 Welcome to lecture number 27 on measure and integration. In the previous lecture, we had started looking at how to compute product measure of a set in the product sigma algebra. We had shown part of a theorem and we will continue looking at the proof of that theorem in this lecture. So, let us just recall what we have been doing. So, we were looking at computing the product measure. So, we will continue that study today. So, let us just recall the settings. We have a set E contained in the product set x cross y and for any element x in x and y in y, we define what is called the x section E x and E y in the previous lectures. Then, we claimed that for every set E in the product sigma algebra, set sections E x is an element of a sigma algebra B and the section at y is an element in the sigma algebra A. So, this we had proved. So, I am just recalling them and then we proved that the functions x going to the new measure of E x, E x is a subset of is an element in the sigma algebra B and new is a measure defined there. So, we can compute what is new of E x and the claim is that the function for every x, the image being new of E x, this is a function defined on x and the claim is it is a measurable. Similarly, function y going to the measure of the y section is a measurable function on the set y with respect to the sigma algebra B. So, these two we had proved and we wanted to prove finally, the third one that if we integrate these functions with respect to mu and with respect to new, these are non negative measurable functions and we can integrate them. So, the claim is that the integral nu E x d mu x is same as the product measure mu cross nu of E and it is same as the integral of the y section with respect to y. So, this is the step we were trying to prove in the previous lecture. So, to prove this what we said, let us look at the class of those subsets E in the product sigma algebra for which this is true. So, we constructed the class P all those subsets in the product sigma algebra such that the previous two claims namely this claim 2 and claim 3 both hold namely x going to nu E x and y going to mu E y are measurable functions. So, P is the family of all subsets A cross B such that the property 2 and 3 hold. So, let us recall what are the properties 2 and 3. Property 2 is that x going to nu E x and y going to mu E y these are non negative measurable functions and the property 3 says that the integrals of nu E x with respect to mu is same as the integral of mu E x with respect to nu and both are equal to the product measure of E. So, the both these properties holds for a set E then that set is in the collection P. So, our aim is to prove that P is equal to the product sigma algebra A cross B. We had already observed in the previous lecture to show this for the first step is to prove that this class P is includes the rectangles. So, that is one so that we had proved and also we had proved that this class P is closed under finite disjoint unions. So, once this class is P is closed under finite disjoint unions and includes the rectangles rectangles form a semi algebra. So, the algebra generated by it looks like the class of sets which are finite disjoint union of rectangles and P being closed under such operations will get that as a consequence of this that the algebra F r generated by these rectangles is also inside P. So, as a consequence of step one we get that the algebra generated by the rectangles in the inside the class P. So, the second step we want you to prove that this class P is a monotone class and the reason for that to prove that is a monotone class is the following. This class P it is directly it is difficult to show that it is a sigma algebra because if you could show directly that P is a sigma algebra it includes algebra generated by rectangles. So, then it will include the sigma algebra generated by it that direct root is not possible. So, we follow the monotone class result namely if we are able to show that P is a monotone class and F r being inside it the monotone class generated by F r will be inside P and F r being a algebra the monotone class generated by an algebra is same as the sigma algebra generated by that class. So, we will get the sigma algebra generated by rectangles will be inside P and that is precisely what we want to show and that is a times b because the sigma algebra generated by rectangles is the product sigma algebra a cross b. So, to complete that proof we have to only show that the class P is a monotone class. So, let us start proving that P is a monotone class. So, P is the class of all those subsets e belonging to the product sigma algebra a times b such that if we look at the set x going to take the say e take its section x that is the subset of the set y in the sigma algebra b. So, nu of that makes sense. So, we get this function. So, this is measurable and the function y going to mu of e y that is is measurable. So, both these functions are measurable and the property that if we integrate nu of e x with respect to mu. So, we are integrating over x this is same as the integral over y of the second function mu of e y with respect to nu e y and both of them are equal to the product sigma algebra mu cross nu of e. So, this is the collection of all those sets e in the product sigma algebra this holds and we want to show that P is a monotone class. .. So, let us look at the first property. So, let E n belong to P, E n be a collection of sets in the class P says that E n is increasing. So, to show that the set e which is equal to union of E n's also belongs to P. So, this is what we have to the first to show that P is a monotone class. We have to show it is closed under increasing unions and decreasing