 Welcome to today's lecture on measure and integration. This is the eighth lecture on measure and integration. Today, we will be looking at a problem called the uniqueness of measures on algebras and sigma algebras. So, for this, we will need to define some term terminology. So, let us look at the uniqueness problem for topic for today's lecture. So, the problem is as follows. We are given c an algebra of subsets of a set x and S of c is the sigma algebra generated by c. So, c is an algebra and S of c is the sigma algebra generated by c. We have got two measures mu 1 and mu 2 defined on the sigma algebra generated by c such that mu 1 of a is equal to mu 2 of a for every a belonging to c. So, for all elements in c, mu 1 and mu 2 agree. The question is, can we conclude that mu 1 of e is equal to mu 2 of e for every element in the sigma algebra generated by c? So, this is a general uniqueness problem which plays a role later on when we extend measures to general settings. So, to answer this question, let us make some definitions. So, first of all, this is not true in general for all measures. We have to put some conditions on the measures. So, let us look at what is called a totally finite measure. A measure c is called totally finite if mu of every subset a is finite in that domain of mu. So, c is a collection of subsets and mu is a set function. So, we say mu is totally finite or sometimes we also say it is finite if mu of a is less than plus infinity for all a belonging to c. Then note that in case c is an algebra and mu is finitely additive, then mu is finite if and only if mu of x is finite. So, to look at that, let us assume c is an algebra and mu from c to 0 to infinity is finitely additive. So, suppose mu of the whole space is finite. Note the whole space x belongs to c because c is an algebra. Then we had seen earlier finite additivity of mu implies mu is monotone. We had seen this property earlier. So, we will not go into the details of this again. So, because mu is monotone, thus for every a contained in x, mu of a will be less than or equal to mu of x, which is finite. So, thus implies mu of a finite for every a subset of x and converse is obviously true. So, converse is true because x belongs to c and so mu of x is finite. So, whenever we are dealing with finitely additive set functions, saying mu is totally finite, it is enough to say that mu of x is finite and as a consequence, mu of every subset will be finite. So, this is a easy consequence of saying for a finitely additive set function on an algebra, mu of the whole space finite is same as saying mu of every subset a of x, a in the algebra of course is finite. Let us look at the property that let us take a set function mu and we say that it is sigma finite. So, we define what is totally finite and now we define what is called sigma finite. So, mu is said to be sigma finite if we can write the whole space as a union of sets x n and 1 to infinity such that these sets are pair wise disjoint. So, we want these sets to be pair wise disjoint and mu of each x n to be finite. So, each x n should be element in the domain of mu in the class c and mu of x n should be finite. So, essentially saying that for totally finite, we said mu of the whole space is finite and sigma finite means what that x can be cut up into pieces x 1, x 2, x n so on and mu of each x n is finite. So, this is what is called sigma finite set functions. So, what we are going to let us look at some examples of set functions. So, the length function lambda on the class of all intervals is sigma finite. So, that is easy to see. Why is the length function sigma finite? So, here is the length function lambda on the class of all intervals taking values in 0 to infinity. The whole space that is real line of course is the interval minus infinity to plus infinity and length of r, we know it is not finite, it is equal to plus infinity. But we can write r as a disjoint union of the intervals n to n plus 1, for example, n belonging to integers. So, the real line is cut up. So, here is the real line. So, here is 0. So, take 0, 1, 1 close to, open to close 3 and so on and on this side minus 1 and so this is minus 2 and so on. So, we have cut up the real line. We have divided the real line into countable many disjoint pieces. Each one is an interval and length of n to n plus 1 for every n is equal to 1, which is less than of course infinity for every n. So, the whole real line is written as a countable disjoint union of intervals, each one having finite length. So, lambda on the class of intervals is sigma finite. Of course, it is not finite because the measure of the whole space, length of the whole space is equal to plus infinity. So, this is an example of a set function, which is sigma finite. We will give another example of a set function, which is totally finite. So, for that, let us look at the length function restricted to any finite interval. So, let us look at, say for example, let us look at the interval a to b and let us look at all subinterval, so let us look at i tilde a b to be all subintervals of a b and define of course the length function as before. So, a b to, so this will be a function from 0 to b minus a, length of i is equal to the usual definition