 Welcome to lecture 31 on measure and integration. In the last two lectures, we have been looking at the properties of the Lebesgue measure on this space R 2 with respect to Porel measurable subsets of R 2 and Lebesgue measurable subsets of R 2. We will continue studying these properties of Lebesgue measure on the space R 2 a bit more today. If you recall, we had looked in the previous lecture how does the Lebesgue measurable sets and the Borel measurable sets behave with respect to the group operation and the topologically nice subsets of the plane. So, we showed an important aspect of the Lebesgue measure namely up to a constant multiple Lebesgue measure is the only translation invariant measure on the set of all Lebesgue measurable and of course, in particular the Borel measurable subsets of the plane. So, we will be making use of that property today, but we will begin with looking at some more transformations on the plane with respect to which Lebesgue measure can change and how does it change. So, let us begin by looking at today's topic is going to be Lebesgue measure and its properties, further properties. So, let us just recall what we had shown in the previous lecture was that if E is a Borel measurable subset of R 2, then its translate E plus X is also a Borel measurable subset in R 2 and the Lebesgue measure of the translated set is same as the Lebesgue measure of the original set very much similar to the properties on the real line. And we also showed that for a non-negative measurable Borel measurable function on R 2, if you look at the integral of the translated function that means f of X plus Y d lambda X is same as integral of f X with respect to X and also we also showed that this is also equal to actually the integral of f of minus X. So, under reflections and translations the integrals do not change. So, today we will start looking at another transformation on the plane namely for any vector X with components X and Y in R 2. Let us see how does this change if you multiply every element of a set with this vector. So, let us define the vector X dot E to be equal to A X comma B Y for every element A B in E that means the component the coordinate X coordinate is multiplied by X and the Y coordinate is multiplied by Y for every element A B in E. So, the X the first coordinate changes by A. So, the first coordinate A changes by X so it goes to A X and the second coordinate which is B goes to B times Y. So, the claim now we want to prove is that for every Lebesgue measurable set E this multiplied set X dot E is also a Lebesgue measurable set and the Lebesgue measure of this transformed set is equal to absolute value of X Y where X is the X component of the vector X and Y is the second component of the vector X. So, absolute value of X Y times Lebesgue measure of the set E. So, let us see how do we prove this property. So, we are given a subset E which is Lebesgue measurable and we are given a vector X with components X comma Y and we look at the vector X dot E which is all elements of the type A X comma B Y where A and B is an element in E. So, we want to check that for every subset E which is Lebesgue measurable in the plane implies that the vector X dot E is also Lebesgue measurable. So, that is first part of the claim. The proof of this is on the similar lines as for the translation namely we will collect together all the sets for which this property is true and show that rectangles come inside and show that this class for which this is true forms a sigma algebra. So, everything will be inside. So, that is a sigma algebra technique essentially we are going to use. So, let us look at that. Let us form the collection A to be equal to all subsets E belonging to L of R 2 such that X dot E belongs to L of R 2. So, let us look at that. So, let us observe. So, first observation is that this collection A is a sigma algebra. That is easy to check. I will just, I will not write, I will just orally discuss this. So, let us take if a set E belongs to it, then the observation is that we want to show that complement also belongs to it. So, that is easy because X dot E complement is nothing but, so X dot E complement is same as X dot E complement. So, if X, if E belongs to this collection A, then X dot E belongs to this collection. So, it is complement belongs to the Lebesgue measurable sets and the complement is equal to X dot E complement and that implies so this belongs to L of R 2 will imply that X implies that E complement belongs to L of R 2. And a similar computation will show that this is also A is closed under countable union. So, that will prove it is a sigma algebra. Let us check the second property namely if I take a rectangle. So, if I take a set E cross F, if I take a set E cross F where E and F both belong to L of R, then what happens to X dot E? So, let us compute that. So, in that case