intersections. So, that is the two properties we have to check. So, let us take a sequence E n in P which is increasing and let us say E is the union of this E n's. So, the claim is that E belongs to E n. So, what we have to do? We have to look at the corresponding. So, what is the first property we have to check? So, to check that E belongs. So, we have to look at nu of E x. So, the first thing we have to show is that this is measurable is A measurable function. .. So, to do that let us observe the following. So, this is what we have to show. Now, when each E n belongs to P, so implies that x going to nu of E n, its section at x is measurable for every n. So, this is what is given to us and we want to come to nu of E n. But for that, so let us observe that as E n is increasing to E, the sections E n x is an increasing sequence of sets increasing to E of x. So, this is a sequence of sets in the sigma algebra B. So, that we have already seen that if A is a subset of B, then the section of A is a subset of section of B. So, that will prove that this is the sections are increasing and the increase to the union. So, union of the sections, so union of E n at x is same as union of each E n and hence this is increasing to x. So, this is a simple observation using the properties of the sections. So, E n x is increasing and now you recall that nu being a measure, if a sequence of sets increases to another set, so that implies that nu of E n x, the sections that will increase, that will converge to nu of E x. So, that proves that nu of E n x increases. Now, each one of them is a measurable function. So, nu of E x is a limit of measurable functions, so that implies that x going to nu of E x is measurable. So, basically what we are saying is because the nu of E x, so the function x going to nu of E x is a limit of the functions nu of E n of x and that comes from the fact that because E n is increasing to E, so the sections E n x increase to the section E x and that means in the sigma algebra b and nu being a measure nu of E n x must converge to nu of E x and each one of them being measurable because it is in the collection p. So, each is a measurable function, so limit of measurable functions is measurable. So, that proves one part that x going to nu of E x is measurable. So, next what we have to check is the following, we have to check that integral of nu of E x d mu x over x is equal to mu cross nu of E. So, this is what we want to check. So, now once again let us go back to the earlier fact that we saw that nu of E n x, the sections these measurable functions, these are actually non-negative measurable functions and they are converging to the function nu of E x and that is the increasing sequence of measurable functions. So, this is nu of E n x is a increasing sequence of non-negative measurable functions converging to a measurable function nu of E x. So, we can apply our monotone convergence theorem. That says so by, so once again this property star star and monotone convergence theorem theorem apply and apply and they give us as a consequence that integral of nu of E x d mu x over x because nu of E x is the limit of increasing sequence of non-negative measurable functions. So, integral of nu of E x must be equal to limit n going to infinity of the integrals of the corresponding sequence of non-negative measurable functions and they are nu of E n section at x d mu x. So, this is application of monotone convergence theorem. Let us observe that E n belongs to the class p. So, the property 2 of that says that if I integrate nu of E n, so the sections with respect to mu, this integral is equal to the product measure mu cross nu of E n. So, that is because E n belongs to the class p. So, by the third property of that collection of sets in p that means nu of mu cross nu of the product measure of E n is the integral of the sections with respect to x. So, we can say that this integral is equal to limit n going to infinity of mu cross nu of E n. So, once that is true, we want to look at this limit. Once again, let us observe that E n is a increasing sequence of sets in the sigma algebra A cross B and mu cross nu is a measure. So, once again using the property of measure that if a sequence of sets is increasing, then the measure of limit of the measure of the sequence is equal to measure of the limit. So, that is equal to mu cross nu of E. So, once again we have used the fact that E n is increasing to E and mu cross nu is a measure. So, this limit must be equal to mu cross nu of E. So, what we get is that this limit is equal to this. So, that means we get that mu of integral over x nu of E x d mu x is equal to mu cross nu of E. So, we have proved that if E n is increasing to E, so this implies that E belongs to the class p because we showed that if E n is increasing to E, then both the properties hold for this. Now, we want to do the similar thing for decreasing. So, next let us consider E n belonging to p, n bigger than or equal to 1 and E n is decreased to E. That is, E is equal to intersection E n's n equal to 1 to infinity. So, we want to claim that E belongs to p. So, this is what we want to check. So, we can try to copy the proof for the increasing case. So, let us go back to the proof of the increasing case and let us see, can we carry over the proof by saying similarly. So, now we have got E n's decreasing. So, because E n's belong, so what we said first thing was that because E n's belong to p, so this is a measurable function. So, that is the property of the set E n being in the class p. So, increasing or decreasing is not coming into picture. So, this step will carry over and then if E n is increasing to E, so now we have got E n is decreasing to E. So, this