y belongs to F of x also. So, this is the definition of E, F over x namely y in y is that x comma y belongs to F. So, this implies that E lower x is a subset of F lower x and the other property is similar. So, similarly we will have that E section at y is a subset of the section of F at y. So, these two properties hold. So, that is property 1. Here, we do not use any effect that E cross F, these such subsets are of in the sigma algebra E A cross B. This is true for any subsets. The next property says that if we look at the difference E difference F and take its section, it is same as first taking the sections and then taking the differences whether at a point x or at a point y. So, let us analyze this namely that if we E and F are subsets of x cross y. So, we are looking to look at the difference and the sections x is a point in x. So, let us look at the section of E minus F at the point x. So, by definition this is all points y belonging to y such that x comma y belongs to E minus F. So, what does that mean? So, that means y belonging to E minus F section implies that x comma y belongs to E minus F. But what is the meaning of saying that x comma y belongs to E difference F? That is same as saying that x comma y belongs to E belongs to E, but x comma y does not belong to F. That is the meaning of this, but that is same as saying that x comma y belongs to E that means y belongs to E x and x comma y does not belong to F that means y cannot belong to the section of F at x. So, that implies that y belongs to the section of E at x, but does not belong to the section of F at x. So, that means y belongs to E x difference of F of x. So, that says that E lower x. So, that says that E difference F section x is a subset of E lower x difference F lower x. But notice in this all the arguments are reversible. So, supposing y belongs here, so that is same as implying the earlier statement that y belongs to E x and y does not belong to F section at x, but that is same as saying x comma y belongs to E and x comma y does not belong to F and that is same as saying that the earlier statement and that is meaning of that is applied. So, all the arguments are reversible. So, the other way around inequality also holds. So, these two sets are equal. So, it says that if you take the difference of E with F and then take the section at x that is same as taking the sections first and then taking the difference. So, that is at the point x a similar proof will work for the differences at if you take the difference of E with F until the section at y the corresponding property says it is first taking the section and then taking the difference or the corresponding section at y. So, basically we are saying the properties of subsets are preserved under taking sections that is part one and the properties of taking sections is preserved also undertaking differences of sets. So, whether you first take the sections and then take the difference it is same as first taking the difference and then taking the sections. So, these are two elementary properties of the sections. Let us look at some more general properties of the sections. Once again these are true for any sets not necessarily sets in A cross B, but we will be using them only for A cross B, but they are true. So, let us look at a sequence of sets not a sequence actually arbitrary family of sets E i's in subsets of x cross y where i is any indexing set. Then if you look at the intersections of the sets E i's and then take the section the claim is it is same as taking the sections first and then taking the intersections and similarly at the point y. So, let us section whether at x or at y. So, let us look at this property. So, E i's are subsets of x cross y i belonging to some indexing set i. So, we want to look at take the intersections of the sets E i belonging to i. Look at the intersections of this family E i's and then take its section at a point x. So, let us take a point x belonging to x. We want to compute and show that this is same as first take the section of every set E i at x and then take the intersection i belonging to y. So, to show this let us take a point. So, y belonging to intersection i in i of E i at x. So, let us take a point y in this set on the left hand side. So, that is if and only if the definition says that means x comma y belongs to intersection of i in i of the sets E i. But if a point belongs to the intersection that is that means that the point belongs to each one of the sets. So, it belongs to E i for every i belonging to i and once that is true. So, that is but the saying that x comma y belongs to E i that is same as saying y belongs to E i x for every i belonging to i. So, y belonging to the intersection section is same as y belonging to intersections. So, for every i that means y belongs to intersections of the sections of E i at x i belonging to i. So, that proves that this property is true. So, what we have shown is that this property is true. Namely, for any arbitrary family of subsets of x cross y, if you take the intersections of these sets and then take the section at a point x it is same as first taking the sections and then taking the intersection of those sections. So, this is at a point x in x a similar proof will work for the sections at y. Namely, for every y in y the section of the intersection is same as intersection of the sections. A corresponding result also is true for the unions. So, let us prove that also. Once again the proofs are similar in all these cases is only a matter of is pure set theory actually. So, what we want to do is E i's are subsets of x cross y where i belongs to i. So, what we want to do is we want to look at the union of these sets E i i