the vector X dot E cross F will look like, so that is the X component multiplied with E cross the Y component multiplied by F. So, it is X E cross Y F and now if X is a Lebesgue measurable set, we know that X dot E is a Lebesgue measurable set and Y times F also is a Lebesgue measurable set in the real line. So, this means this belongs to L of R 2, this is a set in L of R 2. So, that means if I take a rectangle sets of the type E cross F where E is in, so that E is in L of R and F is in L of R, then the rectangle satisfies the property of being in the class A. So, that means the L R cross L R is inside L of R 2. So, just now we observed that this is a sigma algebra, so that will imply that L of R, the product L of R is also in L of R 2. So, all the elements in the product sigma algebra L R cross L R are inside L of R 2 and now L of R 2 is just the completion of this space. So, that is easy to show that if I take an L set then X dot E also is an L set. So, also if E is subset of R 2 and lambda R 2 of star of E is 0, then it is easy to check, I leave it as an exercise that X dot E is again an L set, then X dot E lambda star R 2 is again an L, so lambda outer measure of that is again measure 0. So, that will imply that also implies that such sets, so lambda sets we hope for which, so the sets for which the Lebesgue measure is 0, Lebesgue outer measure is 0 are also in that class. So, that will imply this along with the earlier fact. So, this fact and this fact together imply that the A is equal to L of R 2. So, this is basically the sigma algebra technique, which is used to prove that the class for every to prove the fact that for every set E, this property is true. So, X dot E belongs to L of R 2 whenever E belongs to L of R 2 and next we want to check that the Lebesgue measure also is preserved. So, that again is a proof which is similar to the earlier proof. So, let us so we want to check the claim, the second part of the claim is that the Lebesgue measure, the Lebesgue measure of the set X dot E is equal to XY product Lebesgue measure of the set E. So, this is what we want to check. So, let us observe the following. Basically, we are going to apply the monotone class technique here as in the case of the translation. So, let us define M to be the class of all sets E belonging to L of R 2 such that this required claim. So, let us put that as a star. So, this property claim holds for the set E. So, the technique is to show that one that this class M is a monotone class. So, that is one step to show it is a monotone class to show that the sets L R cross L R, the rectangles are inside this class M and third this M is closed under finite disjoint unions is closed under finite disjoint unions. Because these three steps are approved, this will imply that M is equal to L of R 2 essentially. So, the idea is the following because these rectangles are inside M. So, second step is rectangles are inside M and first one says is a monotone class. So, the monotone class generated by these rectangles will be inside M. Now, third property says that this is also closed under finite disjoint unions. So, once it is a monotone class and it is closed under finite disjoint unions that will imply that the semi-algebra generated by the algebra generated by L R cross L R also is inside M. So, the monotone class generated by L R by L R will be in this. So, let me just write these steps that why this will imply this. So, basically the reason is the following. So, here is the reason why this will happen. So, L R cross L R, the rectangles inside M will imply by step 3, finite disjoint unions that the algebra generated by these rectangles will also be inside the class M because it is closed because the algebra generated by a semi-algebra is nothing but the finite disjoint unions. Now, this will imply by one that M is a monotone class and this algebra is inside it. So, that will imply that the monotone class generated by this algebra, so that is L R cross L R will also be inside M. But, this our monotone class theorem says that this is nothing but the sigma algebra generated by this class. So, this will imply this is L R, the product sigma algebra, Lebesgue measurable sets cross Lebesgue measurable sets is inside M. Now, once again one shows that this is also true for null sets. So, it will hold for completion. So, that will imply that L of R 2 which is the completion of this is also inside M and that inside M. So, that will prove the required result. So, the steps that this collection M is a monotone class and includes rectangles and is closed under the finite disjoint unions. This proof is similar to that of the proof when we had the translation of a set E by a vector. So, I will suggest that you try this as an exercise yourself and on the same lines as the earlier proof because it is a repetition of the same idea again and