 Welcome to lecture number 32 on measure and integration. In the previous few lectures, we have been looking at the Lebesgue measure on the space R 2 and its various properties. In the previous lecture, we had started analyzing how does a Lebesgue measure of a set change when we apply a linear transformation to it. So, we had started analyzing it. Let us recall what we had done and then we will continue analyzing this problem and some more properties of Lebesgue measure under other transformations. Let us recall what we had done last time was that we started looking at the theorem namely if T is a linear transformation from R 2 to R 2 such that and E is a Lebesgue measurable subset, then we showed that we wanted to show that the transform set T of E, the image of E under T is again a Lebesgue measurable set and the Lebesgue measure of the transform set is obtained by multiplying the Lebesgue measure of the original set with the constant called determinant of E. So, Lebesgue measure of T of E is equal to determinant of T times Lebesgue measure of the original set E. So, this property, this theorem we had started analyzing. We had analyzed the proof of this theorem in the first case when T is a singular transformation and there we argued that if T is a singular transformation, then T of E, the image set is going to be a Lebesgue null set and for a singular transformation determinant of E, T is also equal to 0. That is how singular transformations are characterized. So, in that case both the terms, the Lebesgue measure of the translated set is equal to 0 is same as determinant of T times the Lebesgue measure of the original set. In the case when T is a non-singular transformation, when T is non-singular, so non-singular means T is invertible and here are a few facts about linear algebra saying that T is non-singular is same as saying that determinant is not equal to 0 and it is equivalent to saying that as a map T is a 1 to 1 onto map and the inverse of course is also a linear transformation. In analyzing the proof of that, we had already shown that if E is a Borel set, so we first restrict ourselves to Borel subsets of R2. We showed that if E is a Borel set, then for every non-singular linear transformation T of E, the transformed set T of E also is a Borel set. That basically follows from the fact that every linear transformation is a continuous map and if it is non-singular, then the inverse also is a continuous map. So, essentially saying that for every Borel set E, T of E is Borel. One analyses that when E is open set, T of E is an open set and n set is a Borel set and then one shows that the collection of all sets for which this property is true namely the image is a Borel set, is a sigma algebra including open sets and hence one can includes that for every set E, the transformed set, the image set T of E is also a Borel set. We also analyze that if you consider this as a measure for all Borel sets, then it is a translation invariant because T is linear and that means by the uniqueness property we got that for every linear transformation T, which is non-singular, the Lebesgue measure of the transformed set namely T of E must be a constant multiple of the original measure lambda R 2 of E and that constant will depend on the transformation T. So, this is the stage we had reached and then we wanted to analyze further and the claim we want to prove is that this C of T is nothing but determinant of T. So, this is the stage we had reached. So, let us continue the proof. Let us observe that this map T to C of T, see for every transformation T, we are associating a number C of T, which is non-negative. So, we get a map T going to C of T. So, this is a map for every non-singular transformation T, we are associating a number C of T. This association, this map has the following properties namely if T is a diagonal transformation then we observed that was the beginning of our analysis of this theorem that C of T is nothing but the determinant of T. So, for diagonal transformations C of T is equal to determinant of T determinant. And the second property is that if the linear transformation is a orthogonal transformation then this C of O is equal to 1 is equal to determinant of T. And this is because of the fact that a orthogonal transformation on R 2 leaves the set namely the unit circle, you can think of it as all x in R 2 such that norm of x is less than or equal to 1 invariant. So, let us just look at this property, a slightly more sum of you may not be knowing about linear transformations. So, T is a linear transformation from R 2 to R 2. And let us, every linear transformation is given by a matrix A. So, which is a 2 by 2 matrix, so it is A, B, C and D. So, saying that T is orthogonal is same as saying that if you take A and look at it as A transpose that is same as A transpose A and that is equal to identity. So, saying that T is orthogonal is characterized by this property about the matrices, matrix of the linear transformation namely A times A transpose is same as A transpose times A and that is equal to identity. And which is also equivalent to the property that if you look at the row vectors and the column vectors, so the row vector is A, B and C, D. So, these two vectors are orthogonal. Namely, the dot product is equal to 0 