of length i for i contained in a b and we know length function is finitely additive, it is countably additive and so on. So, on these intervals also it is countably additive and lambda of the interval a to b we know is equal to b minus a, which is finite. So, lambda, so one says lambda restricted to subintervals in a b is finite or totally finite for every a and b. So, this is the example of a measure, which is totally finite. Let us look at another example of a set function, which is not, so let us look at example of, let x be any set and for any subset a of x, let us define mu of a to be equal to plus infinity, if a is non-empty and mu of 0 to be equal to 0. Then obviously, this is the measure on p of x, this is a simple property that one can easily check, because mu of empty set is 0 is given and if a is any set, which is a countable, this is not union or not, so mu of the union will be equal to again plus infinity, which is equal to sigma mu of a i's, at least one of them has to be non-empty. So, I hope it is clear that this mu is countably additive, this is a measure on p x. So, if not, let us look at supposing a is equal to union of a i's, i equal to 1 to infinity. If a is equal to empty set, then a i is equal to empty set for every i. So, implying mu of a, which is 0 is same as sigma mu of a i's, i equal to 1 to infinity. The second possibility, if a is not empty and a is equal to union, so that implies there exists at least one i, i such that a i is not empty. Then, so let us say that such that i says that, so let us say that is a i not. There is at least one i, so let us say that is i not. Then, union a i not is not empty and implying that not empty. So, that is a is not empty anyway, so this is not required. Then mu of a is equal to plus infinity is equal to summation mu of a i's, because at least one term here is not empty, so this is equal to plus infinity. So, mu of a i not is equal to plus infinity, so that says this is also plus infinity, so they are same. So, this is a measure on the class of all subsets. So, mu of a plus infinity, if a is not empty and mu of empty set is equal to 0 is a measure and this obviously is not sigma finite, because there are no subsets anyway whose mu is finite. So, this is an example of a non-sigma finite measure. .. So, the theorem we want to prove today is the following, namely let us take see a semi algebra of subsets of a set x and S of c be the sigma algebra generated by c. Let mu 1 and mu 2 be to finitely additive set functions on S of c, such that mu 1 of e is equal to mu 2 of e. For all e belonging to c, then we want to show that mu 1 of a is equal to mu 2 of a. For all a belonging to first a of c, where a of c is the sigma algebra generated by c. So, we are saying as a first step we are going to prove that if two measures mu 1 and mu 2 defined on a semi algebra agree, then they also agree on the sigma on the algebra generated by that semi algebra. So, this is what we want to prove. So, let us see the proof of that. So, c semi algebra a of c that is the algebra generated by c and we are given mu 1 of e is equal to mu 2 of e for every e belonging to c to show mu 1 of a is equal to mu 2 of a for every a belonging to algebra generated by c. So, how do you prove it? So, let us start. Let us take a set a which belong to a of c, then that implies. Recall, we had shown a characterization of elements of the algebra generated by a semi algebra. So, we showed that if a is an element of the algebra generated by a semi algebra, then this a must look like a finite disjoint union of elements c i i belonging to n, where c i is belong to the semi algebra c. So, every element a in the algebra generated by a semi algebra we had shown must have a representation like this. But then mu 1 of a is equal to mu 1 of this finite union and we know mu 1 is finitely additive. So, that implies this must be equal to sigma i equal to 1 to n mu 1 of c i, but each mu 1 is equal to mu 2 on each element of c and c i are elements of c. So, that implies that this must be equal to 1 to n mu 2 of c i, but again by using mu 2 is finitely additive I can write this as mu of a, because a is a finite disjoint union of elements of this. So, mu 1 of a is equal to mu 2 of a whenever a belongs to a of c. So, this proves theorem that whenever two measures finite, whenever two finitely additive set functions mu 1 and mu 2 agree on a semi algebra, then they also agree on the algebra generated by it. . So, let us go to the next step of this uniqueness problem. So, that is saying that c be a semi algebra of subsets of s at x once again and s of c be the sigma algebra generated by c. So, this we have already proved. So, let mu 1 and mu 2 be sigma finite measures on s of c such that mu 1 of e is equal to mu 2 of, so this is a misprint, mu 1 of e should be equal to mu 2 of e for all e in c, then mu 1 of e is equal to mu 2 of a for all a belonging to s of c, whereas of c is the sigma algebra generated by it. So, let me state and we will divide the proof into steps of course. So, let us look at the statement of the theorem. Once again we were saying that let mu 1 and mu 2 be two measures which are sigma finite defined on the sigma algebra generated by a semi algebra c measures c