 Welcome to lecture 12 on measure and integration. If you recall, last time we looked at the extension of a measure from an algebra to the sigma algebra generated by it and slightly beyond the class of all outer measurable subsets. Today we are going to look at some special applications of this, a particular case of that extension theory for the real line and that is the topic for today's discussion namely Lebesgue measure and its properties. So, for the extension theory, we are going to apply it for the case x is equal to real line, the set is the real line, the algebra a which is algebra generated by all intervals in the real line and mu on this algebra is the length function that we have defined and we have seen that the length function on the algebra generated by all intervals is a countably additive set function. The outer measure induced by this length function which is denoted by lambda star is on all subsets of the real line and that is called the Lebesgue outer measure. So, the outer measure induced by the length function is called the Lebesgue outer measure. Let us just look at what is the Lebesgue outer measure for a subset of the real line. So, if you recall, we defined it as outer measure of a set E is look at all possible coverings of the set E by elements in the algebra, but here the algebra being algebra generated by intervals, it is finite disjoint union of intervals. So, we can write this outer Lebesgue measure as the infremium over summation lambda of the intervals i i, where the intervals i i's form a covering of the set E and these intervals are pair wise disjoint. So, lambda star of E is the infremium of the sums of the lengths of the intervals which form a covering of E and then we can take these intervals to be disjoint, because if not then we can make them disjoint. So, that is the Lebesgue outer measure for a set E. The class of all Lebesgue outer measurable sets, so lambda star measurable sets is called the sigma algebra of Lebesgue measurable sets. So, the sets which are outer measurable with respect to lambda star is called the sigma algebra of outer measurable or Lebesgue measurable sets and is denoted by L suffix R, just to indicate L for the Lebesgue and R for the real line. In case there is no confusion, we will just denote L R by simply L. So, this is the class of all Lebesgue measurable sets. If you recall, we had also defined the sigma algebra of Borel subsets of real line and that was the sigma algebra generated by all intervals and a being the algebra generated by intervals. So, the sigma algebra generated by finite disjoint union of intervals is same as the sigma algebra generated by all intervals and that is same as the definition of the Borel sigma algebra of the real line. So, these properties we had already seen. So, the length function in particular is also defined for all Borel subsets because the sigma algebra generated by A is inside the class of all outer measurable sets that is L. So, we have got that S of A that is the Borel sigma algebra is inside the class of all Lebesgue measurable sets. So, for all Borel subsets the notion of length is defined. So, this is called the Lebesgue measure. So, let us just summarize what we were saying. We were saying that the extension theory when applied to the particular case of the real line gives us the notion of length for a class of subsets of the real line which are nothing but the class of outer Lebesgue measurable sets and that includes the class of all Borel subsets. So, that is also gives us the notion of length for all Borel subsets of the real line. So, the triple R Lebesgue measurable sets the length function as extended by the extension theory this triple is called the Lebesgue measure space. So, the extension theory applied to the real line gives us the notion of the Lebesgue measure space and it extends the notion of length from intervals to the class L of all outer Lebesgue measurable sets. Let us just recall that the sets Borel subsets form a subset of the class of all Borel measurable sets is a sub class of the class of all Lebesgue measurable sets and of course Lebesgue measurable sets is a sub class of all subsets of real line. So, the question is can we say something more regarding these three classes namely Borel subsets, Lebesgue measurable sets and P R. So, let us observe which we have done during outer measures that the Lebesgue measurable sets are characterized by the Borel subsets of the real line union the null sets. So, what are the null sets? Sets in the R subsets of R such that N is contained in a Borel set of measure 0 or equivalently one can also define it as sets of outer Lebesgue measure 0. So, B R is a subset of L we know that outer measure 0 sets are also measurable. So, this and we said this class is nothing but this form the sigma algebra and that is equal to the