and each is a unit vector. So, that is called orthogonal, but we are not going to use this property. Let us look at this property which says A A transpose is equal to A transpose A. So, this property gives us the following fact namely let us look at the dot product. So, let us take any vector x belonging to R 2 and look at the dot product of, so let us look at the image, the image is T of x. So, x goes to T of x. Let us look at the dot product. So, norm of T x square that is given by the dot product of T x with itself. So, that is the definition of the magnitude, the dot product in R 2, but this dot product can also be written as T times, this T can be written as T transpose T of x. So, T transpose is same as A transpose basically. So, you can think it as matrices and that being identity, so this is same as x comma x. So, orthogonal transformations are also characterized by the property that the norm of the image of any vector is equal to norm of the original vector. So, that is another way of characterizing orthogonal transformations. We can take that as the definition of the transformation if you like. Now, from both these properties that A transpose A is equal to identity. So, that implies the following fact that A transpose A equal to identity. This implies that the determinant of A transpose determinant of A is equal to 1 and that implies determinant of A transpose is same as determinant of A. So, that says determinant of A times determinant of A. So, this square is equal to 1. So, that implies that the absolute value of determinant of A is equal to 1. So, T orthogonal implies that determinant of T absolute value is equal to 1. So, that is one fact. Also, the fact that norm of T x is equal to norm of x implies that if norm of x is less than or equal to 1, so that implies norm of T x is also less than or equal to 1. So, that means if in the plane we look at the unit circle, this is the set where norm of x is less than or equal to 1 and if you take the transformed set, the transformed set is same. So, under T, this goes back to the same thing. So, norm of T x is equal to norm x less than or equal to 1. So, that is saying that T leaves, if T is orthogonal, then it leaves the unit circle, the region inside the unit circle invariant. And that means, that essentially means that the Lebesgue measure of this set is same as the Lebesgue measure of that set. So, that implies that the Lebesgue measure of the transformed set, so that unit circle, so mod x less than or equal to 1 is same as the Lebesgue measure of mod x less than or equal to 1. And this being equal to determinant of T, this is being determinant of T times, this is same as that, so this is same as, so that implies that, but this can be written as determinant of T, because that is equal to 1 times Lebesgue measure of norm x less than or equal to 1. So, this implies, this is a constant, so C of T, which is determinant of T is equal to, so C of T is 1, which is same as determinant of T. So, that means, for orthogonal transformations, so T orthogonal implies C of T is equal to 1. So, that is the second fact that we wanted to prove. Namely, if O is an orthogonal transformation, then the C of O, the constant, this typo here, this should have been determinant of O is because the unit circle, region inside unit circle is left invariant by the orthogonal transformation. .. Now, and the next property we want to analyze is that, for all non-singular transformations T 1 and T 2, C of T 1 T 2 is equal to C of T 1 times C of T 2. That means, this map is multiplicative. The map T going to C of T is a multiplicative map. Namely, if I look two transformations T 1 and T 2, then C of T 1 T 2 is same as C of T 1 times C of T 2. Now, let us look at a proof of that. So, T 1 and T 2 are transformations from R 2 to R 2. So, we are going to look at, so let us write T is equal to T 1 T 2. Then, to compute C of T 2, we are going to look at the measure mu of T T, which is equal to, so let us take any set. So, let E be any set, which is a Borel set. Let us look at lambda of lambda R 2 of T 1 T 2 applied to E. Let us look at this. This by definition of the constant C of T is C of T 1 T 2, because T 1 T 2 is a linear transformation applied. So, this composite T 1 T 2 is applied to E. So, by definition of C of T 1 T 2 that should be equal to C T 1 T 2 of Lebesgue measure of the set E. On the other hand, we can also think of this as lambda Lebesgue measure of R 2 of T 1 applied to T 2 of E. So, this composition T 1 T 2 is same as saying the linear transformation T 1 is applied to the set T 2 of E. But, if you do that, then we know that this is equal to lambda R 2 of, it is lambda R 2 of T 1 of a set, so it is C T 1. So, this is equal to C T 1 times lambda R 2 of T 2 of E. Now, that once again lambda R 2 of C of T 2 of E gives you C T 2 and the original is C T 1 into lambda R 2 of E. So, we get that C of T 1 T 2 times the Lebesgue measure of any set E is same as C T 2 into C T 1 of Lebesgue measure of E. So, this happens for every set E, so this implies that C of T 1 composite T 2 is same as C of T 1 times C of T 2. So, this map is a multiplicative map. So, this is the property we wanted to prove. Now, we need another fact from linear algebra, namely what is called a singular value decomposition for