 Welcome to lecture 9 on measure and integration. As you recall, we have been looking at classes of subsets of a set X called semi-algebra, algebra, sigma algebra and so on. And then we also looked at set functions defined on these classes with properties. So, in particular we define the concept of a measure. A measure is a set function defined on a collection of subsets such that the measure of mu of the empty set is equal to 0 and mu is countably additive. Today, we are going to start the process which is called extension process. So, the topic for today's discussion is going to be extensions of measures. Basically, the question arises from some properties on the real line. Let us look at mathematically the question. We know that notion of length is defined for all intervals. So, the question is, can the notion of length be extended to arbitrary subsets of the real line? That means, can we define the notion of the length for an arbitrary subset of the real line? Of course, it should be compatible with the definition of the length for the interval. So, the need for such an extension, one of course, is purely a mathematical curiosity that we have the notion of length for an interval, can we define it for a arbitrary subset? Another reason which is more important is that it arises from some problems in Riemann integration. The concept of Riemann integral which is defined for a class of functions fails to satisfy some nice properties, like if a function is the fundamental theorem of calculus does not hold for Riemann integrable functions. So, in order to remove those difficulties, one started for looking for an extension of Riemann integral and that led to the problem of extending the notion of length from a class of subsets that is intervals to all subsets if possible. If you are interested in looking at more details about that, why Riemann integral should be extended to a wider class of functions and how that leads to the concept of extending the notion of length to arbitrary subsets, read chapter 1 and 2 of the text book that we have mentioned earlier, namely an introduction to measure and integration by myself Inderke Rana. . So, let us start with the question of what is an extension. So, let C 1 and C 2 be two classes of subsets of a set X and let us assume C 1 is a subset of C 2. We have two measures or two set functions mu 1 and mu 2, mu 1 is defined on C 1 and mu 2 is defined on the collection C 2. So, mu 1 and mu 2 are set functions mu 1 defined on C 1 and mu 2 defined on C 2 with the property that mu 1 on C 1 is same as mu 2 for subsets of C 1. So, mu 1 and mu 2 agree on subsets of C 1 because C 1 is a sub collection of C 2. So, in such a case we call mu 2 is an extension of mu 1. So, on C 1 which is a smaller class mu 1 and mu 2 are same. So, and mu 2 is defined on a bigger class that is C 2. So, we say C 2 and mu 2 is an extension of the measure of the class set function C 1. So, the problem is given a measure mu, we start with a measure mu on a semi-algebra C of subsets of a set X. We want to show that there exists a unique extension to a measure mu tilde on A of C the algebra generated by it. So, this is going to be our first step of extension theory namely given a measure on a semi-algebra. We are going to extend it to a measure on the algebra generated by that semi-algebra. So, let us see how this process is carried over. So, recall that a set E in the algebra generated by a semi-algebra where characterize such sets can be given by a representation E is equal to union i 1 to n E i. So, every set in the algebra generated by a semi-algebra is a finite union of sets form the semi-algebra and in addition they are pair wise disjoint. So, this was the result we had proved that the algebra generated by a semi-algebra is nothing but all finite disjoint union of sets in the semi-algebra. So, let us take any set E in the semi-algebra. So, we define mu tilde of E to be sigma i 1 to n mu of E i and the claim is that this is the unique extension which we are looking for. So, let us see how do we do it. So, we have got mu on C and this is a semi-algebra. So, we define mu tilde on the algebra generated by C. So, this is algebra generated by C. So, we want to define a function here a set function which should look like an extension. So, if E belongs to F of C then we know this set E looks like a disjoint union of element C i i equal to 1 to n for some n C i belonging to C. Now, why we defined it the way we have defined mu tilde? See, if mu tilde of E is going to be defined and it is going to be a measure on the algebra F of C then we know that every measure is also finitely additive. So, by the finite additive property of mu tilde which we have not yet defined, but the finite additivity property will say that this should be equal to mu tilde of the union C i i equal to 1 to n and this being finitely additive we should have i equal to 1 to n mu tilde of C i. But mu tilde is going to be an extension. So, that means mu tilde on C i is same as mu on C i's. So, this is same as 1 to n of mu of C i. So, that actually fixes what is going to be the definition of mu tilde of E. So, if E is a finite disjoint union of elements