Lebesgue measurable sets. So, that means the B R union N is equal to L. So, all null sets are part of L, but we want to characterize what is the relation between B R and L and what is the relation between L and B R. So, at present we only know that the Borel sets are subsets of all Lebesgue measurable sets which is a subset of B R. To say something more we need to look at what is called a special subset of the real line called Cantor's ternary set. So, we are going to discuss spend some time on a special subset of real line which is called Cantor's ternary set and Cantor's ternary set is an example of a set which has very nice properties and it is useful both from the topological point of view as well as measure theoretic point of view. So, let us look at what is called Cantor's ternary set ternary. It is called Cantor's ternary set because it was given by the mathematician George Cantor first defined by George Cantor and ternary set which are because it involves ternary expansion of real numbers. So, what we are going to do is it is a construction we are going to construct a Cantor's ternary set. So, as a step one let us look at the interval 0 to 1. So, I will first describe this process of Cantor's ternary set construction and then we will analyze its properties. So, what is the first step? The step is divide this into three equal parts. So, that is 1 by 3 and 2 by 3 and remove the middle open portion. So, this open portion is removed from the interval 0, 1. So, what it gives? It gives us two pieces 0, 2, 1 by 3 and from 2 by 3 to 1. So, it gives us two closed intervals. So, at the first step, at the first stage having removed the middle one-third of the closed interval 0, 1, middle one-third open interval we get these two. Now, we repeat that process again with these two subintervals. So, from each of these subintervals remove the middle one-third portion. So, that is middle one-third is 1 by 9, 2 by 9 and here the middle one-third will be equal to 7 by 9 and 8 by 9. So, this is the middle one-third portion which we are going to remove at the second stage. So, that will give us four subintervals and we continue this process. So, eventually something will be left. So, continue. So, the question is what is left? What is left is called Cantor's ternary set. So, let us analyze and let us denote this set Cantor's ternary set by the letter C. So, how do we mathematically construct this? So, that is the question. So, for that let us start with the first stage that is A0 that is the closed interval 0, 1. After having performed the first stage what is left? I write it as A1. So, that consists of two disjoint intervals 0 to 1 by 3 and 2 by 3 to 1. So, it consists of two disjoint intervals. So, let us write them as at the first stage 1 union the second one, first stage the second one. So, this portion is the first interval and this portion is the second interval. So, that is I11 and this one is I12. So, at the second stage we will be left with four disjoint closed intervals. So, let us write them as union I second stage J, J equal to 1 to 4. So, that is going to be 2 raise to power 2. And let us see what will be at the nth stage. If we continue this process at the nth stage, how many intervals will be here? So, there will be intervals, how many of them? We start with 1 at the next stage 2, at the next stage 4 and so on. So, there will be disjoint intervals J equal to 1 to 2 to the power n. There will be 2 to the power n closed sub intervals of 0, 1. Let us write them as I n J. So, these are the intervals. So, what is A n? A n is the union of those intervals which are left at the stage, at the nth stage. And what do we want? And we continue this process, we want what is C? So, how do we write mathematically C? So, the cantor set we can write it as, so the cantor set C, we can define it as intersection of A n's n equal to 1 to infinity. So, each A n is a subset of the previous one. So, let us write what is left eventually as intersection of all these A n's. So, this is what is called cantors ternary set. So, let us make some observations about this, observations about this cantors ternary set. The first observation is that the end points of the open intervals removed are in C. Say for example, 0 is not removed, is not going to be removed, 1 is not going to be removed. At the first stage, we removed the open middle one third. So, 1 by 3 is not going to be removed, 2 by 3 is not going to be removed and at the next stage, 1 by 9 will not be removed, 2 by 9 will not be removed and similarly, 1 by 3 we have already listed, then 7 by 9 will not be removed, 8 by 9 will not be removed and so on, will not be removed. So, for example, these points will not be removed. They will stay in this process of removing middle one third open interval from each sub interval at every stage. So, that means, so thus the class