 Welcome to lecture 25 on measure and integration. In the previous lectures, we had started looking at measure and integration on product spaces. In the previous lecture, we defined the notion of product sigma algebra and today we will define the notion of product measure. .. So, let us recall. We will fix for today's discussion two measure spaces X, A, mu and YB mu. So, X is a set, A is a sigma algebra of subsets of X and mu is a measure defined on the sigma algebra A. Similarly, for the measure space YB mu, B is a sigma algebra of subsets of Y and mu is a measure on the sigma algebra B. So, we have already defined the notion of the product measure namely A cross B. So, if you recall, so we defined the notion of A times B. So, this is the sigma algebra generated by all rectangles and rectangles were defined as the sets A times B, where A belongs to the sigma algebra A and B belongs to the sigma algebra B. Now, we are given a measure mu on the sigma algebra A and given a measure mu on the sigma algebra B. So, that is a measure on B. So, our aim or the problem is to define a measure eta on the product sigma algebra A times B using the measure on A and using the measure mu on B. Why such things are important? So, let us just recall that on the real line and the Lebesgue measurable sets, we had defined the notion of the Lebesgue measure. So, that extended the notion of length. So, that extended the notion of length on R for subsets of R, which are not necessarily intervals. So, we want to do the corresponding thing on R 2. So, on R 2, given a set E, we would like to define the notion of area of E and if E is a nice set, for example, if E looks like a rectangle i cross j, then we know that its area, so let us call it as area of E is defined as length of i times length of j. So, this motivates the notion of the product. So, if we have sets in x cross y, so in general, if E is a set in x cross y and we have a notion of size here and notion of size here, then we will like to define the notion of size for subsets E in x cross y and for sets which are nice, which are very simple to describe, we will like to put it as a product of the length into breadth. So, the abstract question, the problem, abstract problem is the following that we want, given two measure spaces x a mu and y b mu, we want to construct a measure, let us call it as eta on the product sigma algebra a times b such that for sets which are rectangles, so what are the rectangles on abstract measure spaces, they are the sets of the type a times b, where a belongs to the sigma algebra a and b belongs to the sigma algebra b. So, for such sets, we want that the notion of the size for subsets namely, so our notion of size is the measure, so measure of a set a times b should look like mu of a, so something like length of a into mu of b length of the set b. So, this is the abstract problem, given two measure spaces x a mu and y b mu, how to define a measure in a nice way on the product sigma algebra such that on rectangles, it looks like the product of the corresponding measures, so eta times eta of a times b should look like mu of a times mu of b. In fact, this requirement that eta of a cross b is mu of a times mu of b itself says a way of doing this, so that means this fixes the notion of the measure for rectangles which are of the type a cross b. So, if we can show that this set function eta which is defined by this equation for measurable rectangles a times b by this equation and if you can show that is a measure it is countable additive, then we know that measurable rectangles form a semi algebra and they generate the sigma algebra a times b. So, we can take advantage of our extension theory and then extend this eta if it is a measure on the semi algebra of all rectangles to the sigma algebra a times b, so that is roughly the root we want to follow. So, to do that to implement this possibility, so let us write that eta defined on rectangles, so we are defining first eta on rectangles a times b, so eta of a cross b is defined as mu of a times mu of b. Obviously, it is a well defined set function and we want to claim that this is actually a measure, so eta of empty set is 0 that is because if a or b are empty set then this is 0. So, we want to show that it is a measure on R that means we have to show that eta is a countable additive set function. So, to show that let us take a rectangle a cross b a times b and suppose it can be represented as a union of rectangles a n cross b n which are pair wise disjoint. So, a cross b is written as union n equal to 1 to infinity of rectangles a n cross b n where all the sets a n's are all in the sigma algebra a the sets b and b n's are all in the sigma algebra b and these rectangles are pair wise disjoint that means a n cross b n intersection with some a m cross b m is empty whenever n is not equal to empty n is not equal to m. So, if a rectangle is written as a countable disjoint union of rectangles then we want to show that eta of a cross b is equal to summation n equal to 1 to infinity of eta a n cross b n. So, this is what we have to show. So, to show that let us proceed as follows. So, let us write, so here is a rectangle a cross b which is written as a union of rectangles a n cross b n n equal to 1 to infinity and these rectangles are pair wise disjoint. So, the disjointness we will represent it by putting a square cup. So, the notion of union instead of putting the u we will put it as square just to indicate that. So, we do not have