which is C i's then mu tilde of E must be given by this and that also shows the uniqueness of the definition of mu tilde. So, mu tilde should be defined by this it is necessary and we will show that actually this definition works also. So, let us prove this property that mu tilde. So, first we want to show that mu tilde is well defined. So, what does that mean? Suppose E is a set which is in F of C then we know that E can be written as a finite union of set C i's finite disjoint union of set C i in C. But it is possible it can have some other representation. So, it is possible that is also representable as j equal to 1 to m of some sets D i, where C i's belong to C and D j's also belong to C. So, to show then because our definition dependent on the representation. So, we should show that mu of C i summation i equal to 1 to n is same as summation mu of D j, j equal to 1 to m. So, this is we should show then only we can claim if we can able to show this then only we can claim that our function mu tilde i is well defined. So, let us show this. Now, note because E is given by these two different representations. So, I can write union C i i equal to 1 to n also as union C i intersection union D j's j equal to 1 to m. So, that is equal to union i equal to 1 to n union j equal to 1 to m C i intersection D j. Similarly, union D j's j equal to 1 to m is also representable by the same way because the two sets are same. So, it is same representation. Now, let us compute sigma mu of C i i equal to 1 to n. I can write it as sigma i equal to 1 to n. Now, this mu of C i is this is what union of C i intersection D j. That is where we are using this representation that we just now wrote j equal to 1 to m. This is a disjoint union C i's belong to the semi-algebra D j's belong to the semi-algebra. So, this intersection belongs to the semi-algebra and their unions is C i which is also in the semi-algebra and mu is a measure on the semi-algebra. So, it so this is also finally additive. So, it is i equal to 1 to n. So, I can write this sigma equal to j equal to 1 to m mu of C i intersection D j. Similarly, we can also write j equal to 1 to m mu of D j to be equal to this is summation j equal to 1 to m and mu of D j. So, that I can write as union of D j intersection C i i equal to 1 to n. Now, again by finite additive property this is j equal to 1 to m sigma i equal to 1 to n mu of D j intersection C i. Look at this equation 1. Look at this equation 2. 1 and 2 imply that sigma i equal to 1 to n mu of C i is equal to sigma j equal to 1 to m mu of D j. So, that says implies mu is well defined. So, what we have shown is the following that we take any set in the algebra generated by the semi-algebra. So, that has got a representation in terms of the elements of the semi-algebra. So, any element E in the algebra generated by the semi-algebra can be represented as a finite this is not union of elements in the semi-algebra say C i. So, pick up any such representation and define mu tilde of E to be equal to sum of mu's of this P C i sigma i equal to 1 to n. It does not matter which representation you choose you will always get the same sum. So, that means mu tilde of E is well defined. Now, let us look at the next property namely that mu tilde which is defined on the algebra generated by the semi-algebra is finitely additive. So, let us prove that property. So, we want to prove that mu tilde. So, mu tilde is finitely additive. So, to prove that what we have to show? So, let E be written as a union of E j's j equal to 1 to n where each E j belongs to the algebra generated by C. And of course, E also belongs to the algebra generated by C. So, we want to show to show that E mu tilde of E is equal to summation j equal to 1 to n mu tilde of E j. So, this is what is to be shown. Now, to show any such property we have to go back to the definition of mu tilde of any set. So, since E belongs to the algebra generated by C that implies let us write each E j. E j belongs to the algebra. So, each E j can be written as a disjoint union of E j and say k, k equal to 1 to n j, where E j k belongs to C for every j and k. So, every element E j is in the algebra generated by C. So, it must be a finite disjoint union of elements of C. So, that implies that the union E j, j equal to 1 to n is equal to union j equal to 1 to n union k equal to 1 to n j of E j k. Then this is my, so this is my over set E. E is equal to union. So, we have represented E as a finite disjoint union of elements of C. So, that implies that mu of E, mu tilde of E, I can choose any representation. So, in particular this, so it is equal to summation j equal to 1 to n summation k equal to 1 to n j of mu of E k j. And now using the finite additive property of mu, we will write this. So, this is equal to, look at this sum. So, look at this sum. That is nothing but j equal to 1 to n mu tilde of E j. That is by definition, because E j is a union of E k j over k. So, by definition I can take that representation and write this is equal to this. So, that says mu tilde of E is equal to this. So, hence mu tilde is finitely additive. So, we have proved that mu tilde is finitely additive. The uniqueness