 Welcome to lecture 4 on measure and integration. I recall, we have been looking at classes of subsets of a set X with various properties. We will start with the collection called as semi-algebra of subsets of a set X. Then, we looked at what is called the algebra of subsets of a set X. Today, we will start with looking at some more classes of subsets of a set X. We will start with what is called a monotone class. Then, we will look at the monotone class generated by a collection of subsets of a set X. Then, we will go over to describe what is the monotone class generated by an algebra. That is an important relation, which we will be using again and again. Let us start with describing what is a monotone class. A monotone class is a collection of subsets of a set X. Let us denote that collection of subsets by M. M is a collection of subsets of X. We say it is a monotone class, if it has the following two properties 1. Whenever there is a sequence of sets A n belonging to M, the collection M says that the sequence is increasing. That means for every n, A n is a subset of A n plus 1. Then, we demand that the union of these sets A n's also belongs to M. The first property is that the collection M of subsets of set X is closed under unions of increasing sequences. The second property that we expect from this collection is that whenever a sequence A n is in M and A n is decreasing, that means A n is a subset of A n plus 1 for every n, then the intersection of this sequence of sets A n should also be an element of M. Let us recall once again what is a monotone class. A monotone class is a collection of subsets of a set X with the two properties 1. For every sequence of sets A n in M such that A n's is increasing, their union also belongs to M. Secondly, whenever A n is a sequence of sets in M such that A n is decreasing, then the intersection of these sets is also in M. That is why the name monotone comes. That means this class M of subsets of M is closed under monotone sequences. Whenever a sequence A n is increasing in M, the union belongs to M. Whenever a sequence A n is decreasing in M, then their intersection also belongs to M. Such a collection of subsets of a set X is called a monotone class. Let us look at some examples of such collections. Firstly, let us observe every sigma algebra is also a monotone class. Why is that true? Because a sigma algebra is a collection of subsets of X which is closed under any countable unions. Because it is closed under countable unions, it will also be closed under increasing unions. First property will be true. Secondly, if a sequence A n is decreasing in a sigma algebra, then look at the complements of that sequence. So, that sequence of complements of these sets will be an increasing sequence of sets. Because it is a sigma algebra, it is also closed under complements. So, A n complements will belong to it. So, union of A n complements belongs to it. That means the intersections of A n's complements belong to it and that means the intersections of A n's belong to it. Let us look at this property. How do we write it? Suppose M is a sigma algebra. Claim M is a monotone class. To prove this, how do we go ahead? So, one, let A n belong to M and A n's increase. So, that is A n is a subset of A n plus 1 for every n bigger than or equal to 1. Then, union of A n's n equal to 1 to infinity belong to M because M is a sigma algebra. Because it is a sigma algebra, it is closed under all unions and n's such types also. Secondly, let us take let A n belong to M and A n's decrease. So, that is A n includes A n plus 1 for every n bigger than or equal to 1. So, in that case, that implies because A n's belong to M and M is so A n complements belong to M for every n bigger than or equal to 1 because M is a sigma algebra. Because it is a sigma algebra, it is closed under a complements. So, once that is true, that implies that A n complements union n equal to 1 to infinity belongs to M. Because A n decreasing implies the sequence of A n complements will be increasing. Just now, we saw that whenever a sequence is increasing, their unions belong to M. But that implies, what is this set? That is intersection of n equal to 1 to infinity A n complements. That is by De Morgan's law. So, this belongs to M and now M is a sigma algebra. So, that implies that intersection n equal to 1 to infinity A n belongs to M. So, we have shown whenever sequence A n is in M and A n's are decreasing, that implies the intersection also belongs to M. So, hence M is a monotone class. So, the first proposition or first observation is that every sigma algebra is also a monotone class. Let us go over to some more properties, more examples. Let X be any uncountable set and let us look at the collection of all subsets A of X, say that A is a countable set. The claim is that this collection M is a monotone class, but it is not a sigma algebra. So, let us look at M. X is uncountable and we are looking at the collection M of all those subsets of X, such that A is countable. So, claim M is a monotone class. So, let us see how do we prove it. So, first property. So, let us take a sequence A n belonging to M, A n's increasing A n subset of A n plus 1 for every n. What we have to prove? We have to check that union of A n's n equal to 1 to infinity also belongs to M. That is what we have to check. But let us note, to check that the union belongs to M, we have to show it is a countable set. Now, each A n is given to be an element of M. That means each A n is countable, A n countable for every n implies union A n is also countable. So, here we are using the effect that the countable union of countable sets is again a countable set. Hence, this