 Welcome to lecture 3 on measure and integration. We had started looking at the concept of algebra of subsets of a set X. We will look at some more properties of that today. And after that we will start looking at what are called sigma algebras of subsets of a set. And then come to sigma algebra generated by a class of subsets of a set X. And then go on to look at what is called monotone class. The monotone class generated by a class and then look at a monotone class generated by an algebra. So, let us just recall what we had started looking at namely the algebra. We said an algebra of subsets. So, a class of subsets A contained in P of X that is a power set of X is called an algebra. If it had the properties 1, the empty set the whole space is a member of A. Secondly, whenever A belongs to A implies its complement is also inside the class A. That is the class A is closed under the operation of complements. And the third property was whenever A and B belong to A that implies A intersection B belongs to algebra. So, these are the three properties that define a class A to be an algebra. And keep in mind this property because of complements this can be equivalently stated as A and B belonging to the class A implies A union B also belongs to the class A. So, this is what we have defined as a collection of subsets of a set X to be an algebra. And then we looked at various properties of algebras. For example, we proved one thing that a class C need not be an algebra, but you can generate an algebra out of it. So, this is the smallest algebra including the given collection C. And then we went on to prove that if you take a collection C and restrict its elements to a set E, which is defined as all elements of the type A intersection E, where A belongs to C. Then if you generate algebra out of this collection C intersection E, which is the algebra of subsets of E generated by this collection C intersection E, we showed this is also equal to the algebra generated by C restricted to E. So, these are the various ways of generating more algebras out of the given algebra. What is the advantage of having an algebra is the following. So, let us suppose that you have got A is an algebra. And then let us take a sequence A 1, A 2, A n inside A, a collection of elements of A. And let us take their union E equal to union of A n's n equal to 1 to infinity. Of course, this E need not belong to the algebra, because algebra is only closed under finite unions. However, there is something nice one can do. Let us define B 1 to be equal to A 1 itself. Let us define B 2 to be equal to A 2 and remove from it the set elements which are in A 1. And similarly, let us define B 3 to be A 1 union A 2 and remove from it the elements which are in A 3 and so on. So, you will define B n in general to be equal to union A i, i equal to 1 to n. Remove from it the elements which are n minus. So, B n is defined as the union of elements up to n minus 1 and remove from it the elements which are in A n. So, we have generated a new sequence out of the given sequence. So, B n, let us observe that B n which is equal to union of this A i's minus A n. What is this look like? It is union A i, i equal to 1 to n minus 1 intersection A n complement, because removing A n is same as taking its intersection with A n complement. Now, observe each A i is an element in the algebra. This is a finite union of elements in the algebra. A n complement is in the algebra, because A n is in the algebra. Algebra is closed under complements. So, this implies that each set B n is an element of the algebra A for every n. So, that is one observation. And secondly, let us observe that B n intersection B n is empty for n not equal to m, because what we are doing? B 1 is A 1, B 2 is from A 2 remove A 1. So, B 1 and B 2 are going to be disjoint and B 3. It is A 1 union A 2 minus A 3. We have removed what is in A 3. So, this B 3 is going to be disjoint from B 2 and B 1 both. So, in general it is quite obvious that B n's are pair wise. This is disjoint. They are elements in A. Further, here is an important consequence. The way we have constructed if I take union of B i is i equal to 1 to n, what is that equal to? It is precisely B 1 is A 1, B 2 is A 2 minus A 1. So, what is B 1 union B 2? That is same as A 1 union A 2. And similarly B 1 union B 2 union B 3 is same as A 1 union A 2 and union A 3. So, that is same as B 1 union B 2 union B 3. So, for every union of B i's i equal to 1 to n is same as union of B i's i equal to 1 to n. So, that implies. So, as a consequence this implies that union of B n equal to 1 to infinity is same as union i equal to 1 to infinity of A i which was our set E. So, what we have shown? We have shown that if we start with any countable union of elements in the algebra, E need not be an algebra, but E can be represented as a disjoint union of sets B n and each B n is a algebra. So, what we are saying is we have proved a theorem which is going to be quite useful and that is advantage of being inside an algebra. So, let us recall once again if A is an algebra of subsets of a set X and a set E is union of A n's n equal to 1 to infinity where each A n belongs to A. Then there exist disjoint sets. So, there exist sets B n's belonging to the algebra which are pair wise disjoint and their union is equal to B n's. So, any countable union in algebra can be represented as a countable disjoint union. So, that is the advantage of being in a class which is algebra. So, that is nice. We will see the applications of this next