 Welcome to lecture 30 on measure and integration. In the previous lectures, we had defined what is called the product measure on product space. In this lecture, we will specialize that construction on the set R 2, which is a Cartesian product of real line with itself and the sigma algebra being that of either Borel sets or Lebesgue measurable sets and the measure being the Lebesgue measure. So, the topic for today's discussion is going to be Lebesgue measure and integral on the space R 2. So, let us just recall. So, we had defined the product measure space. Given measure space is X, A and mu and Y, B and mu, we defined the product sigma algebra A times B on the product space X cross Y and the product measure mu cross mu. So, today we will start looking at the particular case when X is equal to Y equal to the real line and the sigma algebra A is same as the sigma algebra B is same as the sigma algebra of Lebesgue measurable sets on the real line and mu is same as mu, which is same as the Lebesgue measure. So, we are looking at a copy of the real line, the sigma algebra of Lebesgue measurable sets and lambda the Lebesgue measure and taking its product with itself. So, that will give rise to the product measure space R 2, the Lebesgue measurable sets times the Lebesgue measurable sets the sigma algebra and the product measure lambda cross lambda. If you recall, we had mentioned that even if the original measure spaces are complete, the product measure space need not be complete. So, this product measure space R 2 lambda r cross lambda r and lambda cross lambda is not complete. So, we can always complete it and the completion is denoted by R 2, Lebesgue measurable subsets of R 2 and lambda of R 2. So, this is called the Lebesgue measure space. So, Lebesgue measure space is obtained from the sigma algebra Lebesgue measurable sets times Lebesgue measurable sets completed with respect to the product Lebesgue measure on R 2. So, this is normally called the product called the Lebesgue measure measure space on R 2 and the sets in the sigma algebra lambda of R 2 are called Lebesgue measurable sets in R 2 and the measure lambda of R 2 defined on this completed space is called the Lebesgue measure on R 2. So, whenever one refers to the Lebesgue measure space, it is the complete measure space obtained via completing the product measure on the product sigma algebra. So, today we will start looking at properties of Lebesgue measurable sets and Lebesgue measure. So, let us denote by i tilde as we have done for the real line, the collection of all left open right closed intervals in real line. So, let us look at the rectangles obtained by such intervals. So, that we denoted by i tilde 2 upper script 2 as i cross j rectangles whose sides are left open right closed intervals. Then we claim that this i cross j is a semi-algebra of subsets of R 2 and the sigma algebra generated by this is equal to the Borel sigma algebra of R 2. So, to prove this we already know that i tilde, the left open right closed intervals form a semi-algebra of subsets of real line and we have already shown that if you take rectangles consisting of elements of the semi-algebra, then itself form a semi-algebra, namely the product of semi-algebras is always a semi-algebra. So, that general construction will tell that this space, the set of all rectangles with left open right closed intervals is a semi-algebra. To show that this sigma algebra generated by the rectangles i tilde 2 is the Borel sigma algebra, we observe few things. First of all, if you recall, we had shown, so let me just recall the effect that we have shown in the beginning of defining product sigma algebra, namely if we take a set x and take a set y and here we have got a collection of subsets c and we have got a collection of subsets, so these are collections of subsets. Then we can form c cross D, that is a collection of subsets of x cross y. So, this is equal to all sets of the type e cross f, where e belongs to c and f belongs to d. Now, one can generate a sigma algebra out of this collection c cross d. On the other hand, we can generate a sigma algebra by the collection c. We can also generate a sigma algebra by the collection d of subsets of y and take the sigma algebra generated by the rectangles of this type. So, let us call it as S of c cross S of d. Then the claim is that these two are equal, not always whenever, so we showed that these two are equal if x can be represented as a union of sets partition 1 to infinity and d can be written as a union of some sets d j's in the collection d of 1 to infinity. So, whenever this x can be represented at a disjoint union of sets from c and y can be represented as a disjoint union of elements of d, then whether you take the rectangles first and generate the sigma algebra or generate the sigma algebras and then take rectangles and generate the sigma algebra, both will be equal to same. So, this result we had proved in the beginning of the topic. So, as a consequence of this we obtained, so this implied one observation that if you take the Borel sigma algebra cross the Borel