we have already shown. So, thus we have shown that a measure which is defined on a semi-algebra can be in a unique way extended to the algebra generated by it. And basically the idea is, because every element intuitively keep in your mind that mu of asset is the size. So, any element in the semi-algebra, in the algebra generated by the semi-algebra is a union of disjoint pieces in the semi-algebra and size of each of them is known. So, the size of the union must be equal to some of the sizes of the individual pieces, because they are disjoint. So, that was the idea and that helped us to extend a measure from a semi-algebra to the algebra generated by it. So, that is the first step of the extension theory. So, as a consequence the length function can be extended. That we have already shown, length function can be extended from the collection of all intervals to the collection of finite disjoint union of intervals that is algebra generated by it. So, now we will go to the next step of the extension. So, we will start with a measure which is defined on an algebra and we want to try to extend it to the sigma algebra generated by it. So, the next step in the extension theory is we can the length function. For example, we would like to say can the length function be extended to all subsets of the real line. We have done it from intervals to the algebra generated by intervals. There is a theorem by a mathematician called S. M. Ulam and that theorem was proved in 1930. It says that under the assumption of continuum hypothesis it is not possible to extend the notion of length to all subsets of real line. And this is a very important theorem. So, it uses two things namely one is what is called continuum hypothesis. I will not go into the discussion of what is called continuum hypothesis at this stage. I would say that one should read about this theorem from the text that we have just now mentioned an introduction to measure and integration. So, this is very nice and important theorem which says as a consequence that it is not possible to extend the length function to all subsets of real line. So, the question comes if we cannot extend. So, in general a given a measure mu on an algebra of subsets of x, we would like to extend it to a bigger class than a. It cannot be done it for all subsets anyway. But let us try to intuitively follow our idea of measuring the size of an object. So, intuitively given a measure mu on an algebra a collection of subsets of a set a of a set x, mu of a is the size of the set a which you can measure. And given an arbitrary set e one may not be able to measure its size exactly using the mu, but we can at least try to approximate. So, let us define what is called the outer measure induced by a measure. So, let us take mu and algebra of subsets of a set x, a and algebra of subsets of a set x and mu a measure defined on it. For any subset e in x let us define what is called mu star of e. So, what is mu star of e what we do is given the set e. So, here is a set e you cover it by sets a i is in the algebra you cover it by the sets in the algebra. Take a covering of e by the sets a i's in the algebra and you know what is the size of the set a i. So, let us take the size of the set a i and add up all these sizes. So, what do you think this sum will represent? This sum will represent in some sense the approximate size of the set e of course it depends on the covering a i. Now, what we do is we take the infimum of all these approximate sizes. That means, we take the infimum of these numbers over all possible coverings of the set e and define that number as mu star of e and we will try to analyze what are the properties of this mu star of e. So, first of all let us give it a name this mu star of e is called the outer measure induced by mu. Why outer? Because we are covering e by sets. So, these things cover e we are going may be we are going outside e. So, this is outer and measure because we are trying to measure the size of this in terms of induced by mu because in terms of the known sizes mu. So, once again let us recall and look at carefully what this mu star is given is at e arbitrary subset in x cover it by elements a i whose sizes you know. So, take a covering of e by elements in the algebra look at the sizes of a i's add up all this that is the sum mu a i that is approximate size and take the infimum of all these approximate sizes. So, that we are going to call as the outer measure induced by. So, the first property we want to say is mu star is well defined. Well, what is the meaning of mu star is well defined? Let us go back to the definition this is mu star is infimum of some numbers and infimum of a subset of numbers exists in the real line if it is non-empty and it should be bounded below. Of course, all these numbers are going to be bounded below because all are non-negative numbers. So, it is bounded below by 0. Why is this non-empty? Why is this collection non-empty? Because a is an algebra. So, the whole space belong to it. So, keep in mind a is an algebra and in the definition of an algebra the whole space x is an element. So, e is covered by x itself. So, and x belongs to the algebra. So, at