time. So, let us now start with a class which is slightly more stronger than algebra. So, that is called a sigma algebra. So, let us start with a collection X is a non-empty set and let S be a class of subsets of the set X with the following properties 1, empty set and the whole space are elements of it like in a semi-algebra like in an algebra. A complements belong to S whenever the set A is in the S. That means the collection S is closed undertaking complements as in the case of an algebra. So, these two properties are same as were the case for an algebra. The third property is the one which distinguishes it from an algebra. We want that whenever sets A i's are in S, A i equal to 1, 2, 3 and so on. That means whenever you take a countable collection of sets in S, their union i 1 to infinity A i's also belongs to S. That means the collection S is closed undertaking countable unions also. So, such a collection we are going to call it as a sigma algebra. Sigma indicating that it is closed under a sequence of unions. So, let us just emphasize once again a sigma algebra of subsets of a set X is a collection which includes the empty set in the whole space. It is closed undertaking complements. So, if A belongs to S, A complement belongs to S and whenever you take a sequence A i of elements of S, their union is in S. That means S is also closed undertaking countable unions. So, such a class is called a sigma algebra. One obvious example every sigma algebra is also an algebra because sigma algebra means it is a collection which is closed under countable unions and algebra only requires finite unions. Of course, both algebra and sigma algebra are closed undertaking complements and empty set in the whole space are always members of both of them. So, every sigma algebra is also an algebra. Let us look at an example of X, an uncountable set and let us look at the collection of all those subsets of X such that either the set is finite or its complement is finite. So, an element E is in this collection F if either the set is finite or its complement is finite. We are already shown that this collection F is an algebra. So, let us recall what we have already shown that if I take this collection F of subsets E of X such that E or E complement finite, then we already observed that F is an algebra. So, the question is, is F a sigma algebra? So, that means, so does it have the property? So, that is E 1 E 2 E n belonging to F does this imply always that union of E n is n equal to 1 to infinity also belongs to F. That is not true for the following reason because, so note X is uncountable. So, as a consequence of this, there exists a subset E contained in X such that E is countable in finite and E complement is not finite because if not, because if this is not true then what will happen? We will have X is equal to E union E complement. This is countable and this is finite. So, that will imply X is countable which is not true, which is a contradiction. So, whenever you have got a set X which is uncountable, there always exists a subset of it such that E is infinite and its complement is not finite that is E complement is infinite. So, we have got a set E which is countable in finite and its complement is not finite. So, that means what? Since E is countable, since E is countable in finite, countable in finite, so I can write E equal to X 1 X 2 X n and so on. So, it is a countable in finite set. So, I can write it as a sequence. I can enumerate the elements of it. So, that is equal to singleton X i union i equal to 1 to infinity. And note, let us observe that singleton X i is an element in the algebra F because singleton is a finite set. So, that implies that and that does not imply E which is a union of these elements belong to F. E does not belong to F. Why E does not belong? Because if E has to belong to this collection F, E should be either finite or E complement is finite. Both of them are not true. So, basically if I take a set E which is countable in finite, then it is a countable union of elements of F and it does not belong to F. So, F is not going to be an algebra. So, what we are saying is, if you look at whenever X is uncountable and look at this collection of sets E contained in X, say that E or E complement is finite, then it is an algebra and it is not a sigma algebra of subsets of X. So, this collection that we have taken, we have proved that it is an algebra of subsets of X, but it is not a sigma algebra of subsets of X. So, every sigma algebra is an algebra, but every algebra need not be a sigma algebra. So, that is the observation that we get from here. .. Let us look at some more examples of sigma algebras. Let X be any set, then obviously the empty set and the whole space put together the two elements that collection is a sigma algebra because there are only two elements, their union belongs closed under complements and so on. And of course, the collection of all subsets of X, the power set of X also is a sigma algebra of subsets of X because it is closed under all kind of operations. So, empty set and the whole space put together is an example of a sigma algebra, power set is an example of a sigma algebra of subsets of any set X. These are called obvious examples of sigma algebras of subsets of X. Let us look at some non-trivial examples of sigma algebras of subsets of X. So, let us take X an uncountable set and let us take S to be a subset, all those subsets of X such that A or A complement is countable and the claim is S is an algebra of subsets of the