 Welcome to today's lecture on measure and integration. This is the second lecture and in this lecture, we are going to cover the following topics. Algebra generated by a class, then we will look at algebra generated by a semi-algebra. We will also look at what is called the sigma algebra of subsets of x and sigma algebra generated by a class of subsets of a set x. Let us just recall what we called as the semi-algebra of subsets of a set x. It was the collection, so let me just recall, a semi-algebra c is a semi-algebra of subsets of a set x. If at the following properties, empty set and the whole space belong to it. And secondly, for a and b belonging to c implies that the intersection of these two sets is also an element of c. That is the class c is closed under intersections. And third property was that if a belongs to c, then this implies that a complement can be represented as a finite disjoint union of elements of the class c. That is, a complement is a union of elements c i, where each c i belongs to c and c i intersection c j is empty for i not equal to j. Such a class was called a semi-algebra of subsets of x. Then we looked at what we called as the algebra of subsets of a set x. So, a class f of subsets of a set x is called an algebra. If it had the following properties, namely empty set and the whole space belong to f, as in the case of semi-algebra. Secondly, a and b belonging to f imply that a intersection b also belongs to f, as was the case for the semi-algebra. And third property, which is different from the semi-algebra, which is a bit stronger than the semi-algebra is, whenever a set a belongs to f, this should imply that a complement also belongs to f. So, an algebra of subsets of a set x is a collection of subsets of x, which includes the empty set and the whole space. It is closed under intersections and it is also closed under complements. And last time, we looked at some examples of algebras and semi-algebras. So, today what we are going to look at is, the question is given a collection c of subsets of a set x, does there exist an algebra, which includes c. So, the given collection c may not be an algebra of subsets of x, it is an arbitrary collection. And we would like to know, can we find a collection of subsets of x, which includes this collection and is an algebra. Of course, there is one obvious answer, namely the power set. For example, the power set of x is always an algebra, because it is collection of all subsets of x. So, it has all the properties, namely it is closed under intersection and complements and includes the empty set in the whole space. And obviously, c is a subset of px. So, in some sense px is the largest algebra of subsets of x, which includes any collection c. So, we should modify our question, namely given a collection c of subsets of a set x, does there exist an algebra, which includes c and is the smallest. So, to answer that question, let us look at the following. So, let us look at the collection all f such that f is a subset of px, it is a collection of subsets of x and such that f is an algebra and c is contained in f. c is the collection, which is given to the universe, which may not be an algebra. So, let us collect all those collections of subsets of fx. So, call them as f such that f is a collection of subsets of x, such that f is an algebra and includes c. So, it has two properties. One, c is a subset of f, this collection c is inside the collection f and f is an algebra. So, first of all let us observe that px is an element of this collection. So, let us call this collection as say i. So, this collection of all algebras, which includes c is a non-empty collection because the power set of x, the collection of all subsets of x is an algebra and includes c. So, this is non-empty. So, implies that this collection is not empty. Now, what we do? Let us define a to be equal to intersection of all this f such that f belongs to i. So, let us take all the collections in this collection, all the algebras, which are members of this collection i and take the intersections. So, keep in mind each f is a collection of subsets and it is an algebra and we are taking the intersection of these collections. So, the claim is that 1, c is a subset of this collection a, which is obvious because the collection i of all the algebras f has that property c is a subset of each member. So, this property is obvious. Secondly, we claim that a is an algebra. So, to prove that a is an algebra, let us observe what we have to do. So, what is a? a is the intersection of all algebras f, which are inside i. So, first we want to show empty set belongs to a. Why? Because we know empty set belongs to each f because each f is an algebra. Similarly, x belongs to a because of the same reason because x belongs to f and f is an algebra. Secondly, let us take a set e belonging to a. Then that will imply by the very definition that e belongs to f for every f inside the collection i. But that implies because e belongs to f and f is an algebra that implies that e complement also belongs to f for every f belonging to i. But then that implies it is equivalent to saying this