implies that union A n's n equal to 1 to infinity belongs to M. The first property we have checked is that if A n's belong to M and A n's is an increasing sequence, then the union A n belongs to M. Let us check the second property namely, let us check the second property that A n belonging to M. A n's decreasing for every n should imply that the intersection A n's belongs to M. What we have to check? We have to check that intersection A n, n equal to 1 to infinity belongs to M. That means it is countable. That is what we have to check. But let us observe that intersection A n's, this is a subset of A n for every n because it is intersection and each A n is countable, A n countable implies and this is a subset of it. So, intersection A n is countable and hence implies intersection A n belongs to M. We have shown that if you take the collection of, if X is a countable or uncountable set does not matter actually, so far what we have we have not used the fact it is a uncountable set. If M is a set which is, if M is the collection of all countable subsets of a set X, then that forms a monotone class. Why it is not a, so claim, so finally we want to prove that M is not a sigma algebra. Let us observe a few things here. So, note here we will be using X uncountable implies first of all X does not belong to, X does not belong to M. So, the very first property of a collection being a sigma algebra, namely the whole space belong to it is violated because X is not countable it is uncountable. Another way of looking at this is the following. So, this is one observation. Secondly, X uncountable implies there exists a subset A in X such that neither a nor a complement is countable and that is obvious because if let us see is that X is a uncountable set then the claim that there exists, this is not actually required and this may not be true. Let us take a subset A of X say that either A or A complement is not countable and that is possible because if for every set A and A complement are countable then X will be a countable set. So, let us choose a set A say that either A or A complement is not countable then suppose A is countable A is countable. So, that will imply that A belongs to M, but A complement is not countable and that implies A complement does not belong to M. So, when X is uncountable you can have in fact for every set which is countable A will belong to M, but A complement will not belong to M. So, this collection M is also not going to be closed under complements. So, X does not belong to it and that it is also not going to be closed under complements and that is because X is uncountable. So, when X is uncountable collection M of all countable subsets of it is a monotone class, but it is not a sigma algebra. So, every sigma algebra is a monotone class, but the converse need not be true. So, this is what we have shown just now that if M is a monotone class, every monotone class is a sigma algebra, but there exist examples of sigma algebra, examples of monotone classes which are not sigma algebra. Let us look at the next scenario. Let us start with a collection C of subsets of a set X. C is any collection it may or may not be a monotone class. So, in we would like to find a monotone class of subsets of X which includes C and is smallest. So, first of all let us observe that all subsets of the set X is a monotone class of subsets of X and C is because it is a subset is a sub collection. So, C is a sub collection. So, given any collection of subsets of a set X C it is always included in a collection namely the power set of X which is a monotone class. So, given a collection there always exists a monotone class of subsets of X including it, but this is too large we want to have the smallest monotone class including C whether such a thing exists or not. So, the proof is something similar to what we have shown is algebra generated by a class, the sigma algebra generated by a class. So, let us look at M of C. So, this is a notation for the intersection of all monotone classes M of subsets of X which includes C. So, look at the collection of all monotone classes of subsets of X which includes C and take their intersection and call this as M of C. So, what we want to prove is that M of C is a monotone class it includes C and it is the smallest. So, let us look at a proof of that. So, C is any collection of subsets of X, M of C is the intersection of all monotone classes M, M including C. Claim 1, C is inside M of C that is obvious because C is inside every collection M and M of C is the intersection. So, this property is obvious. Second, claim that M of C is a monotone class and the proof is goes on the same lines as that of algebra and semi-algebra, algebra and sigma algebra. To prove this, let us look at A ns belong to M of C, A ns decreasing. But this implies that each A n also belongs to M, also belongs to M for every collection M which includes C for every n and that implies that union of is decreasing. So, let us we want to show that intersection A ns n equal to 1 to infinity belong to M for every m and that implies that M is every m and hence that implies that intersection A ns belong to M of C. So, essentially saying that if A n is a sequence in M of C which is decreasing then it is also a sequence which is decreasing in each m and hence the intersection belongs to each m and hence belongs to the intersection M of C. And the same proof, similar proof works for the union. So, let us look at the second part A ns belong to M of C and A ns increasing but that means implies A ns belong to M for every m for every m and A n is increasing that means A ns union because each m is a monotone class. So, this belongs to M, the