set X. So, let us try to prove that this collection S is a sigma algebra of subsets of X. So, what is S? S is the collection of all those subsets A contained in X such that A or A complement is countable. So, first observation, empty set belongs to S because empty set is taken to be a finite set. So, it belongs to S. Does X belong to S? Yes, X belongs to S because its complement is empty set and hence that belongs to S. So, empty set and the whole space both belong to S. Clearly, if A belongs to S, then this implies actually different only if A complement belongs to S because our defining condition is symmetric with respect to A and A complement. So, let us check the third property that if A n belongs to S, n equal to 1, 2, 3 and so on, then this implies union of A n's n equal to 1 to infinity also belongs to S. So, let us check that property. So, obviously like in the case of finite and complement finite, we have to divide it into two cases. The first case is, so case 1, all A i's or all A n's are countable, but that will imply the union of A n's is also countable and hence belongs to S. Why is union of A n's is countable? Because countable union of countable sets is countable that is a set theory property. So, A n, so this set is countable, so it belongs to S. Let us look at the second possibility, the second case. So, what is the possibility? Not all A n's are countable. That means there exist some n naught such that A n naught belongs to S, but A n naught is not countable. So, that means what? A n belongs to S not countable, that means A n naught complement is countable by the very definition of S. Now, observe that union of A n's n equal to 1 to infinity includes the set A n naught because that is one of the members. So, that implies that union of A n's n equal to 1 to infinity, their complements is contained in A n naught complement and this is countable. So, this set, its complement is subset of a countable set, so that implies union A n, n equal to 1 to infinity is countable and hence, so implying that this belongs to the class S. So, this is a set whose complement belongs to, so this is a set which is, whose complement is countable, so that means this set must belong to S. So, this is contained in a, and this is countable that means this set is countable and hence, because the complement of this set is countable, so this belongs to S. So, what we have shown is, let us look at the set collection S of all subsets of A, that A or A complement is countable. Then, this collection is a sigma algebra of subsets of X and let us observe whether we have not used anywhere the fact that X, the underlying set is a countable set, this is true for any actually. So, what we have shown is that if X is any set and let us take the collection of all those subsets of X, which are either countable or their complements are countable, then that form the sigma algebra of subsets of X. Let us observe one thing that we have not used anywhere the fact that the set X is uncountable. So, this property even remains true when X is any set, of course the collection S still remains a sigma algebra, but its nature will change in the sense that for example, if X is a countable set, you can try to prove yourself, then this collection of those subsets of X, we say that A or A complement is countable, in fact that will be all subsets of the set X. So, it is a non-trivial example only when X is a uncountable set. So, we have given example of a set X of a collection S of subsets of a set X and which is a sigma algebra. So, next let us ask the question given a collection C of subsets of a set X, it may not be an algebra or it may not be a sigma algebra. So, the question arises, can we say that we can find a sigma algebra of subsets of the set X which includes this collection C. So, in some sense this collection C may not be closed under complements or may not be closed under taking countable unions. So, we would like to enlarge it, so that it becomes a sigma algebra. So, obvious examples are, if you take all subsets of the set X, then that itself is a sigma algebra and that includes C, but that is a very trivial example, trivial example of a sigma algebra which includes C. So, we would like to modify our question that given a collection C of subsets of a set X, does there exist a sigma algebra of subsets of X which includes C and is the smallest. So, let us and the answer is yes and it is something similar to what we have done in case of algebras. So, let X be any set and C be any class of subsets of set X, let S of C denote the intersection of all the algebras S of subsets of X which includes C. Then the claim is that this collection S of C is a sigma algebra of subsets of X, it includes the class C and it is the smallest. So, that will show that given a collection C of subsets of a set X, you can always find a sigma algebra of subsets of X which includes C and which is the smallest. So, let us check these three properties one by one. So, S of C is defined as the intersection of all the algebras S say that S algebra and S includes C. So, the first property we want to check that the empty set and the whole space belong to S of C. So, that is obvious from the fact that empty set and the whole space will belong to every algebra S which includes C. Because it is S is an algebra, so empty set and the whole space belong to it, every element S in this collection whose intersection we are taking. So, the intersection also will have that property. So, that is the obvious property like observed