implies that e complement belongs to the intersection of all these f in i and that is precisely our a. So, we have shown if e belongs to a, then e complement also belongs to a. So, the class a is closed under complements and let us finally show that it is closed under intersections also. So, let us take two sets e and f belonging to a. Then that implies that e and f both belong to each f, f belonging to i. But that implies that e intersection f belongs to the collection f because f is an algebra. So, that is crucial. So, that is being used again and again. e and f belong to the collection f which is an algebra. So, the intersection also belongs and hence e intersection f belongs to a. So, what we are showing is this collection a which is intersection of all the algebra which includes c is an algebra. So, we have shown the second property namely that this collection a of intersection of all the algebras which includes c is an algebra of subsets of the set x. Also, c is inside a and by the very nature because it is the intersection of all the algebras which include a, it should be obvious. So, obvious property namely a is the smallest algebra of subsets of x such that c is inside a. What we mean by that? That means that if, so this is same as saying if f is any algebra and c is inside f that implies that this f have to include a. So, what we have shown is that given any collection of subsets of a set x, if we define a by this a is the intersection of all the algebras including i then this such an algebra then this collection exists because there is at least one algebra which includes c namely the power set and it is a smallest. So, what we have shown is given any collection of subsets of a set x there is an algebra of subsets of x which is smallest and includes a. So, we have answered this question that given a collection c of subsets of a set x does there exist an algebra which includes c and is smallest. Yes, the answer is yes. So, let us state this as a theorem namely let x be any set and let c be any class of subsets of a set x. Let f of c denote the intersection of all the algebras a which include the collection c. So, f of c is the intersection of all the algebras that include the collection c then what we have just now shown is that c is a subset of x that means f of c includes c and f of c is an algebra of subsets of x. So, f of c is an algebra which includes x and it has additional property it is a smallest with that property that is if a is any other algebra of subsets of x is that a includes c then a must include f of c. So, this is not s of c it is f of c. So, f of c is called the smallest sigma algebra of subsets of x containing c and is called the algebra generated by the class c. So, what we have shown is that given any collection of subsets of a set x we can always find an algebra of subsets of x which is smallest and includes it. Let us look at some examples. So, let us look at the collection x is any non-empty set. Let us look at the collection of all singleton subsets of this collection x. So, c is the collection of all singleton sets where x belongs to x. We want to know what is the algebra generated by it and the claim is that the algebra generated by this c is nothing but all those sets e in x say that either e or e complement is finite. So, let us prove this how do you prove such kind of a assertion. So, we have got x any set and c is the collection of all singleton sets belonging to which are subsets of x and then we are looking at we claim. So, this is our claim that f of c is nothing but all sets a contained in x say that a or a complement is finite. .. So, let us observe. Does empty set belong to f of c? Yes, because by definition empty set is a finite set. Does x belong to f of c? So, for x to be an element of f of c either x should be finite which we do not know because it is an any set. It may or may not be finite, but we know it is complement which is empty set is finite and in f of c. So, by the second criteria a or a complement. So, x may or may not be finite but it is complement is empty set which is finite. So, this property is true. Let us check the second property namely if e belongs to f of c does this imply e complement belong to f of c? Is this true? Well for e complement to belong to f of c either e should be finite or e complement complement should be finite. So, this will be true if and only if e complement or e complement complement finite which is same as e complement or e complement complement is e finite which is same as saying. So, this is if and only if this is if and only if this is if and only if e complement belongs to f of c. So, e belonging to f of c is true if and only if e complement because our definition of f of c is symmetric with respect to a and a complement. So, the collection f of c of all those sets for which a or a complement is finite includes empty set the whole space it is closed under complements. Let us show the third property namely that if e and f belong to f of c then that implies the intersection also belongs to f of c or equivalently this is equivalent to saying that e union f belong to f of c. Because we have