 My name is Inder Kumar Rana. I am professor in the department of mathematics IIT Bombay. I will be taking you through this course on measure and integration. This is a course which is normally taught at master's level, MSc in mathematics and sometimes in departments other like physics, electrical engineering also. So, let us go through the basic objectives of this course. This course as I said is called measure and integration. This also goes by various names such as real analysis, advance real analysis and so on. The aim of this course, the objectives that we will be covering in this course are as follows. So, the aim is to generalize the concept of length, area, volume, etcetera to abstract spaces. That leads to the notion of Lebesgue measure on Euclidean spaces and general concept of measures on general spaces. Then also we will extend the notion of integration which is normally done in UG levels called Riemann integration to more general settings. That leads to the notion of Lebesgue integral and other notions of abstract integration. So, these are the basic sort of outline of the course that we are going to follow. We will be following the text book, an introduction to measure and integration written by me. This is published jointly by graduate text in mathematics by American mathematical society and an Indian edition of this is available through Neroza publishers in New Delhi. So, why Lebesgue integration is needed and what is the need for extending the notion of Riemann integration? There are some problems, drawbacks of Riemann integration. To study about them, you should look at chapter 1 of the text book that I have mentioned. We will not have time to go through these drawbacks of Riemann integration and how efforts which were made to remove these drawbacks led to the development of Lebesgue measure, Lebesgue integration and so on. So, for these we refer chapter 1 of the book. Historically, this was developed by the French mathematician Henry Lebesgue who published as a part of his PhD thesis in 1902, integral, longer, and then this was developed further into the abstract spaces by various mathematicians in 19th and 20th century, some of them being Emil Borrell, Carrethe-Durie, Radon and Freshé among others. So, we will assume to start with, there are some prerequisites for this course which we shall assume and we will hope that you have gone through an elementary course in first course in real analysis and you are familiar with the properties of the real line. For example, what is real line, what are called open intervals, closed intervals, what is the topology on the real line, what are called the compact subsets of the real line. So, basic course on real analysis is going to be assumed throughout this course. So, if you have difficulty in this, I would say as look up some elementary book on first course on real analysis and go through these topics, so that you are not left behind and you are able to understand whatever things we are going to discuss, concepts we have to be able to discuss. There is one, the basic space of course is the real line, but there is a notion of what is called extended real numbers which we are going to use in our course and since this is not normally discussed in most of the text books or in courses in real analysis, we will go through some of these concepts on extended real numbers. So, first of all what is this set of extended real numbers, the set of extended real numbers denoted by R star is the set of real numbers to which we adjoin two new symbols, one is called plus infinity and the other is called minus infinity. Now, once we adjoin these two new symbols to the set R, we get the extended set R denoted by R star. Now, as you all know, the set of real numbers has got algebraic operations of addition, multiplication, there is an order on it. So, when we add these two new symbols to them, we would like to define how does these two new symbols, these two new objects behave with respect to the original order structure, the original operations of addition, multiplication and so on. So, we are going to define what are called operations of addition, multiplication and order on the set of extended real numbers. The first one is the order relation. So, we are going to assume or we are going to say that for every real number x in R lies between the two new symbols minus infinity less than x, strictly less than plus infinity. So, this is how the new symbols plus infinity and minus infinity behave with respect to the order structure. So, minus infinity in R star is the smallest element and plus infinity is the largest element in R star as far as the order is concerned. For real numbers, the same original order stays. Next, let us look at the algebraic operations on R star. So, for real numbers x and y, we already know what is x plus y, but for infinity and minus infinity, the two new symbols, how are these operations defined here are the. So, for every x belonging to R, if you add minus infinity to