thing will change. So, if E n's are decreasing, then of course it is true that the sections E n x will be decreasing to the set, the section of the set E at x. So, the sections E n x will decrease to the set E x. So, that step also will be okay. Now, we want to say that when E n's decrease to E, so we want to say that here we use the property that for the increasing case we said whenever a sequence is increasing, nu of E n's converge. The corresponding result we know is not true for decreasing sequences. So, here the proof, try to copy the proof for the increasing thing will fail down because this step will not, this equation star will not hold. To make this star hold, we have to put an extra condition that the measures are finite because if measures are finite, then E n decreasing to E will imply measures converge. So, if E n's mu and nu are finite, so for example if nu is finite, then sections of each E n that is a decreasing sequence, so nu of E n will converge. So, to carry over the proof, in the similar case we have to put an extra condition. So, for this claim to hold, we have to assume that mu and nu are finite. So, let us assume that mu of x is finite and nu of y is finite. Of course, that implies mu cross nu of x cross y is finite. So, under these conditions we want to show that if E n belongs to P, E n's decrease to E that implies E belongs to P. So, to show that we just now we can repeat the steps. So, E n's let me just go through the proof again for the decreasing case also to emphasize where exactly we will be using the finiteness condition. So, E n's decrease to E, so that implies that the sections E n x decrease to E of x. So, that implies that mu of E n x, nu of E n x because E n's x is the subset of B, the subset in B. So, this converges to nu of E x. So, this is the stage where we will be using this condition plus. So, under this condition plus that mu and nu are finite, this holds. Now, each E n belongs to P. So, each one of them is a measurable function. So, that will imply that x going to nu of E x is measurable. So, this is a measurable function and we have got nu of E n x decreases to nu of E of x. So, earlier we used the monotone convergence theorem to conclude that nu of E x, integral of nu of E x must be limit, but here it is a decreasing sequence. So, we cannot use monotone convergence theorem also here, but let us note. So, here is an observation. Note that because mu of x is finite, nu of y is finite. So, this function nu x going to each of the functions nu of E n x is an integrable function. Why is that? Because it is a decreasing sequence. So, let us observe nu of E n x. For every n, if I look at this non-negative function, it is less than or equal to nu of E 1 of x and nu of the section E 1 x integral over x d mu x is less than or equal to nu of E 1, E 1 x is less than or equal to nu of y. So, this is the integral is less than nu of y. So, integral of 1 d mu x is less than mu of x, which is finite. So, nu of E 1 x is an integrable function on the measure space y b nu and each nu of E n x is less than or equal to, so each nu of nu E n of x is integrable. So, we can apply a dominated convergence theorem. So, dominated convergence theorem applied to the fact that nu of E n x is a sequence of non-negative integrable functions and they are decreasing to the function nu of E x. So, this also is integrable. So, implies by dominated convergence theorem and this observation that the function nu of E x d mu x over x, this is this function is integrable and its integral is nothing but the limit n going to infinity of integrals nu of E n x d mu x. So, for the decreasing sequence, the proof differs in both the steps. First of all, when we want to say that E n's are decreasing, the sections decrease. So, the finiteness condition allows us to say that nu of E x is a limit of these functions and that implies that this is a measurable function. So, finiteness says and this function is measurable because of this fact and in fact, the finiteness condition says this is sequence of integrable functions decreasing to the function this. So, dominated convergence theorem can be applied and that gives us this limit is equal to. So, nu of E x integration with respect to mu is limit of and now the proof is as before this E n being in the collection p. So, this integral is nothing but measure of mu cross nu of the set E n. So, that is limit n going to infinity of measures of the sets E n and once again E n's are decreasing to E and mu cross nu is a finite measure. So, that will imply. So, this is equal to mu cross nu of E again using the fact that mu and nu are finite. So, we get the conclusion that again using finiteness condition namely that the integral nu of E x d mu x over x is equal to. So, we have already shown it is the class p is closed under increasing sequences. Now, we have shown it is closed under decreasing sequences. So, p is a monotone class. So, that proves that p is a monotone class. So, as a consequence of the fact that p is a monotone class the consequence of this would be namely that we already have f of r is inside the class p and p is a monotone class. So, that will imply that the monotone class and rated by f of r will also be inside p, but this is nothing but the sigma algebra generated by the class r of rectangles or same as the sigma algebra generated by rectangles and that is same as the product sigma algebra. So, that will prove that the product sigma algebra is equal to p namely that the required conditions hold for the corresponding. So, that proves step 2 namely that p is a monotone class and that implies that the monotone class generated by f of r is inside p and hence and that we will prove that a monotone