semi algebra s of c, this s of c is equal to sigma algebra generated by c. Then mu 1 of a is equal to mu 2 of a for every a belonging to the semi algebra to show mu 1 of a is equal to mu 2 of a for every a in the sigma algebra generated by c. So, this is what we want to show. So, to show this the first step let us look at the first step. So, we may assume that c is an algebra. So, first of here we are given that c is a semi algebra. So, step 1 says we may assume that c is an algebra and that is because of the fact that we have just now shown that if mu 1 and mu 2 agree on the on a semi algebra then they also agree on the algebra generated by it. So, by the given hypothesis mu 1 of a is equal to mu 2 of a for every a belonging to the algebra generated by c. So, we already have mu 1 and mu 2 agree on the algebra generated by a of c and we want to show that this implies mu 1 of a is equal to mu 2 of a on s of c the sigma algebra generated by c. But note this is same as the sigma algebra generated by a of c that also we have shown that given a semi algebra you can directly generate the sigma algebra or you can generate the algebra first and then generate the sigma algebra both are same and just now we showed whenever two measures agree on a semi algebra they agree on the algebra generated by it. So, mu 1 and mu 2 agree on the semi algebra they show they agree on the algebra generated by it and we want to show that they agree on the sigma algebra generated by it and which is nothing but s of c. So, that proves the first step. So, as a first step in our proof we are saying that we can assume that the given class c on which mu 1 and mu 2 are defined is actually an algebra. So, that is the first simplification in the proof that without loss of generality we may assume that c is an algebra. So, next step let us look at next step says that we may assume that both mu 1 and mu 2 are totally finite we are given mu 1 and mu 2 are sigma finite. So, next step is that we may assume that we may assume mu 1 mu 2 are totally finite. So, what is the meaning of saying we may assume that if the statement. So, this is same as saying if the statement if the statement mu 1 of a equal to mu 2 of a for every a belonging to s of c is true when mu 1 mu 2 are totally finite then it will also be true when mu 1 mu 2 are sigma finite. So, that is the meaning of saying that we may assume that mu 1 and mu 2 are totally finite. So, let us see why that is the case. So, let us take a set A contained in. So, what we are given is mu 1 and mu 2 are sigma finite. So, mu 1 mu 2 sigma finite imply I can write x mu 1 is sigma finite. So, I can write x as union of x i's i equal to 1 to infinity where x i's belong to c and mu 1 of each x i is finite. Similarly, mu 2 is sigma finite. So, I can write x as union some j equal to 1 to infinity y j where y j's are subsets in c and mu 2 of each y j is finite. But then from both of these statements I can write x as. So, this implies we can write x as union over i 1 to infinity of x i's, but that I can decompose into union of y j's. So, x i intersection y j, I can write that. So, I can write this as a decomposition of x into subsets x i intersection y j. Now, what we have achieved is the following, mu 1 of each x i was finite, mu 2 of each y j was finite, but now this implies note that mu 1 of x i intersection y j is finite and mu 2 of x i intersection y j is also finite. So, now both mu 1 and mu 2 are finite on this piece. So, in the picture you can think of this as x. So, you divide these are sets x 1, x 2, x i and so on and then you also have sets which are, you also have sets y j's. So, they are decompositions like this. So, this piece is nothing but x. So, this piece is y j. So, this is this piece here is x i intersection y j. So, the whole space is cut up into pieces. So, this is what this statement means, where each one of them is finite. Now, let us note, so here is observation. So, note, mu 1, mu 2 restricted to x i intersection y j are totally finite. So, what is the meaning of this statement? They are restricted. That means, if we look at the subsets, that is, for every a contained in x i intersection y j, a belonging to C mu of a is finite, mu 2 of a is finite. And for totally finite measures, we have already proved that we have already assumed that this statement is true and we are trying to show it for our sigma finite. Now, for any set a contained in x, mu 1 of a can be written as summation over i summation over j mu 1 of a intersection x i intersection y j. So, that is because a is equal to union over i, union over j, a intersection x i intersection y j. So, this is a countable disjoint union, mu 1 is a measure. So, this must be true. And now, note that this a intersection this is a set in the set x i intersection y j, where mu 1 is finite and then that we know that there the statement is true. So, here a is contained in x. Of course, a belonging to S of C. So, the statement is true. So, that means what? So, by the assumption that statement is true for finite measures, we conclude that this is same as a intersection mu 2 of a intersection x i intersection y j and once again that is equal to mu 2 of a. So, here we use. So, a basic idea is for any set, we can bring it to