sigma algebra of r that is equal to Borel sigma algebra of the space r 2. So, that is one observation because the real line can be represented as a countable union of say open sets or intervals. Similarly, the same argument also implies that if I take the sigma algebra generated by this left open right closed intervals cross the sigma algebra generated by left open right closed intervals and then look at the product sigma algebra, then that will be same as the product sigma algebra of left open right closed intervals cross left open right closed intervals because the whole real line can be written as a countable union of left open right closed intervals. So, these two facts follow from our earlier construction. So, we will keep that in mind and now what we want to show is that the Borel sigma algebra of r 2, so that we know it is Borel sigma algebra of real line times the product Borel sigma algebra of real line and Borel sigma algebra, we know from our construction of real numbers that is the same as the sigma algebra generated by left open right closed intervals. So, left open Borel sigma algebra is generated by the sigma algebra of left open right closed intervals. So, and this just now we observed is the sigma algebra generated by i cross i, so that is same as the sigma algebra generated by i 2. So, that completes the proof of the fact that the sigma algebra generated by rectangles which are left open right closed intervals is same as the Borel sigma algebra of r 2. So, that is one observation. So, that is very much similar to the result in the real line where the left open right closed intervals generated the sigma algebra Borel subsets. The same result is true if we replace intervals by rectangles which are left open right closed. So, that is the proof we have just now said. So, si is equal to si cross i. So, Borel sigma algebra B r cross B r is the sigma algebra generated by intervals left open right closed cross left open right closed intervals which is same as this rectangles. So, now let us look at the next property that the Lebesgue measure that we have defined for a rectangle is lambda of i cross lambda of j is same as lambda of i into lambda of j. So, that is obvious because we obtained the product measure where the extension of the measure on the rectangles. So, what we are saying is the Lebesgue measure on r 2 is the natural extension of the notion of area in the plane. So, this is property is obvious built in the definition of the product measure. And third observation is, so recall we just now said that the Lebesgue measure space r 2 Lebesgue measures subsets of r 2 and Lebesgue measure the Lebesgue measurable subsets. So, this space which is the space Lebesgue measure space on one hand we defined it as the completion of the Lebesgue measurable sets course Lebesgue measurable sets. And this is also the completion of the measure space of real line with Borel subsets of r 2. And that is once again by the effect that the Borel subsets of r 2 are inside this and the Borel subsets of r 2 and the Lebesgue measurable sets they differ only by sets of measure 0. So, that is also the completion. So, one way of looking at is look at the Lebesgue measurable subsets r 2 it being the completion. So, it is the is the class of all outer Lebesgue measurable subsets in r 2 with respect to the product measure. And on the semi algebra i 2 the of rectangles it is given by the product. So, this is obviously the completion of the measure space r 2. So, these are obvious facts. So, we should keep in mind which are very much similar to that of the real line. They plays a role later on when we want to look at null sets in r 2. So, basically the sets which are going to be of importance are going to be the Lebesgue measurable sets cross Lebesgue measurable sets in r 2 or Borel subsets in r 2. Here is another useful fact about Lebesgue measure in r 2 which connects it with the topological in eyes sets namely the Lebesgue measure of r 2 of any open non-empty open set is always bigger than 0. So, that follows from the fact that if u is contained in r 2 is open and u is not equal to empty set then so here is the set u. So, there is always a rectangle left open right closed rectangle inside it. So, there is a rectangular neighborhood. So, implies there exists a rectangular so non-empty. So, there is a point x belonging to u. So, there is a rectangular neighborhood of x. So, let us call that neighborhood as n. So, this is the rectangle n which is contained in u. But the Lebesgue measure of so that means the Lebesgue measure of u will be bigger than Lebesgue measure of n which is always going to be bigger than 0 because it is a non-empty neighborhood. So, for every non-empty open set the Lebesgue measure is always positive if the set is non-empty. The second important thing is supposing you take a set k which is a compact subset of r 2. So, let us look at a compact subset of r 2. So, k is contained in r 2 and k compact. If a set is compact that implies it must be bounded. So, k compact implies k bounded and that means so saying the set is bounded implies