again. So, it is better to get used to it by doing it yourself. So, prove that. So, that will prove the required claim namely, so this property from the Lebesgue measure will be proved. Namely, if E is a Lebesgue measurable set, then the product X times E is also a Lebesgue measurable set and its Lebesgue measure is equal to the Lebesgue measure of the set E multiplied by absolute value of the components of X that were X comma Y. So, this property relates to something about multiplication and now we will like to rewrite this property in a slightly different way. So, let us before that let me just state the corresponding result for integrals namely, that the if F is a non-negative measurable function on R 2, then integral of X times T, this multiplication has defined now is equal to the absolute value of X Y times the integral of the function F. So, that is how if I multiply the function F each by a vector X, then the integral changes by the value mod of X minus Y and the proof of this is once again an application of sigma algebra monotone class there simple function technique I am sorry and I would like to leave it as an exercise once again. So, copy the proof try to copy the proof for the translation X plus T saying that the integral is invariant and where the multiplication comes. So, the steps are essentially first take F to be the indicator function of a set measurable set and that is just now we proved that result and once it is true for indicator functions this being a equality involving integrations. So, the for it will hold for finite linear combinations that means this will be true when F is a non-negative simple measurable function and for a general non-negative measurable function one takes limits of non-negative simple measurable functions and essentially applies monotone convergence theorem to get that this result is also true. So, the standard simple function technique will give you a proof of this. So, I will says that you have a look at that proof yourself. Let us go to some more properties of Lebesgue measure. So, just now we proved that for a Lebesgue measurable set E and a vector X in R 2. So, this should be R 2 if you multiply. So, lambda R 2 of X E is equal to the absolute value of the product of the coordinates of the vector X with which you are multiplying into the Lebesgue measure of E. One can reinterpret this result as follows. Namely, let us look at a transformation T from R 2 to R a linear transformation whose matrix is given by the diagonal matrix X 0 0 Y. So, here is some knowledge of linear algebra is required. So, let me state a few things about some facts about linear algebra that we are going to use. So, here is some facts about linear algebra that we are going to use. So, the first thing is that a linear transformation is in linear algebra. So, let us look at I will be looking at linear transformations on R 2 only. So, let us look at linear transformations on R 2. So, what is a linear transformation? A linear transformation is a function T from R 2 to R 2 which has the following property that T of if I take alpha times a vector X it is alpha times T of X for every alpha belonging to R and the vector X belonging to R 2. That is one property and the second that for any two vectors X and Y if I look at image of T of X plus Y that is equal to T of X plus T of Y for every X Y belonging to R 2. So, such a map is called a linear transformation and any such linear transformation T is also given by a matrix 2 by 2 matrix. So, let us call it as A, B, C and D. So, what is the meaning of this? That means that T applied to a vector X is same as the if you call this as the matrix A, that is the matrix A applied to the component of X. So, let us call it as A, B, T applied to A, B where X is equal to A comma B. So, this is very standard thing in linear algebra that linear transformations are described by matrix multiplication where A is called the matrix corresponding to the linear transformation that is obtained via basis of R 2. So, I will not go into details. So, that is what essentially we will require. So, let us look at a special linear transformation namely T which comes from the matrix which is a diagonal matrix. So, A, B, 0, 0. So, where will this be? So, what is the effect of the linear transformation T? T applied to a vector with components X, Y that is same as the matrix A, 0, 0, B applied to the column vector X, Y and the matrix multiplication says it is just A, X and B, Y. A, X plus 0, Y plus 0, X and B, Y. So, that is same as in our notation that the vector X multiplied with the vector with components X, Y. So, in our notation when we had this matrix we had this for any set E in subset in R 2 and we had a vector X with components X, Y. So, saying that we are multiplying X with E, the property that we studied just now is same as looking at the image of T, image of the set E