to write every time that they are pair wise disjoint. The symbol itself indicates that they are pair wise disjoint. So, we want to compute eta of a cross b and show it is equal to summation of eta a n cross b n. So, to do that let us proceed as follows. Let us fix any point x belonging to a. So, for any x belonging to a, if we look at y belonging to b, then x comma y belongs to a cross b. We have fixed a and take any point y in b then the ordered pair x comma y belongs to a cross b. So, that is a cross b. So, it will belong to one of the sets here. So, which set it will belong? It will belong to some a n and b n. So, then x cross y belongs to a n and y belongs to b n. So, this implies there exist a n such that x comma y will belong to a n cross b n. So, that is a possibility. Now, if this weapon implies that x must belong to a n and y must belong to b n. If the ordered pair x comma y belongs to a n cross b n, then x belongs to a n and y belongs to b n. But what does that imply? That implies that if y belongs to b, that means implies. So, thus let us write what we have. y belonging to b implies y belongs to b n, where what is this n, where x belongs to a n. So, whenever x belongs to a n, y will belong to some b n. So, that implies that I can write the set b as union of our sets b n, where n belongs to, let me write S of x. So, what is S of x? S of x is the set of all those indices n such that it will belong to b n. That x, which is fixed, belongs to a n. So, out of the indices 1, 2, 3, so on, look at those n for which x belongs to a n. So, if x belongs to a n and then y will belong to some b n, that means y belongs to those b n such that x belongs to a n. So, this is what we want to claim. And not only that, we want to claim that these b n which are involved here, they are pair wise disjoint. That means this union is a pair wise disjoint union. Why is that? Because if this is not disjoint, that means if b n, if a point y belongs to b n intersection b n, where both n and m are in S x x, that means that will imply that x y belongs to a n cross b n and also it belongs to a m cross b m, where n and m are in the set. So, n and m are in the set of S of x. So, n and m belong to S of x. Suppose, I want to show this union is disjoint. Take two elements here, b n and b m. That means for n and m belong to S of x, look at the intersection. Suppose, there is a y in that intersection, that will mean what? That x belongs to b n and y belongs to a n and y belongs to b n. And similarly, x belongs to a m and y belongs to b m. That means x y belongs to both of this, which is a contradiction because a n and b m are disjoint. So, what we are saying is the following, namely we are saying the following that for any x fix, I can write my set b as a disjoint union of sets. So, this is what the conclusion is. I can write the set b as a disjoint union of sets b n's, which are coming in this union. But, which ends? Those ends such that x belongs to a n. So, to look at a pictorial view of this, let me just take a very simple example, illustration of this. So, this is the set a. This is the set a and this is the set b. So, we have got a rectangle a cross a cross b. We have got a rectangle a cross b and it is written as a disjoint union of rectangles. So, here are the disjoint union of rectangles. What are those rectangles? One is this rectangle. The second rectangle is this. The third rectangle is this. The fourth rectangle is this. The fifth rectangle is this and the sixth rectangle is this. So, this a cross b. So, the set a cross b is written as a disjoint union of six rectangles. So, let me call this first rectangle R 1. This is R 2. This is rectangle R 3. This is rectangle R 4, R 5, R 6 and R 7. So, these seven rectangles and their sides of each one of them, we can write down a 1 cross b 1. So, this rectangle, this side is a 1 and this side is b 1. For this rectangle, this side is this portion and this is the width and so on. Now, I wanted to illustrate that point. So, let us take a point x belonging to A fixed. So, when x is fixed, what are the points y in B n? So, to find those, let us go vertically. If we go vertically, for any y belonging to this set B, y belonging to B, either y will belong here or y will belong here. So, that means, this set B can be written as a disjoint union of this portion and this portion of B. So, that will be b 1 and this will be b 7. So, if x belongs to A 1, then y belongs can belong to B 1 or B 7 or it can belong to B 7. So, that means, B will be equal to B 1 union B 7. So, B 1 and union B 7. B 1 is the width, height of R 1 and B 7 is the height. For example, let us take a point x here. This is a point x. Let us take a point x here, which belongs to A 2. So, when I go above, for any, if I fix this, then how does the, how does the y split? So, if x belongs to A 2, if that is fixed, then to be inside the rectangle, I can be here, I can be here, I can be here. So, it will belong to B 2, B 6 and B 5. So, y can belong to B 2 or B 5 or B 6. So, that means, in that case, B will be equal to B 1, B 2, union B 5, union B 6. So, what we are saying? So, depending on where the point x is, the set n x will be 1 and 7. If x is in A 2, then n x will be 2, 5 and 6. So, B, in either case, B is a disjoint union of rectangles, some of the B i's. So, this is the important thing, which I wanted to convey. So, this is what is the conclusion of this argument, that if A cross B is a disjoint union of rectangles A n cross B n and I fix any