 Welcome to lecture number 13 on measure and integration. If you recall in the previous lecture, we had been looking at the Lebesgue measure, Lebesgue measurable sets and its properties. We will continue that study of Lebesgue measurable sets and its properties today itself. We will be especially looking at the translation in the various properties of the Lebesgue measure and then Lebesgue measurable sets vis-a-vis topologically nice subsets of the real line. Let us recall that we had defined what is called the Lebesgue measurable sets and that gave us the space, the real line, the Lebesgue measurable sets and the Lebesgue measure. This was called the Lebesgue measure space and the Borel sigma algebra of real line, the Borel subsets of the real line form a sub sigma algebra of Lebesgue measurable sets. This properties we had seen and now today what we are going to look at is the following. Recall that on the real line, there is a binary operation of addition. You can add real numbers. This operation can be used to transform subsets of the real line. Let us take a set E contained in real line and define what is called E plus x. E plus x is defined as all elements y plus x such that y belongs to E. It is the set E which is translated by an element x. The question is if E is Lebesgue measurable, does this imply that E plus x is Lebesgue measurable? Similarly, we will also look at the second question namely if E belongs to Br, if E is a Borel subset of real line, does it imply E plus x belongs to Br? These are the two questions we will start analyzing. The importance of these two questions is the class of Lebesgue measurable sets. Is it invariant under translations and is the class of Borel subsets invariant under the group operation of translation on the real line? These two questions we will answer in the first to start with. To answer the first question, let us recall that Lebesgue measure is nothing but the restriction of the outer Lebesgue measure. The Lebesgue outer measure for real line is defined as the infremum of sigma lambda of intervals i i, where these intervals form a covering of E is a subset of union i i's i equal to 1 to infinity and i i's are all intervals. Keep in mind the remark, we said you can choose these intervals i i's i j's to be open if necessary. Whether you take all possible coverings of E by intervals or all possible coverings of E by open intervals, both will give you the same value namely the Lebesgue measure of the Lebesgue outer measure of the set E. Let us start with a set E, which is Lebesgue measurable to show that E plus x is also Lebesgue measurable. This is the question. To show that, recall what is the definition of a measurable set? To show that for every subset y of real line, we should have Lebesgue outer measure of y is equal to Lebesgue outer measure of y intersection E plus x plus Lebesgue outer measure of y intersection E complement plus x, E plus x complement. That is what we have to show. Let us start observing that we are given that E is Lebesgue measurable. So, E Lebesgue measurable implies for every subset y of real line, the Lebesgue measure of y is equal to outer Lebesgue measure of y intersection E plus outer Lebesgue measure of y intersection E complement. Now our aim is to transform this E to E plus x. That means we should be looking at the properties of the outer Lebesgue measure of a set in terms of translation. Let us observe. Let us note that if a set E is covered by a union of intervals i j 1 to infinity, that is if and only if E plus x is covered by union of the translated intervals i j plus x j equal to 1 to infinity. That means every covering of the set E gives a corresponding covering of the set E plus x by the intervals i j plus x. Note that if i j is an interval i j plus x also is an interval. So, if E is covered by intervals i j, we get a corresponding covering of E plus x by the intervals i j plus x. Conversely, given a covering of E plus x, we can construct back a covering of E by looking at by translating by minus x. So, this property obviously implies that Lebesgue measure of a set E is same as the Lebesgue measure of the set E plus x. So, this observation implies that the Lebesgue measure of a set remains invariant under translations. So, this is a property. So, this is an important property of the Lebesgue outer measure we are going to use to conclude that if E is Lebesgue measurable then E plus x also is a Lebesgue measurable. So, as we said E Lebesgue measurable implies for every subset y, we have got lambda star of y is equal to lambda star of y intersection E plus lambda star of y intersection E complement. And now observing that Lebesgue measure is invariant, outer Lebesgue measure is invariant under translations, we can write this as lambda star of y intersection E plus x plus lambda star of y intersection E complement plus x. So, here we are using the fact that lambda star is translation invariance. And now, a simple observation I will tell you that y intersection E plus x is same as y intersection, it is same as y intersection y plus x intersection E plus x. So, meaning that if you take intersection and translate, it is same as translating and intersectionist. So, that is simple set theoretic property. So, the first term here is equal to lambda star of y plus x intersection E plus x. And similarly, the second one will give you it is lambda star of y plus x intersection E complement plus x. So, that is using the fact that lambda star is translation invariant and intersection translation is same as translation intersection, they commute with each other. And now, this property is true for every