x, you should get minus infinity. So, that is the rule we are specifying. How does minus infinity behave with respect to addition of real numbers? Similarly, plus infinity plus x is equal to plus infinity. Whatever be x positive or negative, when added to minus infinity, you get minus infinity and when added to plus infinity, you will get plus infinity. Now, how does plus infinity added to itself, what is the outcome? So, it says plus infinity plus plus infinity is plus infinity and minus infinity plus minus infinity is minus infinity. Let us specify that plus infinity plus minus infinity is not defined. So, these are the only four relations among addition of plus infinity minus infinity with respect to x plus infinity with itself and minus infinity with itself. Next comes the rules for multiplication. For every real number x, x into plus infinity is equal to plus infinity into x is plus infinity if x is non-negative. Similarly, x multiplied by minus infinity is same as minus infinity multiplied by x is equal to minus infinity again under the condition that x is bigger than 0. More or less, we are following the rules of multiplication for real numbers. Similarly, if x is negative, we have x multiplied by plus infinity or plus infinity multiplied by x is equal to minus infinity. The sign changes of infinity and similarly, x multiplied by minus infinity is equal to minus infinity multiplied by x is equal to plus infinity if x is less than 0. So, depending on whether x is bigger than 0 or x is less than 0, the rules for multiplication are as specified. Of course, if x and y are real numbers, the multiplication between x and y is same as that of real numbers. So, these are the rules for multiplication and of course, there is a specific element. There is a particular element called 0 in the real numbers. How does that behave with respect to plus infinity and minus infinity? Here are the rules for plus infinity into 0 is same as minus infinity into 0 is equal to 0. That is the same as for real numbers also x multiplied by 0 whether positive or negative is always equal to 0. Of course, if I am multiplying plus infinity with itself, the answer is plus infinity and if minus infinity is multiplied with plus infinity, the answer is minus infinity. Plus minus infinity multiplied by plus infinity is plus minus infinity and similarly, plus minus infinity multiplied by minus infinity is minus plus infinity. The sign changes of the outcome. So, these are the rules for addition, multiplication and order structure on the set r star which is nothing but the real numbers along with two new symbols plus infinity and minus infinity. So, once again let us specify that the relations minus infinity plus plus infinity and plus infinity plus minus infinity are not defined. So, with these rules we get the set r star of extended real numbers which is also denoted by this square bracket minus infinity comma square bracket plus infinity. So, that is essentially something like saying like the real numbers are denoted by the open kind of interval minus infinity to plus infinity. If you close it on both sides, that is a notation used for extended real numbers. So, once you are familiar with the order, familiar with the addition and multiplication on the extended real numbers, we can look at the notion of sequences in real numbers and also the notion of supremum and infremum on subsets of extended real numbers. So, let us first look at a is a subset of extended real numbers and let us assume a is a non-empty set. Now, there is a possibility that a is a subset of real numbers only. Then we know that the completeness property of real numbers is if the set a is bounded above, it must have least upper bound or namely the supremum. Now, in the case a is a subset of r star is a subset of extended real numbers. That means there is a possibility of minus infinity or plus infinity being a part of it. Suppose if it is bounded above, then it has to be a subset of real numbers and supremum will exist. If it is not bounded above, that means plus infinity is going to be a part of it. So, we will define the supremum of a to be equal to plus infinity, if a as a part of r is not bounded above. Similarly, we will define the infremum of the set a to be equal to minus infinity, if a intersection r is not bounded below. What we are saying is in the subset of extended real numbers, a set which is bounded above or bounded below, we do not have to say that. So, every subset, non-empty subset of extended real numbers will always have supremum and will always have infremum. Of course, this supremum will be equal to plus infinity, if the set a is not bounded above and it will be equal to minus infinity and the infremum will be equal to minus infinity, if it is not bounded above. So, it is a very nice situation. Every subset has supremum as well as infremum. Now, similar conditions will hold or similar results will hold for limits of sequences in r star. Let us look at a sequence xn, which is