the finite pieces. There we know it is true and go back to the original piece. So, this is the proof of the second step that we may assume without loss of generality that our measures mu 1 and mu 2 both are finite. So, we have made two simplifications in our proof. The first one being we may assume that C is an algebra and second one that mu 1 and mu 2 are totally finite. So, what we want to prove now? So, we only are left with the case to prove that if C is an algebra mu 1 and mu 2 are totally finite, defined on the algebra C and if they agree on C then they will agree on the sigma algebra generated by C. So, that is the next step we want to show. So, for that let us write. So, to prove the final step let us write m to be the class of all those elements of S of C where mu 1 and mu 2 agree and what is the aim to prove? Our aim is to prove that this collection m is nothing but S of C. We are picking up subsets of S of C. So, m is a subclass of S of C. We want to prove that this is equal to S of C and that is proved as follows. First we will observe then m is a monotone class. We will prove that once we have proved m is a monotone class. We will also observe that we are given that mu 1 and mu 2 are equal on C. So, C is a subclass of m and C is an algebra and C is contained in m. m is a monotone class. So, that will mean what? That the monotone class generated by C must be inside m, but C is an algebra and the monotone class generated by an algebra is the sigma algebra generated by it and that also we have proved. So, that will prove as a step 4 that m is equal to S of C. Let us prove step 3 and then conclude from it step 4. So, step 3 we want to prove. We are given that mu 1, mu 2 totally finite. We are in this case C algebra mu 1 of A equal to mu 2 of A for every A belonging to the algebra C to show that mu 1 of A is equal to mu 2 of A for every A belonging to the sigma algebra generated by C. So, that is the question. So, we are saying the proof define m to be the class of all subsets belonging to S of C for which this property is true. So, that means mu 1 of E is equal to mu 2 of E. So, claim that this m is a monotone class. So, what is a monotone class? Recall a monotone class is a collection of, so a monotone class is a collection of subsets of a set X which is closed under increasing unions and decreasing intersections. So, these two properties have to be checked. So, let us check that. So, let us take, so let E n be a sequence in m such that E n is increasing. So, E n is inside E n plus 1 for every n to show union of E n's belong to m. Now, let us note that E n is an increasing sequence, E n increases to E which is union of E n's. E n's belong to m. So, keep that in mind. So, and we want to show E belongs to, so that is the set E, we want to show that E belongs to m. That means mu 1 of E is equal to mu 2 of E. So, now note, so let us note what is mu of E? How do we compute it? Let us observe that mu 1 of E is nothing but limit of mu 1 of E n's and why is that? That is because mu 1 is a measure, it is countable additive. So, we had proved that countable additivity of the set function implies whenever a sequence E n increases to a set E then mu of E must be limit of mu 1 of E n's. That was the characterization property for countable additivity. So, this go back and refer that was because of countable mu, countable additive. So, that is the property being used here or equivalent form of it. Now, each E n belongs to m. So, that implies that E 1 mu of 1 of E 1 is equal to mu 2 of E n. So, this is equal to limit n going to infinity mu 2 of E n. So, that is we are using mu 1 equal to mu 2 on mu 1 equal to mu 1 equal to mu 2 because sorry this is because E n belongs to m. And once again mu 2 is countable additive. So, there mu 1 was count and that implies that this is so this is mu 2 of E using the fact that mu 2 is countable additive. So, let us once again we have used lot of things from which we have proved earlier mu 1 is a measure E n is increasing to E. So, by countable additivity mu 1 of E must be equal to limit n going to infinity mu 1 of E n by countable additivity. Now, each E n belongs to m. E n is a sequence E n is a sequence in m. So, that means mu 1 of E n is equal to mu 2 of E n. So, this is equal to this. So, this is the second step equality. And now once again mu 2 is countable additive E n increases to E. So, by countable additivity this limit must be sorry this limit must be equal to mu 2 of E. So, it says mu 1 of E is equal to mu 2 of E. So, that implies that E belongs to m whenever. So, E belongs to m whenever E n is a sequence which is increasing to m. So, this is for increasing and the corresponding thing we have to prove when it is decreasing. And that is where we are going to use the fact that mu and mu 2 are totally finite. So, for the second case let E ns belong to m. E n include E n plus 1 for every n decreasing and E b equal to intersection of E ns and n equal to 1 to infinity. So, we want to show that E also belongs to m. So, for that once again mu 1 of E is equal to limit n going to infinity mu 1 of E n because mu 1 totally finite mu 1 of x finite and mu 1 countable additive. And that as above earlier is same as mu 2 of E n because each E n belongs