observed that from union you can go to intersection by complements because the class is already closed under complements. So, let us but e and f belonging to f of c means e or e complement finite and f or f complement finite. So, various possibilities arise let us take case one both e and f are finite. In that case that will imply that e union f is finite because union of finite sets is a finite set and that will imply that e union f belongs to f of c cover collection. So, whenever e and f are finite there. So, what is the case two what is other possibility either e or f is not finite. So, let us for the sake of definiteness suppose e is not finite, but e belongs to the class f of c. So, that implies e belongs to f of c means e complement is finite. Now, e complement is finite let us look at now and e is contained in e union f, but that implies that e union f complement is contained in e complement and e complement is finite. So, that implies so is finite. So, that implies e union f complement is finite and hence by definition this implies e union f belongs to f of c. So, we have shown in either case hence whenever two sets e and f belong to f of c that implies e union f belongs to f of c. So, thus we have shown that the collection so f of c is an algebra. So, f of c is an algebra added. So, f of c which was defined as all sets a contained in x such that a or a complement finite is an algebra and c is contained in f of c. Is it clear why c is contained in f of c? Because for every x belonging to x the singleton x is finite. So, implying the singleton x belongs to f of c. So, implying that c is a subset of f of c. So, thus f of c is an algebra and includes f of c claim finally that f of c is smallest. So, that is let a be any algebra such that c is inside a then that should imply so that is the question that should imply that a includes f of c. So, let us show how is that true. So, to prove this let a belong to f of c. So, two possibilities one a is finite. So, let us write a as x 1, x 2, some x n. That means this is equal to union of singletons x 1, x i, i equal to 1 to n and each singleton x i is an element of class c and c is inside f of c is inside a, c is given to be inside a. So, this implies that implies that a belongs to c. So, let me just go through again because a is finite, a is written as x 1, x 2, x n. So, I can write as a finite union each x i belongs to c. So, each x i belongs to f of c. So, this implies here is because each x i belongs to c which is subset of f of c. So, each one is in f of c, f of c is an algebra. So, a belongs to f of c. This is not what we wanted to prove. So, this is not true. This is of course I am proving whatever was required. So, let a be finite. So, this is we want to show that a is inside the class a. So, each x i belong to c, c is contained in a. That means each x i belongs to a and a is an algebra. So, that implies that the union, so a belongs to a. Same argument basically using that c is inside a and a is an algebra. So, that will prove that a is if a belongs to f, then this implies that a belongs to a. So, that proves that a is a subset, a a always includes f of c. So, hence what we have shown is the following that if c is the collection of all singleton sets x belonging to x, then the algebra generated by it is nothing but all subsets a in x say that a or a complement is finite. So, this is the algebra generated by c. So, we have computed, we have described the algebra generated by a collection c of singletons of a set x. In general, it may not be possible to give a description of the algebra generated by a collection of subsets of a set x. In a special case, when the starting collection c is at least and semi-algebra of subsets of x, it is possible to describe the algebra generated by it and that is our next theorem. So, we want to describe the algebra generated by a semi-algebra. The theorem states the following. Let c be any semi-algebra of subsets of a set x, then f of c the algebra generated by c is given by the following collection. So, it is the collection of all subsets e in x such that e can be written as a union of sets finite number of sets c i's i say 1 to n with each one of the elements c i is in the collection in the semi-algebra c and these c i's are pair wise disjoint. So, what we are claiming is that the algebra generated by a semi-algebra is nothing but the finite disjoint union of elements of c. We have seen an illustration of this in the previous lecture when we described c was the collection of all intervals and we took f of c the finite disjoint union of intervals and showed that that was an algebra. So, that is a typical model example or illustration of describing the algebra generated by a semi-algebra. Let us prove this that the algebra generated by the semi-algebra is nothing but as described above. .. So, given c a semi-algebra f of c the algebra generated by c is equal to all sets such that I can write e as a finite disjoint union of elements belonging c i's belonging to c and note this square bracket square union means that the sets involved are pair wise disjoint. So, let us observe first a few things first of