C, the set C is a non-empty set. It is non-empty, is non-empty. So, that is the first observation. There is something left behind and the second observation we want to show that in fact, C is uncountable, that it is an uncountable set. So, how do we prove C is uncountable? What we are going to do is, we are going to define a map from the closed interval 0, 1 to C. So, to prove this, we will define a map which is 1, 1. We will define a 1, 1 map from 0, 1 to the cantoristary set and that we prove that the cardinality of the set C is at least as much as 0, 1 and C being a subset of 0, 1 it cannot be more than that of 0, 1. So, cardinality will of C will be same as cardinality of 0, 1. That may seem a very strange observation to you that from C we have removed from the interval 0, 1. We have removed so many pieces and still what is left is as much as the points in 0, 1. So, these are properties of infinite sets. Actually, they are the characterizing properties of infinite set. The interval 0, 1 is an uncountable set and from that we are removing sub intervals and still what is left behind is as much as 0, 1. So, let us prove this fact namely that there is a 1 to 1 map for this. For this, let us start. Let us take a point x belonging to 0, 1 and consider its binary expansion. So, what is the binary expansion? The binary expansion of a point in 0, 1 is written as, so x can be written as point a 1, a 2, a 3, a n and so on, where each a n is equal to 0 or 1. So, that is the binary expansion of every point. Essentially, the idea is that the interval 0, 1 can be divided into two parts. Name first part as 0, second part as 1 and see at each stage where it lies. So, that is 0, 1, 1. And let us assume there are two different ways of writing. For some points, there are two different ways of writing binary expansions. So, we will fix one of the ways and say there is a unique binary expansion for every point in 0, 1. So, we will fix that binary expansion process. And now, what we do is the following. Construct a point y with ternary expansion b, with ternary expansion. So, y is equal to point b 1, b 2 and b n, where for every point b n, b n is nothing but 2 times a n. So, even the binary expansion, look at the nth place. Either it will be 0 or 1, double it and call that as b n. So, b n is twice as much as a n. So, each b i is either going to be 0 or it is going to be 2. So, this is the ternary expansion. So, note y belongs to 0, 1 because it is dash is dot b 1, b 2, b 3, so on. So, no integral part. So, it is going to be part of, it is a point in 0, 1. And it has in the ternary expansion, the only numbers that come are 0, 2 times a n, a n is 0 or 1, it is 0 or 2. So, in the ternary expansion of y, which is in 0, 1, only 0 or 2 appear and that implies that y belongs to c, because in the construction of the counter ternary set, we have removed the middle one-third. So, in the ternary expansion, the number 1 is not going to appear. So, each one is, so this is a part of, so this is the observation we make, that starting with a point x belonging to 0, 1 with binary expansion a 1, a 2, a n construct a point y. So, send it to the point y. So, this x is sent to the point y, which is again in 0, 1. In fact, it belongs to, so let us write more specifically, it belongs to c. So, we have got a map from 0, 1 to c and the claim is that this map, this is it is 1, 1. And that is obvious, because for every point x, we have got this binary expansion a 1, a 2, a 3, a n, the unique binary expansion. So, if you take two different points x 1 and x 2, so let us write, try to write this mathematically, that this is, so let us take a point x 1 with binary expansion a 1, 1, a 1, 2, a 1, n and so on. Let us take another point with binary, unique binary expansion, that we have fixed in the methodology, so a 2, 1, a 2, 2 and a 2, n and so on. And x 1 not equal to x 2. So, that implies, if x 1 is not equal to x 2, that implies, there exists some stage n not such that a 1, n not will not be equal to a 2, n not. And that implies that 2 times a 1, n not will not be equal to 2 times a 2, n not. And that is, this is b 1, n not and this is called that b 2, n not. That means, y 1, if we have y 1, that is point b 1, 1, b 1, 2 up to b n, n and so on. And y 2 is the other point, the image of x 2, so that is b 2, 1, b 2, 2, b 2, n and so on. Then, so if this is so, then y 1 is not equal to y 2. So, that means, this process of sending x, taking x with binary expansion is this. And constructing y with, constructing y with ternary expansion is this. So, if we send x to y, this gives us a map from 0, 1 to c, which is 1, 1 and hence, so this implies as a consequence. So, hence the cardinality of c is same as cardinality of 0, 1. And if you recall, cardinality of 0, 1 that is, so let us write this implies that c is uncountable, because 0, 1 is