that so this is the set k which is compact. So, that means it is bounded so it must be inside a rectangle. So, k bounded implies so k bounded implies k is inside some i cross j with lambda of i. So, implies lambda r 2 of k will be less than lambda of i cross lambda of j which is finite. So, finite intervals compact implies bounded so there is a finite rectangle including it so that means it is finite. So, these are two relations about open sets and compact sets. There are more relations which relate like in the real line that one can prove a result. For example, a set E is Lebesgue measureable if and only if for every epsilon you can find a open set which includes it and the difference as measure small. So, that is very much similar to the real line and the proof is also very much similar to the real line. So, we will not prove this result. An interested reader if is somebody who is interested should try to copy the proof of the real line and extend that proof to the case of r 2. So, this will give us that another result is that for the Lebesgue measure of r 2 you can approximate it from inside by compact sets. So, supremum of lambda of r 2 where k is compact. So, these results basically are of importance. So, these are called regularity conditions for the Lebesgue measure in r 2. So, we will not prove these results just for the sake of knowledge I am mentioning these results here. So, that later on if you come across you can look at proofs of these results. So, the next result we want to look at is how are the Lebesgue measureable sets related with the group structure of the space r 2. So, let us take a subset E of r 2 and let us look at is a point x vector x in r 2. So, we will define the translate of the set E by x to be as in real line y plus x where y belongs to E. So, take the set E and shift every element of E by the vector x. So, it is y plus x. So, the claim the first claim is that if E is a Borel set and the point x belongs to r 2 then E plus x also is a Borel set that is one property and the Lebesgue measure of r 2 of the set E is same as a Lebesgue measure of the set E plus x. That means the Lebesgue measure is once says it is translation invariant on the class of all Borel subsets of r 2. So, the proof of this fact that for every set E E plus x belongs to B r 2 and the fact that the Lebesgue measure of the translated set is equal to Lebesgue measure of the original set are standard applications of the techniques that we have been using namely the sigma algebra monotone class theorems. So, let me illustrate this once again so that this idea of using monotone class convergence theorem monotone class sigma algebra technique settles down in the mind. So, we first want to prove namely that we want to show that for every E a Borel subset of r 2 if I look at E plus x that is also a Borel subset of r 2. So, the technique is as follows let us collect together all sets A. So, form the collection A of all those subsets E belonging to B r 2 all Borel subsets say that the required property is true E plus x belongs to B r 2. So, look at all sets having this property. So, claim so that is a sigma algebra technique claim one all open subsets of r 2 are inside this collection. So, we will prove two claims one and secondly that the class A is a sigma algebra. So, if you prove these two facts about the class A then that will imply because it includes open subsets of r 2. So, it will include the smallest sigma algebra generated by. So, these two facts will imply these two facts will imply that the sigma that the sigma algebra generated by sigma algebra generated by open sets will be inside the class A and that is equal to the Borel sigma algebra. So, that will prove that Borel sigma algebra is equal to A. So, let us first show that the open subsets of r 2 are inside A. So, let us take a open set. So, to prove the first fact we have to show that if a set. So, to show the first one. So, let U be open in r 2. So, we want to show that this implies U plus x belongs to B r 2 and this follows because if U is open implies U plus x is open. So, that is a simple fact because how do we show that U plus x is open? Basically, saying that U is open. So, let us take a point y plus x belonging to U plus x. If y plus x belongs to U plus x, where y belongs to U and U open implies there is a neighborhood. So, let us call it as B delta y. So, y belongs to a neighborhood which is contained in U, but then that implies that y plus x belongs to the translation of the neighborhood that is contained in U plus x. So, that means for every point y plus x there is a neighborhood when you shift a ball that remains a ball in the plane. So, that is a basic fact we are using. If you translate a neighborhood that remains a neighborhood in U plus x. So, that implies that if U is open in r 2 then U plus x is also an open set and hence belong to B r 2. So, that proves the first fact namely open subsets belong to A. So, let us now to show that. So, this proves the first fact that open subsets belong to r 2. So, to show that A is a sigma algebra that is this very standard technique. We