 Welcome to lecture 5 on measure and integration. If you recall in the previous lectures, we have been looking at the various classes of subsets of a set x with various properties. We looked at what is an algebra, what is a sigma algebra and a monotone class. Today we will start looking at functions defined on classes of subsets of a set x. We will look at first what is called set functions and then we will look at a very important example of a step function namely the length function. So, let us start defining with what are called set functions. Let us start with c, a class of subsets of a set x. Any function mu, so this is a Greek symbol called mu, so a function mu defined on the class of subsets c of a set x and taking non-negative extended real value functions. So, this interval 0 plus infinity both included denotes the set of all non-negative extended real numbers. So, a function mu defined on this collection c of subsets of a set x taking values in non-negative extended real numbers is going to be called a set function. So, it is a function whose domain is a collection of sets, that is why it is called a set function. Next we will be looking at some special properties of, we will be analyzing such functions. So, let us define a set function mu of course, where c is a collection of subsets of a set x and 0 to plus infinity is the non-negative extended real numbers. So, a set function mu is said to be monotone, if it has the following property namely for any two sets a and b in c mu of a is less than or equal to mu of b whenever a is a subset of b. So, it is some kind of a and monotone property that whenever a is a subset of b and both are in the collection c, we want that mu of a should be less than or equal to mu of b. So, this is called the monotone property. . Next, we will look at what is called finite additive t property of a set function mu. So, a set function mu is said to be finitely additive. So, I am emphasizing the point finitely and additive, if it has the following properties that mu of union of sets a i i equal to 1 to n. So, given any finite collection of sets a 1 a 2 a n in c, we want mu of the union of these sets is equal to mu of a i's. Of course, this will be whenever a 1 a 2 a n, this is the finite collection of sets in c such that their union also belongs to c for otherwise this number on the left hand side of this equation will not be defined and we want further these sets are pair wise disjoint. So, a i intersection a j is empty for i not equal to j. So, once again let us see what is finite additivity. Finite additivity means for any finite collection of sets in c a 1 a 2 a n in c such that their union is also an element in c and these sets are pair wise disjoint. For any such finite collection of sets, we want that mu of the union is equal to summation of mu of the individual a i's. Intuitively keep in mind mu in some sense is denoting the size of a set a. So, we are saying mu of the union is equal to sum of the individual sizes whenever the sets a i's are disjoint and this is we are requiring it for any finite collection i 1 to n. So, in a 1 a 2 a n is any finite collection of sets in c which are pair wise disjoint and such that their union is element in c. We want mu of the union is equal to summation of mu of the individual a i's. So, such a property is called finite additivity property of mu or one says mu is finitely additive. We can extend a generalization of this definition. We will say mu is countably additive. So, from finite we are going to countably additive if mu of union a n's 1 to infinity is equal to summation of mu n a n's. Of course, whenever a 1 a 2 a n is a sequence of sets in c such that the union is also an element of c and they are pair wise disjoint. So, countable additivity is a property about a sequence of sets a 1 a 2 a n so on in c which are pair wise disjoint and their union is an element in c. Then we want for any such sequence of pair wise disjoint sets mu of the union must be equal to summation of mu of a n's n equal to 1 to infinity. There is another notion of countable is sub additive if mu of a is less than or equal to summation 1 to infinity mu of a n's whenever a is a set in c and a is contained in union of a n's where a n's also in c for every n. So, in some sense if a is covered by a union of sets a n's then we want the size. So, that is mu of a to be less than or equal to summation mu of a n's n equal to 1 to infinity. So, this is called countable sub additivity because here we are just saying that mu of a is less than or equal to and we are not requiring that a n's are pair wise disjoint. So, this is called countable sub additivity property of the set function mu. A set function mu is called a measure on c and c is a collection of subsets on c. If mu has a property that it is countably additive. So, it should be countably additive and we want that empty set belongs to c and with the property that mu of empty set is equal to 0. So, mu is defined on a collection c of subsets and we want the properties that empty set should belong to c, mu of empty set should be 0 and mu on this collection should be countably additive. Such a set function is going to be called a measure on c. .. So, some examples of set functions. So, let us start with a very simple one. Let us look at a set x which is a countable set. So, its elements are x 1, x 