least there is one number in this collection over which you are taking infimums namely mu of x. So, it is a non-empty collection of extended real numbers. So, its infimum always exists and hence mu is a well defined number. Of course, it could be equal to plus infinity. Keep in mind the numbers here they are all extended real numbers. So, this is this set is a collection of non-negative extended real numbers and their infimum always exists and infimum could be equal to plus infinity. So, we have shown that mu tilde is mu star the induced outer measure is well defined. The next property. So, mu star is a well defined set function on the class of all subsets of the set x and we want to show some properties of it. So, the first property is mu star of empty set is equal to 0. That is true because empty set belongs to the collection A in the algebra and mu star of there is equal to mu of A and that is equal to 0. And for any set that is a non-infinimum of non-negative numbers. So, this infimum has to be bigger than or equal to 0. So, that first property is obvious. Second property we want to check that mu star is monotone. So, let us check that mu star is a monotone function. So, let us take let A and B be subsets of x and A a subset of B to show mu star of A is less than or equal to mu star of B. Now, what is mu star of A? This is in all these properties we are going to use the definition of infimum critically. So, what is mu star of A? mu star of A is defined as by our definition. It is the infimum over sigma mu of E i's say 1 to infinity, where this set A is contained in union of E i's disjoint union. Of course, E i's belong to the algebra. And what is mu star of B? That is the infimum i equal to 1 to infinity of mu of say f i's, where B is contained in union of f i's i equal to 1 to infinity disjoint union, where f i's also belong to the algebra A. Now, note if A is given to be a subset of B, if A is subset of B and B is covered by a union j equal to say 1 to infinity, then that implies A is also inside. So, this is also inside f of j. So, what we are saying is every covering of B is also a covering of A. So, and this is the infimum over all possible coverings of B and this is the infimum over all possible coverings of A and every covering of B is also a covering of A. So, here we are taking infimum over a larger set and here we are looking at the infimum over a smaller collection of numbers. And whenever you take infimum over a smaller collection of numbers, that is always bigger than or equal to infimum over a larger collection of numbers. So, that is a simple property about infimums. If you are taking a infimum of a larger collection, then that tends to be smaller than the infimum over a smaller collection. So, that property implies that mu star of A has to be less than or equal to mu star of B. So, that is purely a property of the infimum over what collection you are taking. Every covering of B is also a covering of A. So, coverings of B form a subset of coverings of A and hence this property is true. So, that is the monotone property namely mu star is monotone. Let us look at the next property namely mu star is countably sub-additive. So, we want to prove mu star is countably sub-additive. So, that means what? To show, so whatever to show that if A is a subset of X and A is contained in union of Ais, Ais also a subset of X, then we want to show that mu of A is less than or equal to summation mu of Ais. So, this is what is to be shown. Now, let us observe. So, note we want to show one number mu of A is less than or equal to sum of these numbers. If one of these numbers is equal to plus infinity, then obviously this property is true. So, note if mu of A i is plus infinity for sum i, then clearly mu star of A is a number which is less than or equal to plus infinity which is at least one of the mu Ais. So, that is less than or equal to mu of mu. So, everything about star, so it is mu star we are looking. So, let us just write mu star. We are trying to prove that mu star is countable. So, mu star of A i is equal to 1 to infinity. So, what we are saying is this inequality is obvious if one of the terms in this sum is equal to plus infinity. So, let us take the case when all of them are finite. So, let us assume, suppose mu star of each A i is finite for every. Now, what is mu star of A i? mu star of A i is infremum of a certain collection. So, here we are going to use the property of something being infremum and that being finite. So, let epsilon greater than 0 be arbitrary, of course, fixed. You choose arbitrarily and fix it. Then mu star of A i is the infremum over all summations approximate sizes. So, then there exists at least one covering. So, there exists sets say A ij, j equal to 1 to so on in the algebra A such that this A i is contained in this disjoint union of A ij. So, and mu star of A i, which is an infremum, if to this I add the small number epsilon, this becomes bigger than summation mu of A ij, j equal to 1 to infinity. So, this is let me stress here. This is the kind of definition or this is the kind of analysis we will be coming across and we will be doing again and again. So, let us be