linear transformations. So, in case we have not come across this theorem called singular value decomposition for linear transformations, please look into the text book that we have suggested, namely an introduction to measure and integration and look at the appendix of that book, you will find a proof of this singular value decomposition for linear transformations. So, let us state what is a singular value decomposition. It says that every linear transformation T can be represented as a product of three transformations, where the first one P and the last one Q are both orthogonal transformations P and Q are orthogonal transformations and this D is a diagonal transformation. So, every linear transformation T can be represented as P times D times Q, where P and Q are some orthogonal transformations and D is some diagonal transformation. So, this is a theorem called the singular value decomposition in linear algebra. So, please have a look at a proof of this in case you have not come across this theorem in the text book mentioned. So, once we know that for every linear transformation T can be represented as P times D times Q. So, and the property 3 just now stated says that the constant C of any composite is the product. So, we apply that property through this. So, we get C of T will be equal to C of P times D times Q, which is nothing but the product. So, the C of a transformation T will be C of P into C of D into C of Q. So, we get that the constant for any linear transformation T is equal to the constant for some orthogonal transformation P times the constant for a diagonal transformation D and the constant for another orthogonal transformation Q. But just now we observed that for orthogonal transformations the C of orthogonal transformation the constant associated is 1. So, C of P is 1 and C of Q is 1. So, that gives you C of T is equal to C of D because both first and the last multiplicative things are 1. So, it is C of D and for a diagonal transformation we have already shown this is equal to determinant of D. So, for the linear transformation T the constant C of T is equal to determinant of D, where D is the diagonal transformation which appears in the singular value decomposition T is equal to P D Q. But on the other hand we can also look at the determinant of T from this. So, determinant of T recall that determinant is also a multiplicative map. So, determinant of T will be equal to determinant of P into determinant of D into determinant of Q, but determinant of P and determinant of Q both are equal to 1. So, that saves determinant of T is equal to determinant of D and just now we said determinant of D is equal to C of T. So, combining these two we get determinant of T is equal to determinant of D. So, this should be C of T is determinant of D. So, that says that C of T should be equal to determinant of D. So, here it should be this is redundant. So, C of T is determinant of D and determinant of D is equal to determinant of T. So, this two combined together gives you C of T is equal to determinant of T. So, that completes the proof of the fact that for a linear transformation. So, we have completed the proof that for a linear transformation if you take this at E and change it, transform it by a linear transformation that is same as determinant of T times the Lebesgue measure of E. So, the Lebesgue measure of the transform set is determinant of T times the Lebesgue measure of the set E. So, that proves the theorem for all sets which are Borel measurable sets. So, till now we have proved the theorem only for Borel measurable sets. We will like to extend this theorem for Lebesgue measurable sets. So, for that let us observe the following. How are the Lebesgue measurable sets in R 2 obtained from Borel measurable sets? What is the relation between Lebesgue measurable sets and the Borel measurable sets? So, the first property is that if let us take any set N which is in R 2 and such that its Lebesgue outer measure is 0. Look at sets of Lebesgue outer measure 0 in the plane. So, we first claim that under any linear transformation T, the image is also a Lebesgue measurable set of measure 0. That means linear transformations take sets of measure 0 to sets of measure 0 in plane. So, to prove that let us observe the following thing. Let us take N a subset of R 2 and Lebesgue measure of N equal to 0. But, saying Lebesgue measure of N is equal to 0 is same as saying for every epsilon bigger than 0, there exist rectangles. There exist rectangles say R i bigger than or equal to 1 such that the set N is contained in the union of these rectangles R i and the Lebesgue measure of the rectangle R i added together is less than epsilon. So, saying a set is a null set is same as saying it can be covered by rectangles such that the total measure of the rectangles put together is less than epsilon. But note now so each R i is a rectangle so it is a Borel set is a Borel subset of R 2. So, that implies that T of R i also is a Borel subset of R 2 for a non-singular linear transformation T if T is non-singular and the Lebesgue measure of T of R i by what we have proved just now is equal to determinant of T times Lebesgue measure of Lebesgue measure of R i. So, that is just now we have proved for Borel sets this property holds. .. So, now the fact