to m and that is equal to mu 2 of E. Once again mu 2 is finite and E n is decreasing to E. So, that proves this also. So, this proves that m is a monotone class. So, the class m which was equal to all subsets E belonging to S of C such that mu 1 of E is equal to mu 2 of E is a monotone class. And we are given that mu 1 of A equal to mu 2 of A for every A belonging to C. So, what does that mean? An equivalent way of stating that is saying that the collection C is inside the collection m. So, that is what it means by very definition. So, m is a monotone class, C is inside it so that implies that the monotone class generated by C must be inside m. So, because recall what is monotone class generated by a collection of subsets of C? It is the smallest monotone class of subsets of x which includes C. During the smallest it must be inside it. But note, C algebra implies m of C is equal to S of C. So, this is an important theorem which we have proved that if you take an algebra and generate a monotone class out of it that is same as generating the sigma algebra out of it. So, this is same as saying that S of C is contained in m. But m is a collection of subsets of S of C. So, that is inside S of C. So, that is same as saying that m is equal to S of C, the sigma algebra generated by C. That means what? For all elements in S of C, mu 1 is equal to mu 2 of A. So, that proves the theorem, the uniqueness theorem namely f 2. So, we have finally proved in these four steps the theorem that if mu 1 and mu 2 are two measures defined on a semi-algebra of subsets of a set X and mu 1 and mu 2 are both sigma finite then and they agree on the semi-algebra then they also agree on the sigma algebra generated by C. So, this is an important theorem which we are going to use quite often. So, with this we come to an end of part, a part of our course. So, this is probably the right stage to revise what all we have done till now. So, let us revise what we have done till now. So, we started with looking at collections of subsets of a set X. We defined what is a semi-algebra. So, what was a semi-algebra? Semi-algebra was a collection of subsets of a set X with the properties. The whole space belongs to it, the empty set belongs to it. It is closed under intersections and the complement of a set is inside, not inside, not necessarily inside it, but can be represented. So, semi-algebra C, semi-algebra meant that empty set, the whole space belongs to it, one property. The second one A and B belonging to C should imply A intersection B belong to C and the third property that A belonging to C implies A complement can be written as a finite disjoint union of elements of C. So, N for some c i is belonging to C. So, that is a semi-algebra. Then we defined what is called an algebra. So, a collection C is called an algebra. The first property as it is empty set, the whole space belongs to it. A and B belonging to C should imply it is closed under intersections that also belongs to C. Now, we have something stronger, instead of just saying that A belongs to C, this complement is representable. Actually, we want that this complement also belongs to C. So, this is something stronger. So, we said this is a stronger property. So, semi-algebra implies semi-algebra and the converse need not be true, that we had said. Then we defined what is called a sigma algebra. A collection C is a sigma algebra, if of course it is an algebra first of all. So, it is phi x belong to C. It is closed under intersections A and B belong to C, if A and B belong to C. But this is not enough. So, actually we want a countable, this should be true for any countable collection. So, let us write whenever A i's belong to C that should imply that intersection A i's belong to C and because it is going to be closed under complement. So, this is the property 1, this is second and the third property is that whenever A belongs to C, it should imply a complement belong to C and that automatically implies that C is also closed under. So, this property of countable intersections can be equivalently stated as because of the complements that A i's belong to C imply union A i's also belong to C. So, a sigma algebra is a collection which is closed under countable unions and complements and of course empty set under the whole space belong to it. Then we define what is called a monotone class. So, what was a monotone class? A m was called a monotone class, if whenever a sequence E n belong to m, E n's increasing, E is equal to union E n's should imply that E also belongs to m. So, this is one property and second property we want that whenever E n's belong to m, E n's are decreasing and E is equal to intersection of E n's should imply that E also belongs to m. So, a monotone class is the collection of subsets of a set X with the property it is closed under increasing unions and decreasing intersections. Of course, sigma algebra implies monotone class, the converse is not always true. Then we looked at, so this was the first basic concepts of properties of collection of subsets of a set X we looked at and then we looked at what are called the algebra generated by a collection of subsets or the sigma algebra generated by a collection of subsets or the monotone class generated