in the case of algebra generated. The second thing let us take a set A which belongs to S of C. So, that implies that A belongs to S every, sorry, this is a sigma algebra. So, we are looking at the case of sigma algebras. So, S is that we are taking the intersection of all sigma algebras which includes C. So, if A is inside the class S of C, then A belongs to S and S is a sigma algebra that implies that A complement belongs to S for every S and that implies that A complement belongs to the intersection of all this collection. So, hence A complement belongs to which is nothing but S of C. And similarly, let us take the collection, the third property. Let us take a sequence A n which belongs to S of C. So, that implies A n for every n. So, that implies A n belongs to S for every S and that implies because this is a sigma algebra union of A n's n equal to 1 to infinity also belongs to S and that implies for every S and that implies that union n equal to 1 to infinity A n also belongs to the intersection of all this algebras S over S and that is nothing but S of C. So, what we have shown is that if you look at S of C, the intersection of all sigma algebras which includes C, then they themselves form a sigma algebra. So, the out here shown is S of C is a sigma algebra that C is contained in S of C is once again obvious because S of C is the intersection of all sigma algebras which includes C. So, the intersection also will include. So, that also is an obvious property. And why it is the smallest? The smallest property also is true once again by the very fact that S of C is the intersection of all the algebras all the sigma algebras which include C. S of C being the intersection is the smallest anyway. So, that proves the fact that S of C is a sigma algebra of subsets of X, C is inside S of C and if S is any other algebra of subsets which includes C, then S of C must include must be inside S because S of C is the intersection of all. So, intersection is inside every element. So, what we have shown is given a collection of subsets of a set X, C a collection of subsets it may not be an algebra, but we can put it inside a sigma algebra of subsets of X denoted by S of C and such a thing exists because of this construction. So, such a thing is called the sigma algebra generated by the class C. So, S of C we are going to call it as the sigma algebra generated by the class. Let us look at some examples of sigma algebras generated. So, let us look at X the collection of all, actually X is an only non-empty set and let us look at all singleton subsets of this set X. So, let us call that collection as C. So, C is the collection of all singletons where singletons are elements of the set X. So, the claim we want to find out what is the sigma algebra generated by it. If we take a sequence of elements say X 1, X 2, X 3, X n in X and look at those singletons then their union is going to be an element in C. So, that means all countable sets must be elements of C and similarly all sets whose complements are countable also must be elements of the set E. So, as a consequence we expect that this answer is nothing but the algebra. So, f of C the sigma algebra generated by. So, this is not correct. So, what we should have is that the sigma algebra generated by C. So, this is nothing. So, this is a correction here that then the sigma algebra generated by this must be equal to this collection f of C where either E or E complement is countable. So, let us prove this fact. So, X is any set we are taking S of C to be equal to all those subsets A contained in X say that A or A complement is countable. So, this is what we want to prove. So, let us call this collection as S. So, sets whose which are countable or their complements are countable that collection is called S and the claim is S of C the sigma algebra generated by the singletons is nothing but this collection. So, let us observe one of the first observation that S is a sigma algebra that we have just now proved this collection S is a sigma algebra. Second that C is inside S because what are the elements of C they look like singletons and singleton is countable. So, this belongs to S. So, because of this reason C is a subset of X. So, this is a sigma algebra which includes and third we want to show it is a smallest S is smallest. So, let us take let us take any other algebra. So, let us look at let F be any sigma algebra such that C is a subset of such that C is a subset of F. So, what we want to show we have to show. So, to show that F includes this collection S. So, let us take let A belong to S. So, either two cases arise either A is countable. So, in that case A can be written as X 1, X 2 and so on. So, that can be written as a union of singletons X i i equal to 1 to infinity and each X i is an element in C. So, and C is and C is inside F and C is inside F. So, every singleton belongs to F. So, and F is algebra. So, this is sigma algebra. So, that means this implies this belongs to F. So, if A is countable then it belongs to F. If not what is the second possibility that A complement is countable then by the same argument this will imply that A complement belongs to F and F is a sigma algebra. So, this implies that A belongs to F. So, in either case we have shown that if F is any sigma algebra which includes C then this F must include S. So, that proves the fact that S of C the sigma algebra generated by the singletons is nothing but all those sets say that either the set is countable or its complement is countable. So, we are