that N is covered by union of R i's is implies so this fact star implies that T of N is covered by union of T of R i i equal to 1 to infinity. So, this is contained by T of R i so that implies by the countable subordinary property of the Lebesgue measure outer Lebesgue measure R 2 of T N is less than or equal to summation i equal to 1 to infinity Lebesgue outer measure of T of R i and Lebesgue measure of T of R i is determinant so this is equal to determinant of T absolute value times summation of i equal to 1 to infinity Lebesgue measure of R i and that is less than epsilon so it is less than or equal to absolute value of determinant of T times epsilon and since this property holds so this holds for every epsilon bigger than 0 so let epsilon go to 0 so that will imply Lebesgue measure of outer measure of T of N is equal to 0 when T is if T is non singular transformation we know T of R 2 itself is 0 so T of N will be 0 so this proves the fact that for every set N which is of Lebesgue outer measure 0 lambda the image T of N under any linear transformation is again a set of measure 0 and the second fact between Lebesgue measure and Lebesgue measure and Lebesgue measurable sets and the Borel measurable sets is the following every Lebesgue measurable set E can be represented as a union of two sets 1 a Borel set A and NL set N and the Lebesgue measure of E is same as the Lebesgue measure of the set A N is a set of measure 0 so that is the second fact one has so let us take so let let A be a Lebesgue measurable let us take Lebesgue measurable set in R 2 then this set A can be written as E union N where E is a Borel set in R 2 and the Lebesgue measure so lambda R 2 of A is same as lambda R 2 of E now if we look at T of A so if it is look at the set T of A then that will be equal to T of E union T of N if T is say non singular you know this is a Borel set and this is again NL set so Lebesgue outer measure of T of A is equal to Lebesgue outer measure of T of E which is equal to determinant of because this is a Borel set so determinant of T times Lebesgue outer measure Lebesgue measure of the set E and which is same as the Lebesgue measure of the set A so this is same as Lebesgue measure of T times Lebesgue measure of A so we have used the fact that if A is a Lebesgue measurable set then A can be written as E union N where E is a Borel set and N is a NL set that means the Lebesgue measure of R 2 of the set A is same as the Borel Lebesgue measure of the Borel component of it that is R 2 of E so now if I apply transformation T to it and say T is non singular then T of A will be equal to T of E union T of N and just now we observed that T of N is a NL set and T of E is a Borel set so Lebesgue measure of T of A will be nothing but the Lebesgue measure of T of E which by the earlier case is determinant of T times Lebesgue measure of E and Lebesgue measure of E is same as the Lebesgue measure of A so that proves that for a Lebesgue measurable set the Lebesgue measure of the transform set T of A is same as determinant of T times the Lebesgue measure of the set A itself so that proves the theorem completely namely that if E is any Lebesgue measurable set then T of E is also Lebesgue measurable and the Lebesgue measure of T of E is equal to the determinant of T times the Lebesgue measure of E. So this is how the Lebesgue measure of a set E in the plane changes with respect to linear transformations. Now we will give some applications of this now because there are many nice linear transformations in the plane so let us look at the first application of this namely so let us take two vectors A and B A 1 B 1 and A 2 B 2 in R 2 and look at the set which are all sets set P of all vectors in the plane of the type where the first component is alpha 1 A 1 plus alpha 2 A 2 and the second component is alpha 1 A 1 plus alpha 2 B 2 where alpha 1 and alpha 2 are numbers between 0 and 1. This is called the parallelogram determined by the vectors A 1 B 1 and A 2 B 2. In the picture it is the following so if we have the plane let us take two vectors one vector is this vector and other vector is this. So this is the vector which is A 1 B 1 and this is the vector which is A 2 B 2 then they determine a parallelogram the geometric object so let us see what is that parallelogram so that is nothing but this parallelogram. So that is the parallelogram P determined by these two vectors A 1 B 1 and A 2 B 2 and any vector in between so this vector so this parallelogram P is characterized by that this any vector inside here so call it as x y so P is then x looks like alpha 1 A 1 plus alpha 2 A 2 and the second look looks like alpha 1 B 1 plus alpha 2 B 2 where this alpha and alpha 2 have the property their numbers between 0 and 1 so alpha 1 alpha 2 between 0 and 1. So this is the point this is the point with components alpha alpha 1 and alpha 2 so if I take so this is how this is what the parallelogram so one can check geometric fact that if I take two vectors A 1 and B 1 and look at the geometric picture of this parallelogram then if I take any alpha 1 alpha 2 between 0 and 1 and look at this then this is nothing but the parallelogram given by these two vectors. The claim we want to is that we want to show that the Lebesgue measure of this parallelogram is same as the