uncountable. Thus, c is an uncountable set. So, this is follows from our construction that c is uncountable set. In fact, let us try to now calculate. So, note that c, which is equal to intersection of a n's implies that for every n, c is a subset of a n. And what was a n? That was a disjoint union of intervals i n j, j equal to 1 to 2 to the power n. And at the nth stage, what will be the length of each where? The length of each i n j, so what is the length of the intervals, which are left at the nth stage? That is 1 over, so let us just look at the construction. At the first stage, when we removed 2, at a 1, 2 intervals were left each of length 1 by 3. So, this is 1 by 3, this is 1 by 3, this is 1 by 3 and this is 1 by 3. So, 4 intervals at the second stage of length 1 by 3. So, at the nth stage, how many, 2 n intervals of each of length, how many will be left? They are 2 to the power n intervals each of length, how much? So, here the length of each i n j, so at the second stage it is 1 by 3. First, second stage is 1 by 3 and nth stage will be 1 over, so it will be 1 over 3 raised to power 2 n minus 1. So, that will be the length of each one of them and there are 2 to the power n of them. So, what is the total length? So, sigma lambda of i n j, j equal to 1 to 2 to the power n, so that is 2 to the power n intervals each has got the same length. So, divided by 3 raised to power 2 raised to power n minus 1 and observe that this number goes to 0, it is 2 to the power n by 1 over 3 raised to power n that goes to 0 as n goes to infinity. So, that means what? That means that c can be covered by for every n by 2 to the power n intervals whose length is this and that can be made as small as they want. So, that means that the outer Lebesgue measure, so lambda star of c is equal to 0 because what is lambda star of e? It is the infremium of the sums of the intervals which cover the set c and here we have just now shown that c is contained in an which is a finite disjointed line of intervals and the total length of these intervals is becoming smaller and smaller. So, that is length of c. So, that implies that c is a lambda star null set. Hence, c belongs to is a Lebesgue measurable set and not only that c is in fact what we know is something more that if e is any subset of c, then that implies that lambda star of e also equal to 0 because lambda star is monotone and that implies that e also belongs to l. So, hence all subsets of c, power set of c is a subclass of the Lebesgue measurable sets and of course, Lebesgue measurable sets are a subset of power set of real line. Now, but c is uncountable and r is uncountable. So, what does this imply? That means this implies that l has as many elements as p of r. So, what is the meaning of this? It has as many elements as p r that is same as saying the cardinality if you know what is cardinality. Cardinality of l is same as the cardinality of the power subset of real line and if you know that the cardinality of real line which is what we call as cardinality of continuum is denoted by the small letter c. So, this is denoted by 2 to the power c. So, what does that prove? So, that proves that if you look at from the cardinality point of view, if you look at how many elements are there in the class of all Lebesgue measurable set, then it says cardinality of Lebesgue measurable sets is as much as the cardinality of all subsets. So, if you look from the cardinality point of view, you cannot say that the class of all Lebesgue measurable sets is a proper subset of the class of all subsets of the real line. So, but that does not also imply that all subsets of real line are Lebesgue measurable. So, the question still remains undecided whether the class of all Lebesgue measurable sets is a proper subclass of all, proper subclass of all subsets of real line. So, to decide this question is a bit difficult and that relates to some fundamental questions in set theory. So, let us look at, so what we have shown just now, let us just recapitulate that lambda star of c is equal to 0 and that says that power set of c is a subset of L and hence there are at least as many elements in L as 2 to the power c. So, that is the cardinality of the continuum. So, we get cardinality of L and power set is same, both have got same cardinality. So, question still remains is L a proper subset of P R. So, if you recall the answer to this question is related to some of the fundamental questions in set theory. So, if you recall we proved what is called Ulam's theorem, we did not prove it really, we mentioned what is called Ulam's theorem and I said that one can read a proof of this in the text book that we have mentioned and that is the statement of the Ulam's theorem says, assuming continuum hypothesis, Lebesgue measure cannot be extended to all subsets all of real line. There is something called