union belongs to M for every m and that implies that the union A ns n equal to 1 to infinity belongs to M of C and that implies. So, hence M of C is a monotone class. So, it is a monotone class that includes the collection C and third M of C is smallest such that C is M of C, smallest monotone class and that is obvious because it is the intersection of all monotone classes. So, obviously it is going to be the smallest. So, what we have shown is that given a collection C of subsets of a set X there exists a monotone class M of C which includes C and which is the smallest. So, this monotone class is called. So, this is what we have proved. So, thus M of C is the smallest monotone class of subsets of X such that C is the entire M of C and this collection is called the monotone class generated by C. So, every collection C has got the smallest monotone class in which includes it. So, that is called the monotone class generated by it. So, given a collection C of subsets of a set X we are able to generate an algebra out of it. We are able to generate a monotone class out of it. We are able to generate a sigma algebra out of it and the next question that we want to analyze is what is the relation between these collections. So, we want to prove a theorem which relates these concepts. So, first of all let us start with any collection of subsets of a set X. Then the first observation is if C is an algebra which is also a monotone class then C is a sigma algebra. So, let us first prove this fact that if C is C an algebra plus monotone class implies C is a sigma algebra. So, what we have to prove? To prove C is a sigma algebra, we have to prove first empty set the whole space belongs to C. That is true because C is an algebra. So, this property is true. Second, we should show that if a set A belongs to C implies A complement belongs to C and that is again obvious because it is the collection C is a algebra. So, these two properties are true. The third property is the only property to be checked that if A n belongs to C then that should imply union of A n is n equal to 1 to infinity also belongs to C. So, C is closed under countable unions that is what we have to prove and what we are given is C is a monotone class. So, a monotone class is a collection which is closed under only increasing and decreasing. So, let us look at union of A n's. Can we represent this as a union of increasing sets? Well, the one possibility is let us take union of A i's i equal to 1 to n then this collection as n increases will be an increasing sequence and the union n equal to 1 to infinity will be an increasing union. It will be a union of increasing sequence of sets which are namely union A i's and now observe that each one of these sets union A i i equal to 1 to n that is a finite union of elements in the algebra. A n's belong to C and that is an algebra. So, that means each one of them belong to C. So, we have written the union A n's as union of sets in C and this is an increasing sequence and C is a monotone class. So, that implies that this right hand side this set belongs to C. So, note we have used we have represented any union as a increasing as a union of increasing sequence of sets and each set here is a finite union of elements of the algebra C. So, hence this belongs to it. So, this becomes an increasing union of sets and hence because it is a monotone class this belongs to C. So, every A n belonging to C implies the union 1 to infinity also belongs to C. In fact, it is closed under countable unions. So, it becomes a sigma algebra. So, this proves the first property namely if C is an algebra which is also a monotone class then it is a sigma algebra. Let us look at the next property that if C is any collection then C is contained in M of C. That is obvious because M of C is the smallest monotone class including C. So, this is obvious and now C is also subset of S of C because S of C is the sigma algebra generated by it. So, C is inside S of C and just now we proved S of C is also a monotone class. So, S of C is a monotone class including C and hence the smallest one must come inside. So, that we prove this C is inside M of C is inside S of C. That means given any collection of subsets of x it is always included in the monotone class generated by it and the monotone class generated by it is also inside the sigma algebra generated by it. So, let me repeat these arguments once again. So, first of all C is contained in M of C that is because M of C is the smallest monotone class including C. Also, C is included in S of C because S of C is the smallest sigma algebra of subsets of x which includes C and this just now we proved is also a monotone class. So, this is a monotone class including C. So, that implies the smallest one must come inside it and the smallest one is M of C that comes inside S of C. So, what we have shown is for every collection C of subsets of x, the monotone class generated by it is a subset of the sigma algebra generated by it. We want to analyze the question when can we say S of C is also a subset of M of C. When is this true? That is same as saying that when is S of C the sigma algebra generated by a collection can I say it is equal to M of C. The answer is given by the next theorem which says that if C is an algebra then this is true. So, this is an important theorem called the sigma algebra monotone class theorem. It says if A is an algebra of subsets of a set X then the sigma algebra generated by it is same as the monotone class generated by it. Now, we have observed that M of, so the first part we have already observed that M of A is a subset of S of A. For that one need not have even A as algebra for any collection