have been using it very often. If a set E belongs to A that means E plus x belongs to B r 2. So, let us write that. So, if E belongs to collection A that implies E plus x belongs to B r 2 and that implies because B r 2 is a sigma algebra. So, that will imply its complement belongs to B r 2. But this is same as the first taking complement and then then taking translate. So, that belongs to B r 2. So, that means E complement belongs to B r 2. So, if whenever A belongs to collection A it is E complement plus x belongs to B r 2. So, that means E complement belongs to A. So, E is closed under complements and similarly E i's belonging to A will imply that union of each E i plus x belongs to B r 2. So, that will imply that union of E i plus x belongs to B r 2, union over i. But this is same as union of E i's plus x belongs to B r 2. So, that will imply that the union of E i's belongs to A. So, A i's belong to union also belongs to A. So, that will prove that A is a. So, A is an algebra. So, A is a sigma algebra. So, A is a sigma algebra including open set. So, it includes everything. So, that will prove that. So, this is a sigma algebra technique. I have been mentioning that sigma algebra technique that we have mentioned says that implies that whenever E belongs to B r 2 implies E plus x belongs to B r 2. So, that is what we have proved. So, to show the other thing to prove that lambda of E plus x is same as lambda of E, everything is in R 2. To show this, once again let us define M to be the collection of all those subsets E. M to be the collection of all those subsets E belonging to B r 2 for which this property is true lambda of E plus x is equal to lambda of E. So, we want to show that B r 2 is inside M because M is already a subset of B r 2. So, that will prove that M is equal to B r 2 and hence this property will hold for all subsets of B r 2. Now, to show this the technique is the monotone class theorem. So, one show M is a monotone class, two M is closed under finite disjoint unions and third the rectangles B r cross B r rectangles are inside M. So, once these three facts are proved, we will be through as follows because these rectangles are inside it and if this is a monotone class. So, the idea is that the step three will imply that the monotone class generated by B r cross B r is also a inside M and this class is also closed under finite disjoint union. So, this collection the sets which are inside M will also be closed under finite disjoint unions. So, that will prove. So, it is a monotone class closed under finite disjoint unions. So, that will imply that so B r the rectangles are inside it. So, the algebra generated by the monotone class generated by finite disjoint unions also will be inside it because this is inside. So, this and this is closed under finite disjoint unions. So, that means the algebra generated by rectangles will also be inside it. But M is a monotone class which is closed under finite disjoint unions. So, that must be a sigma. So, monotone class generated by an algebra is also a sigma algebra. So, sigma algebra generated will come inside it. So, hence we will have everything is equal. So, the idea of the proof is that which one should prove these three things because after these three things are proved. So, what will the proof imply? So, three will imply 3 plus 2. So, this is a semi-algebra because B r cross B r a semi-algebra it is inside M and M is closed under finite disjoint unions will imply that the algebra generated by so, F of B r cross B r will be inside M. So, the algebra generated by this comes inside M, but now implies by one M is a monotone class. So, it includes this algebra. So, the monotone class generated by this algebra is also inside M, but the monotone class generated by an algebra is same as the sigma algebra. So, this is same as the sigma algebra generated by this algebra B r cross B r and that is equal to the Borel sigma algebra of R 2. So, that is the line of argument that will prove that B R 2. So, this is the line of argument which will prove that B R 2 is a subset of M. So, we have to verify these three things namely M is a monotone class, M is closed under finite disjoint unions and rectangles are Borel rectangles are inside M. So, to show that, let us look at the first one that M is a monotone class. So, to show that M is a monotone class, let us look at the proof. So, proof of one. So, let us look at a sequence En which is increasing, increasing to En, increasing to En. Let us say En belongs to M. So, that will imply that lambda of En plus x is equal to lambda of En for every n. Now, if En is increasing, if En is increasing then En plus x is also increasing and lambda being a measure, this converges to lambda of E plus x and by the same thing this converges to lambda of E. So, that says lambda of E plus x is equal to lambda of E. So, if En is increased to E, then that will imply that these two are equal. So, E belongs to M. And similarly, for a decreasing sequence also similar property, if En is decreasing to E and lambda of say lambda of E 1 is finite, then the intersection, the