monotonically increasing and which is not bounded above. If you recall, here as a subset, as a sequence in real numbers, if a sequence is monotonically increasing and is bounded above, then it must be convergent. Now, if a sequence is monotonically increasing and it is in r star, it is a sequence in of extended real numbers and not bounded above, that means plus infinity is going to be an element of it. So, we will say, if it is not bounded above, we will say a sequence is convergent to plus infinity and write this as equal to plus infinity. This is essentially also we would say, when xn is a sequence of real number, which is monotonically increasing and not bounded above. In that case also, we write the limit to be equal to plus infinity. So, for a sequence of real numbers, it is only a symbolic way of saying that monotonically increasing sequence not bounded above converges to plus infinity, but as a sequence in r star, it converges to an element of r star namely plus infinity. Similarly, a sequence xn in r star, which is monotonically decreasing and if it is not bounded below, we say it converges to minus infinity and write this as limit n going to infinity xn is equal to minus infinity. So, this is how we will analyze sequences in r star, which are monotonically increasing or monotonically decreasing. Similar concepts can be developed for series in r. So, let us just say a few things about sequences, because every monotone sequence is convergent. So, if I look at the sequence, given any sequence, look at the supremums of that sequence from the state j onwards. So, supremum k bigger than or equal to j of xk, then that gives me a new sequence and that sequence will always converge. Similarly, the infremum from k bigger than or equal to j xk will also converge, because these are monotone sequences and monotone sequences in r star always converge. So, limit of the supremum k bigger than or equal to j for that is noted by limit superior of xn and similarly, for the infremum k bigger than or equal to j xk is called the limit inferior of the sequence. So, in general, we already know that limit inferior is always less than or equal to limit superior and the sequence will converge when the limit inferior is equal to the limit superior even in the case of sequences in r star. So, this is how sequences behave in. Now, let us look at sequence from sequences. Let us go to the concept of series. Suppose, xk k bigger than or equal to 1 is a sequence in r star, then let us look at the partial sums of the sequence sn that is the sum of the first n terms of the sequence. So, that is denoted by sn which is summation k equal to 1 to n xk. So, for every n this is well defined and one can ask whether this sequence is convergent or not in r star. So, if this series is convergent in r star then we if the sequence is convergent in r star the sum of partial sums, the sequence of partial sums if it is convergent in r star, we say that the series is convergent and the limit is called the sum of the series. So, this is basically same as that of real line only keep in mind how sequences behave in the real line. So, with this basic discussion about what is the basic space of extended real numbers we are going to deal with. We start with a proper concepts in our subject measure and integration. The first few concepts are going to be discussions about class of subsets of a non-empty set. So, we are going to look at some collection of subsets of a given non-empty set x with certain properties. These collection of subsets which we are going to call them as semi algebras, algebras, sigma algebras and monotone classes. These are various classes which play an important role later on in our subject. So, let us start with looking at what is called a semi algebra of subsets of a set x. So, let x be a non-empty set and let c be a collection of subsets of that set x. We say that the class c is a semi algebra of subsets of x. If it has the following properties, this collection c has the following properties. One, the empty set and the whole space are members of this class x. So, the first property desired of c is that the empty set and the whole space x are members of this class. The second one is that this class is closed under intersections. That means, if a and b are two elements of this collection, then the intersection of these sets a and b should also be a member of the class c. So, this class c is closed under intersections. That is the second property. There is a third property which we will describe soon. That is saying that this class need not be closed under complements, but will require additional property. So, let me write that property and explain, because it is best understood when it is written. So, x is a non-empty set, c is a collection of subsets of the set x. So, the first property we said empty set and the whole space belong to c. Second property was if a and b belong to c, then that implies a intersection b belong to c. The third property which is