very clear about this. We have got some number, which is the infremum over some collection and if this infremum is finite, then the infremum plus a small quantity epsilon cannot be the infremum, because that is on the right side of it. So, that cannot be the infremum of that collection. Otherwise, alpha plus if alpha is the infremum, then alpha plus epsilon will be the infremum, which contradicts the definition of the infremum. So, if alpha is the infremum, alpha plus the small number epsilon, any small number epsilon cannot be the infremum. That means what? That means there must be a member of the collection over which you are taking infremum, which so that alpha plus epsilon becomes bigger than that number in the collection over which you are taking infremum. So, that is what we are saying that if because mu star of A i is finite, so given epsilon, the infremum plus epsilon must be bigger than a member of the collection over which you are taking infremum. So, what is the collection that is obtained by taking a covering, a disjoint covering of a disjoint covering, not only disjoint, actually any covering we are taking. So, any covering and such that this is true. So, given epsilon, there exists a covering A ij, j equal to 1 to infinity of A i, say that mu star of A i plus epsilon is bigger than this, and this happens for every i. So, if we add up, so I add these equations over i, so summation over i equal to 1 to infinity, mu star of A i plus sigma alpha over i is bigger than sigma over i equal to 1 to infinity, sigma over j equal to 1 to infinity mu of A ij. That is what we wanted, mu star of A i is bigger than something, we have got that kind of inequality. Now, the problem is this, we are going to add epsilon infinite number of times, so this will tend to become infinity, and we do not want that. So, we go back and refine our estimates. So, given epsilon bigger than 0, this we can do it for any epsilon. So, in particular whenever you are looking at for A i, given epsilon, there should exist a covering, say that we will refine it, we will make it 2 to the power i. So, we will change our epsilon, that is true for every epsilon. So, in particular it should be true for this. So, what we are saying is, given epsilon, there is a covering, say that A i is covered by that collection, and mu star of A i plus epsilon divided by 2 to the power i is bigger than the approximate size, that is summation mu of A ij. This is for every i, now if I add here is epsilon to the power 2 i, so that means we have got, this is now convergent, so that means, so that implies that sigma i equal to 1 to infinity mu star of A i plus epsilon is bigger than this sum. And now note, if i and j both vary, so this is for every i. Now, if I take the union over i's, that will be union over this, so I will get a covering of union A j's, which will be covered by this, and A is inside this. So, what we are claiming is, this is bigger than mu star of A, because A is contained in union over i union over j A ij's and this A ij's belong to C. So, A is covered by this countable union and this is one approximate size for mu of A. So, that is always bigger than or equal to mu star of A, because that is an infimum. So, this quantity implies that this is always bigger than this. So, I can claim that mu star of summation is bigger than this quantity. Now, this epsilon is arbitrary, that was fixed arbitrarily. So, I can let that go to infinity. So, one writes, so now letting, so letting epsilon go to 0, we have sigma mu star of A i i equal to 1 to infinity. This epsilon becomes 0 eventually. Now, I will write bigger than or equal to, because in the limit it can become bigger than or equal to mu star of A. And that shows, so hence mu star is countably sub-additive. So, that we have proved is countably sub-additive. I just want to go through the proof of this once again, because this is an important kind of analysis, we will be doing again and again. Let us just revise the proof once again, that mu star is countably sub-additive. So, to show that mu star is countably sub-additive, we have to show that if A is a subset of x and A is contained in union A i's, A i's contained in x, then I have to show that mu star of A is less than or equal to summation mu star of A i's. Now, to show this, the first observation, which should, should keep in mind that whenever you are trying to show that one number is less than or equal to summation of a collection of numbers, then an obvious case may arise, namely one of the numbers may be equal to plus infinity. So, if mu of A i is equal to plus infinity for some i, then clearly this side is equal to plus infinity and mu star of A is always less than or equal to plus infinity. So, we get mu star of A less than or equal to plus infinity and which is always less than this sum. So, that means that property is true. So, the obvious case is mu star, then mu star of A i is finite for some i. So, what is the other possibility? Other is that mu star of A i is finite for every A i. Now, here is the main construction, part of the construction that we are going to use, namely it is an infremum, which is a real number. So, given epsilon bigger