class generated by f of r is a algebra. So, the monotone class generated by the algebra is precisely the sigma algebra generated by E. So, A cross B will be inside the class p and hence everything is inside. So, A cross B is equal to p. So, this is a theorem where we have used very sensibly the fact that when mu and mu are finite in that case we can extend that argument of the increasing to the case of decreasing also. This also illustrates the technique the monotone class sigma algebra technique. So, we have proved the theorem required claim that p is a monotone class under the conditions mu and mu are finite. So, now with the usual arguments one can extend it to the case when it is sigma finite. So, let us see that, but before doing that let me just go through the proof of the step 2 again to illustrate the basic facts. The first thing we looked at was if E is a product set A cross B. So, I am just revising the proof of step 2 to highlight the important points in the proof. So, A cross B belongs to r is a rectangle then mu of E x. So, that was the first step showing that r is the class in r includes rectangles. So, there we use the fact that if you take a set which is a rectangle then it is section is nothing but either the set A or the set B or the empty set according to the point x or y. So, mu of E x is nothing but mu of B times the indicator function of x because if x does not belong to A then this is 0 and the section is just B. And similarly mu of E y is mu of A times the indicator function of B. So, these two facts prove that x going to mu of E x and y going to mu E y for rectangles are measurable functions. And if we integrate because this so integral of mu will be equal to mu of B into mu of A. So, that is the product measure of the product set A cross B. So, that says the rectangles are inside it. So, that is a straight forward argument which says rectangles comes inside P. Showing that P is closed under finite disjoint unions is also straight forward because that follows from the fact that if E 1 and E 2 are two sets in the class P which are disjoint then the sections are disjoint of these two sets. So, and the sections of the union is equal to union of the sections. So, as a consequence of this the nu of the section of the union. So, E 1 union E 2 section at x nu of that is addition nu of E 1 x plus nu of E 2 x because the sections are disjoint. And E 1 and E 2 both belong to P imply these two are measurable functions and hence the sum of measurable functions is measurable. So, this becomes measurable. So, that is a straight forward proof of the fact that if E 1 and E 2 belong to P then E 1 intersection and they are disjoint then the union also belongs to P. And finally, to look at the integral. So, integral of nu of the section of the union because that splits into two parts. So, nu of E 1 union E 2 is nu of E 1 x plus nu of E 2 x with respect to mu. So, the integral splits into two parts. So, that is mu cross nu of E 1 because E 1 belongs to P. And this is mu cross nu of E 2 because E 2 belongs to P. And now using the fact that mu cross nu is a measure that gives us this equal to mu cross nu of E 1 union E 2. And a similar thing will work for the y sections. So, proving that rectangles are inside the class P and P is closed under finite disjoint unions is a rather straight forward computation. The problem arises when we want to show that P is a monotone class. So, there we first assume that mu and nu are finite. So, once mu and nu are finite we want to show it is closed under increasing union and decreasing intersections. So, take a sequence of sets E n which is increasing. So, a simple fact that if E n's are increasing the sections are increasing. And mu and nu being measures imply mu of the sections E n's will converge to mu of E. So, mu of E x and nu of E y are limits of measurable functions. So, they become measurable. So, straight forward till now no finiteness condition has been used. So, this is true whenever mu and nu are any two measures. But for the decreasing part part where we will need the finiteness condition. So, for the increasing part everything goes straight for a monotone convergence theorem application gives you nu of E x is limit of that and that is equal to the product measure and everything is ok. So, let us look at the part where we find the difficulty arises. So, difficulty arises when we want to show that if E n belongs to P and E n's are decreasing then the set E which is the intersection of E n's also belongs to P. So, here the main step is to conclude that nu of E n x is equal to nu of E x. So, for that we need finiteness condition because whenever a sequence of sets is decreasing to a set then measure of the sets need not converge to measure of the limiting set unless the measures are finite. So, finiteness condition will give us that and then instead of monotone convergence theorem we can apply the dominated convergence theorem to conclude that mu cross nu of E n is equal to corresponding integral. So, that will prove that mu n and nu being finite P is a monotone class. But still we have not concluded the proof for the general case. So, for the general case one can apply the usual sigma finiteness criteria namely whenever two measure is sigma finite the whole space can be cut up into finite number countable disjoint pieces each of finite measure and on each the result holds. So, put them together to get the result holds for the whole space. So, let us see the argument how it works because mu and nu are sigma finite. So, x can be decomposed into a disjoint union of sets A i and y can be decomposed into a union of sets B j such that a disjoint union such that mu of each A i is finite and nu of each B j