absolute value of A 1 B 1 A 1 B 2 minus A 2 B 1 so these are the components A 1 B 1 and A 2 B 2 are the components of the vectors which we started with so the claim is so the claim that Lebesgue measure of P is equal to the absolute value of A 1 B 2 minus A 2 B 1 so to prove this we are going to show that this P is equal to T of is T of a set E where T is a linear transformation T linear and E is a nice set and it is not difficult to guess what is T and what is E so let us just look at that so the claim so let us observe that if T is the matrix with the components A 1 A 2 B 1 B 2 so rows are the vectors which are given vectors are A 1 B 1 and A 2 B 2 so the first column is the vector A 1 B 1 second column is the vector components of the vector A 2 B 2 if I look at this transformation T and look at the set E with components alpha 1 alpha 2 in where alpha 1 alpha 2 are real line and they are between 0 and 1 so what is this set E this set E is nothing but this set E is nothing but a rectangle actually a square in the plane with sides 0 1 2 0 1 and if I look at T of E so any set alpha 1 alpha 2 so what will be this so what we are saying is the following that if I look at the set E so here is the set E which looks like so this is the set E 0 0 this is 1 0 1 1 and 0 1 so if I look at this set and look at the transformation given by T where T is equal to A 1 B 1 A 1 B 1 and A 2 B 2 if I look at this then the image then the image of this under this is precisely is precisely that parallelogram P where this is A 1 and A 2 and this is A 1 B 1 and this is A 2 B 2 so this transforms to this parallelogram so if T is this and E is this set then T of E is equal to P that is because what is so let us look at A 1 A 2 B 1 B 2 and a vector is alpha 1 alpha 2 so what is that so that is A 1 alpha 1 plus A 2 alpha 2 and that is gives you B 1 alpha 1 plus B 2 alpha 2 so that says the vector here alpha 1 alpha 2 goes to the vector they are given by this and that is precisely the parallelogram so under the and this is a linear transformation so under the linear transformation T given by this matrix the unit square changes to the parallelogram and once that is true so this will imply that the Lebesgue measure so this will imply the Lebesgue measure of P is same as the Lebesgue measure of in the plane of T of E and that is equal to determinant of T absolute value times Lebesgue measure of E but Lebesgue measure of E that is the area that is the rectangle so it is the area that is equal to 1 and determinant of T is A 1 B 2 minus A 2 B 1 so that gives us the result so that gives us the result that the Lebesgue measure of P is same as the Lebesgue measure of the transform set T of E which is equal to determinant of T times the Lebesgue measure of E and that Lebesgue measure of E being equal to 1 so that gives us determinant of T which is nothing but A 1 B 2 minus A 2 B 1 so this gives us that Lebesgue measure of the parallelogram is the determinant of that given by the vector so that is A 1 B 2 minus A 2 B 1 so these are this is one of the results that one proves normally in geometric and linear algebra that the determinant is nothing but a measure of the parallelogram area of the parallelogram determined by the vectors. Let us look at another application of this formula how the linear transformation changes so let us look at the Lebesgue measure of the unit circle region enclosed by the unit circle so Lebesgue measure of all vectors x comma y in R 2 such that x square plus y square is less than 1 so let us look at that so let us call the Lebesgue measure of this to be equal to a number pi we are not we are not assuming anything about pi we are just saying that the Lebesgue measure of this unit circle is a finite quantity so let us see why it is finite so here is the unit it is a bounded so this is x square plus y square less than 1 so that is that set so this is a bounded set so for example this is enclosed inside this rectangle inside this square of this is 1 and this is 1 so this is 0 and this is 1 and this is 1 so it is enclosed inside this square of length so that means that the area so this is less than or equal to that it is a subset of the square and square is a bounded thing so that means the Lebesgue measure of the points say that x square plus y square is less than 1 will be less than or equal to the Lebesgue measure in the plane of the square which is finite quantity so that means that the Lebesgue measure of the region enclosed by the unit circle is a finite number and this finite number we are just calling it by the number by we are denoting it by the symbol pi so pi is the Lebesgue measure of the region enclosed by the unit circle then the claim is that if we look at the annulus region if we look at the annulus region that is x square plus y square bigger than a square and less than b square then its Lebesgue measure is pi of b square minus a square so what we want to prove is that if I look at if I look at here is bigger circle and here is the smaller one and this radius is a and this radius is b so we are saying the Lebesgue measure of this portion is nothing but pi b square minus a square that is what we should be expecting from our ordinary geometry that we have been learning in schools namely the area of the circle is equal to pi r square