than 0 arbitrary, we can find a covering A ij of the set A i, such that mu star of A i plus this small number and that small number will make it dependent on i, the stage at which you are doing epsilon divided by 2 to the power i, bigger than the approximate sizes over which you are taking the infremum. So, once again the property of infremum being a real number is used here, nothing more than that. So, once that is done, you add both sides. This is for every i, take the summation on both sides. So, summation mu star of A i plus summation of this over i is bigger than summation of mu of A ij. Now, this is a convergent series, its sum is equal to epsilon. So, this is mu star of A i summation plus epsilon and this quantity on the right hand side is an approximate size of A, that is this is bigger than or equal to mu star of A. Mu star of A is the infremum over all such numbers, because A is covered by union over i union over j, A i is covered by A ij's. So, union over A i's will be covered by this union and A is inside it. So, this implies that summation mu star of mu of A ij's summation over i and j is bigger than mu star of A and once that is done, that means that we have got and let go epsilon go to 0. So, you get this quantity. So, that says that mu star is countably sub additive. So, let us so we have proved this property that mu star is countably sub additive. So, mu star now the only thing left to be shown is that that mu star actually is an extension, otherwise all this process will be a waste. So, we want to claim that mu star is indeed an extension of mu. Mu star is not a countable additive, but at least we should check it is an extension and it is countably sub additive that we have already checked. So, we want to check that mu star of A is equal to mu of A if A is in A. So, to check that let us look at the proper definition. So, we had mu star of A is equal to infimum over summation mu of A i, i equal to 1 to infinity where A is contained in union of A i's of and A i's belong to C belong to the algebra A. Now, so if A belongs to the algebra then A is actually equal to A. So, A is contained inside A. So, this is one of the elements here in this covering A itself covers it. So, it will appear in one of it will be one of the elements over which you are going to take the infimum. So, that implies that mu star of A which is the infimum is less than or equal to mu of A. So, that property is by the sheer fact that A is covered by itself and A is in the algebra. So, that is we want to prove other way around inequality to show that mu of A is less than or equal to mu star of A. Now, once again we want to show that one number is less than the other number. So, there is an obvious possibility case 1 that mu star of A is equal to plus infinity. So, in that case this is plus infinity and mu star mu of A is always less than or equal to plus infinity which is equal to mu star of A. So, that is obvious. So, the obvious case is when mu star of A is equal to plus infinity. So, let us look at case 2, mu star of A is finite. So, in that case again we are going to use the definition of infimum. So, mu star of A is the infimum of all possible approximate sizes, summation so on. So, let epsilon greater than 0 be arbitrary. Then there exists a covering. So, there exists sets A j's belonging to the algebra such that A is contained in the disjoint union of A j's and the infimum says that mu star of A plus epsilon cannot be the infimum that has to be bigger than summation mu of A j's. So, there is at least one such covering possible so that this is infinity. This is not necessarily disjoint. Win can make it, we will see it later on. So, this is finite. Now, note A is contained in union of A j's and all of them are elements in the algebra. We assumed A is in the algebra. So, everything is in the algebra. So, and mu is a measure and we showed every measure implies mu is countably sub additive and that implies that mu of A is less than or equal to summation mu of A j's, j equal to 1 to infinity. So, look at this equation 1. Look at this equation 2. So, what does 1 and 2 apply? mu star of A plus epsilon is bigger than this sum and that sum is bigger than mu of A. So, 1 and 2 imply that mu star of A is bigger than mu star of A plus epsilon. So, plus mu star of A plus epsilon is bigger than mu of A and epsilon is arbitrary. So, let epsilon go to 0 and that implies that mu star of A is bigger than or equal to mu of A. So, that proves the other way round inequality also in the case when mu star of A is less than or equal to mu of A, because A is one of the members which is covering it. So, mu of A is element. So, mu of A is mu star of A is an infinitum. So, that is less than or equal to. That is obvious property and to show that the case when it is finite, mu star of A is finite. We look at once again the definition given epsilon is bigger than 0. There is a covering so that this holds. The infinitum plus epsilon is bigger than one of the elements over which you are taking the covering. Now, using the fact that mu is countable as a derivative, this is bigger than or equal to mu of A and hence that proves the required property. So, what