is finite. So, using that we can write down that mu cross nu of A i cross B j is finite because this is nothing but mu of A i times nu of B j. So, as a consequence on each of these pieces our earlier results were hold the P was a monotone class. So, let us see how that is used to prove for a general set E in 8 cross B for a set in the sigma algebra A cross B. Note that the integral of the measure nu of E intersection A i cross B j x d mu x because each nu of each of these sets has got finite measure. So, we are applying the earlier result on the piece A i times B j. So, for every i and j using the earlier case we have that the integral over x of the x sections of E intersected A i cross B j is nothing but mu cross nu of E intersection A i cross B j and that is equal to the mu integral of the y sections of the corresponding sets. So, this step follows basically from the fact that mu cross nu of A i intersection B j is finite and for any set E, E intersection this rectangle A i cross B j on that rectangle mu and nu are finite. So, this earlier case gives us the result and now we have to only some both sides with respect to i and j. So, let us look at mu of mu cross nu of E is equal to because the whole space is equal to union over i and j of the rectangles A i cross B j that is a partition. So, mu cross nu of E can be written as using countable activity of the measure mu cross nu as summation over i summation over j mu cross nu of the pieces A i times B j and now for each one of this piece we know the result holds. So, I can write this as a integral of the x sections or as integrals of the y sections. So, this term mu cross nu of A i cross B j intersection E is equal to this integral or this integral because of the fact that for the finite case the result holds. And now using the fact that if you look at this section E intersection A i cross B j of x this section is nothing but nu of E x times A i cross. So, this is the small observation that you look at set E and take its piece inside the rectangle A i cross B j and take its section. So, this section is going to be equal to the section of E intersection with B j of course, if x belongs to E j and if x does not belong to E j then there is not going to be any intersection. So, this is going to be empty set. So, this is the observation and that observation can be used in this part that if x does not belong to A i then this thing is going to be 0. So, using that we can write that sum. So, this sum which was integral over x of E intersection this can be written as so this set is nothing but nu of E x intersection B j because that is the only place where the section appears when x belongs to A i. So, this is integral over A i of nu E x intersection B j. So, this integral is equal to this because of this fact and now this summation over i means that this integral is over x. So, this summation you can transform into integral over x and now you can interchange that to integral and the summation again you will be using fact here that this is the integral which depends on j. So, you can push it out and take it inside. Basically, I will be applying implicitly a monotone convergence theorem to say that this is equal to I can take the integral sign x and because this is a sequence of functions which are non-negative measurable and so on. So, here is an application of monotone convergence theorem which helps you to interchange summation and the integral sign. So, summation goes inside and now summation over B j, B j's are disjoint. So, that gives you over the whole space y. So, that is just E x. So, we get that mu cross nu of E is equal to the integral of the section nu of E x d mu x. So, you see at almost every step we are using some theorem or the other to justify the facts. So, this is the case for the x sections and the similar result will hold for y sections. So, that will prove that mu cross nu is also equal to integral over the y sections and that will complete the proof of the fact that one can reduce the result in the case of sigma finite. So, from finite to sigma finite is almost straight forward in the sense that we split the whole space into countable number of pieces of finite measure. So, on each piece we apply and then sum it up to go back to the original piece. .. So, we have proved the theorem namely how to compute the measure of a product set. So, let us observe one thing here. Namely, even if we start with measure space is x a mu and y b nu to be complete, the product measure space which we are denoting y x cross y a times b mu cross nu need not be complete. Because, how do we get this measure mu cross nu on a cross b? We looked at the product mu cross nu on rectangles and extended it and defined the outer measure via that and then looked at the measurable sets mu cross nu and that included the sigma algebra. So, this a times b, the product sigma algebra is not the sigma algebra with respect to which of all mu cross nu measurable sets. So, it may not be complete. So, for example, you can take any set a in x such that a does not belong to the algebra a and take any non-empty set b of measure 0. Then, the outer measure of mu cross nu will be equal to 0, because nu of b is equal to 0. But, the rectangle a cross b does not belong to product sigma algebra, because a does not belong to a. So, in case one wants to look at the completion of this, so that is possible. So, if we look at the sigma algebra a times b bar and denote that to be the sigma algebra of mu cross nu measurable subsets the product space, then of course, the product sigma algebra is inside it and that will be a complete measure space. So, we can say that x cross y and mu cross nu measurable sets as before is the completion of product