subset of y. So, I can replace y by y minus x. So, implies replace y by y minus x, we get this is also equal to. So, lambda, this property is true for every. So, let us replace lambda star of y minus x is equal to lambda star of y plus x minus x. So, that is y intersection E plus x plus lambda star of y plus x minus x. So, that is y intersection E complement plus x. So, in the above equation, we replace y by the set y by y minus x. So, this is true. And now, observe lambda star of y minus x is same as lambda star of y by translating y or y x or minus x does not affect. So, we get for every subset y, lambda star of y is equal to lambda star of y intersection E plus x plus lambda star of y intersection E complement plus x. Now, a simple observation tells you that this set E complement plus x is same as E plus x complement. So, first take the complement and then translate that is same as saying first translate and then take complement. So, it is purely a simple set theoretic exercise, which you should be able to verify easily. So, we get lambda star of y is equal to lambda star of the first y intersection E plus x plus lambda star of y intersection E plus x complement. Hence, this implies that E plus x is Lebesgue measurable. So, we have proved the first property namely if you take a Lebesgue measurable set E and translate, then the translated set also is Lebesgue measurable. So, another way of saying the same thing is that the Lebesgue measurable sets are translation invariant. They remain the class of Lebesgue measurable sets is translation invariant. And we have already seen the Lebesgue outer measure is translation invariant. So, that means that the length function is translation invariant on the class of all Lebesgue measurable sets. So, this proves the first property. So, we have answered the question that if E is Lebesgue measurable, then E plus x is also measurable. Let us look at the next question that if E is a Borel set, can we say that E plus x also is a Borel set. So, to answer this question, we need some topological properties of the real line. So, let us look at the topological properties, what we are going to look at. So, the question is if E is a Borel subset of real line, does that imply E plus x is also a Borel subset of real line? So, for that let us observe consider the map. So, consider the map from real line to real line, where y goes to y plus x for every y belonging to y. So, x is fixed. So, this is translation. So, this is the translation map from real line to real line and the observation is that this map translation is a homeomorphism. So, what does homeomorphism means? That is it is 1 1 on 2 and both ways continuous. That means this is continuous and the inverse map because it is 1 1 on 2 that also is continuous. So, this is an important property very basic, but yet important property that we are going to use to prove that if E is a Borel set, then E plus x is a Borel set. So, to prove our requirement, let us consider the collection say S. So, S is a collection of all subsets which are Borel and which have that required property namely E plus x belongs to B R. So, look at all Borel subsets of real line which such that the translation translated set is also a Borel set. So, we are going to prove two things about this. One that the class of all open sets all open sets are subsets of S and second thing will prove that S is a sigma algebra. So, once these two properties are proved all open sets are inside S and S is a sigma algebra. So, these two properties will imply that the sigma algebra generated by the class of all open sets is inside S and that is nothing but the Borel sigma algebra. So, the Borel sigma algebra will come inside S and that will prove the required property. So, we have to only prove these two facts namely that if you take an open set then it belongs to S. That means if I take an open set and translate that should be a Borel set. But that is again obvious by the fact that translation is a homeomorphism. So, if E contained in R is open is an open set then the set E plus X is also an open set. So, that is also an open set. In fact, that is a very simple property to prove because if you take a if E is open that at every point there is a open interval inside E and the translated one will be inside E plus X. So, that is easy to verify or one can simply verify by saying that E plus X is an open map or the translation is a homeomorphism. So, this is an open set. So, this implies the first property that all open sets are subsets of S. The second thing that it is a sigma algebra. So, for that note empty set the whole space are both open. So, they belong to S and second if a set E belongs to S that implies E plus X belongs to S is a Borel set. But that implies this is a sigma algebra. So, E plus X complement is also a Borel set because it is a sigma algebra. So, must be closed under complements and now a simple observation that this set E plus X is nothing but E plus X complement is same as E complement plus X. So, this is same as this. So, that belongs to B R and that implies that E complement belongs to E complement belongs to S that is same as saying E complement. So, the class S of subsets of Borel subsets of real lines either translates or Borel sets includes empty set the whole space. It is closed under complements and let us prove that this is also closed under countable unions. So, third property namely, so let E n be a sequence of sets in S implies that E n plus X is a Borel set in R and Borel sets being a sigma algebra that implies union of E n plus X also belongs to B R. Hence, now here is a simple observation that this set