E which is the intersection will also belong to M. So, that will prove the fact that E is M is A. So, that will prove the fact that M is a monotone class. So, that is okay. Now, let us show that M is closed under finite disjoint unions. So, for that to show that M is closed under finite disjoint unions, let us take, let E 1 and E 2 belong to M, E 1 intersection E 2 equal to empty set. Now, E 1 and E 2 belong to M. So, this fact implies lambda of E 1 plus x is equal to lambda of E 1 and similarly, lambda of E 2 plus x is also equal to lambda of E 2. Now, E 1 and E 2 disjoint implies that the sets E 2, E 1, the translates of E 1 and translate of E 2 are also disjoint. So, that is a simple thing to observe. So, that will imply that lambda of E 1 plus x union of E 2 plus x, because these sets are disjoint. So, the Lebesgue measure of the union of in R 2 is same as the Lebesgue measure in R 2 of E 1 plus x plus lambda of E 2 plus x. But E 1 and E 2 belong to M. So, this is equal to lambda of E 1 plus lambda of E 2 and that is equal to E 1 and E 2 are disjoint. So, it is lambda of E 1 union of E 2. So, what we have shown is if E 1 and E 2, if E 1 and E 2 belong to M and they are disjoint, then lambda of E 1 union E 2 is same as lambda of E 1 x plus union of E 2 x. But a simple observation will tell you that this is also same as lambda of E 1 union E 2 plus x. So, whether you take translates first and then take the union, the same as taking union and the translates. So, this will imply that E 1 union E 2 also belongs to M. So, whenever E 1 and E 2 are disjoint, their union also belongs to M. So, that proves the second fact, namely M is closed under finite disjoint unions. Finally, we prove the third fact, namely the rectangles are inside M. So, that again is a straightforward simple fact to prove. So, to prove that, let us observe. So, to prove the third thing, let us observe the following, namely, let us take E f belonging to B r cross B r to show E and f both belong to B r. We want to show that the cross product belongs to B r cross B r. So, that is what we want to show. So, to show that, let us observe and E and f belong to B r. So, we know that whenever E is in f, so E plus x. So, let us take a vector x, which is equal to A comma B. Then, what is E plus? Then, we know that E plus A belongs to B r and also E plus B belongs to B r, because E and f are subsets in B r. So, the translates belong and lambda of E plus A is same as lambda of E and that was the set f and lambda of f plus B is same as lambda of f. So, now, look at the set E cross f translated by x. x is A B. So, what is that? So, that is equal to E plus A cross product with f plus B. So, the Lebesgue measure of this set, sorry, E cross f plus x will be equal to, this is a rectangle. So, it is Lebesgue measure of E plus A into Lebesgue measure of f plus B, but that is equal to Lebesgue measure of E into Lebesgue measure of f, because Lebesgue measure on the real line is translation invariant. So, that is equal to Lebesgue measure of E cross f. So, what we have shown is that if E cross f is a rectangle, borough rectangle, then translate of the borough rectangle has the same measure as the rectangle itself. So, that proves the third thing, namely that the borough sets cross the borough sets is inside m. So, all the three facts are proved and that will imply that B r 2 is a subset of m. Hence, for all, so that is what we have shown is that, so what we have shown is that the Lebesgue measure is a measure on the plane, which has the property that Lebesgue measure of every, for every borough set E, its translated is also, translation is also a borough set and the Lebesgue measure of the translated set is equal to Lebesgue measure of the original set. So, this is called the translation invariance properties of the Lebesgue measure on the plane. So, as in the case of real line, in the real line we showed that the Lebesgue measure on the line is a translation invariant measure. And so similarly, we have shown that the product of the Lebesgue measure taken on the r 2 is also a translation invariant measure. Of course, the natural question arises on the real line, we have shown that essentially a Lebesgue measure is the only translation invariant measure and we will show for this Lebesgue measure on the plane also is essentially a unique is the unique translation invariant measure, unique in the sense that a scalar multiple is again translation invariant anyway. So, up to a multiplication by a scalar, we will show that the Lebesgue measure in the plane is also is a unique translation invariant measure on the Borel sigma algebra. But before that, let us prove a property about the integrals of functions on the plane. So, the next property we want to analyze is the following namely, so this proof, so we have already gone through the sigma algebra monotone class technique. So, that sigma algebra monotone class technique that we have already explained, so that is just shown here that show that m includes f of r and hence it will include the sigma algebra generated by it and that will prove. So, the