very crucial is that if a belongs to c, then that implies the set look at the set a complement that need not be in c, but we want to say you can write c as union of elements c i finite number of them i equal to some 1 to n such that c i's are elements of c and they are pair wise disjoint c i intersection c j is empty. So, let us just go through these concepts again. A collection c in p x having the following properties 1, the empty set, the whole space are elements of it. If a and b belong to it, then the intersection of these two sets namely a intersection b is also an element of this collection c. And the third property is if a is a subset of c, then a complement the complement of this set in x of course, should be representable as union of c i's i equal to 1 to n where this c i's are elements of c and they are pair wise disjoint. So, this property that c i intersection c j is not empty for i not equal to j, we just say they are pair wise disjoint. So, such a collection c is called a semi-algebra, semi-algebra of subsets of x. .. So, let us look at some examples to get familiarized with this notion of semi-algebras. So, let us take any non-empty set. Let us take the collection c to be equal to all subsets of x. So, p of x what is p of x that is a power set of x. So, normally this is called the power set of x which is same as all subsets of x. So, c is the collection of all subsets of x. Do you think it is closed under? So, do you think phi and x belong to c? Obviously, it is a collection of all subsets. So, phi and x belong. Obviously, a intersection b also belongs to it if a and b belong to c because if a and b are subsets of it then naturally a intersection b also is a subset and in fact if a belongs to c then that implies a complement also belongs to c because a complement itself is a subset of x. So, a complement belongs to p x it is a subset of x. So, belongs to p x which is c. So, the collection of all subsets of the set x is a example which is obvious example of a semi-algebra subsets of a set x. Let us look at some more examples. This was an obvious example. So, let us look at the second example namely let us look at x equal to real line and let us take the collection c all intervals in R. So, we are looking at the collection of all intervals in R that is a collection c. So, first property is empty set a member of c. Here you will have to understand of course, yes one way of looking at it is empty set can be written as the open interval a comma a for any point a belonging to R and that is the interval. So, that belongs to the collection c. Second property, let us take two intervals i and j belong to c. So, does this imply the intersection i intersection j belong to c? So, that is the question means if i and j are two intervals can I say i intersection j is also an interval. Let us take in the picture, let us take two intervals i and j. So, let us say here is the interval i and here is the interval j. So, one possibility is that they are disjoint from each other. So, here i intersection j is empty. So, hence belongs to c. What is the other possibility? Let us take the intersect. So, here is my interval j and here is my interval i. Then what is the intersection of these two? So, this is i and that is j. So, that is my intersection of i and j and clearly that is also. So, i intersection j is also an interval. So, it belongs. So, two cases, case one when they are disjoint empty set, intersection is empty set and belongs to c. If they overlap, then the overlap itself is an again an interval and that belongs to c. So, it is quite clear that this collection of all intervals is closed under intersection also. Let us look at the third property which is crucial and that says if I take an interval i. So, the third property that we want to verify is if i is an interval, I should look at the complement of that interval and should be able to write it as union of c i i equal to 1 to n, where c i belong to c and c i intersection c j is empty for i not equal to j. So, once again let us look at an interval. So, let us look at an interval say open interval a and b. So, that is my interval. So, what is going to be the complement of this? In fact, the complement of this looks like two pieces. One is this side, other is this side. So, this piece and this piece. So, for this I can write that the complement r minus a b is equal to minus infinity to a in close union b to plus infinity. So, this is this part and this is this part. So, if I take an interval i, so that is i complement, where i is the interval a to b, then its complement is a disjoint union of two elements. Similar cases will follow, if for example, a is left open or right close. So, let us write r, if it is of the type this, then I can write this as this point a is in close. So, minus infinity to a, the complement will be this open here and union b to plus infinity and similarly the other cases. So, I will say that you write down yourself. So, we have verified, see the collection of