continuum hypothesis in set theory, I will not explain at this stage what is continuum hypothesis because will be going slightly off stream, but it is worth mentioning here that the set theory is based on certain axioms. So, whatever modern mathematics we are doing is based on axiomatic set theory and there is a which has some kind of some axioms on which we are we are we can deal with set theory, but there is something called continuum hypothesis which relates to the subsets of real line and so on and that is not part of the axioms of set theory that is why it is called continuum hypothesis. Some people believe in continuum hypothesis and do mathematics according to that and some people do not believe in it. So, if you assume continuum hypothesis and Ulam's theorem says that you cannot extend that means not all subsets of real line are measurable. Another result which one can use which is again not part of the axiomatic set theory is the following which says that supposing you assume what is called axiom of choice. Axiom of choice is another axiom which is not part of the axiomatic set theory and one can either accept it part of set theory and do mathematics or do not accept part of it and do mathematics. So, mathematicians those who accept axiom of choice they are supposed to be doing what is called non-constructive mathematics because there are some existence theorems which assume axiom of choice helps in proving some theorems which are existential in nature. For example, proving that every vector space has a basis requires the need of using axiom of choice. You cannot prove it if you do not assume axiom of choice. There are many results in mathematics which use axiom of choice and which are not true if you do not assume axiom of choice. So, what is axiom of choice is basically saying very statistically saying given a non-empty collection of non-empty sets, you can pick up one element from each set and form a new set. So, it is how sets can be constructed when the sets are not indexed by a family which is finite in number essentially. So, it says given any indexed family of non-empty sets and that indexing set also is non-empty from each one of these sets. You can pick up one element and form a new set. So, using this one can show there exist sets in the real line which are not Lebesgue measurable. So, we will prove this result. So, assuming axiom of choice there exists non-Lebesgue measurable sets in real line. So, let us prove existence of non-measurable sets by assuming axiom of choice. So, let us start. So, what we are doing is existence of non-measurable sets. So, that is what we are discussing. So, we want to construct a subset of the real line which is not Lebesgue measurable. So, to start with consider once again the interval 0 to 1. So, consider the interval 0 to 1. So, this is the interval 0 to 1. On this I am going to define a relation so far x related to y if x minus y is a rational number. So, for x and y take two points x and y in 0, 1 and you say that they are related with each other if and only if x their difference is a rational number. So, the first observation I will just write claims 1 that this x related to y is an equivalence relation. So, what does equivalence relation mean? It means it is reflexive, symmetric and transitive. So, what is reflexive? x related to x that is obvious because x minus x is 0 and that is a rational number. And secondly if x is related to y that means x minus y is a rational number and so the difference so the negative of that that is y minus x also is a rational number. So, that implies that y is related to x. So, if x is related to y then y is related to x that is called symmetry that the relation is symmetric. And the third one is let us suppose x is related to y and y is related to z. So, x related to y means x minus y is rational and y related to z means y minus z is rational. So, if you take the difference that implies that x minus z is a rational so that implies that x is related to z also. So, it is an equivalence relation it is a reflexive, symmetric and transitive. And every equivalence relation given on a set partitions the set into equivalence classes. So, that is the basic idea that 0 1 can be partitioned into equivalence classes by this relation. So, that implies so second that implies so let us write that 0 1 can be written as a disjoint union of equivalence classes. So, let us write it as e alpha, alpha belonging to some indexing set let us call it as a, e alpha equivalence class. And recall equivalence class means e alpha intersection e beta is empty for alpha not equal to beta that is why I have written as a union with a this sign. That means equivalence classes they cover 0 1 and they are disjoint. So, there is a partition of the set on which equivalence classes are defined. .. And the third step is from each e alpha select