able to give a description. So, as a consequence we are able to give a description of the sigma algebra generated by a collection of subsets in this case when the collection C consists of singleton sets. But let us be very careful in general for a set X given a collection C of subsets of X it is not always possible to describe the elements of S of C explicitly in terms of elements of C it is it is not possible always. And remember that was also the case when we looked at the algebra generated by a collection of subsets C when only when C was a semi algebra we were able to describe what is the algebra generated by the semi algebra. We were not able to give a general description of the algebra generated by a class C. Similarly, it is not possible to give a description of the sigma algebra generated by a collection of subsets of a set X. But these are the collection these are the collection of sets such are the collection of sets which are going to play a role in our subject later on. So, we have to study them carefully in detail. Let us look at some more properties of such kind of objects. Let us look at something called a topological space. I hope some of you are aware of what is a topological space. A topological space consists of a set X and a collection of subsets of X which is called tau is called topology. It is a collection of open sets in X and this is a collection which has some properties namely the empty set the whole space belong to it. And if two sets E 1 and E 2 belong to F then that implies E 1 intersection E 2 belongs to F. And if E alpha is a collection of sets in F that implies union of E alphas belong to F. So, a topology is a collection of subsets of a set X such that the empty set belongs to it it is closed under finite intersections and closed under arbitrary unions and such collection of sets are called open sets. So, obviously if tau if is a topology it need not be an algebra or a sigma algebra. And the reason is obviously because this collection need not be closed under complements which is required for a algebra or a sigma algebra. So, there it lacks actually in a topological space the sets whose complements are open are called closed sets. So, if you want the complement also to be in that collection then those are the sets which are both open and closed and they are not many examples of such things. So, let us look at a topological space X F. F need not be is a topology on X and this need not be a sigma algebra. So, the question is can we generate the sigma algebra given by this topology and the answer is yes. So, let us look at the collection of all open sets in this topological space and let us look at the collection of all closed subsets in the topological space. So, you can generate the sigma algebra given by all closed open sets and we can also generate a sigma algebra by the all closed subsets of the topological space. So, question arises is there any relation between these two sigma algebras and let us recall a set is closed if and only if its complement is open. So, using this fact we will prove that the collection of all the sigma algebra generated by open sets is equal to the sigma algebra generated by closed sets. .. So, u is open sets, c is closed sets and the claim is the sigma algebra generated by open sets is same as the sigma algebra generated by the closed sets and to prove this we will follow a technique which is going to be used again and again. So, let us observe. Let us take a set E which belongs to U that means that is same as saying that E is open and that implies which is same as equivalent to saying E complement is closed and that implies that E complement is in the collection C which is inside S of C. So, what we have shown is if E is an open set then E complement belongs to S of C and as a consequence of this E belongs to S of C because S of C is a sigma algebra. So, this implies that all open sets are inside the sigma algebra generated by all closed sets. Now, this is a sigma algebra which includes open sets. So, this sigma algebra must include the smallest sigma algebra including containing U that means once U is inside the sigma algebra S of C the sigma algebra generated by it also must come inside S of C by the very definition. So, this implies that S of U the sigma algebra generated by open sets comes inside S of C. So, this is a technique which is used very up. So, to prove S of U is inside S of C what we have done is we have shown that U is inside S of C and hence S of U is inside S of C. So, this technique is going to be used very often in our course of lectures. We want to show certain collection of sets as required property. So, we show that that collection of sets includes a set of generators and hence will include the sigma algebra generated by it provided that collection forms a sigma algebra. So, by this technique we have shown S of U is inside S of C and by same technique we can show that S of C is also inside S of U because if I take a set A which is in C that means A is closed that means A complement is open. So, it belongs to U which is in S of U. So, that will imply that A belongs to S of U because that is a sigma algebra. So, hence this implies that the sigma algebra generated by C must come inside S of. So, the collection S of C is same as S of U. So, this is a very important collection of subsets of a topological space and this is given a name such this collection is called the Borel subsets of the set X. So, the sigma algebra generated by all open sets or by all closed subsets of