is same as you first take the union and then take the translation. It is same as first translating and then taking the unions. So, that belongs to B R. So, basically translation commutes with all set theoretic operations that is observation we have been using again and again. So, this belongs to B R so that implies union n equal to 1 to infinity E n is a set in S. So, we have proved so hence open sets are inside S a sigma algebra. So, implies that the Borel sigma algebra generated by open sets which is the Borel sigma algebra is inside the smallest sigma algebra generated by open sets namely the Borel sigma algebra must also come inside S and that is a subset of B R. So, hence all are equal. So, this proves the second fact namely if E is a Borel set then it is translated translation is also a Borel set. Once again it emphasizes the use of the technique that we had called as the sigma algebra technique namely we wanted to prove that for every Borel set E the translation is a Borel set. So, note we collected all the sets which have this property and we proved two facts namely the open sets are inside this collection S and S is a sigma algebra. So, that implied that the smallest sigma algebra generated by open sets which is nothing but the Borel sigma algebra also comes inside S. So, this technique we will be using we have been using and will be using quite often is called the sigma algebra technique. So, you want to prove some property about subsets of a set X collect them together and try to show that those sets form a that collection forms a sigma algebra and includes the generators of the required sigma algebra. So, we have proved that under translation the measurable sets the collection of Borel sets are very well behaved and we get a translation invariant measure on real line that is the length function. So, this gives us the fact that lambda the length function on B R the Borel sigma algebra of subsets of real line is the unique translation invariant measure such that the length of the interval 0 comma 1 here comma is missing. So, 0 comma 1 is equal to 1 and this is a very important property of the length function. So, if you observe on the real line there is a notion of addition and we just now pointed out that the group operation X comma Y goes to X plus Y is a continuous map and one can also easily check that X going to minus X that is the inverse and the group operation is also a continuous map that is summarized by saying that the real line is a topological group. So, on the real line there is a topological structure there is a metric there is a topology on the real line there is a group structure on the real line and the two behave very well with respect to each other saying that the group operations X comma Y goes to X plus Y and X goes to X inverse both are continuous maps. So, one says such a thing is called a topological group. So, the real line with addition and the usual metric the normal metric forms what is called a topological group and we have shown on this topological group there exists a translation invariant measure on the sigma algebra of subsets there exists a translation invariant measure. This is a very important fact and that can be generalized to what are called locally compact topological groups that on every locally compact topological group there exists a invariant measure because the group may not be abelian. So, one has to make a specific thing there exists a left invariant or a right invariant measure on every locally compact abelian group and that plays a important role in doing analysis on such groups. So, just a pointer that you may come across in your other courses in higher studies of higher mathematics that on every locally compact topological group there exists a invariant measure and that actually is called Haar measure on the topological group. So, Lebesgue measure on the real line is an example of Haar measure on the locally compact topological group real line under addition and the usual multiplication the usual metric space topology. So, these were the properties of Lebesgue measurable sets vis-a-vis group structure. Now, let us look at properties of the Lebesgue measurable sets with respect to topologically nigh subsets of the real line namely topologically nigh sets are open sets and closed sets. So, we will actually analyze and characterize measurability of sets in terms of open sets and closed sets. So, this is the precise theorem that we are going to prove namely if E is any subset of the real line then the following statements are equivalent saying that a set E is Lebesgue measurable is equivalent to saying that for every epsilon bigger than 0 there exists an open set G lower epsilon such that E is a subset of G epsilon that means that open set includes E and the difference between the two sets has got outer measure small. So, that means there is a little little difference between a Lebesgue measurable set and open set which covers it. So, we will show that for every a set E is Lebesgue measurable if and only if for every epsilon bigger than 0 there exists an open cover of it such that the difference between the cover and the set. So, lambda star of G epsilon minus E is small and we will show that this is equivalent to saying that there exists a G delta set G such that the set includes E and the difference has got measure 0. A set G delta set what is a G delta set? A G delta set is nothing but a intersection of open countable intersection of open sets. So, subsets of the real line or in any metric space which are countable intersections of open sets are called G delta sets. So, let us prove this theorem. So, saying