point x belonging to A and analyze the points y in B, then x comma y belongs to A cross B. So, it will belong to some A n cross B n. So, y will belong to some B n's. So, which B n's it will belong? It will belong to only those B n's for which x belongs to A n. So, B can be written as a disjoint union of B n's, where n belongs to S of x. So, this is a disjoint union. So, that was the first important observation, once we have. So, once this is a disjoint union and all the B i's, the set B n's are all in the sigma algebra, where nu is defined. So, this being disjoint that implies, what we get is that nu of, so this star, so star implies namely that nu of B is equal to summation nu of B n's, where n belong to S of x. So, now I would like to transform this equation slightly. So, what was x? x was a point in A. It was, when x is in A, so for every x fix, that means for every x fix in A, we had this. Now, suppose if x belongs, so suppose x does not belong to A, if x does not belong to A, then obviously x does not belong to B n's. x does not belong to B n for every n, sorry. Then if x does not belong to A, then x does not belong to, sorry, not B n's, x does not belong to A n for every n. So, that implies that chi A n of x will be equal to 0. So, if x belongs to A, then x will belong, then it will belong to some of the A n's and in that case for those n, it will be equal to 1. So, if x does not belong to A and if x belongs to A and x, that means x will belong to, x will belong to some A n and that means this n will belong to nx and that means we will have chi A of x will be equal to 1. So, what I am saying is this equation, nu of B equal to this, I can write it as nu of B times the indicator function of A x is equal to summation over all n equal to 1 to infinity chi of A n x times nu of B n. So, let us understand this once again that why is this so. So, if x belongs to A, then the left hand side is this value of the indicator function is 1. So, the left hand side is nu of B, so that is here and if x belongs to A, then it will belong to some A n. So, if x belongs to A n, this value is 1, so the right hand term is nu of B n. If x does not belong to A n, then this value is going to be 0. If x does not belong to A n, then this value, so if x belongs to A and x belongs to A n, this value is 1, otherwise this value is 0. So, on the right hand side in this summation, only those terms will be non-zero for which x belongs to A n and that means the value of the right hand side terms will be indicator function will be 1 and nu of B n. So, this will be this equation, otherwise both sides are equal to 0, so that holds. So, what we are saying is that, so from the earlier equation, we have come down to the second conclusion namely this holds for every x belonging to A and so for every x belonging to x. So, for every x, this equation holds. So, that is what we have proved, because when x belongs to A, it is the earlier equation star. When x does not belong to A, both sides are equal to 0, so this equation holds. So, this is the second crucial step in the arguments namely, if A times B is a rectangle which is written as a countable disjoint union of rectangles A n and B n, then indicator function of A times x into nu of B is summation over n indicator function of A n times nu of B n. Now, this is the equation involving non-negative, this is the equation involving non-negative simple non-negative measurable functions, so on the measure space xA mu. So, this is a sequence, so now I can apply monotone convergence theorem. So, this apply monotone convergence theorem on xA mu. So, this is a non-negative function which is a limit of, so the sum means, it is a limit of the partial sums. So, it is a limit of non-negative measurable functions. So, monotone convergence theorem will give me that the integral of this is equal to limit of integrals of this. So, an application of monotone convergence theorem to this equation star gives me that integral chi of B indicator function of A x d mu x can be written as summation n equal to 1 to infinity integral of indicator function of chi A n nu of B n d mu x. So, this is a straight forward application of the monotone convergence theorem. Left hand side is a non-negative measurable function which is a limit of non-negative measurable functions on this measure space. So, integral of the left hand side with respect to mu must converge to the integral is equal to the limit of the integrals on the right hand side. So, this is an application of monotone convergence theorem. Now, let us compute the right hand side. The left hand side, the integral of nu of B, nu of B is a constant, so that goes out. Integral of the indicator function of A with respect to mu, so that is mu of A is equal to summation n equal to 1 to infinity and this integral again nu of B n is a constant. So, nu of B n goes out of the integral sign and integral of A n with respect to mu, so that is mu of A n. So, what we have gotten is nu of B into mu of A is summation nu of B n into mu of A n, but this thing is nothing but eta of A cross B and this is nothing but each term is nothing but eta of A n cross B n. So, what we get is eta of A cross B is equal to summation n equal to 1 to infinity eta of A n cross B n and that proves that the, so hence eta is countably additive. So, that proves that the eta is a measure. So, this is how one proves that eta which is defined, so eta which is defined as eta of a rectangle equal to A cross, rectangle A cross B to be mu of