so we will first prove that the area of a circle of radius r is equal to pi r square and from there we will reduce this fact so let us observe that so the first thing is let us take the linear transformation T which is diagonal which is given by a 0 a so we are looking at the diagonal transformation a 0 0 a and look at the unit circle area enclosed by the unit circle so that is e so e is the set of all x vectors x comma y in r 2 say that x square plus y square is less than 1 then if we look at any point here and transform it according to this T that will look like so let us look at what will that look like so let us look at the transformation a 0 0 a and let us look at a vector x y so that gives us the vector a x a y so if this vector had the property that x square plus y square is less than 1 then the transform vector a x a y has the property that a x square plus a y square is equal to a square times x square plus y square so which is less than a square so that shows that the unit circle so if e is the unit circle that is x y x square plus y square less than 1 then T of e is the circle the region enclosed by the circle x square plus y square less than a square so that is what we know so that means now we apply our apply our theorem of linear transformations so look at the Lebesgue measure r 2 of the transform set e so that is equal to by the property of the theorem it is determinant of T times Lebesgue measure of the set e but determinant of T is equal to it is a diagonal transformation so that is a square and Lebesgue measure of e which is a unit circle is pi so Lebesgue measure of the transform set is equal to pi a square so that means the Lebesgue measure of all the points x y say that x square plus y square is less than a square is nothing but pi a square so that is the magnification that we are getting and so as a consequence of this let us deduce for the annulus region the area is the required Lebesgue measure is pi of b square minus a square so let for that we have to just observe that if I look at the circle so the set of points x y x y such that x square plus y square is less than b square is a set which is a bounded set and Lebesgue r 2 of that is finite so the set of finite outer measure finite Lebesgue measure so I can write that the Lebesgue outer measure of x square plus y square bigger than a square and less than b square is nothing but the Lebesgue measure of the set x square plus y square less than b square minus the inner circle so that is x square plus y square less than a square and now everything being finite I can write this as the Lebesgue measure of the region enclosed by the outer circle so that is x square plus y square less than b square minus the Lebesgue measure of x square plus y square of Lebesgue measure of x square plus y square less than a square this is possible because everything is a finite quantity so the measure of the difference a minus b measure of a minus b is measure of a minus b whenever b is a subset of a and everything is finite so that property gives you this and this is pi of b square just now we saw pi of a square so that is pi of b square minus a square so that proves so what we are saying is this simple properties help us to confirm that the Lebesgue measure on the plane that we have defined is essentially the extending the notion of area in the plane to a bigger class of subsets and the usual formulas for the areas that we have been using a priori without any justifications are now being justified by the Lebesgue measure. I will just point out one more extension of this result namely that the annual Lebesgue measure of an annular region is pi of b square minus a square to something in integration which is looks like the change of variable formula in multiple integrals namely if you have a double integral then and you change to Cartesian to polar coordinates then the d x d y normally we have that formula that when you change d x d y it should be r d r d theta d theta. So, a more rigorous we are saying that for a particular class of functions I want to state and give an outline of the proof we will not be proving it fully we will give an outline of the proof. So, let us go to the next application of so this is what I just now said that the Lebesgue measure of the annular region is Lebesgue measure of the outer circle minus the Lebesgue measure of the inner circle that is pi b square minus a square. So, this is another application or extension of this result just now we proved it is called the integration of the radial functions. So, let us look at if the theorem says let us look at a function defined from 0 infinity on the positive on the non negative part of the real line taking values in non negative values so that is 0 to infinity. So, it is a non negative measurable function defined on the non negative part of the real line. Then the claim is that if I look at the double integral over r 2 of f absolute value of x, x is a vector so absolute value means the norm so the magnitude of the vector x. So, look at this so this is like a composite function the vector x goes to the magnitude that is a non negative real number and f evaluated at that. So, the double integral integral with respect to r 2 is given by 2 pi times f r d r d theta. So, that is the claim that this integral