that these three statements are equivalent is saying that if one of them holds then the other one also holds. So, what we are going to do is we will assume one and show that one implies two and then we will show that statement two if you assume statement two that implies statement three and if you assume statement three then that implies one and that will imply that all these statements are equivalent. So, if one of them is true then the other two statements are also true. So, that gives you a characterization of Lebesgue measurable sets in terms of open sets or G delta sets. So, let us prove this theorem. So, we will start with looking at. .. So, let us assume. So, let us suppose the statement one holds. So, that is E is Lebesgue measurable. To show two and that is for every epsilon bigger than 0 there exists a set G epsilon open such that G epsilon includes the set E and the Lebesgue outer measure of G epsilon minus the set E is less than epsilon. So, this is what we have to show. So, let us start with something this is regarding outer measure. So, let us start looking at the set E. So, let us first suppose E is such that E is Lebesgue measurable. So, let us suppose that Lebesgue measure of E is finite. So, what is Lebesgue measure of E? Recall, Lebesgue measure of E is same as its outer Lebesgue measure, which is same as infremum over sigma lambda of i j 1 to infinity where the set E is covered by union of i j's intervals and each i j open. So, recall we had made a observation that in the definition of Lebesgue outer measure, you can assume that all the intervals involved are open. So, let us fix and since this number is finite and it is infremum. So, by the definition of infremum there exists a covering. So, there exists intervals i j's such that E is contained in union of i j's, each i j open and we have got the property namely lambda star of E, which is same as lambda star of E, which is same as lambda of E because it is measurable plus the small number epsilon is bigger than sigma lambda of i j, j equal to 1 to infinity. So, and that implies, so this implies, so note lambda star of E is finite. So, this is finite quantity. So, that implies sigma lambda of i j, j equal to 1 to infinity is finite. So, all the sets I have got. So, this implies that if I take a look at lambda star of union i j, j equal to 1 to infinity, that will be less than or equal to summation j equal to 1 to infinity lambda of i j by the sub additive property of the length function and that is finite. So, this set union of i j's is a set of finite outer measures. So, let us define, so put G epsilon to be the set, which is union of i j's. Now, let us note that first of all G epsilon is open. Why it is an open set? Because each i j is an open interval. So, a countable union of open intervals is an open set is open and E is inside union of i j's. So, E is contained in G epsilon. So, we have got the required property that we have got a cover of E by an open set and let us note and what is the difference? So, lambda star of G epsilon minus E. Now, note G epsilon is an open set. So, it is a Borel set. So, it is a Lebesgue measurable set and E is given to be Lebesgue measurable and it is subset of it and everything is finite that we have just now observed. So, we can say this is equal to lambda star of G epsilon minus lambda star of E. So, here we are using the finite additivity property of the length function and lambda star of G epsilon that is here is that is same as lambda star of union i j because G epsilon is union i j minus lambda star of E and that is less than or equal to sigma by sub additive proper countable sub additive property 1 to infinity lambda star of i j minus lambda star of E and that by our choice of our intervals i j's if you recall this was the choice. So, that means lambda star of this summation minus this is less than epsilon. So, which is less than epsilon. So, lambda star of G minus G epsilon minus E is less than epsilon. So, that means what? That means we have proved the required property when lambda star of E is a finite set. So, thus if lambda star of E is finite then 1 implies 2. Let us remove this condition and that condition here is a step where is the important step which would keep in observation that we first proved a property about the length function for finite sets of finite measure and now we are going to extend this using the fact that lambda is sigma finite. So, whenever you want to one wants to prove a property about the length function or about sigma finite measures many a times it is easier to prove it when the underlying set is of finite measure and then extend it to sets general sets of sigma finite measure. So, that is what we are going to do now. So, let us in the general case. So, in general the set E which is Lebesgue measurable may not have a finite Lebesgue measure, but the Lebesgue measure being sigma finite we can write E as a disjoint union of sets E j, j equal to 1 to infinity such that lambda of E j is finite. It is a measurable set and it is finite. So, now by the earlier case because lambda star of E j is finite. So, for fix epsilon bigger than 0 there exists open set call it G j which includes E j such that lambda star of G j minus the set E j is less than the small number epsilon, but we are going to write less than epsilon to the power 2 to the power j and you will see soon why we are doing that because for each piece we are going to make it small and we are going to add up these pieces. So, now define the set G epsilon to be equal to union of G j, j equal to 1 to infinity. So, then G j is open. So, it is an open set because it is a countable