some element x alpha and form the set call let us call it as e which is x alpha x alpha belonging to the indexing set a. So, what we are saying is using this equivalence relation partition thus interval 0 1 into equivalence classes and from each equivalence class pick up one element exactly one element x alpha select one element x alpha choose one element x alpha from each equivalence class and put them together in a box call that e and claim is that e is a set. And this is a place this is the place we are using axiom of choice. So, that means e alpha is a collection of non-empty collection of non-empty sets from each we can pick up one element and form this set this is possible only if we assume axiom of choice. So, here is the place where we are using axiom of choice. So, from each equivalence class we have picked up one element and constructed a set e. So, obviously this set e e is a subset of 0 1 because each equivalence class is a subset of 0 1 and from each we have picked up one element. So, this is a subset of 0 1. Let us write so, let rationals in minus 1 to 1 be written as r 1 r 2 r n and so on. Rationals in the interval minus 1 to 1 is a countable set. So, they can be enumerated they can be written in the form of a sequence. We are not saying r 1 is smaller than r 2 or anything. We are just giving a enumeration of the rationals. They are countable in many. So, we can write them as a sequence and construct define a set en which is e plus r n n bigger than or equal to 1. Construct a set en this. So, let us observe where is the set en? e is in 0 1 and each r n is between minus 1 to 1. So, what can you say about the set e plus r n? So, e can be 0 to 1, r n could be minus 1 to 1. So, that means each one of them is a subset of minus 1 to minus 1 to 2. At the most this sum can become minus 1 where elements of e are smaller the smallest ones 0 and possibility here is a minus 1 and the largest possible is r n is equal to 1 and e also element is 1. So, 1 plus 1 2. So, for every n en is a subset of 0 1 of minus 1 to 2. So, this implies that the union of en is also contained in minus 1 to 2. So, that is one observation. Also, if I take x belonging to 0 1, if I take an element x in 0 1 that implies x is related to x alpha for some alpha because the equivalence classes cover 0 1. So, every element x in 0 1 has to belong to one of the equivalence class. So, say it belongs to e alpha. So, that means it is related to x alpha the element that we have picked. So, that implies that x minus x alpha is a rational x minus x alpha related means the difference is a rational and so where will that rational be? x is in 0 1 x alpha is in 0 1 this is a rational is a rational in minus 1 to 1 because both could be 1 and that means that is x minus x alpha belongs to en because if it is a rational in minus 1 to 1 that must be equal to some R n and that means x is equal to x alpha plus R n and that means it is in en. So, what we are saying is for every x in 0 1 x minus x alpha belongs to is in R n. So, sorry that not that to that implies that x is equal to x alpha plus R n and that belongs to en. So, x belongs to en. So, the second observation is that 0 1 is inside the union of en's. So, that is what we have gotten. So, this construction we have got is the following that 0 1 is contained in union of e plus R n that is en n equal to 1 to infinity and that is contained in minus 1 to 2 and in this construction of the set e we have used this axiom of choice. Now, here is one observation that let us move on to observe claim that these sets e plus R n intersection e plus R n are disjoint sets for n not equal to m. To prove this, so let us take an element x which is common. So, if not x belongs to e plus R n that means x is equal to x alpha plus R n is also equal to it is also in e plus R m. So, it is also equal to some e beta plus R m and that implies that so that implies x is related to x alpha and x is related to x beta that is x is related to x element that implies either x alpha is equal to x beta if that is there should be same and that is possible implies that alpha is equal to beta. So, if alpha is not equal to beta then this is not possible. So, that says that means that these two sets are disjoint. So, this is what we have gotten. So, as a consequence let us write this as that 0 1 is contained in a disjoint union of e plus R n n equal to 1 to infinity and that is contained in minus 1 to 2. So, till now we have not done anything except we defined a equivalence relation and using axiom of choice we constructed a set e and this as this property. Now, suppose assume that e is Lebesgue measurable then there are two possibilities one Lebesgue measure of e is equal to 0 but that implies Lebesgue measure of e plus R n is equal to 0 for every n because Lebesgue measure is translation invariant and that implies that