by a collection of subsets of a set X. In all these cases, basically given a collection in C, so let us just recall what was the meaning of saying generation. So, C any collection of subsets of a set X, so algebra generated by C generated by C is the smallest one. So, it was the intersection of all the algebras which include C and we showed such a thing exists. Similarly, the sigma algebra generated by C, we said it is nothing but look at all sigma algebras of subsets of X which include C and take their intersection, so that is a sigma algebra, so that is called the sigma algebra. So, another way of saying is the algebra generated by C is the smallest algebra of subsets of a set X which includes C. Similarly, S of C is the smallest sigma algebra of subsets of C which includes C and similarly, we have monotone class generated by C. It is the smallest monotone class of subsets of X which include C and we showed by these properties that such a object always exists. Then, we proved a very important theorem namely the monotone class generated by C is equal to the sigma algebra generated by C if C is an algebra. So, this was an important theorem that we had proved. So, these concepts were basically about collection of subsets of a set X. Then, we looked at functions defined on such collection of subsets of this set X and we called them as set functions. So, set functions are functions defined on a collection of subsets of a set X and the important class of set functions was the length function and we showed that the length function had important properties namely the length function which is defined on the class of all intervals in the real line was shown to be a countably additive set function which was also invariant under translations. So, that was an important property. Then finally, we had proved some equivalent conditions for countable additivity and these conditions are very useful. So, let us just recall these equivalent conditions. We have used one of them today also. So, one of the important conditions for example, if mu is defined on a algebra. So, this is a algebra and it is finitely additive. Then, we said that mu is countably additive if and only if mu is countably sub-additive of course plus. So, let us write plus mu finitely additive. So, let us remove this condition. Let us write mu to be an algebra and mu of empty set equal to 0. So, let us put. So, mu is a set function defined on an algebra and mu of empty set is 0. Then, we proved that saying that mu is countably additive is equal into saying that mu is finitely additive and countably sub-additive and this is quite useful in proving the countable additivity of set functions. So, this was one and second we proved that mu finitely additive we assume that then mu countably additive if and only if whenever e n's decrease of course, under the condition a is algebra to e should imply mu of e n's is equal to the decrease. So, let us write decrease to e then limit n going to infinity mu of e n is equal to mu of e. This provided we had to put an extra condition mu of x is finite. So, mu of x is finite e n's decrease to e and implies that limit of e n's is equal to mu of e. So, this condition is equivalent to saying mu is countably additive when we have this. If we do not put this condition then this may not be true, but then I can also one can equivalently prove another thing that if e n's increase to e then that should imply mu of e is equal to limit n going to infinity mu of e n's. So, this was the property of saying that when is something countably additive and finally today we proved the uniqueness theorem saying that if mu 1 and mu 2 are two finite countably additive set functions defined on a semi algebra. So, this is a semi algebra mu 1 and mu 2 are sigma finite measures and mu 1 of a is equal to mu 2 of a for every a in the semi algebra. Then this implies mu 1 of a is equal to mu 2 of a for every a belonging to the sigma algebra generated by c. This is mu 1 and mu 2 should already be defined. So, we should say they are already defined in S of c, c is a semi algebra. So, c is a semi algebra. Let me state it once again c is a semi algebra mu 1 and mu 2 are defined on the sigma algebra generated by c. Both mu 1 and mu 2 are sigma finite and mu 1 and mu 2 agree on the semi algebra. Then they agree on the sigma algebra also. So, they agree on the whole domain. So, if they agree on the part of the domain which is a semi algebra, then they agree on the whole of the sigma algebra also. That is a uniqueness result and that we proved under the condition that mu 1 and mu 2 are sigma finite measures. We will see this how that is used in extension theory in the next few lectures when we come to them. So, that is what we stop today here. So, in the next lecture, we will start a new topic called extension theory. So, we would like to extend a set function defined on a class to a bigger class. For example, on the real line, we have the notion of length defined on a collection of intervals on the collection of all intervals. We would like to define the notion of length for any set. So, that is the aim of that is a motivating thing for extension theory. So, we will use that and prove the theorems in the next lectures. Thank you.