M of C is contained in S of C. So, in particular if A is an algebra then M of A is a subset of S of A. So, we have to prove the second the other way around inclusion namely M of A includes S of A when A is an algebra. To prove the other way around inclusion let us observe it is enough to prove that M of A is an algebra because why it is enough to prove this? Let us observe enough to show M of A is an algebra will imply because M of A is an algebra also a monotone class and just now we proved every algebra which is a monotone class will imply M of A is a sigma algebra because M of A is a monotone class and if we are able to show it is an algebra then it will be also a sigma algebra. Then we have A is inside M of A and now if M of A is a sigma algebra that even imply the smallest one must come inside it. So, S of A will be inside M of A. So, that will prove S of A is a subset of M of A and will be through. So, we have to only prove to show M of A is an algebra when A is an algebra. So, this is what we have to prove. So, let us start looking at a proof of this. So, first of all to prove that M of A is an algebra we should show that it is closed under complements. So, let us try to prove it is closed under complements. So, that means what to show it is closed under complements I have to show that for every subset in M of A its complement is also in M of A. So, this is a technique which we are going to use very often. So, let us collect together all the sets B which have the property that whenever B is the collection of all those subsets say that E complement belongs to M of A. So, to prove M of A is closed under complements what we have to show is that M of A is a subset of B. So, M of A is a subset of B we have to show that. So, let us try to prove that. So, claim let me repeat that we want to show that M of A is closed under the operation of complements that is A belonging to M of A should imply A complement belong to M of A. So, to show that consider all those subsets. So, let us consider the collection B of all those subsets E contained in X such that E complement belongs to M of A. So, to prove this this will be true. So, the required claim will be true. So, let us so this will be true if we can show M of A is contained in B. So, that is what we want to prove. We want to show because then for every set A belonging to M of A it will belong to B that means its complement will belong to M of A. So, this is what we have to show. Now, let us observe we are trying to show that M of A is inside a collection B and what is M of A? M of A is the smallest monotone class including A. So, suppose we are able to show that A is inside B and B is a monotone class then this claim will be true. So, for this enough to show. So, to prove this claim it is enough to show one that A is inside B and second we should show that M of and secondly B is a monotone class because once B is a monotone class including A the smallest one will come inside. So, let us try to prove these two facts that A is so proof of one that A is a subset of B. So, let set A belongs to A that implies A is algebra. So, that implies A complement belongs to A because A is algebra and note implies A complement belongs to A which is inside M of A because A is always inside M of A. So, what we have shown that if A belongs to A then its complement belongs to M of A. Hence that is same as saying that A belongs to the collection B. So, we have proved that B includes A. So, first property is true. Let us look at the second property. So, what is the second property we want to prove? The second property we want to prove is that B is a monotone class. So, let us take a collection A n, a sequence belonging to B such that it is decreasing or increasing. So, let us say A n is increasing. But A n belonging to B means what? That means what A n complements belong to M of A. That is the definition of the class B. So, saying that we have got a set A n in this that means A n complements belong to A. Now, M of A is a monotone class. So, that implies that A n complements intersection will belong to M of A. Provided we can say A n complements are decreasing and that is true because A n's are increasing because A n complements are decreasing and M of A is a monotone class. So, that means this intersection belongs to it. So, that means union of A n's n equal to 1 to infinity, complement of this belongs to M of A. So, whenever a sequence A n belongs to B and A n's are increasing, we have got the complement union. The complement of the union belongs to it. So, that means union of A n's n equal to 1 to infinity belongs to B. So, the collection B is closed under increasing unions and let us finally prove that it is also closed under decreasing sequences. .. Let us take a sequence of sets which is decreasing. So, let A n belong to B and A n's decrease. We want to show that the intersection of A n's belong to it, but A n's belong to B implies that just now we observed A n by definition A n complements belong to M of A. By definition A n belongs to B means A n complements belong to A for every n and that implies now A n complements is a sequence because A n's are decreasing that is same as A n complements are increasing. So, union of A n complements belong to M of A because M of A is a monotone class and that implies that if the intersection of A n's n equal to 1 to infinity complement belongs to M of A. So, whenever A n's belong to it and take the intersection of A n's there complement belong to it that means intersection of A n, n equal to 1 to infinity belongs to B. So, what we have shown is the collection B is closed under increasing unions is closed under decreasing intersections that means