belonging to i and then it take its section at a point x. So, let us take a point x in x. We want to show this is same as for each one of the E i's take the section x at a point x that x we have fixed and then take its union over i belonging to y. So, section of the unions is equal to union of the sections. So, that is what we want to prove. So, to prove that so note so y belong to the left hand side of the set. So, that is i belonging to i union E i and its section at x. But y belonging to the section at a point x is same as saying that the point x y belongs to the union of E i's i belonging to i. But the definition of saying that a point belongs to the union means at least it should belong to one of them. So, x comma y belongs to E i for some i belonging to i. But that is same as saying x y belongs to E i that means y belongs to the section E i at x for some i. But that is same as saying it belongs to the union. So, it belongs to at least one of the sections of E i's that means y belongs to union i belonging to i of E i at a point x. So, y belonging to the section of the union if and only if y belongs to union of the sections. So, that proves that this property is true. Namely section of the unions is union of the sections at a point x in x and a similar property holds section of the unions at a point y in y. So, basically what we are saying is all the set theoretic operations behave nicely with respect to taking sections. And this is true for all subsets E i's of x cross y. So, now using this properties we will prove namely if E is a set in the product sigma algebra A times B and x and y are elements x is in x and y is in y. Then the claim is that the section E x belongs to B and the section E y belongs to A. So, that means for every x in x look at the subset of y which is the section of E at a point x that belongs to the sigma algebra on B whenever E is element in the product sigma algebra. And similarly the section at y is a subset of x and our claim is that this belongs to the sigma algebra A. So, these are the two properties we want to check for every set E belonging to A cross B. Now, here is the technique of proving all these results in the product sigma algebra. Basically we will apply the monotone class sigma algebra techniques namely we will whenever we want to show a property holds for A cross B elements in A cross B. We will collect together all subsets for which this property is true and try to show that we will collect sets for which this property is true in A times B and show that that collection includes rectangles and this collection is a sigma algebra. So, once this collection is a sigma algebra and includes rectangles it will include the product sigma algebra A times B. So, that is what I had called as the sigma algebra technique. So, we will apply that technique here. So, let us define the collection S to be all subsets in A times B such that this property which we are calling as star. So, E x the section at x belongs to the sigma algebra B and the section at y belongs to the sigma algebra A. So, what we want to prove? We want to prove that this S is equal to A times B. So, to prove that S is equal to A times B we will prove two things namely S is a sigma algebra and it includes rectangles and that will prove that it actually is equal to A times B. So, let us prove this properties. So, what we are given is we are given that the set E belongs to the sigma algebra. So, set E belongs to the sigma algebra A times B. So, let us so S is the collection of all the subsets E belonging to A cross B such that the section E x belongs to B and the section E y belongs to the sigma algebra A. So, to show S is equal to A product sigma algebra B. Note it is already a subset of A cross B. So, we will follow two things one. So, let us check the properties of this. The first is the rectangles are inside S. So, to check this property let us take a rectangle. So, let A belong to A and B belong to B and let us take the rectangle E which is equal to A cross B. So, if you recall we had calculated what is the section of E at x. So, that is all y belonging to y say that x comma y belongs to B x comma y belongs to A cross B. So, now x comma y can belong to A cross B only when x belongs to A and in that case y should belong to B. So, this set is equal to, so if x belongs to A, so for all x belonging to A this set is equal to B. So, the section is equal to B if x belongs to A and if x does not belong to A then in no way x comma y is going to belong to B. So, this is empty set if x does not belong to A. So, for a rectangle we have already seen I am repeating the steps which we had done earlier namely for a rectangle A cross B the x section is either B or empty set. So, in either case this belongs to the sigma algebra B. So, the property that E x belongs to B is true. A similar argument will show that E y also belongs to A. So, this proves that the rectangles are inside the sigma algebra S. So, the next step we want to check is the following. So, second step we want to check is that this collection S S is a sigma algebra. So, this is what we want to check. .. So, for that the first property look at the empty set the sections of the empty set either x section is same as the section y and that is the empty set and that belongs to both A and B. Similarly, if I look at the whole space that is x cross y that is actually a rectangle which is inside A cross A times B which is A times B and already rectangles are inside S. So, both the whole space and the empty set are inside S and A and B and hence it is also a rectangle. So, actually we should say that this belongs to a rectangle and which is part of S. So, empty set and the whole space both belong to