is equal to this integral. So, and what is the meaning of a non negative radial function f is a non negative I should have said here it is a radial function that means it depends only on the absolute value of the function it does not. So, f is a non negative so this is a radial function. So, this you can think it as f composite the magnitude is the radial function is a radial function it the value of this composite function depends only upon the magnitude of the vector and not on the position of the vector. So, to prove such a result the proof is a typical application of simple function technique. So, one tries to prove that for simple measurable functions this is true and then apply monotone convergence theorem and so on. So, I will just outline the steps for a detailed proof you may consult the text book. So, let us look at the first step. Let us look at the first step when this function f is the indicator function of a interval a b when f is the indicator function of a interval a b where a b is the interval in the non negative part of the real line. So, a less than b bigger than 0. So, when f is the indicator function let us compute these both sides and what is the look like. So, when f is the indicator function of the a b so here is the indicator function so this is 0 to infinity. So, 0 to infinity means this will give you indicator function will give you only a to b. So, this will be a to b of the function f r the function is indicator function this value is 1 so x r d r. So, when you integrate r d r you get r square by 2 between a and b. So, when you put the values you get b square minus a square by 2 so that is equal to pi of b square minus a square. So, this side is nothing but pi of b square minus a square and what is this f so the indicator function of a b evaluated at the absolute value means you are integrating in the annulus region between the limits a and b. So, it is pi b square minus pi a square so in that then it is just equal to pi of b square minus a square. So, this both sides are nothing but the result we did discuss just now that the area of the annulus region is equal to pi b square minus a square. So, step 1 is for indicator function it is that result. The next thing is you look at a sets which are either sequence is e n's which are sequences of sets which are Lebesgue measurable of course either or pair wise disjoint or increasing sequence and supposing for the indicator function of each set e n this result holds then the claim is it also holds for the union of e n's. So, if for each e n the result holds then the result for each indicator function of each e n it holds then it also holds for the indicator function of the set e and that essentially is an application of the monotone convergence theorem to the earlier result. So, and then the step 3 is from such things one comes to open sets by the fact that each open set is a countable disjoint union of countable disjoint union of intervals. So, for intervals that property holds so this will hold for every open set and from the open sets and null sets. So, one shows the corresponding property also holds for null sets. So, open sets and null sets one goes to the indicator function of any Lebesgue measurable set because any Lebesgue measurable set can be written in terms of open sets and null sets. So, and then from this one applies the usual monotone convergence theorem technique from the indicator function to non negative measurable function. So, these are the steps one follows to prove the theorem of this kind. I just want to conclude today's lecture by saying what we have done it for product spaces of two product spaces can also be extended to any finite number of product spaces. So, namely if you are given a finite number of measure spaces x i, a i, mu i we define the product of two of them can be extended by sort of doing one at a time iteratively you can define the product of the space this x space is x i is 1 to n you can define the product sigma algebra i i is 1 to n and you can also define the product measure inductively one can do that. So, this is called the product space will I will not go into the details of it, but this is useful and one can also show that if you take product of say some finite number m number and take product of some n number of copies and then take the product again that is same as the product of them put together. So, it is same as the product of x i is from n 1 to m plus n. So, these are same. So, one can same with usual identity. So, basically saying that what we have done it for two product of two measure spaces can be done for a finite number of them. So, as a consequence one can define the notion of Lebesgue measurable subsets in R n and the notion of Lebesgue measure in R n. So, this can be done. So, this again those who are interested should refer the text book for more details. So, what we have done today is we have completed the study of product measure spaces and with that we have essentially completed what is called the basic concepts in measure theory namely we have done the extension of measure then integration of measure then measure and integration on product spaces. This is the core of the subject and from now onwards I will be looking at some special topics in our subject of measure theory. Thank you.