B is a monotone class. So, A is inside B, B is a monotone class and that will prove that M of A is a subset of B because this is a monotone class including A. So, it must include the smallest one. So, that proves the first step of our claim namely that the collection M of A is closed under complements. We wanted to show it is a algebra. So, what is the next step? Next step should be to show that M of A is closed under unions. So, that means whenever two sets E and F belong to M of A the union must belong to M of A. So, let us fix one of them. Let us fix the set F in M of A and let us look at the collection L of F such that it is a collection of all those sets say that A union F belongs to M of A. So, what we have to prove? So, in this we have to prove that M of A is closed under unions. We have to prove that M of A is a subset of L of F. So, once again the required property that M of A is closed under unions we are translating into a property of a collection of subsets. So, let us try to show that M of A is contained in L of F. So, that is first we should try to show. So, to show that the collection M of A is closed under unions. So, this is what we want to show. So, let us fix a set F belonging to M of A and consider the collection which is let us call it L of F. What is this collection? It is a collection of all those subsets in X such that E union F belongs to M of A. So, saying that M of A is closed under unions so to show. So, we should show that M of A is a subset of L of F. So, that is what we should show. So, once again we want to show M of A is a subset of L of F. M of A is a monotone class generated by A and we want to show it comes under L of F in some other collection L of F. So, that means we should try to show that A is inside this collection and this collection L of F is a monotone class. So, we should try to show that A is inside L of F for every F belonging to M of A and second L of F is a monotone class. Let us just observe second one which is quite obvious. So, let us observe L of F is a monotone class. So, for that what we have to show let us take a sequence En belonging to L of F, En's increasing. But that will mean if En's are in L of F that will imply that En union F belongs to M of A. That is by definition of L of F and M of A is a monotone class. En's are increasing. So, En union F is also increasing. So, implies that union of En union F belongs to M of A because En's are increasing. En union F is increasing and belongs and that means it is same as saying that union of En's union F belongs to M of A and what does that mean? That means union of En's belong to the class L of F. So, whenever En's belong to L of F, En's increasing this implies that union En's belong to L of F. So, this is what we have just now proved. A similar proof will work for decreasing also. So, saying that L of F is a monotone class is a straightforward argument because M of A is a monotone class. Let us try to check that A is inside L of F for every F belonging to L of M of A. So, we want to check the first property namely. So, this is a property we want to check that this collection algebra is inside L of F for every F belonging to M of A. Let us note for the time being, we want to check this property for every F in M of A. Note if F belongs to A, then for every E belonging to A, E union F belongs to A because A is a algebra. If E and F are two sets in A, A is algebra that will mean that the union belongs to algebra and that is included in M of A. So, what does this imply? This means for every F in A, E union F belongs to M of A. That means that the set E belongs to the collection L of F. Once again, we are starting with a very simple observation. If a set F belongs to A and E belongs to A, then the union belongs to A and A is always inside M of A. So, that means E union F belongs to M of A for every E belonging to A. That means for every E belonging to A. So, that means we have shown that A is inside L of F. Now, A is inside L of F for every F in A implies L of F is a monotone class. This now proved implies M of A is inside L of F for every F belonging to A. So, what we have shown is M of A is a subset of L of F for every F belonging to A, but we want to check that M of A is inside F for every F belonging to M of A. We have got only for F belonging to A. Now, here is a very simple observation which helps us note a set E belongs to L of F if and only if F belongs to L of E. So, this is an observation which is going to be very important and very useful for us. A set E belongs to L of F means what? E union F belongs to M of A, but if E union F belongs to M of A, that is same as F union E belongs to that means F belongs to L of E. So, saying that E belongs to L of F is same as F belongs to L of A. Now, let us translate this property. Here, it says M of A is inside L of F for every F belonging to A. So, that means for every E belonging to M of A implies that E belongs to L of F and that is if and only if F belongs to L of E. Here F was in the algebra, F in the algebra. So, what we have got? For every F in the algebra L, it belongs to L of E whenever E belongs to L of A. So, that means what? That means A is inside L of F for every F belonging to M of A. See, how nicely we have turned the tables. Earlier we had M of A is inside L of F for every F in A. So, that means every element here E is element in L of F, but here E belongs to L of F means F belongs to L of E. Now, F is in A that means A is in L of F for every F belonging to A, but that means once that is true, now A is inside L of F for every F in M of A and that implies that M of A is inside L of F for every F belonging to M of A. Because L of F is a monodon class, it