S. The next property let us take a set E belonging to S and show that its complement also belongs to it. But, E belongs to S implies the sections E x belongs to B and E y belongs to A that is the definition of S. So, let us just recall so what was the definition of the set S? The definition of the set S is all subsets A cross B. So, that E x belongs to B and E y belongs to A. So, by the definition this is true, but E x belongs to B and B is a sigma algebra E y belongs to A and A is a sigma algebra. So, that implies that E x complement belongs to B and E y complement belongs to A because of the properties of sigma algebras that A and B are both sigma algebras. So, they must be closed under complements, but on the other hand this set taking section and the complement as now we just now we observed it is same as I can take the complement first and then take the section. So, that should belong to P and similarly here the section y and then complement is same as taking the complement first and then taking the section that should belong to A. So, this set is same as this. So, for every set E in S if I look at the set E complement its section at x belongs to B and its section at y sorry this is E complement. So, at y belongs to A so that implies that E complement also belongs to S. So, S is closed under taking complements and finally to show it is a sigma algebra I have to show it is also closed under say countable unions. So, to show that so let E i is belong to S say S bigger than or equal to 1, but each E i belonging to S implies. So, for every I E i belongs to S implies that E i section at x is in the sigma algebra B and E i section at y belongs to the sigma algebra A. So, this property is true, but once again A and B both are sigma algebras. So, that implies that the union of E i's sections at x i equal to 1 to infinity belongs to B. Similarly, the corresponding one the union i equal to 1 to infinity of E i's section at y belongs to A. So, this is true, but that implies by the fact that this set taking the sections and taking a union is same as first taking the unions and then taking the sections. Let us now we have observed that. So, that belongs to the sigma algebra B and similarly this set first taking the first taking sorry this was E i's at y because union belongs to A. So, that is same as now I can write as the same as union 1 to infinity of E i's at section at y belongs to A. So, that implies that union of the set union of E i's its section at x belongs to B and its section at y belongs to A that means this belongs to the collection S. So, that proves so hence S is a sigma algebra. So, S is a sigma algebra and we know that rectangles are inside S. So, that implies that A cross B is inside S, S is the subset of already A times B. So, all these are equal that means the property that for every set in the product sigma algebra. So, this property that for every set in the product sigma algebra the x section belongs to B and the y section belongs to A is true. So, let me once again emphasize the fact that we are looking at this proves which are nothing but application of the technique called the sigma algebra technique. So, now let us go to the next property namely we want to check the property that we already know that for every x E x is a section of E at x. So, if E is in the product sigma algebra this set E lower x E section of E at x is in the sigma algebra B. So, nu of that set makes sense because nu is defined on the sigma algebra B and similarly the section of E at y is in the sigma algebra A. So, measure of this section mu of this section makes sense, but both nu of E x depends on x and mu of E y depends on y. So, this gives us two functions x going to nu of E x and y going to nu of E y. The first one is a function on the set x and the second one is a function on the set y. So, we want to prove that both of these are measurable functions and clearly these are non-negative functions. So, they are non-negative measurable functions on x and y. So, their integrals make sense with respect to this is a function on x. So, it is integral with respect to mu make sense and this is a non-negative measurable function with respect to y. So, it is integral with respect to the measure nu make sense. So, we want to claim that the integral of the first function with respect to mu is same as the product measure of the set E and which is same as the integral of the second function with respect to nu. So, that will give us a nice way of computing the product measure namely the product measure of a set E can be computed either by taking its sections with respect to x, finding the size of those sections that is the nu measure of the sections with respect to x and then summing it up that is taking integrals with respect to mu or we can interchange the roles of x and y. We can take the sections of E with respect to y first, take its measures with respect to mu and then add up. So, take integrals with respect to nu. So, we want to prove that this property 2 and property 3 hold for every subset E of product sigma algebra A cross B. So, once again this proof is going to be an application of the sigma algebra monotone class technique and you will see how effective these techniques are. So, what we will do? We will collect together all the subsets of A cross B for which these two properties are true and we will try to show rectangles are inside it and hence everything is inside it. .. So, let us look at the collection P of subsets of E cross elements in A cross B. So, that property 2 and 3 both hold. So, what is going to be our technique? So, what is the problem to be proved? So, the problem is to show that this P is equal to A times B. So, to show that