under this linear transformation E. So, that is what the interpretation is. So, we can say that this multiplication by the vector X is transforming the set E by the map, linear transformation T, which is a map from R 2 to R 2. And our result says that the result we proved just now says that the Lebesgue measure of R 2 of the set X dot E is equal to the absolute value of X, Y times the Lebesgue measure of the set E, where the vector X is equal to with components X and Y. So, we said X dot E is the image of E under the linear transformation T and also this T is given by the diagonal matrix A00B. So, for this diagonal matrix there is a notion of what is called determinant of this transformation T. So, what is the determinant? Determinant for 2 by 2 is cross multiplying and subtracting the value. So, it is A times B for this determinant. So, absolute value of X, Y when you are multiplying by X. So, here our vector is with component X and Y. So, this result can be interpreted as lambda R 2 of X dot E is equal to this absolute value of X, Y is nothing but the determinant of T times lambda R 2 of E. So, our result that under multiplication that is how the value changes of the Lebesgue measure can be interpreted in terms of the linear transformations that if we take the linear transformation T from R 2 to R 2, which gives this multiplication as interpreted earlier. Then the Lebesgue measure of the translated set. So, this is T of E. The Lebesgue measure of the transformed set is determinant of T times the Lebesgue measure of the original set. So, this is the property interpreted in terms of maps. So, let us also observe something called when T is singular in that case determinant of T is 0 and basically saying that if X or Y are 0, then this both sides are 0. So, if neither X nor Y is 0, then determinant is not 0 and T is non singular. So, let us just look at these facts a bit more seriously. So, let us we have a map T from R 2 to R 2 and this is given by the diagonal matrix A 0 0 B. So, this is the matrix. So, that means any vector X Y goes to. So, X comma Y, a vector X goes to T of X, which is nothing but A X comma B Y. Now, let us observe a thing here that if either A is 0 or B is 0, if either A is 0 or B is 0, then determinant of T is equal to 0, because determinant of T is equal to equal to A times B. So, determinant of that means that is T is singular. Whenever determinant of a linear transformation is 0, T is singular that means it is in terms of functions, it is not 1 1, it is not 1 1. So, T is a linear transformation which is singular, so it is not 1 1. So, that implies that the image T R 2 of the whole space is a subspace of R 2 because under linear transformations, the image is always a subspace of dimension of T R 2 has to be less than or equal, it cannot be 2 because then it will be 1 1, because it is not 1 1, so it is less than or equal to 1. So, here I am discussing a bit of linear algebra because that will be required here. So, the question arises, what are subspaces S of R 2, S a subspace and dimension of dimension of S less than or equal to 1. So, one possibility is dimension of S is equal to 0, so that implies that S is just the vector 0 vector or secondly, dimension of S is equal to 1 and that will imply that geometrically S is a line through the origin or mathematically I can write S as all x comma y, where it is all x comma y, where a subspace S should look like y is equal to m x for some m. That means, in R 2, a subspace has to be nothing but a subspace is just a line through the origin, so S, a subspace has to be a line through the origin. So, once it is a line through the origin, what is going to be the Lebesgue measure of this line? So, let us look at the Lebesgue measure of R 2 of this line S. So, obviously the guess is this is going to be equal to 0. There are various ways of proving this, what one can do is to prove that Lebesgue measure of any line is equal to 0, what one can do is try to approximate this line by small rectangles or so I think this is a good exercise to leave. So, I leave it as an exercise to check that the Lebesgue measure essentially saying that the area of the line is equal to 0. So, let us use these facts. So, if T is singular, then for every subset E contained in R 2, T of E is going to be a subset of dimension 1. So, it is a subset of S, dimension 1 of S less than or equal to 1 implies that the Lebesgue measure of R 2 of this set E is going to be equal to 0. On the other hand, we also know that determinant also of T is also equal to 0. So, that implies for singular transformation T singular, then the Lebesgue measure of T of E equal to 0 equal to determinant of T, which is again 0 times Lebesgue measure of E. So, that property holds when T is singular. And if it is non-zero, if T is non-singular, if T is non-singular that means, where T is given by a 0 0 b