all intervals in r is a semi algebra of subsets of real life. Let us look at some more examples. So, let us look at example number 4. Let us take the set x to be equal to r 2 and c is the collection of all rectangles in r 2. So, let us just look at the picture and try to understand. So, here is r 2. So, let us take a, can I say empty set is a rectangle? Of course, empty set can be written as a comma a cross a comma a, if you like. It does not matter or you can also write as a comma a cross c comma d. In both cases it is the empty set or any other such representation. So, empty set is an element of the collection of all rectangles in the plane. Let us look at the whole space x that is r 2. Of course, that is r cross r and r is a rectangle, sorry r is a interval. So, r is an interval which we write normally as minus infinity to plus infinity. So, empty set is an element of it. The whole space is an element of it. Let us take two rectangles and see whether the intersection of these two rectangles is also. So, let us take one rectangle here and another rectangle. The possibilities are they do not intersect. In case they do not intersect, then there is nothing to prove because the intersection is an empty set which is already a rectangle. So, let us take a rectangle which intersects with the earlier rectangle. So, we are taking two rectangles r 1, say r 1 and r 2 and r 1 intersection r 2. We want to check whether that is a rectangle or not a rectangle, but let us from the picture it is quite clear that r 1 intersection r 2 is this rectangle. So, is a rectangle. One can write down a formal proof by writing this to be equal to a, b, c and d and so on, but that is not necessary once we understand from the picture that intersection of two rectangles is again a rectangle. Of course, let us verify the third property namely can I represent the complement of a rectangle as a finite disjoint union of a rectangle. So, let us take a rectangle. So, let us take a rectangle in the plane, this one. So, this is a rectangle r and I want to write r complement. I want to see what does it look like. Can I represent this as a finite disjoint union of rectangles again? Let us obviously in the picture I can try to do is the following. I can draw lines passing through the sides and I can draw another line passing through this. Then it is quite clear this complement of r. So, this was the set rectangle r and its complement is nothing but a rectangle r 1, a rectangle r 2, a rectangle r 3, a rectangle r 4, a rectangle r 5, a rectangle r 6, a rectangle r 7 and a rectangle r 8. Of course, these rectangles r 1, r 2, r 3, r 4, r 5, r 6, r 7 and r 8. So, this is r 5, this is and this part is. There are many ways of. So, I am looking at this whole infinite. I can look at this whole infinite, this side and this corner as a rectangle and this part as a rectangle. So, I can write it as union of r i i equal to 1 to 8, where r i intersection r j is empty. So, it is a matter of writing down the details that depending on in r, whether which part of the boundary is included or excluded, accordingly I can make this rectangles r i is to be disjoint. So, this is true that the complement of a rectangle in the plane is also a rectangle. So, that means what? That says that the collection of C of all rectangles in r 2 is also a semi-algebra, is also a semi-algebra of subsets of x. So, we have given a lot of examples of objects which are semi-algebras of subsets of x. Now, let us go to a next stage of understanding, extending this concept of semi-algebra to what is called an algebra of subsets of a set x. So, let x be a non-empty set and a collection f of subsets of x with the following properties. One, like semi-algebra, the empty set belongs to it, the whole space belongs to it and of course, there is another property. So, let us write, better let us write this. So, x is a non-empty set, a non-empty set, f is a collection of subsets of x with the following properties. One, if empty set and the whole space belong to f, like that in the semi-algebra and secondly, we are going to look at the intersection property. If a and b belong to f, they are elements of f, then that implies their intersection also belongs to f and of course, third property, namely in the case of semi-algebra, if I take an element f in f, then its complement need not be in f, but we were able to represent it as a finite disjoint union of elements of that class. In an algebra, in the new concept, we are demanding, this implies a complement also belongs to f. In this case, we say f is an algebra of subsets of x. So, this is called an algebra of subsets of x. So, how does the algebra differ from a semi-algebra? This property is all true for algebra as well as semi-algebra. This property is true for algebra as well as semi-algebra. This property may not be true for a semi-algebra. Namely, in that case, we recall, we said a complement is a finite disjoint union of elements of f and here we are saying a complement itself