the Lebesgue measure of the union e plus R n is equal to 0 and that implies because 0 1 is inside this that means 0 1 equal to 0 which is a contradiction because Lebesgue measure of 0 to 1 is equal to 1. The second possibility is that the Lebesgue measure of e is strictly bigger than 0 then that implies Lebesgue measure of minus 1 to 2 this close interval is bigger than or equal to Lebesgue measure of this union because that is a subset of it and that is equal to sigma lambda of e plus R n and that is equal to sigma lambda of e because for every n it is same and this being a positive quantity added in infinite number of times that is equal to plus infinity which is again a contradiction because lambda of minus 1 to 2 actually is equal to 3 and 3 equal to infinity is a contradiction so either case this assumption cannot be true. So, this is a set which is in 0 1 and which is not measurable. So, what we have shown is the following that if we assume axiom of choice then there exist non-Lebesgue measurable sets in the real line without axiom of choice or without continuum hypothesis it is not known that you can construct subsets of the real line which are not measurable non-Lebesgue measurable. In fact, there is a theorem which says that the condition that assume axiom of choice actually if you put this as an axiom in set theory that every subset of every subset of the real line is Lebesgue measurable if you take that as an axiom and if your set theory axioms are already consistent then adding this new axiom to your set theory will not make any difference it will still leave it consistent. So, existence of non-measurable sets get related to fundamental questions in set theory. So, on this side we will leave it as it is saying that if you either assume continuum hypothesis or you assume axiom of choice then there exist sets which are not Lebesgue measurable. Let us turn to the other side can we say that the Borel sigma algebra the Borel subsets of real line they form a subset of for this form a subclass of Lebesgue measurable sets what is the relation between these two can we say that the Borel sets form a proper subset of the class of all Lebesgue measurable sets. So, one can show will not prove most of the things here because they are slightly technical. So, first observation is that the Borel sigma algebra of the real line which is a sigma algebra generated by all intervals is the same as the sigma algebra generated by all open intervals because one can show that every open set in real line is a countable union of open intervals actually that is using the basic topology in the real line. So, topological properties of real line come into play and not only that in fact you can take open intervals with only rational end points and if you generate the sigma algebra by them that is same as the Borel sigma algebra. So, this needs a proof so we will not prove it, but just indicate what is involved here. So, the Borel sigma algebra so this is a countable family open intervals with rational end points. So, you take a countable family of intervals and generate the sigma algebra and that is B R and one can show that the cardinality of this process of generating is exactly equal to C. So, one using these properties one shows using this construction one shows that the cardinality of the sigma algebra Borel sets is same as that of C that of the continuum and that is of the real line. Whereas, the cardinality of the Lebesgue measurable sets was 2 to the power of C so that means there exists sets. So, cardinality looking at the cardinality says that there exist sets which are Lebesgue measurable but which are not Borel sets. But actual construction of these sets is not very easy it is possible to construct such sets which are Lebesgue measurable but which are not Borel sets they are called analytical sets, analytic sets and for that we refer the our textbook for more details those who are of you are interested they should refer the textbook for more details. So, what we have shown today is that in the special case of the extension theory we get the notion of length function on a class of sets which are called Lebesgue measurable sets which include the Borel sigma algebra of subsets of the real line and the cardinality of the Lebesgue measurable sets is same as the cardinality of all subsets and if you make some assumptions like continuum hypothesis or axiom of choice you can show existence of sets which are not Lebesgue measurable. Otherwise you cannot show there is no such proof known and on the other side the Borel sigma algebra has got cardinality C which is much stricter strictly less than the cardinality of Lebesgue measurable sets. So, we will continue looking at the properties of Lebesgue measurable sets vis-a-vis open sets closed sets and the group sector on real line in the next lecture. Thank you.