A into nu of B is countably additive. So, let me slowly go through the proof once again. There is only one small idea involved in it and other is rest is straightforward applications of the earlier results. So, let us write A cross B. So, I am going through the proof once again that A cross B is written as a countable disjoint union of rectangles A n cross B n and what we want to show that eta of the rectangle A cross B is equal to summation of the measures of the rectangle, each rectangle. So, summation over n of eta A n cross B n. So, to prove this what we do is as follows look at the set A cross B. So, fix any element x belonging to A, then for any y belonging to B we know that x comma y belongs to A cross B which is nothing but union of A n. So, that means x comma y will belong to exactly one of them, but which one of them. So, x comma y will belong to that A n comma B n whenever this x belongs to that A n because x comma y belonging to A n cross B n implies x must belong to A n and y must belong to B n. That means what we are saying is x comma y belongs to A cross B if and only if x belongs to A n and for that x, the y should belong to B n because x is fixed. So, that n is fixed. So, what are those n's which are fixed? So, y belongs to B n provided x belongs to A n. So, for a fixed x collect together those n's. So, find the set S of x all those indices n say that x belongs to A n. See, A n's are not disjoint. So, x can belong to more than one of the A n's. So, look at those if x belongs to A n, but for a fixed x it will belong to only one of them. So, if x belongs to A n then y will belong to B n. So, as x varies over A, for every fixed x you will get a collection of B n's. So, what are those B n's? Those B n's are indexed by n belonging to S of x say that x belongs to A n and this union is a disjoint union. So, for every x fixed in A we can decompose B into a disjoint union of B n's over those n's say that n belongs to S of x. This being a disjoint union because A n's B n's are disjoint we get our first inequality that for any fixed x in A nu of B is summation nu of B n's over those n's which belong to S of x. And now we observed that equivalently this thing we can write it as nu of B I can multiply it by the indicator function of A because x belongs to A. So, this will be equal to 1. And this nu of B n I can multiply if x belongs to A n that means n will belong to S of x. So, I can multiply here by the indicator function of A n if n belongs to S of x. And if x does not belong to A n that means it cannot belong to anyone of the A. So, this all the remaining terms here will be 0 and all the remaining this side is also equal to 0. So, what we are saying is for any x in A I can write this is equal to this and this equation makes sense whenever x does not belong to A also because if x does not belong to A this side is equal to 0 and that side x does not belong to A. So, it does not belong to anyone of the A n's. So, all the terms are 0. So, this equation first we can write it as indicator function of A times x nu of B is equal to summation of chi A n's. And now we realize that not only this equation is valid for x belonging to A this equation is valid for all x in x. So, once that is observed so that is what is observed here. So, what we get is that the equation chi of A x nu of B is equal to the summation chi of A n nu of B n for all x. And now this is the equation about non-negative measurable functions. So, left hand side is a non-negative measurable function which we can realize as a limit of non-negative measurable functions namely the partial sums of this series and apply monotone convergence theorem. So, that will give us that the integral of the left hand side is equal to summation of the. So, I can take the integral sign inside by monotone convergence theorem and say that integral of chi A x nu B d mu y is nothing but integral of this summation. And so here is the application of the monotone convergence theorem I can take this integral inside. So, that is equal to summation of integral of indicator functions. And now it is just a matter of writing down the values of this nu of B n is a constant so goes out. So, this integral is nothing but mu of A n so and that nu of B n. And the left hand side this was integral of chi A of A nu of B nu of B is a constant. So, that is integral of mu of chi of A with respect to mu so that is mu of A. So, that gives us that eta of A cross B is equal to summation eta of A n cross B n whenever A cross B is a disjoint union of rectangles. So, that proves that this is eta is a countably additive function. So, what we have gotten is eta is a countably additive function so let us just observe. So, what we got is we got eta defined on A cross B 0 to infinity by eta of A cross B equal to mu of A nu of B is a measure. We got that this is a measure on the semi algebra. So, this is important on the semi algebra A times B. So, implies by our general extension theory via outer measures and so on we can extend we can define eta tilde on A times B define eta a measure eta tilde a measure and eta tilde of A cross B to be equal to eta of A cross B. That means this eta can be extended via outer measures to the sigma algebra generated by A cross B the semi algebra A cross B. And if you recall we had said that this extension will be unique provided this eta is a sigma finite measure. So, we claim eta is sigma finite if mu and nu are sigma finite. So, we want to show next that if A and B if mu and nu