includes the algebra A. So, it must include the smallest one. So, M of A is inside L of F for every F belonging to M of A. So, we have proved the required thing, namely M of A is inside L of F for every F belonging to M of A and hence that means M of A is also closed under. So, here is what we wanted to prove. So, M of A is closed under unions that we translated into the property that M of A is inside L of F for every F in M of A and that we just now proved finally and you see again and again whenever we want to show something is true, we convert it into a property of a collection of objects and show generators come inside and everything comes inside. So, that proves that M of A is an algebra, M of A is an algebra, it is already a monodon class. So, it must be a sigma algebra and S of A is a sigma algebra. So, that will prove that the required theorem. So, this step proves the required theorem namely. So, for this, let us just go through this proof again to show that M of A is closed under unions. We fix F in M of A and look at this collection and we want to show M of A is inside L of F. For this, we have to show that L of F is a monodon class and A is inside L of F whenever F belongs to A. So, that says M of A will come inside for F belonging to A and now reverse the roles of these two that M of A for F in A that means E belongs to L of F. So, F belongs to L of E and reverse the roles and that comes. So, that gives you the property that M of A is inside L of F. So, it is closed under unions and hence it is a sigma algebra. So, it must include the smallest one and that proves the fact that the sigma algebra generated by algebra is same as the monodon class generated by the algebra. So, let us just recall what we have done till now. We started with this at X, looked at a collection C of subsets C contained in Px. The first thing we have looked at what is called a semi-algebra. Then, we looked at this collection C to be an algebra. Then, we looked at this collection to be a sigma algebra and then it to be a monodon class. So, a semi-algebra, every algebra is a semi-algebra, every sigma algebra is also an algebra. Every sigma algebra is also a monodon class. So, this is something here. This way around implication may not be true, this way around implication may not be true and this way around implication may not be true. Finally, we proved monodon class generated by an algebra is the sigma algebra generated by algebra, if A is an algebra. So, that finishes our study of collections of subsets of X with special properties. I just want to leave you with some exercises which you should try, which are important. The first property or first exercise we would like to, that whenever a collection is an algebra, it is equivalent to saying that empty set in the whole space, the collection is closed under complements, it is closed under unions and that is equivalent to saying whether it is closed under complements because of this de Morgan laws. Another property that whenever something is an algebra, a collection algebra, it is also closed under symmetric references and any finite union in an algebra can be represented as a finite disjoint union whenever you are inside algebra. That property we have seen, but you should try to prove this exercise yourself. Another property about semi-algebras is you can or sigma-algebras is that you can restrict. So, take a collection C of subsets of a set X and restrict it to a set E. That means, take intersection of all sets in C with E and this is C restricted to E and the property we want to prove is that if C is a semi-algebra, then C intersection E is a semi-algebra of subsets of E and similarly, you prove the property that the algebra generated by the restricted sets is same as the algebra generated by generate the algebra and restrict. So, f of C intersection E is equivalent to f of C intersection E. So, you restrict and generate, it is same as generate and restrict and the same property is true for algebras is true for sigma-algebras. So, that is the property about sigma-algebras and these exercises you should try to prove yourself. The steps are outlined here for you to prove. We have already proved these things in our lectures, but I will strongly advise that you prove these things yourself. Here is another example of generating new sigma-algebras, semi-algebras. F is a function from x to y. If you take sets in y and take inverse images that give you a collection of subsets of x. So, try to show that whenever if C is a collection of subsets of y, which is a semi-algebra or a sigma-algebra, then the pullback sets also form a semi-algebra or sigma-algebra. Here is another example of taking two collections of subsets f and g of two sets x and y. Look at the addition products of these collections show that in general it is not a algebra. Of course, if you take a collection of subsets f alpha, which are all algebras or sigma-algebras, the corresponding intersection also is a algebra or a semi-algebra. So, these properties unions may not be true. So, show that for union this property need not be true. So, that sigma-algebra technique, that same thing is inside, then the generated sigma-algebra comes inside that we use. So, we use that to prove that if you take the collection of all intervals in i and left open or right closed intervals, then the sigma-algebra generated by all intervals is same as the sigma-algebra generated by all left open right closed intervals and same as the boron sigma-algebra. So, I would strongly advise you to try these properties to get used to this concept of algebra, semi-algebra, sigma-algebra and monotone class. So, let us stop here today. Thank you very much.