we have shown is that the mu star is indeed an extension of mu of A. So, let us go back and have a look at what we have done is the following. We started with a measure mu on the algebra A. Measure means it is mu of empty set is equal to 0 and mu is countably additive. We are trying to extend it. So, we try to find out the size of any set by looking at sizes of sets in A. So, take any set E, cover it by elements in the algebra A and look at the sizes of mu, call it as mu of A i. So, take the summation. So, this gives an approximate size of the set E. Look at the smallest possible of these numbers, call it the infremum. So, mu star of E, the induced outer measure is defined as the infremum over all these summations and these summations arise from coverings of E. So, this is called the outer measure and we showed it is well defined. We showed it is, it has the obvious property namely mu of empty set is equal to 0, mu star of A is bigger than or equal to 0. It is monotone and so that means mu star of A is less than or equal to mu of B and mu star is countably sub additive. And finally, it is an extension. So, let me point it out that we have taken mu star of A as the infremum over those summations and we have taken the coverings which are countable in number. One can ask the question, cannot we take only finite coverings instead of countable coverings of E? So, let us give an example to show that that is not possible to do that. The finite coverings will not suffice. So, let us look at the set E, the case is the real line. We will look at the set E which is rational's intersection with 0 to 1. So, we are looking at all the rational's in the set in the interval 0, 1. Clearly lambda star of E we expect it to be equal to 0. Why we expect size of this? Because it is a countable set and the length of each singleton is equal to 0. So, we expect the length of each when added together this also should remain small or lambda star of E is equal to 0. It is quite natural. Now, suppose we define. So, this is when lambda star is defined by taking countable coverings. Now, let us take a finite covering of E by interval. So, E is covered by finite number of intervals union E i. We claim that in that case this number the approximate size of E will always be bigger than or equal to 1 because of the following reason. What is E? E is rational's inside 0, 1 and suppose E is covered by union i, j is j equal to 1 to n. So, if possible let sigma lambda of i, j, j equal to 1 to n be less than or equal to 1. Now, these are finite collection of. So, here is 0, here is 1 and i, j's are intervals of course intervals in 0, 1 which are covering. So, i, j's are in 0, 1 which are covering the side E which is rational's in 0, 1. So, now let us say that i, j say for the sake of just definition it is a, j, b, j. It does not matter whether open or closed. You can just assume to be open. It does not matter much actually. Then we have got these numbers between a, j's and b, j's. So, look at all the left end points and look at the smallest of them. Let us say the smallest is here that is a. So, what is a? a is the smallest of the numbers a 1, a 2, a n and b, j. Look at the largest of b, j's and call it as b. Then this a, b is equal to, may be closed does not matter, is equal to union of i, j's or at least it will cover the union of i, j's, j equal to 1 to n and they cover E. Now, so and that covers E. Now, if this is less than or equal to, this is the smallest and that is the largest ones which is covering. Now, this number a, so that means what? That means b minus a is less than or equal to summation length of i, j's, j equal to 1 to n and if that is less than or equal to 1, that means b minus a is strictly less than 1. That means it has to be like this, but then there is a rational here between 0 and 1 which belongs to E and E is inside a, b. So, that is not possible, so that will be a contradiction. So, this situation is not possible that means whenever these are covering, we have to have lambda of i, i is bigger than or equal to 1, but that means all approximate sizes of E is bigger than 1. So, that will imply that lambda star of E is bigger than or equal to 1. That means, but that is not possible because we just now said lambda star of E is equal to 0. So, in the definition of the outer measure, we cannot limit ourselves to only finite coverings. We have to allow all countable coverings also. So, today we have tried to go beyond an algebra. So, we started with a semi-algebra and a measure on it. We extended it to a measure on the algebra generated by it as a first step. As a next step, we started with a measure on an algebra and we showed that by an example on the real line, Ullam's theorem that you cannot extend it to all subsets of real line. So, let us try to go as much as far as possible. So, we define given a measure mu on an algebra, we defined the notion of an outer measure for any subset A of that set X and we showed this outer measure has some nice properties. It extends one given measure, it is monotone, it is countably sub additive. So, in the next lecture, we will see how to get from it an actual extension which is a measure. Thank you.