is an element of f. Let us look at some examples of this again to understand. First observation, of course, every algebra is also a semi-algebra because of the third property that we looked at, namely in a semi-algebra one would like to have a complement to be a disjoint union of elements of f. In an algebra, it is itself in f. So, it is much stronger. So, every algebra is also a semi-algebra. .. Let us note, when x was equal to r and the c is equal to class of all intervals, we showed that c is a semi-algebra. That question is c and algebra. Obviously, the answer is no. For example, I can take an interval, any non-degenerate interval, say a to b. Let us take this interval a to b, that is my i. So, i belongs to c, but i complement is not an interval because i complement is nothing but minus infinity to a union b to plus infinity. So, when i belongs to c, i is an interval, its complement need not be an interval in general. So, that implies that this collection c is c is not an algebra. .. So, let me emphasize again, property here, one says every algebra is also a semi-algebra and this says every semi-algebra need not be an algebra. Means, there are examples of collection of subsets of sets. For example, in the real line, the collection c of all intervals is a semi-algebra, but it is not an algebra. So, a collection of subsets is an algebra is a much stronger property, is a concept than that of a semi-algebra. .. Let us look at the example, same example of x is real line, c is all intervals. We know that this collection is not an algebra, it is a semi-algebra. Let us write f to be the collection of all subsets of the real line, such that look at this collection of all intervals was not an algebra, because the complement of an interval need not be an interval, but it looked like a union of two disjoint intervals. So, let us write where e such that e complement is equal to a union of intervals i i, i equal to 1 to 2, where i i intersection i j is equal to m t. So, what we are saying, look at all those subsets of real line, which can be represented as fine as disjoint union of two intervals. So, claim f is an algebra, we claim that this is an algebra of subsets of r. So, let us first observe, make some observations, namely c is a subset of f, because if I take an interval, if I take an interval, then it has this property, namely it is a disjoint, its complement is a disjoint union of two intervals. So, c is part of subsets of x. So, this as a consequence implies m t set and the whole space belongs to f. Let us look at the second property. Let us look at two elements. So, let us call e 1 and e 2 belong to r, sorry e 1 and e 2 are elements of c. We want to check whether e 1 intersection e 2 belong to c or not. So, what is e 1? Because e 1 belongs to c implies e 1 complement can be written as a disjoint union. So, this square bracket normally indicates I am writing something as a disjoint union. So, i 1 1, i i equal to 1 to 2 or let us just simply write it as this is union of two intervals. So, i 1 1 union i 1 2, where both of these are disjoint. Similarly, e 2 complement can be written as a disjoint union, let us call j 1 union j 1 2, where j 1 and j 2 are disjoint. Now, we want to look at e 1 intersection e 2. We want to look at e 1 intersection e 2 and we want to check whether this belongs to c or not. That means what? I should look at e 1 intersection e 2 complement and try to represent that as a union of two intervals. So, let us look at this is equal to e 1 complement. Here is intersection. So, by demarcall laws that becomes e 2 complement. Now, this is same as e 1 complement is nothing but i 1 1 union i 1 2 union e 2 complement that is j 1 1 union j 2 2. Now, from here i 1 and these two are disjoint, these two are disjoint, but all of these four may not be disjoint. So, this set of ideas seems not leading us to a claim that f is an algebra of subsets of x. So, let us modify our arguments. So, instead of checking this second one, whether the intersection belongs to it, let us look at the property, the third property that we want, whether that is true or not. So, what was the third property? That property said, if a set e belongs to c, we want to check an algebra. So, let us look at e belongs to f. Does that imply e complement belongs to f? So, let us take a set e belonging to f. Now, what is e complement? e complement looks like a finite disjoint union of elements because e belongs to it. So, it is a disjoint union of two elements. So, e complement is equal to e complement is i union j. So, it seems to suggest that if I can show that the collection of finite disjoint unions, this f is closed under union, then I may be through. So, we modify all our arguments again and see how do we proceed. This is how one does not get all the time a polished proof in mathematics. One has to modify the arguments. So, I modify my arguments to prove that f is an algebra as follows. The first step, let us keep in mind what are we trying to do. Here is the