that implies that neither a nor b is equal to 0. And determinant of T is equal to a b. So, that again once again implies that our earlier result implies that lambda of R 2 of T of E is equal to determinant of T times Lebesgue measure of this set T. So, when T is diagonal, so for diagonal transformations, we have this result that if we take a Lebesgue measurable set E and transform it according to a linear transformation, then the transformed set has got Lebesgue measure, which is determinant of T absolute value times the original Lebesgue measure of R 2. So, this is for non-singular determinant. So, the question arises, can we say that this result is true for all linear transformations. So, we have got a result for diagonal transformations, namely if T is a diagonal transformation from R 2 to R 2 and we transform a Lebesgue measurable set according to this, then the Lebesgue measure of the transformed set is absolute value of the determinant of T times the original measure. And the question is, can we say this result is also true for arbitrary linear transformations of the plane. And we are going to prove, yes that is true, it is this result holds for all linear transformations in R 2. So, that is what we want to prove the theorem. So, the theorem says for all linear transformations in R 2, one can say that the Lebesgue measure of the transformed set is determinant of the absolute value of the determinant of T times the Lebesgue measure of E. So, as before we will let assume T is singular and that we have just now observed, when T is singular this result is true, because T of R 2 is a subspace of R 2 of dimension less than 2. So, it will be either a line or a single point the origin vector. So, in either case the Lebesgue measure of the transformed set is equal to 0. So, it is a outer measurable set of measure 0. So, it will belong to the Lebesgue measurable set. So, what we are saying is that if E is a Lebesgue measurable set and T is a singular transformation, then T of E is a set of outer Lebesgue measure 0 in the plane and hence it is Lebesgue measurable and since determinant of T is also 0. So, the required property holds for required property holds when T is a singular transformation. Now, second case is when T is non-singular. So, let us look at the case when T is non-singular and we want to prove. So, T R 2 to R 2 non-singular. So, claim for every E belonging to Lebesgue measurable set T of E is also Lebesgue measurable and the Lebesgue measure of the transformed set T of E is absolute value of determinant of T times Lebesgue measure of E. So, that is what we want to check. We will first do it. So, let us assume for the time being that the set E is a Borel subset in R 2. When E is a Borel subset of R 2, we want to check that T of E is also a Borel subset of R 2. So, that is the question we want to first analyze. So, for that we observe that if T is a linear transformation from R 2 to R 2, T linear and non-singular that implies the non-singularity implies T is bijective. Every linear transformation which is invertible of course has to be bijective that is a and secondly and on the plane we have got the notion of topology convergence. One can easily check that every linear transformation is continuous. So, we are using two things here namely a linear transformation is non-singular if and only if actually it is bijective and every linear transformation on R 2 to R 2 is continuous. Actually it is true for R n to R n, but we are only considering it for R 2 to R 2. So, T is continuous. So, not only T is continuous, the inverse map bijective T inverse is also linear and hence continuous. So, any non-singular linear transformation T from R 2 to R 2 is a continuous map and the inverse also is a continuous map. Because that is true, let us take a set U contained in R 2 and U open. So, that will imply T of U is open because T inverse is continuous. So, that means this implies that T of U is a Borel set in R 2. So, what we are saying is if I look at the collection of all subsets all Borel subsets in R 2 such that the image is a Borel subset in R 2, then this collection A includes. So, open sets are inside this collection A and it is easy to check that. So, easy to check that this collection A is a sigma algebra basically because B R 2 is a sigma algebra and T is a bijective map. So, A is a sigma algebra. So, that will imply that for that this collection A is actually equal to B R 2. That means, so that is for every set E which is a Borel set in R 2 implies T of E is also a Borel subset of R 2. So, T preserves the collection of all Borel sets. Now, let us T is a map from R 2 to R 2. So, let us define. So, for every set E which is a Borel subset of R 2, let us define a new measure mu of T as follows, mu T of E is equal to Lebesgue measure of the set T times E. So, let us define. So, the claim one that mu T is