union of open sets G j's and G epsilon includes G j includes E j. So, includes union of E j's which is equal to E. So, it is an open set which includes E. Now let us look at the difference further. So, let us observe further that G epsilon minus E what is that equal to that is union of G j's minus union of E j's that is the definition that is how we constructed. Now here it is a simple set theoretic property namely that this is a subset of union j equal to 1 to infinity G j minus E j. So, this is a simple set theoretic curve that union of G j's minus union of E j's is a subset of union of G j minus E j and once that is verified which is easy to verify we get that lambda star of G epsilon minus E is less than or equal to sigma j equal to 1 to infinity lambda star of G j minus E j which by our choice is less than epsilon to the power j. So, this is less than or equal to summation j equal to 1 to infinity epsilon by 2 to the power j which is equal to epsilon. So, that proves the second property completely in the general case also. So, hence what we have shown is that 1 implies 2 namely if E is Lebesgue measurable then I can find given epsilon we can find a open set G epsilon such that the outer measure of this is small is less than epsilon. So, now let us go to the second step of the verification. So, we have verified the first step namely 1 implies 2. Now, let us verify that 2 implies 3. So, let us assume 2 holds. So, 2 implies 3. So, we are given a set E. So, given 2 means let E be a subset of real line such that for every epsilon bigger than 0 there exists a open set G epsilon which includes E and lambda star of E is just a set. So, we cannot write lambda now it is lambda star of G epsilon minus E is less than epsilon. So, that is what is given to us and we have to construct a set say that the difference has got measure Lebesgue measure of 0. So, the obvious way is make this epsilon small and small. So, in particular so that says for every epsilon equal to 1 by n there exists G n including E G n open such that lambda star of G n minus E is less than 1 by n. So, we have specialized this given condition for each epsilon equal to 1 by n and we got an open set G n which includes E. Now, because we want it small s. So, we want to let this become smaller and smaller. So, it says the following. So, define G equal to intersection of G n n equal to 1 to infinity. So, what is G? G is intersection of open sets. So, note G is what is called a G delta set. So, G delta set is by definition an intersection of open sets and this set G includes E because each G n includes E. So, G also includes E. Further lambda star of G minus E note G minus E is a subset of G n minus E because G is the intersection. So, G minus E is a subset of G n minus E. So, by monotone property lambda star of G minus E is less than lambda star of G n minus E which is less than 1 by n. So, lambda star of G minus E is less than 1 by n for every n. So, that implies the fact that implies the fact that lambda star of G minus E is equal to 0. So, hence we have shown 2 implies 3. Now, let us conclude the proof by showing. So, let us show we show that 3 implies 1. So, what is 3? So, 3 says that by 3, so 3 implies that for E E a subset of R, there exists not an open set, there exists a G delta set G including E and lambda star of G minus E equal to 0. So, here is the set E that is the set E inside and this is the set G which covers it, so that the remaining part has got measure 0. But note what is E? E is same as you take the set G, this is the full set and intersect it with the complement of the outer portion. So, this is the complement. Look at the complement of this. So, intersection G minus E complement. So, this is a simple observation because what is G minus E complement? So, this inside portion is, so let me just draw a picture again. So, this is inside is E and outside is G. So, this shaded portion is G minus. So, what is complement? Complement is the outside portion here and E. So, when you intersect it with G you get E. So, E is nothing but G intersection G minus E complement and this is a G delta set and a G delta set is an intersection of open sets. So, this set belongs to B R and hence this set also belongs to L. So, it is Lebesgue measurable and this set G minus E has got outer measure 0. So, G minus E belongs to is a Lebesgue measurable set because all sets of outer measure 0 are Lebesgue measurable. So, this set also belongs to L and so intersection of two Lebesgue measurable sets is Lebesgue measurable. So, this implies that E belongs to L. So, 3 implies, so this implies 1. So, this proves completely the fact that the three properties that E is Lebesgue measurable is equivalent to saying for every epsilon bigger than 0, there is an open set G epsilon such that E is a subset of G epsilon and lambda star of G epsilon minus E. The difference is outer measure small and that is equivalent to saying that for the set E there exists a G delta set covering it such that the difference has got measure 0. So, this gives us a characterization of Lebesgue measurable sets in terms of open subsets of real line. A corresponding characterization of Lebesgue measurable sets is obtained in terms of closed sets. Let us state that also and prove it. So, let us look at the next that for any set E in R the following is true namely E is Lebesgue measurable is equivalent to saying for every epsilon bigger than 0. Now, there is a closed set inside E such that Lebesgue measure of the difference is small and that is equivalent to saying that there is a F sigma set. So, what is the F sigma set? A F sigma set is nothing but a set which