we will do the following. First, we will show that rectangles are inside A cross B. So, that is one that the set of all rectangles are inside the class this collection P and we will show the second step namely this collection P is closed under finite disjoint unions. So, what will that prove? You recall we had shown that R is a sigma algebra and if the collection P which includes R is closed under finite disjoint unions that means finite disjoint unions of elements of the rectangles also will be inside P, but finite disjoint union of rectangles. So, is nothing but the algebra generated by this semi algebra R. So, that will prove. So, this step will imply that the algebra generated by the rectangles is inside the class P. So, this first step is to conclude that the algebra generated by rectangles is inside P and the method is to show that R is inside it and f of R. So, it is closed under finite disjoint unions. So, let us prove this step one first. So, we have got the collection P. So, P is the collection of all subsets E belonging to A times B such that those two properties hold. Now, the two properties the properties were that x going to going to nu of E x and y going to mu of E y. These two are measurable functions and that the integral of nu E x with respect to mu is same as integral of mu E y with respect to d nu. So, this is over x and this is over y and both of them are equal to the product sigma algebra namely mu cross nu of E. So, E is essentially what we are saying is we are looking at the sets E in the product sigma algebra for which the required properties hold. So, we want to show the first thing is we want to show that the rectangles are inside P. So, to prove this let us take a rectangle E. So, E is equal to A cross B where A belongs to the sigma algebra A and B belongs to the sigma algebra B. So, let us look at we recall what was the sections. The section E x was equal to empty set if x does not belong to A and it is equal to B if x belongs to B. So, that means this E x is nothing but when x does not belong to A it is empty set. So, what is going to be nu of that? That is going to be 0 E x is going to be set B. So, it is going to be so it is nu of B into the indicator function of A at x. So, this is what is important that for a rectangle A cross B we have already computed the sections x section was empty set if x does not belong to A and it is B if x belongs to B. So, nu of E x is going to be nu of empty set which is 0 if x does not belong to A and if x belongs to A. So, this should be x belongs to A. So, if x belongs to A then it is nu of B and here nu of A is 1. So, this equality whole because if x belongs to A this value is 1 and indicator function of A at x does not belong to A is 0. So, we have got this equation namely nu of E x which we want to show is measurable is nothing but the indicator function constant times the indicator function of a set in the sigma algebra A is in the sigma algebra. So, that implies that x going to nu of E x is A measurable. So, this is a measurable function and similarly if we take the corresponding section with respect to y. So, let us write that also. So, if we look at E of y so that is we are writing it up actually. So, let me write follow the same rotation the section of E at y. So, E superscript y is equal to it is empty set if x does not belong to A and it is equal to A I am sorry let me write it properly. So, the section E y is equal to empty set if y does not belong to B and it is A if y belongs to the set B because y is a point in x. So, that means that mu of E y is going to be equal to mu of the set A times the indicator function of the set B at the point y. So, as a function of y it is just the indicator function of the set B at the point y multiplied by A constant. So, that will imply that y going to mu of E y is B measurable. So, that proves the first thing namely the rectangle we wanted to show that rectangles are inside. So, what we have shown there for a rectangle the first property namely x going to nu of E x and y going to nu y are measurable with respect to the corresponding sigma algebra and let us compute the integrals of these things. So, nu of E x is this function. So, what is its integral with respect to mu this is a constant and this is the indicator function. So, it is nu of B into mu of A. So, from this equation star. So, let us write that from the equation star it follows. So, from star integral of nu of E x d of. So, this is the property star d of mu x is equal to integral of this quantity. So, that is nu of B times mu of A. So, that was the property star and similarly let us look at the integral of the other function .So, we want to compute integral of the function mu of E y. So, but mu of E y is equal to this quantity let us call it as double star. So, once we integrate this what we will get is integral of mu of E y with respect to y d nu y is equal to mu of A into nu of B. So, from the equation double star we will have integral mu of E y d nu of y is equal to mu of A into nu of nu of B. So, in either case these integrals are which is nothing but the product. So, that says so this is equal to the product sigma algebra product measure mu cross nu of the rectangle A cross B and similarly here this is the product mu cross nu of A times B. So, this proves. So, hence what we have shown is that rectangles are inside the collection P of the sets for which we wanted to prove the required claim holds. So, what was the second step we wanted to prove? We want to prove that this is the collection R P. So, claim so the second collection so here is the second thing in the step one. So, we have proved part of the step one namely we have proved that P includes R P includes