collection f of subsets of x or subsets of the real line, which are union of two of them. But that seems to complicate the issue. So, instead of this, let us modify this definition of f itself. Let us look at the modified version of this example. x is real line. Let us look at f. The collection of all those subsets e of r such that instead of seeing e complement is a union of two disjoint intervals, let me just write e complement is equal to a finite disjoint union of intervals i j, j equal to 1 to n, where i j's are intervals and they are pairwise disjoint. So, that is already indicated by writing this square bracket. Now, let us observe keeping in mind our previous arguments that see the collection of all intervals is a part of f. So, that implies that empty set and the whole space are members of f. Second, now if e and f belong to this collection, then what is e? e is a disjoint union of intervals. So, let us write e 1 union e 2. I can write this as e 1 is a disjoint union. So, let us write e 1 as union of i j, j equal to 1 to n, union of disjoint union of some i j k, k equal to 1 to m. Now, these collection of sets, intervals are disjoint. This collection of intervals are disjoint, but all of them may not be disjoint, but that does not matter much. I can write this as union over j equal to 1 to n, union over k equal to 1 to m of i j intersection j k. So, what I am doing is I am intersecting. So, the basic property is if two intervals are not disjoint, then I can write them as a union of disjoint pieces. So, this collection of intervals is a union of intervals which may be overlapping, but I can intersect one with another other and write this as a disjoint union. Now, these pairs of intervals will be disjoint. So, this will imply that if e and f belong to f, then that implies e union f also belongs to f. So, this collection of finite disjoint union of intervals is closed under unions. Let us write finally that if e belongs to f, then that implies by definition e complement is a disjoint union of intervals and each i j is an interval. So, it belongs to f and just now we proved it is closed under unions. So, this implies this also belongs to f. So, the collection of sets of the real line which are finite disjoint union of intervals have the property c is a subset of it. If e and f belong to it, then it is closed under unions and also closed under complements. Now, it is a simple matter for us to check that these two properties whenever a collection of sets is closed and the unions e f belonging to f, we showed implies e union f belongs to it and also e belonging to f implies e complement belongs to f. But these two properties imply that e intersection f also belongs because I can write this as because I can write e as e complement, complement union because what is the comp, if you like this intersection of e f complement, complement and that I can write as e complement union f complement, complement. Simply it is just saying that because these are De Morgan's laws, this will give you e complement, complement intersection. Now, e belongs to f, so this belongs to f, this belongs to f, this union belongs to f and complement belongs to f. So, this belongs to f. So, basically in an algebra, if f is an algebra, then saying that it is closed under unions and complements is equivalent to saying it is closed under intersections and complements. So, let us just summarize what we have done today. We started by looking at our course measure and integration and I said the underlying set of real numbers need to be extended to a larger class namely the set of extended real numbers and there we defined the notion of order, addition and multiplication and analyze how do sequences, series, supermums of sets behave there. Then we started at looking at collection of subsets of a set x with some properties. We started at the first thing was we looked at what is called a semi algebra of subsets of x namely it is a collection of objects, subsets of the set x with the property empty set and the whole space belong to it. It is closed under intersection and the complement of any set in this collection is representable as finite disjoint union of elements of that collection again and a typical example was that of all intervals in the real line. Then we looked at a slightly stronger concept namely the algebra in algebra of subsets of a set x which we defined as the collection with the properties empty set and the whole space belong to it. It is closed under intersections and also closed under complements and then we made a remark every algebra is a semi algebra. The collection of intervals in the real line form a semi algebra, but they do not form an algebra and we showed how to construct an algebra out of these intervals namely we looked at the collection of all finite disjoint unions of these intervals and that collection we proved it is an algebra of subsets of it. So, we will continue analyzing such collections of objects in our next lecture. Thank you.