can be expressed as a countable union of closed sets such that there is an F sigma set F inside E such that lambda star of E minus F is equal to 0. So, let us quickly prove this and this will use the earlier characterization. So, suppose one holds E belongs to Lebesgue measurable sets. Now, E belongs to Lebesgue measurable sets implies there is an open set which covers E with difference of measure small but we want closed sets. So, let us observe E belongs to L also implies that E complement belongs to L because this Lebesgue measurable sets is a sigma algebra. So, if E is Lebesgue measurable complement also measurable. So, this implies by just now what we proved. So, E complement is Lebesgue measurable. So, for every epsilon bigger than 0 there exists an open set G epsilon such that this includes E and outer Lebesgue measure of G epsilon minus E is less than epsilon. This is just now we proved this fact. So, this is of E complement. We are applying the previous just now proved result for E complement and now so if E complement is inside G that means E will be including G epsilon complement and note G epsilon is an open set. So, it is complement is a closed set. So, let us call it as C epsilon. So, C epsilon is closed. It includes E and we want to find what is the Lebesgue outer measure of E minus C epsilon. So, what is that? That is E intersection. What is this? This is just C epsilon complement by set theory and that is same as E intersection. This is G epsilon. So, we want E complement. We want to find out what is the difference between the closed set inside E complement. So, that is sorry not not not the complement E. So, E minus C epsilon is E intersection C epsilon complement and C epsilon complement is nothing but G epsilon G epsilon. So, this is G epsilon intersection E. So, we want to conclude. .. So, now note that Lebesgue outer measure of E complement. So, Lebesgue measure of we had E minus. So, what is this? So, this we can also write it as G epsilon minus E complement because E complement complement will be E. So, E minus C epsilon C epsilon is same as outer measure of G epsilon minus E complement and that is less than epsilon. So, what we have shown if E is Lebesgue measurable then there is a closed set inside it such that the measure outer measure is less than epsilon. So, implies two holds. So, one implies two we have proved and here we have just used the fact that a set is open if and only if it is complement is closed and applied the previous criteria. So, one implies two we have proved. Let us look at two implies three and that is once again for every epsilon what is given that there exists a closed set C epsilon closed inside E, C epsilon closed with the property that outer measure of E minus C epsilon is less than epsilon. So, in particular for every epsilon equal to 1 by n there exists a set C n which is inside E, C n closed such that Lebesgue measure of E minus C n is less than 1 by n. So, put so technique is same. So, put C equal to union of this sets C n. So, this 1 to infinity. So, this is a F sigma set because it is a union of closed countable union of closed sets and each C n is inside E. So, this set C is also inside E with the property that lambda star of E minus C. C is the union of all the C n. So, if I take only one of them C will be a subset of E minus C will be a subset of that. So, it is less than or equal to lambda star of E minus C n which is less than 1 by n. So, implying lambda star of E minus C is equal to 0 and that proves hence three holds. So, for given for every epsilon there is a closed set inside it we have shown there is a closed set inside E with difference there is a not a closed set a F sigma set with the measure 0. Finally, let us put that prove that 3 implies 1. So, E is a subset of real line E E is a subset of real line with the property that there exists a F sigma set C contained in E with the property that lambda star of E minus C is equal to 0. So, here is the set E and I have got a set C inside it. So, this is a set C inside it say that the difference has got measure 0, but now note the set E can be written as C union E minus C. So, this is this portion outside and union the inside portion and this is F sigma set. So, it is a Borel set and hence it belongs to is a Lebesgue measurable set the set E minus C is a set of measure 0. So, that is a Lebesgue measurable set. So, E is a union of two Lebesgue measurable sets. So, implies 1 that is E is Lebesgue measurable set. So, we have proved the third property namely 2 implies 3 and now we have just now proved 3 implies 1. So, the Lebesgue measurable sets are very nicely connected with topological nice sets namely open sets and closed sets. So, for every Lebesgue measurable set can be covered by an open set such that the difference has got Lebesgue measure small and which is equivalent to saying a set is Lebesgue measurable that is also a characterization and similarly a set E is Lebesgue measurable if and only if we can find a closed set inside it such that the difference has got measure small. So, Lebesgue measure is a very nice measure on the real line and it is defined on all Lebesgue measurable sets. So, in particular it is defined for all Borel sets and it is translation invariant it gives nice is compatible with the group structure and it also has nice properties with respect to the topological structure namely with respect to open sets and closed sets. So, with that we conclude our analysis of Lebesgue measure and Lebesgue measurable sets. In the next lecture we will start looking at functions on measurable spaces and their properties. So, thank you very much.