all c is a subset of f of c that should be obvious because every set in c is union of itself single. So, e is has representation as union of itself. So, only one set is involved so e is a subset of and that implies that the empty set and the whole space belong to f of c. So, as a consequence it implies the first property required for f of c to be an algebra and generated we will see it later. So, second thing we are trying to show that f of c is an algebra. So, first property we have checked let us take a set e belonging to f of c that means that implies that e can be written as a finite disjoint union c i equal to 1 to n where c i's belong to f of c. But recall saying that c i's belong to c sorry recall saying that c i's belong to c and c is a semi algebra this is important implies by the property of the semi algebra that each c i can be written as c i complement can be written as a disjoint union of elements of c again. So, let us write it as a i j j equal to 1 to some k i where a i j's belong to c again and they are disjoint. So, this together with the representation for this implies. So, look at e is a union of c i's each c i complement is equal to this. So, that means first of all e complement can be written as intersection of i equal to 1 to n of c i complement and each c i complement is a finite disjoint union. So, this is i equal to 1 to n of finite disjoint union j equal to 1 to k i of a i j. So, here we have used the property namely that e complement by demarcola's is intersection of c i's because e is union of c i's complement and each c i complement being an element of the semi algebra can be written as a disjoint union of elements a j i. So, that is the representation. Now, we use the fact that intersection distributes over union. So, what kind of this set is this can be written as a union of finite number of sets which will involve a i j. So, sets of the following type intersection a k l because when I distribute it over I will be getting one set a i j and some other set a k l and this intersection of that and unions of those and these unions. So, these sets belong each one of them belong to c. Why does it belong to c? Because the collection c is a semi algebra both of them are elements of the semi algebra. So, this intersection is element of the semi algebra and this sets are pair wise disjoint because if i k j and l if any one of the pairs is different then those sets will be disjoint. So, I can say this is a disjoint union of sets of the following type where i belong between one and k i k between one and some k some index and j and l between one and l. So, basically we are saying is e complement can be represented again as a finite disjoint union of elements of c again and that is following because of the fact that e complement which is a union of sets is a intersection and that is intersection of distributive property and this. So, it is because the index is involved are so many it is difficult to write all these things, but it should be clear that e complement is a finite disjointed union of elements of c. So, that means e belonging to f of c implies e complement belong to f of c and let us look at the third property namely if e and f belong to f of c does this imply that e intersection f belong to f of c. That should be obvious because e belonging to f of c means e is a disjoint union of some a i's 1 to n, a i's belonging to c, f is a union of some j's 1 to m of some sets b j's j equal to 1 to m where each b j belong to c. So, that implies so let us conclude from here. So, e intersection f will be equal to this union a i's intersection of union b j's i equal to 1 to n j equal to 1 to m and once again using the distributive property we got this intersection of a i's intersection b j i n j. So, the intersection and each one of them belongs to c, this is union over i and j. So, this by the distributive property will be union over i and j of a i intersection b j and these elements are elements of c because c is a semi algebra a i belong to c b j belong to c and these are again disjoint because they are a i and b j. For disjoint pairs, for different pairs they will be disjoint by the property that these unions are disjoint. So, implies that e intersection f also belongs to f of c. So, what we have shown is the following namely if we take, if I look at c is any and is algebra if I look at this collection of then subsets which are finite disjoint unions then this includes c and this is an algebra. To show that it is algebra generated we have only to show that if a is an algebra and a includes c that should imply a includes f of c. To prove that is obvious because let us take let e belong to f of c but that will imply that e is a disjoint union of elements a i where a i belong to c and c is a subset of a. So, c is inside a so that implies that e is a union of some elements in a and a is a algebra that means a belongs to a e belongs to a. So, hence f of c is a subset of a. So, what we have shown is that this collection f of c of finite disjoint unions of elements of the same algebra is indeed the algebra generated by the semi-algebra. So, we are able to describe completely