rectangles. So, the next part of the proof requires first to show that P is closed under finite disjoint unions. So, let us prove that so P is closed under finite disjoint unions. So, for that let us take two sets E and F which belong to P which belong to the collection P. So, that means what? So, that implies that for E and F the corresponding results are true and E and F are disjoint that is also given to us intersection is equal to empty. So, to show we want to show that E union F belongs to P. So, that is what we want to show. So, let us start with looking at the sections of E union F its section at a point x by the definition properties of the sections. The section of the union is union of the sections. So, it is union of E x union of F x. So, what is going to be these are the sections. So, what is going to be nu of the union E union F section? So, what is nu of that? So, that is now as E and F are disjoint these sections are going to be a disjoint. So, it is nu of the disjoint union of the sections E x union F of x and these being disjoint. So, that means this is equal to nu of E x plus nu of F x it is nu of E x plus F of x. So, here is something for you to think and confirm that if E and F are disjoint sets then their corresponding sections are also disjoint and hence this property is true. Now, E and F both belong to P that means this is a measurable function of x and this is also a measurable functions of x and we have proved that some of measurable functions is measurable. So, this will imply that x going to nu of E union F section at x is A measurable. So, this is a measurable function. Similarly, y going to nu of mu of E union F section y is B measurable. So, to check that P is closed under finite disjoint unions we have checked the first property namely if E and F are two disjoint sets in P then x going to nu of E union F section and y going to mu of the section E union F at y are both respectively measurable functions. Now, let us check the next property namely that the integral property is true for the union. So, for that what we want to do is the following. So, we want to integrate. So, let us integrate nu of E union F section this is a measurable function. So, with respect to mu we can integrate this. So, this over the set x of course this we know whether just now we proved that nu of. So, here is nu of E union x E union F at x is nu of E x plus nu of F of x. So, let us use that property and so this we can write as x. So, the integral nu of E union F of x is equal to nu of E x plus nu of F of x d mu x. So, now using the properties of the integral this we can split it as integral over x of nu E x d mu x plus integral over x of nu of F of x with respect to d mu x. Now, because E and F both are inside the class inside the collection P for which this property integral of the section nu E x d mu x is nothing but mu cross nu of E and the second integral is nothing but mu cross nu of F. Now, by the fact that E and F are disjoint and mu cross nu is a measure this is nothing but mu cross nu of E union F. So, what we have shown is that if I integrate nu of E union F section with respect to mu that is the product measure of the set E union F. A corresponding result will also hold when I take y sections namely we can show that integral of mu E union F at y. So, similarly so let me just write the argument that is the corresponding result will be similar. Similarly, if I integrate over y and mu of E union F section at y d nu of y. So, if I take the section of E union F with respect to y take its mu measure. So, that is a measurable function and its integral with respect to nu that we just now we observed that this section is nothing but mu of E y plus mu of F y and that was because this section E union F section is same as E section union F section and they are disjoint some measures add up and that is equal to d of nu y. Now, once again as before we can write this as mu cross nu of E plus mu cross nu of F and by again using the property of that mu cross nu is a measure E and F are disjoint. This is mu cross nu of E union F. So, that proves the second part of the property namely that not only P includes rectangles in fact P is closed under finite disjoint unions. So, as a consequence of this because finite disjoint union of elements of a semi algebra give us the algebra generated by the semi algebra. So, as a consequence of step one we have gotten that the algebra generated by rectangles is inside the class P where F r is the algebra generated by rectangles. So now, so our next step should be that trying to show that this P is actually a sigma algebra, but once one tries to do that one tries to show that P is a sigma algebra one lands into problem one is not able to show that it is closed under arbitrary unions. So, that will be a problem. So, one modifies the arguments and instead of showing that P is a sigma algebra one tries to show that P actually is at least a monotone class. So, once one tries to show that P is a monotone class it includes an algebra. So, it will include the monotone class generated by the algebra which is the sigma algebra. So, that is a root we will follow. So, from here onwards our technique will be the monotone class technique. So, we will try to show that P is a monotone class. So, it will include the monotone class generated by the algebra F of r which is same as a sigma algebra generated by r and that will complete the proof. So, the second step we will do it in the next lecture. So, today's lecture we have just concluded that the class P for which we want to prove the required claim holds includes the algebra generated by rectangles. So, we will continue the proof in the next lecture. Thank you.