2, x 3 and so on. So, x is equal to x n, n 1, 2, 3, so on and let us fix p n a sequence of non-negative real numbers. So, x is a set which is a countable set with elements x 1, x 2, x 3 and so on and we are fixing arbitrarily some sequence of non-negative real numbers. .. So, let us define for any set a contained in x, for any subset a contained in x, let us define mu of the empty set to be equal to 0 and for the set a if it is non-empty, let us define mu of a to be equal to summation over those p i's such that x i belongs to a. So, a is a subset of x, so some of the x i's will belong to a. So, look at those indices i such that x i belongs to a, pick up those p i's from the given sequence p n and add them up and that is called mu of a. So, mu of a is defined as summation over those p i's such that x i belongs to a. We want to check that this is a measure on the collection of all subsets of the set x. Well, that is quite obvious because mu of empty set is defined to be equal to 0 and let us observe that mu of the singleton, if a is a singleton set, then mu of the singleton set is going to be the number p i. So, if a set a is a countable disjoint union of sets, so let us check that this mu is a measure. So, we are defining mu of a to be equal to summation p i, where i is such that p i sorry i is such that x i belongs to a. So, mu is countably additive. So, let us take a set a. So, let us take a set a, which is union of a i's i equal to 1 to n, a i's belonging a i, a subset of a, a subset of where a i is any subset of x. To check that mu of a is equal to a i's belong to and we want that a i intersection a j to be empty. So, we want this to be equal to mu of a i's i equal to 1 to infinity. Let us observe enough to check when each a i is a singleton x i. So, let us check that case first. So, what is a? a is let us when a i's are some not x i, let us because x itself is x 1, x 2, x n. So, this will be the whole space. So, let us look at the special case when say a i is equal to some x k i, i bigger than x k i, i bigger than or equal to 1. Then the set mu of a is going to be equal to summation p of k i, i equal to 1 to infinity which can be written as limit n going to infinity of which can be written as n going to infinity i equal to 1 to infinity p i up to n. So, sum of a series these are non negative numbers. So, this sum is nothing but the limit of the partial sums because they are non negative. So, there is no problem in writing it that way. So, that means mu of a is equal to limit n going to infinity of sigma i equal to 1 to n of p i. But that means what we want to check that it is summation of mu of each a i. So, this is limit n going to infinity of summation i equal to 1 to n mu of a i because each one is p of k i. This is summation p of k i. So, this is k i and this is mu of a i. So, that is equal to i equal to 1 to infinity mu of a i. So, mu of a is equal to summation mu of a i whenever they are whenever a i is a singleton set. If not it is a finite set then each finite is a union of finite sets and then because for a non negative series you can add it in any way you like it is easy to check that mu of a is also equal to. So, it is easy to check that mu of a is equal to summation mu of a i i equal to 1 to infinity whenever a i is are contained in x and a i intersection a j is empty. So, that says that this set function mu that we have defined is countably additive. So, this is what is called a discrete measure because it is given by a sequence and p i is called the mass at the point x i. So, as we observed mu of the singleton x i is equal to p i for every i and mu of the whole space which is equal to summation mu of the singletons mu of the singletons that is p i. So, mu of x is equal to summation of p i. So, obvious consequence of this is that mu of x is finite whenever this series is convergent. So, the mu is one says this discrete measure mu is finite that is mu of x is less than infinity mu of the whole space is finite if and only if summation mu of p i is less than infinity. If this summation of p i is if the series p i is convergent and the sum is equal to 1 then this measure mu is called a discrete probability distribution on the set x which is x 1, x 2, x n. A very special case which plays important role in the theory of probability and so on x is the set of the numbers 0, 1, 2 and so on. Let us fix any number p which is between 0 and 1 and define p k to be equal to n choose k. So, this is the binomial coefficient n choose k p to the power k 1 minus p raise to power n minus k k between 0 and n. This is called the binomial distribution because of this binomial coefficient appearing in the definition of p k. It is quite easy to check the summation of this p k is equal to 1 that is the summation of p k is summation k equal to 0 to n and this side is nothing but the sum is equal to of p plus 1 minus p raise to power n and that is equal to 1. So, this is a distribution which plays a very important role in probability. This is the probability distribution supposing you have got a coin and you are tossing a coin with probability p for head appearing then in n tosses this p k represents the probability that in n tosses you will get k heads. Another special case of this discrete distribution is called the binomial distribution is called the Poisson distribution which is characterized by the definition