are sigma finite. So, let us assume so if mu sigma finite so that implies I can write x as a disjoint union of sets x i 1 to infinity each x i in the sigma algebra A and mu of x i finite for every i. And similarly, nu sigma finite implies I can write y as disjoint union of sets y j, where each y j is an element in the sigma algebra B and nu of B j is finite. And then this implies we can write x cross y as disjoint union of x i cross disjoint union of y j's. And now it is just a simple matter to set the equality namely this is same as the unions over i unions over j of rectangles x i cross y j. Because if x comma y belongs here that means x belongs to the union x i's and y belongs to union y j's. So, that means x belong to only one of x i and only to one of y j's. So, it will belong here and conversely. So, this is a disjoint union and now we only have to observe the fact that x cross y has been decomposed into a disjoint union of sets x i cross y j and we only note that eta of x i cross y j it is a rectangle. So, its measure is mu of x i times mu of y j and both of them being finite. So, this is a finite quantity. So, x cross y is written as a disjoint union of sets x i cross y j and each piece has got finite measure. So, that implies eta is sigma finite. So, the measure eta is sigma finite on the rectangles and hence has a unique extension to the sigma algebra. So, this is what we wanted to prove that eta that extension is also sigma finite. So, general extension theory gives me a unique. So, that mu and mu are sigma finite. So, that implies x is a disjoint union y is a disjoint union. So, we can write x cross y as a disjoint union of the rectangles x i cross y j. That is what I just now illustrated and each piece has got a finite measure. So, by that process we get a eta is sigma finite on rectangles. So, by extension theory eta can be extended uniquely. So, that is the important thing eta can be extended uniquely to a measure on a on the product sigma algebra. So, that for rectangles it is the product. So, this is the measure eta which is defined on a times b on the product sigma algebra is called the product measure and is normally denoted by mu cross mu. So, let us summarize what we have done. We started with the two measure spaces x a mu and y b mu and for the product set x cross y we first define the rectangles namely sets of the type a cross b where a belongs to the sigma algebra a and y belongs to the sigma algebra b. So, that gives us sets of subsets of x cross y called measurable rectangles they only form a semi-algebra. So, we extend here, we generate the sigma algebra by this semi-algebra of rectangles and call that as the product sigma algebra denoted by a with circle cross a times b and now given measures mu on the sigma algebra a and a measure mu on the sigma algebra b we want to define a measure on the product sigma algebra. So, that is done by defining the product for a defining the new measure first on rectangles. So, eta of the rectangle a cross b is defined as the product of mu of a and mu of b and we show that this is a measure. So, this becomes a measure on the semi-algebra of rectangles and if it is sigma finite that means if we assume that the given measures mu and mu are sigma finite then this extends uniquely to a measure on the product sigma algebra a cross b and that measure is called the product measure product of the measures mu and mu. So, given two measure spaces x a mu and y b mu which are sigma finite we get the product measure space x cross y the sigma algebra a cross b generated by the rectangles and the product measure mu cross mu go obtained via the extension theory. So, this is the product measure space constructed as just now said. Now, the next problem we want to analyze is the following namely this product measure mu cross mu that we have gotten is obtained via extension theory, but it does not tell us how does one compute the product measure mu cross mu of a set in a cross b. So, that is not indicated because we are making use of the extension theory. So, next problem that we want to analyze is the following. So, namely so we have got the product measure space x cross y the product sigma algebra a times b and the product measure mu cross mu. So, let us take a set E contained in x cross y which is of course E is an element in a times b. So, mu cross mu of this set E is defined. So, the question is can we compute this quantity mu cross mu of E using mu and mu? So, that is the question. And there let us just recall something from our elementary calculus. Supposing in the plane we have got a set which looks like the following. It looks like this is a set. So, this is a set E which looks like the following namely here is a point A and here is a point B. So, the set E looks like, so let us just write what does the E look like? E is equal to all x cross x comma y so that x belongs x belongs between A and B and y. So, at any point x if I look at y this is the portion of y. So, it starts with the green boundary. So, y is bigger than or equal to some function f of x that is the green curve and less than or equal to here is g of x. So, this is what we call in calculus or elementary analysis sets of type 1. And for such sets for such sets we can find out what is the area. So, area of the set E if you recall from calculus it can be obtained as you look at this difference height what is this height. So, that is nothing but g of x minus f of x and integrate that from A to B dx. So, Riemann integral