the algebra generated by a semi-algebra and note one thing we have described the algebra generated by a semi-algebra explicitly and as remarked earlier that the algebra generated by a semi-algebra is described but in general the algebra generated by any collection of subsets we cannot describe it explicitly one may not be able to say what are the elements of that algebra generated by collection of subsets. So, that is not possible always it is only in the case when it is a semi-algebra c we start with we are able to describe the algebra generated by it. So, we have looked at a semi-algebra we have looked at an algebra of collection of subsets of x and then we have looked at the algebra generated by a semi-algebra of subsets of x. Next we go to the next level of collection of subsets which are slightly stronger and that is called the semi-sigma algebra of subsets of x. But before that probably let us look at another property namely how do we generate more algebras or sigma algebras out of a given collection. So, let us take c any collection of subsets of a set x and let e be a fixed set in x. Let us write c intersection e to be the collection of sets of the type c intersection e c belonging to c. Note that c intersection e c is a collection of sets and e is a set. So, c intersection e is just a notation for all sets of the type c intersection e and note these are all subsets of the given set x. So, the claim is that given any collection c and given any set e let us look at c intersection e and generate the algebra by this collection of subsets of. So, f of c intersection e is same as the algebra generated by c intersected with e. So, this is a very useful thing of restricting in some sense restricting the sigma algebras because restricting the algebras. So, what we are saying is c any collection of subsets of a set x e contained in x fixed define c intersection e to be c intersection e where c belongs to c. So, what are these sets? So, note that it is c intersection e. So, this is a subset of, so this is a collection of subsets of the set e and now I can look at f of c intersection e. So, what will be f of c intersection e? It is the algebra of subsets of e generated by c intersection e generated by the collection c intersection e and the claim is that f of c intersection e is same as you first generate the algebra by c. So, this will be the algebra generated by the collection of subsets in c. This is subsets of x. So, this is algebra of subsets of x take its restrictions to e and that is same as this. So, this is the theorem or this is the result that we say is true. So, let us prove this. So, to prove this let us first observe note, so first of all we observe that c is contained in f of c. Given any collection c of subsets of a set x, f of c is algebra generated by c. So, by the very definition c is a subset of f of c. So, that implies that c intersection e is a subset of f of c intersection e. Is that clear? So, if I take a set in c intersection e that is going to look like here c intersection e and c belongs to e. So, that is also an element in f of c. So, it is f of c intersection e. So, this is, so let me write because if a belongs to c intersection e implies a intersection e belongs to c which is in f of c. So, and a belongs to and a is an element in c. So, c intersection e. So, a belongs that means a belongs to c. So, this is in c. So, that is in f of c. So, implies this is an element of f of c intersected with a e and that is precisely the meaning of this. So, we have got c intersection e because of this. This is true and now let us observe that f of c intersection e is an algebra of subsets of e. This is an algebra of subsets of e. Obviously, empty set belongs to this because empty set can be written as empty set intersection e. So, that belongs to f of c intersection e and second. Now, note these are subsets of e. So, what is the whole space? That is e. So, e is equal to e intersection e. So, that is x. So, that is x intersection e and that belongs to f of c intersection e. So, I can write e as x intersection e, x in element of f of c and e is here. So, that belongs to f of c. So, the empty set and the whole space that is e, they belong to this collection. What is the second observation? So, let us look at the second observation from here. Let us take two sets. So, let us take two sets a and b belonging to f of c intersection e. That means this a is equal to… So, if a belongs to that means c intersection e, that means a can be written as some set, let us write g intersection e where g belongs to g belongs to looking at f of c and b is also in this. So, b is written as some h intersection e where h belongs to f of c. So, that implies a intersection b is written as g intersection h intersection e. Now, g and h both belong to f of c. So, this belongs to f of c and that is in a. So, implies a intersection b belongs to f of c intersection of e. So, and finally, let us look at the third. If a belongs to f of c intersection e, that means a can be written as some g intersection e where g belongs to f of c. So, now let us look at a complement, but keep in mind the a complement is in