a set X is called the Borel sigma algebra of subsets of the set X. So, let us observe a few more things. So, we recall we started with a collection C of subsets of a set X and now we can generate we said that given a collection of subsets of a set X we can generate a algebra out of it which is something algebra has a property which the class A may not C may not have. Now, given this collection of C we can also generate a sigma algebra out of it and we can also generate the sigma algebra by the algebra generated by that class. So, the question comes what is the relation between these three things. So, the observation we are going to prove is that given a collection C you can directly generate the sigma algebra by this collection or you can generate first the algebra and then the sigma algebra by that algebra both processes are same. So, let us give a proof of this obvious fact because such kind of proofs are going to be useful. So, let us look at the proof that if C is any collection of subsets of a set X and I take the collection C I generate the algebra by this collection and then generate the sigma algebra by this collection that is same as the sigma algebra generated by C. So, this is what we want to prove. Let us observe. So, note C is contained in A of C and which is contained in S of A of C by very definition. So, that implies C is the collection which is contained in the sigma algebra. So, that means the sigma algebra generated by C is contained in the sigma algebra generated by the algebra generated by C. So, that proves one way to prove the other way around what we have to show. So, let us observe that C is contained in S of C also C is contained in S of C and S of C is a sigma algebra. So, hence it is also an algebra. So, this implies that C A of C must be inside S of C. So, here we have used the fact every sigma algebra is also an algebra. So, this is an algebra including C. So, the smallest one must be inside it and now this collection is inside S of C. This is a sigma algebra. So, that means the smallest one. So, the sigma algebra generated by A of C must come inside S of C. So, that proves the other way around inequality and this is already proved. So, these two together imply that the sigma algebra generated by any collection C is same as you can first generate the algebra and then generate the sigma algebra. It does not matter both are same and the proof illustrates the use of this technique again and again. If C is inside something and that something is algebra, then algebra generated comes inside and so on. So, these are going to be techniques which are going to be used again and again in our course of lectures. So, this is one observation that the sigma algebra generated by any collection C is also the sigma algebra generated by the algebra generated by that collection. Let us observe another thing, another way of generating more examples of sigma algebras. If we take any collection C and take a subset Y of it, then that gives us C intersection Y is the collection of subsets of Y which are inside C and then we want to show that the sigma algebra generated by this. So, we are restricting C to Y and then generating the sigma algebra by it and the claim is it is same as the sigma algebra generated by C first and then restricting it to Y. So, let us give a proof of this fact. So, let us start with, so X is any set and C is a collection of subsets of P of X. So, and Y is a subset of X. So, we want to show that the sigma algebra generated by C intersection Y is equal to the sigma algebra generated by C restricted to Y. So, this is what we want to prove. So, let us observe, so note C is contained in S of C by the very definition C is inside. So, if I look at C intersection Y, so look at the intersection of these sets that is going to be inside S of C intersection Y. And if I can show that this is a sigma algebra, if I can show this is a sigma algebra, then I would have that S of C intersection Y will be inside S of C intersection Y. So, one should try to show that this is a sigma algebra. So, let us try to show that that is a sigma algebra. So, does empty set belong to S of C intersection Y? Yes, because empty set can be written as empty set intersection Y. And empty set belongs to S of C. Similarly, the whole space, so what is the whole space? The whole space here is Y. We are S of C intersection Y. So, claim is that this belongs to S of C intersection Y. That is true, because Y can be written as the X intersection Y and X is in the sigma algebra S of C. So, both these things belong. So, second thing let us look at is at E belonging to S of C intersection Y. We want to look at the complement of this, but what is the complement of this? If this set belongs to S of C intersection Y, that means this set E must be equal to some element. So, some A intersection Y, where A belongs to S of C, that is by the definition. So, what is E complement? But E complement in keep in mind, we are looking at subsets of Y. So, what is E complement in Y? That is same as E complement in Y, that means E intersection Y and that is same as A complement intersection Y. So, and that again belongs to S of C intersection Y. So, this collection S of C intersection Y is closed under complements. Finally, let us show it is closed under countable unions. So, if E ns belong to S of C intersection Y, let us assume E n because it belongs here. So, it will be some A n intersection Y, where A ns belong to S of C. So, that will imply union A ns is union A n intersection Y and this