as an application of Riemann integration we do that we define it equal to this. But now let us rewrite this. This I can write it as this is Riemann integral. So, Riemann integral I can write integral over AB of D lambda with respect to the Lebesgue measure. And what is g x minus f x that is precisely the Lebesgue measure of this Lebesgue measure of this height. So, Lebesgue measure of let me write as E x what is E x? E x is equal to all y such that x comma y belongs to E which is same as all y such that y is between f x and g x. So, that set I am writing it as follows. So, I am writing as Lebesgue measure of a notation called E x. So, you can think of that look at this set x let look at that look at that set E to find its area we are just adding up the areas of these small strips. So, I can think it as that way that is what this integral seems to indicate. So, we would like to generalize this in the case of our construction the same idea we want to generalize it. So, here is what we want to do. So, given a set E in x cross y for x belonging to x fix let us look at E x that is so here is abstract now x is abstract set y is some abstract set. So, look at all those points y belonging to y section is the part of the horizontal line and the x section is part of the vertical line. So, as I said E x is called the section of E at x or just the x section of E and similarly E y this set E y is called the section of E at y or just the y section of E. So, here are some simple properties we want to verify for this sections. So, first of all we want to verify and let us look at some examples first. Let us take a set E which actually looks like a rectangle. So, in this x cross y let us take actually a rectangle A cross B where A belongs to A and B belongs to B. So, then for any x in A if for what are the points y such that x comma y will belong to E that is means y must belong to B. So, E x the x section of E for a rectangle A cross B is nothing but B the set B itself if x belongs to A and if x does not belong to A then the point x comma y is never going to belong to E. So, the x section is empty set. So, here is a simple observation that for a rectangle A cross B the x section is equal to the set B if x belongs to A and it is empty set if x does not belong to A. Similarly, the y section of E or the section of y at a point y in y. So, all x say that x comma y belongs to B then for all x in A x comma y is going to belong to E. So, that means the y section of E is equal to A if y belongs to B and it is empty set if y does not belong to E. So, for rectangles these are very easy to compute what are the sections. For a rectangle A cross B the x section for x belonging to A is B otherwise empty. Similarly, the y section is equal to A if y belongs to B otherwise it is empty. Now, let us look at another example. So, let us take a measurable space x A and look at the ordered pairs x comma t. So, t belongs to R such that this t lies between the evaluated indicator function of A at x. So, we are looking at the ordered pairs x comma t such that for every x t lies between 0 and A and x belongs to x. So, what are the sections of this set E? This is a subset of A cross B and where B is y is the real line. So, it is subset of x cross R we want to find its sections. Let us observe that for a point x in A if x belongs to A then this indicator function of A the value will be equal to 1. So, t will be between 0 and 1. So, if x belongs to A then t will be between 0 and 1. So, the section is going to be the interval 0, 1 and 1 not included and if x does not belong to A if x does not belong to A then this is going to be 0. So, t is going to be the singleton 0. So, the section if x belongs to A so the section depends on whether x belongs to A or not. So, the section of E at a point x is equal to the interval close interval 0 open at 1 in R if x belongs to A otherwise it is the 0 set or another way of looking at this is the following that the set E I can write it as A cross the interval open close at 0 and open at 1 union A complement of this A complement cross the singleton 0. This is another way of writing the same set E as I explained just now. So, the section now is a union of two disjoint rectangles. So, section in the first case when x belongs to A section is going to be 0, 1 and in the second case the section is going to be the singleton 0 if x does not belong to A. So, these are the x sections. We can similarly find the y sections. So, for y belonging to 0 to 1 that means y is the real line. So, for a real number between the closed at 0 and open at 1 interval it is going to be A at 0. So, and if y is equal to 0 then this is going to be the whole space x and empty set. So, this is easy computation from this it follows. So, this is how one computes the sections of these sets. So, these sections are going to play important role in computing the measure of a set E in the product space. So, in the next lecture we will analyze the x sections, the y sections, various properties of these sections under complements, intersections and unions and then show that each section for a set E in the product sigma algebra, each section is again a appropriately measurable set whose measure can be defined and then you can take the measure of that and define the functions and compute the integral of the product set, compute the product measure of the set E. So, we will continue this study of sections and their implications for product measures in the next lecture. Thank you.