a measure, that is easy to check because if I take the disjoint union of sets E i, one to infinity and then look at mu T of that. So, that is going to be lambda of T of the disjoint union of the sets E i of the sets. Now, let us observe that T is a one to one on to map. So, T of the union is going to be union of, so it is a disjoint union of T of E i, i equal to 1 to infinity and lambda being a measure, this is equal to sigma i equal to 1 to infinity lambda R 2 of T of E i. And that is same as saying, this is same as the summation i equal to 1 to infinity lambda R 2 of T of E i is mu T of E i. So, that proves the fact that mu T is a measure. The second property we want to check for this measure mu T is that mu T is translation invariant. So, let us check that second property that mu T is translation invariant on R 2. So, that means what? Let us take a Borel set in R 2 and let us take a vector x in R 2 and look at the vector E plus, let us look at the set E plus x. So, that is a translated set. So, mu T of this set is equal to by definition lambda R 2 of T of E plus x, but T is a linear transformation. So, that implies T of E plus x is just T of E plus T of x. So, that is a consequence of the fact that T is linear. So, that will imply and Lebesgue measure being translation invariant that says this is lambda R 2 of T of E and that is same as mu T of E. So, that is same as mu T of E. So, that proves that mu T is a translation invariant measure. Another fact about this measure, let us write it as the third property. Let us take the set 0 1 cross 0 1 which is a Borel set. Let us call this set as s the square in the plane that is of course, a close set cross a close set. So, that is a close set. So, it is a Borel subset in R 2 and mu T of this set s is equal to lambda R T of the Lebesgue measure of T of this mu T of 0 1 cross 0 1. Now, let us observe T of 0 1 cross 0 1 and let us observe that T of 0 1 cross 0 1. The observation is that this is a bounded s is bounded and hence T of s is also bounded. It is also a bounded set and of course, its Lebesgue measure is positive with Lebesgue measure of R 2 T s being positive and being bounded it has to be finite. So, that implies that the measure mu T of s the measure of mu T of s is the measure mu T of s is positive is bigger than 0 and less than infinity. So, these three properties of the measure mu T, let us look at what are the three properties of the measure mu T that we have proved 1. So, we defined the measure mu T of E to be equal to lambda R 2 of T of E and we said first of all it is a measure that is one property that we proved. The second property we proved it is translation invariant and the third property we proved that there is a set of finite positive measure with respect to mu of T. So, all these three properties by the uniqueness of the Lebesgue measure in R 2 implies that mu T of every set E has to be equal to a constant multiple. So, that is C of T times constant multiple of the Lebesgue measure of E for every. So, implies there exists a constant C of T bigger than 0 such that for every set E which is a Borel set. So, this is where we are using that the Lebesgue measure is essentially the only translation invariant measure on the plane. So, there is a, so any other translation invariant measure has to be a constant multiple of the Lebesgue measure. So, what we have gotten is for every T non-singular, for every T non-singular there exists a constant C of T such that Lebesgue measure mu T which is nothing but the Lebesgue measure of the transform set T of E is equal to the constant multiple C times T of the Lebesgue measure of the set E. So, this is the property that we have established that for every linear transformation the transform set T of E will be a Borel set and its Lebesgue measure will be a constant multiple of this. So, that means this gives us a map. So, hence we have for every non-singular linear transformation we have got a constant C of T for every T non-singular and what we want to do is to prove, so to show that C of T is equal to absolute value of determinant of T for every T non-singular, so that is what we want to show. So, once we do that we will be through with our construction because C of T being determinant that will prove that it is non-singular. So, at this stage to prove that I need some more facts about linear algebra and it will, so to prove that we need some more facts about linear algebra and we will not be able to complete the proof in the remaining part of today's lecture. So, I will continue the proof next time. So, we will start from here saying that we have got for every non-singular linear transformation T a constant C of T that means there is a map of T going to C of T and we want to show that this map does not this map actually is nothing but the determinant of T. So, we will prove this next time. Thank you.