belongs to S of C. So, it is intersection Y. So, it belongs to. So, that implies that union also is an element. If E ns belong to this, then union also is an element of this. So, we have proved that this collection is a sigma algebra. This collection is inside this. So, this will prove one way inequality. We have to prove the other way around inequality, namely. So, let us try to prove the other way around inequality, namely. So, claim that S of C intersection Y is also a subset of S of C intersection Y. So, this is what is to be proved. The proof once again uses a similar technique. .. Let us write A. Look at all those subsets E such that E intersection Y is an element in S of C intersection Y. So, look at all those subsets in X such that their intersection with Y is inside this sigma algebra. So, one shows A is a sigma algebra and C is inside A. That means what? So, this will imply because C is inside A, it is a sigma algebra that will imply that S of C is inside A. That means for all elements in S of C, if I take as an intersection that is going to be inside it, so that will prove this required inequality. So, this claim is equivalent to proving these two things, namely A is a sigma algebra and C is inside A. So, let us observe why it is a sigma algebra? That is once again by the similar properties, namely if E E. So, let us show that this is a sigma algebra because if we look at, so let us try to show that E is a sigma algebra. So, one does empty set belong to A? Yes, because empty set intersection Y is empty set which belongs to this because this is a sigma algebra. So, that is similarly the whole space belongs to A, that will be second property. Let us take a set E which belongs to A. So, what does that imply? E intersection Y belongs to S of C intersection Y, but this is a sigma algebra. So, it must be closed under complements and complements in Y. So, that means E complement intersection Y also belongs to S of C intersection Y and that implies that E complement belongs to A. So, the collection A is closed under complements and similar arguments will show that it is closed under unions also. So, E n belonging to A will imply E n intersection Y belongs to this sigma algebra and hence implies union of E n intersection Y also belongs to this sigma algebra and hence this set union E n belongs to A. So, this proves clearly that A is a sigma algebra and clearly C is inside A because E intersection Y in that case will belong to this. So, this will prove the required fact that what we were trying to show that if I take a set Y inside X and restrict the collection C to subsets of Y and generate the sigma algebra that sigma algebra is same as the first generate the sigma algebra and then restricted to Y. So, this is another way of generating more sigma algebras out of a given sigma algebra for a given collection. So, this is the kind of technique which we are going to use later on in our subject. For example, X will be the real line, we will look at Y will be an interval. So, we will look at the open sets in the whole space of real line, generate the sigma algebra that is the Borel sigma algebra of subsets of real line and then we can restrict them to the interval and it will say it is the same as looking at the open sets in the interval and generating the sigma algebra out of it. So, these are various ways of generating sigma algebras. So, what we have done today is the following. We started with looking at algebras and described a special property of the algebras namely we said that in an algebra any countable union can be represented as a countable disjoint union a very important aspect of an algebra. Then we moved on to looking at restricting the algebra to a set. So, we said if E is a subset of the set X and C is a collection then you can restrict the class C to E and generate the algebra out of it that is same as first generating the algebra and then restricting it to it. So, that is another way of restricting of algebras and then we define what is called a sigma algebra of subsets of it. It is a collection of sets which is closed under complements, includes empty set and the whole space and is closed under countable unions. In unlike algebra, algebra is closed under complements and finite unions only, finite unions. So, every sigma algebra is also an algebra and we give examples of an example of a collection of subsets of a set X which is an algebra, but which is not a sigma algebra. So, sigma algebra is something more stronger than being an algebra and then we finally, looked at how we one can generate a sigma algebra out of a given collection of sets and that is called the sigma algebra generated. The process is similar to the algebra generated like the intersection of all sigma algebras that include that collection and that will be the sigma algebra generated by it. The sigma algebra generated by the open subsets of a topological space is an important sigma algebra called the sigma algebra Borel subsets of that topological space and that is same as the sigma algebra generated by all closed subsets of the topological space and that is an important thing. And finally, we proved that you can take any collection, restrict it to a set and generate the sigma algebra that is same as generating the sigma algebra first and then restricting it to the set. So, next time in the next lecture, we will continue this study of classes of subsets of a set X and we will look at some another important class called the monotone class of subsets of a set X. Thank you.