that p k is equal to lambda to the power k e raise to power minus lambda divided by k factorial. This is called Poisson distribution. This is another important distribution in the theory of probability. Finally, when we take only the finite number of points 0 1 up to n and p k is 1 over k and each point is given the same mass 1 over k then this is called the uniform distribution. They are special cases of discrete probability distributions. .. Next we give an important example of a measure which is defined on the collection of all intervals in the real line. So, to do that let us fix our notations. So, we will denote by i the collection of all intervals on the real line. For a interval i with n points a and b the left end point being a the right end point being b we will write it as i a comma b. So, a will denote the left end point and b will denote the right end point. We are not saying that this is a open interval a comma b. We are just saying that it is a interval with left end point a right end point b where the left or the right may or may not be or both way or may not be included in that interval. So, it is just an interval with n points a comma a and b left end point is a the right end point is b. So, on this collection of all intervals we are going to define a function. For example, recall that the open interval a comma a is the empty set and the interval 0 to plus infinity. So, the square brackets indicate that we are including 0 and we are including plus infinity. So, this is the closed interval in r star x belonging to r star the extended real numbers x bigger than or equal to 0 which is same as the open interval closed on the left 0 open on the right infinity in the real line union the special symbol plus infinity that we had added in the extended real numbers. So, with these notations we define what is called the length function on the class of intervals. So, it is a set function lambda defined on i taking values in 0 to infinity and is defined by take any interval i with left end point a right end point b. So, we define it as the absolute value of b minus a if a and b are both real numbers that means if a if the interval i is a finite interval with n points a and b then it is length is defined as b minus a and we define it equal to plus infinity in case either the left end point a is minus infinity or the right end point b is equal to plus infinity or both. So, length of i for a unbounded interval is defined as plus infinity. So, this function is called a length function on the class of all intervals and this length function going is going to play an important role for in our subject. So, let us study its properties. So, next we will be studying properties of this length function. The first property is the length function has the property that lambda of the empty set is 0 because empty set is an interval is an open interval with left end point say a right end point a. So, both is open interval a comma a which is empty set. So, by very definition that is equal to a minus a which is equal to 0. Next let us check that this is a monotone set function. So, namely length of i is less than or equal to length of j if i is a subset of j. So, to prove that so we want to check whenever we have got intervals i comma j and i is a subset of j this should imply that length of i is less than or equal to length of j. Since the intervals are defined characterized by the end points. So, let us case 1, let us say i is say infinite say i is equal to minus infinity to say a. Let j be equal to now since i is a subset of j obviously j has to start with minus infinity and can go up to some point say c where c is bigger than or equal to a. So, essentially what we are saying is if this is a and this side is the interval and this side is the interval all of it is the interval i and if j is to contain i then j must be somewhere ending somewhere here. So, that is c. So, clearly both are infinite. So, length of i is equal to plus infinity is length of j. So, that case is obvious. Let us look at the next case i is a subset of j. j is infinite whether i is infinite or not does not matter because the length of i is always less than or equal to plus infinity which is equal to length of j. j being infinite its length is always going to be plus infinity. So, this is obvious if this is the case. So, finally let us look at the case when both i and j are finite. So, here is let us say i has got the end point a and b. Now, i is subset of j that means the end points of j has to be somewhere here and it has to be somewhere here. So, if i is with end points a, b and j is with end point c, d then we should have c is less than or equal to a less than or equal to b is less than or equal to d. So, this implies this and that is same as saying that d minus c is bigger than or equal to b minus a and that is saying that the length of j is bigger than or equal to length of i. So, the monotone property is checked. So, the length function lambda is monotone property namely if i is a subset of j whenever interval i is contained in another interval j, length of i is less than or equal to length of j. Next, let us look at another property is called finite additivity property. So, what should be finite additivity property? We want whenever a interval i is a disjoint union of some other intervals then the length of i should be summation of