next property I wanted to illustrate is the following namely, for every non-negative Borel measurable function f on r 2 and any vector y in r 2, the integral of the translated function, so integral of f of x plus y with respect to the Lebesgue measure is same as the integral of the function itself and it is also same as integral of the negative of the function namely f of minus x. That means the Lebesgue integral for non-negative functions is invariant under translation and this is what is called deflection x goes to minus x. So, a proof of this is basically applications of the simple function technique, so let me just illustrate one or two steps of this proof that this is true. So, let us look at the first one, so let us try to, let us prove that if f is a non-negative measurable function on r 2, then we want to prove that the integral of f of x plus y d lambda r 2, so this is over r 2 is equal to integral of f of x d lambda r 2 of x, so this is what we want to prove. The simple function technique as you recall is the following first step, let us take f to be the indicator function of a set E, where E is a Borel subset of r 2. So, in that case the left hand side, so this left hand side is integral of the indicator function of E x plus y d lambda r 2, which is nothing but, so you are integrating with respect to x, so it is same as x plus y belonging to E means it is x belonging to E minus y, so this is integral of the indicator function of E minus y, so it is lambda r 2 of the set E minus y, but that is same by the translation invariant property, it is lambda of the set E, so and this thing f is the indicator function, so indicator function of x d lambda r 2, which is same as lambda r 2 of E. What we are saying is that as a first step the required claim namely integral of f of x plus y is integral f holds, whenever f is the indicator function of a set E. Now, both sides being integrals, so implies step 1 implies step 2 namely required claim holds for f equal to non-negative simple measurable function r 2 to r. So, this claim holds because any non-negative simple measurable function is a finite linear combination of characteristic functions of the indicator function, so for each indicator function we have shown this, so that will imply that the required claim holds for non-negative simple functions. So, in the third step if f is non-negative measurable then we know implies there exists a sequence S n of non-negative simple measurable functions S n increasing to f and integral of f to be equal to limit n going to infinity integral of S n d lambda. So, saying that f is non-negative measurable means that f is limit of non-negative simple measurable functions and the integral of f can be defined as the limit of the integrals of non-negative simple measurable functions. But for non-negative simple measurable function each S n, so for every n we know that the required claim holds by step 2. So, by step 2 we know that S n of x plus y d lambda r 2 of x is equal to integral of S n of x d lambda r 2 of x, so that is by step 2. Now, as S n is increasing to f, so clearly the translates this will increase to the translate of the function f. So, this implies that in the limit by monotone convergence theorem, so an application of monotone convergence theorem will say that as n goes to infinity this will converge to integral of f of x plus y d lambda of r 2 of x and on the other hand we know this converges to integral of f x d lambda of r 2. So, this must be equal, so that means for a non-negative measurable function this required conclusion holds. So, that is how one proves the claim namely f of x plus y is equal to f of integral of the translate is equal to the integral of the original function. So, basically what is what we call as the simple function technique applied to it. So, a similar argument will show that integral of f of x is same as integral of f of minus x. So, for there one has to use the fact that the Lebesgue measure of S at E in R 2 is same as the Lebesgue measure of. So, for the step 2, so let me just indicate what we need for step 2 to show that integral of f of x d lambda r 2 of x is equal to integral of f of minus x d lambda r 2. When f is equal to indicator function of the set E, that means we need the fact that lambda r 2 of a set E is equal to lambda r 2 of minus of E. What is minus of E? So, minus of E, so minus of E is the set minus the vector x where x belongs to E. Now, so to prove that this is so once again one has to go to the sigma algebra technique. So, consider define A to be the collection of all those sets E belonging to B R 2 where for which you can say that lambda of E is equal to lambda of minus E. So, look at all those collection of these sets. So, claim so one will show that show rectangles are inside it. That means if I take sets E cross f belonging to B R cross B R then E cross f belongs to A and A is a sigma algebra. Once again if these two sets are proved that will prove that this claim holds for every Borel subset also and hence for the indicator function of a set E. So, that is we leave it as a exercise once