measure space x cross y a times b mu cross nu. So, this is just a small observation, which we should keep in mind that the product sigma algebra, which is a sigma algebra generated by the rectangles need not be giving you a complete measure space. However, one can always complete it and the corresponding result holds for sets in a times b. That is a small technical result, which we can prove that we had proved this result for sets in the product sigma algebra. Namely, you can integrate the sections and get back the product measure. So, this also applies to any set E in the product sigma algebra. That means in the completion space also the corresponding result holds. So, this is the way we can compute the product measure of a set in the sigma algebra. I want to go over to an interpretation of this result, which leads to a very important result in integration of product spaces. So, what we had was the result namely, so what we have shown is for every set E in the product sigma algebra a times b, we can take it section with respect to every point x that gives us a set in the sigma algebra b. So, we can define nu of that and that becomes we show it is a non-negative measurable function. So, I can integrate this over x with respect to mu. On the other hand, I can also take the section of E with respect to every point y and then take its measure. We showed that these sections belong to A, take its measure mu of E y and we showed that that is a non-negative measurable function and I can integrate it over y d nu of y. We showed that these two are equal and in fact both of them are equal to the product mu cross nu of E, but a simple observation that the measure of a set is the integral of the indicator function. So, what is this? I can write it as integral over x, this nu of E x, I can write it as integral over y of the indicator function of E x y d nu y. And similarly, this thing I can write it as integral over y mu of E y. So, that I can write as integral of over x of the indicator function of E y x d nu of y and then we should have d mu of x. So, integral, sorry, this is E y. So, this is d nu of y. So, this is E y. So, there should be d mu of x and then d nu of y. And this product thing, I can write it as integral over x cross y of the indicator function of E d, the product measure mu cross nu. So, we get an integral representation of this result namely that I can take the indicator function of the set E. So, but note that this function, the indicator function of E x y is nothing but, see this is non-zero when y belongs to E x that means x comma y belongs to E. So, this is just the indicator function of E x comma y. So, and similarly this is also the indicator function of E x y. So, everywhere it is the indicator function of E. So, what we are saying is look at the indicator function of the set E and integrate it with respect to y. So, keep x fixed and integrate with respect to y. That depends on x, integrate with respect to x or take the indicator function of E then integrate with respect to x. So, keep y fixed. So, that integral depends on y and integrate it over y. So, that is another number that you will get and it says both of them are equal to integral of the indicator function of the set E with respect to the product measure mu cross nu. So, let me just rewrite and show it to you in the form of in this in the slide. So, what we are saying is the result that we proved just now for every set E in the product sigma algebra A cross B. I can rewrite the result in the form of integrals that namely it is same as saying that the integral of the indicator function of E with respect to the product measure mu cross nu is same as look at the indicator function, it is a function of two variables. For this function of two variables I can fix an x. If I fix an x and vary only y then this indicator function becomes a function of one variable y. So, it says let me integrate this function indicator function of E for a fixed x with respect to y. So, this integral is can be computed and this integral depends on x and says that is a measurable function and its integral can be taken with respect to x with respect to the measure mu and that is same as that integral. And similarly instead of fixing the first variable x I can fix the second variable as y I can fix this as y. So, then this becomes a function of x I can integrate it with respect to x I get a number which depends upon y and that function is integrable with respect to y and that integral is also equal to the original one. So, the result of computation of product measure of a set E in the set A cross B can be written in terms of the integrals of indicator function over the product set. .. So, basically this illustrates that to integrate the indicator function which is a function of two variables I can integrate it as one variable at a time. So, this is an important result which leads to an important result in integration that given a function of two variables if you want to integrate it with respect to the product measure then this gives a hint then possibly what one can do is fix one variable of the two variable function. So, it becomes a function of one variable integrate it out the one variable and then it becomes a function of the other variable integrate out that variable also you get the integral with respect to the product measure. So, we will prove this in the next lecture namely that this result can be extended to non-negative measurable functions on product spaces and eventually it can be extended to integrable functions. So, that leads to important theorems in the theory of integration on product spaces called Fubini's theorems. So, we will continue looking at that in the next lecture. Thank you.