e. So, this is nothing but g complement intersection e. See, this complement is in e. We are taking the complement inside set e because we are looking at the collection of subsets of e. So, this belongs to f of c intersection e. So, what we have shown is that c intersection e is subset of this and this collection is an algebra. So, that implies the property. So, this implies, so let us observe. So, this implies that the algebra generated by c intersection e is a subset of f of c intersection e because this collection is inside this and this is an algebra. So, the algebra generated being the smallest must come inside. So, we get this property. Now, we have to prove the other way in equality. So, claim that f of c intersection e is a subset of f of c intersection e. So, this is what we have to prove. So, to prove this, let us define a collection a. To be all sets, a contained in x such that a intersection e belongs to f of c intersection e. Let us look at this collection a. Note, c is contained in a. That is obvious because if I take an element in c, then that is at a intersection e belongs to c intersection e which is inside this because a belonging to c implies a intersection e belongs to c intersection e which is inside f of c intersection e. So, c is inside a. So, if a is an algebra, what will this imply? c is inside a, a is an algebra that will imply f of c is inside a. But what is the meaning of f of c is inside a? That means for every element f belonging to f of c, if we look at f intersection e, that is belonging to f of c intersection e. So, that means that f of c intersection e is f of c intersection e is a subset of f of c intersection e. So, to complete the proof, we only have to show that a is an algebra of subsets of x. Now, let us look at this and show this is an algebra of subsets of this at x. So, let us start observing. So, we want to show this is an algebra of subsets of a set x. So, to show that, let us observe one. Empty set belongs to a because empty set intersection e is empty set which belongs to f of c intersection e. The whole space x belongs to a because the whole space intersection with e which is e that belongs to f of c intersection e. Keep in mind this is the algebra of subsets of e generated by this collection. Secondly, let us take a set e belonging to a that implies, let us take a set g belonging to a that means g intersection e belong to f of c intersection e and this is an algebra of subsets of e. So, this will imply that g complement intersection e also belongs to f of c intersection e because it is an algebra of subsets of e. So, its complement should also be inside that and that implies that g complement belongs to a. So, g belonging to a implies g complement belongs to a. Finally, let us conclude that if g and h belong to a that imply that g intersection e and h intersection e belong to f of c intersection e and that implies g intersection h intersection e belongs to f of c intersection e. Once again by the fact that this is an algebra f of c intersection e. Once again the fact this is an algebra. So, the collection a of subsets of x is an algebra includes c. So, that must include f of c and that proves the theorem that the algebra generated by a collection c that is f of c when restricted to a the set e is same as first restrict the collection by which you are generating and then restrict and generate. So, what we are saying is given a collection c of subsets of x if we restrict the class c to the subsets of e and then generate the algebra of subsets of e that is same as first generating the algebra and then restricting it that class of sets to that of e. This is going to be very useful later on when we want to restrict collection of sets to subsets of it. .. So, let us just conclude what we have done today. We started with the recalling what is an algebra what is a semi algebra what is a algebra and then we started by observing that in general if a collection c of subsets of a set x is not an algebra we can always generate an algebra out of it. That means we can show the existence of a smallest algebra of subsets of the set x which includes this collection c and basically that is the proof is by showing that if I take the intersection of all the algebras which includes c then that is the smallest one and that also had the important property namely observation namely intersections of algebras is again an algebra. So, that was a crucial property a crucial observation that helped us to prove that the intersection of all the algebras that includes c are also is also a algebra and that includes because the intersection so it has to be smallest and hence that is the algebra generated by. .. And then we gave examples of how to find algebra generated by a collection for example if you take singleton sets of any collection of for any set x then the algebra generated is the collection of all sets which are either finite or the complements are finite and in general if c is a semi algebra then the algebra generated by it is nothing but the collection of all finite this joint unions of elements of that semi algebra. So, thank you so we will continue next time.