again is a straightforward verification. So, do that. So, once that is done so that will prove the second equality also. . Now, in this proof one more observation we want to make here is the following. If I replace lambda R 2 by any see in this proofs of these two things we have not use anywhere the fact that we are on the we are on lambda is especially the Lebesgue measure. Essentially we use the fact that this measure lambda of R 2 is translation invariant. So, if mu is if we replace this measure Lebesgue measure on R 2 by any translation invariant measure then this result that f of x plus y is equal to integral of f of x will remain true for lambda of R 2 replace by any translation invariant measure. So, this is an observation we should keep in mind for the future reference. So, finally we want to prove the fact that the translation invariance is a unique property for the Lebesgue measure. So, let us take any measure mu which is sigma finite on the Borel subsets of R 2 and assume it is translation invariant. And let us assume that there is some particular set E naught such that the measure of the set E naught is positive and the measure mu of E naught is C times a constant multiple of Lebesgue measure of the set E naught and it is finite. So, there is a set of finite Lebesgue measure finite positive Lebesgue measure such that mu of E naught is a constant C times Lebesgue measure of E naught for some particular set E naught. Then the claim is that this property holds for every subset of Borel subset that means mu of E is constant multiple of the Lebesgue measure. So, that will prove the uniqueness of the Lebesgue measure with respect to translation invariance. So, let us prove this. So, as I observed that the integral of the translate of a function is equal to integral of the function remains true for any translation invariant measure. So, in particular for mu. So, that property will be using. So, now let us so we want to show that mu of E is constant multiple of Lebesgue measure of E for every set E, but C is equal to. So, what is C? Let us just look at C, C I can compute from here C is equal to mu of E 0 divided by lambda of E 0. So, if I put that value so to show that mu of E is equal to C times lambda R 2 of E is equivalent to showing that lambda of E 0, Lebesgue measure of E 0 into measure of E is same as measure of E 0 mu of E 0 into Lebesgue measure of E for every subset. So, this equality we should show for every subset E of R 2. So, that we will show it as an application of Fubinist theorem. So, let us take the left hand side. So, lambda of E 0 mu of E is equal to lambda of E 0 and mu of E is integral of the indicator function with respect to y d mu y. Now, take this lambda R 2 inside and they use the fact that it is translation invariant. So, lambda R 2 of E 0 is same as lambda R 2 of E 0 minus y and I put it in under the integral sign. So, the required quantity is equal to integral of lambda Lebesgue measure of E 0 minus y into indicator function of E and now this Lebesgue measure I will write it as an integral in the form of integral. So, I get Lebesgue measure of E 0 minus y is integral of the indicator function of E 0 minus y same as it. So, it is same as the integral of the indicator function of E 0 of x plus y d lambda R 2. So, here we got double integral, iterated integral and the function involved are non-negative. So, by Fubinist theorem the first part for non-negative functions I can interchange the order of integration. So, let us interchange. So, earlier we had inner integral was with respect to lambda and outer with respect to mu. So, when we interchange mu comes inside and lambda goes outside. So, that is the integral and now once again mu is translation invariant. So, that means in this integral if I shift y to y plus x the integral will remain the same y minus x the integral will remain the same. So, let us do the shifting shift this to y minus x. So, indicator function of y minus x indicator function of E 0 x plus y. So, that becomes y d mu y and now once again we apply Fubinist theorem and go back. So, when I apply some mu goes out and lambda R 2 comes inside. So, that is indicator function of E y minus x lambda of R 2 of x, and that is same as Lebesgue measure of the set E and this is Lebesgue mu of E 0. So, that is equal to this. So, twice an application of the fact Fubinist theorem for non-negative functions and the earlier property gives us the required fact namely the Lebesgue measure is the translation invariant measure unique translation invariant measure on. So, today we have looked at the properties of Lebesgue measure with respect to topologically nice sets namely open sets, compact sets and with respect to the group operation of translation. On the plane there is another transformation possible namely you can take a set E and rotate it not only you can translate you can also rotate it or magnify a set. So, we will next lecture we will analyze how Lebesgue measure changes with respect to what are called linear transformations in the plane and which include rotations and magnifications. Thank you.