length of j's. So, what we are saying if a interval i is written as a finite union of intervals j i, j i 1 to n where this j i's are pairwise disjoint then we want length of i to be equal to summation length of j i's. So, that is going to be called finite additivity property. .. So, let us check the finite additivity property. We want to check if i is equal to union of j i's all are intervals where j i intersection j k is empty then that should imply length of i is equal to summation length of j i's i equal to 1 to n. So, let us assume if i is infinite and i is equal to union of j i's j i's 1 to n then that implies at least one of j i's is infinite because if all of them are finite intervals the union will be again a finite interval. So, i infinite implies i equal to union of j i's implies at least one of this j i's has to be infinite. So, that implies lambda of i is equal to plus infinity is equal to summation lambda of j i's i equal to 1 to n because one of them is plus infinity. So, the case when i is infinite is okay. Let us look at the case when i is finite. So, i finite i equal to union of j i's and j i's are pair wise disjoint. Now, let us say the interval i has got n points a and b and let us say the left end point is a the right end point is b. So, here is a and here is b. We want to compute the length of i. So, note length of i is same as the length of the closed interval a comma b. I can include the end points in the interval i because the length depends only on the values of the end points. It does not matter whether the end points are inside or not. So, what we are saying is without loss of generality let i be equal to a comma b. So, this is the interval a comma b. Now, i is equal to union of j i's. So, the point a belongs to this union. So, it should belong to one of the intervals j i's. So, it belongs to one of the interval j i's and actually it has to be end point of one of the intervals of j i's because the interval cannot start with somewhere else. So, let us say so without we can say we can assume i 1 sorry the interval j 1 is starting at a and ending somewhere let us call it as b 1. The point end point may or may not be included. So, the first interval i 1 we can assume it starts here and ends somewhere here that is b 1. Now, the point b 1 is again in that union. So, either it is already included in the interval i 1 or it should be an end point of another interval in the union j 1 j 2 j n's. So, the second one must start with here and end somewhere here that is b 2. So, what we are saying is i 2 some other interval we can rename it as i 2. So, it should start somewhere again at b 1 and end somewhere here and so on. So, here will be the last one a n and that should be b n. So, what we are saying is we can assume this is. So, going this way we can arrange the end points of j 1 j 2 and j n such that such that a is same as a 1 less than or equal to b 1 is equal to a 2 less than or equal to b 2 and so on. So, a n less than or equal to b n which is equal to b. So, we can rearrange this end points of these intervals because this is a union and that is a disjoint union. So, this is what is possible for us to arrange. So, that means and that clearly says that b minus a is equal to b n minus a 1 that is equal to summation b i minus a i i equal to 1 to n adding and subtracting these terms in between that is same as i equal to 1 to n lambda of j i. So, whenever i is a finite interval i is equal to union of j i is they are pairwise disjoint we have gotten that the length of this b minus a is the length of the interval i is equal to summation length of j i's. So, that means that the length function lambda is finitely additive. So, this is the property of lambda the length function being finitely additive namely if an interval i is a finite union of pairwise disjoint intervals then length of the interval i is equal to summation length of j i's. Next let us look at another property supposing i is a finite interval say that i is contained in union 1 to n i i's where a finite union of the intervals but we are not longer saying that they are disjoint then the claim is that length of i must be less than or equal to summation length of these intervals a i's. So, if you drop the condition that these are pairwise disjoint i is subset of. So, we are saying if an interval i is covered by a finite union of intervals then the length of i must be equal less than or equal to summation of length of this intervals i i's. Let us look at a proof of this the proof of this is once again similar to the earlier property. So, let us say i we are saying i is contained in union of i i's i equal to 1 to n. Obviously, if one of i i's is infinite then clearly this implies length of i is less than or equal to summation length of i i's that is obvious because one of these terms on the right hand side in the summation is plus infinity which is always greater than or equal to length of i whatever be i. So, let us suppose that each i i is finite and this is a finite union. So, that implies i is finite. So, without as before without loss of generality assume without any loss of generality that i is equal to a comma b. So, here is once again the same picture here is a and here is b. Now, the point a belongs to the interval i. So, this is my interval i. So, a belongs to i that means it belongs to this union. So, it will belong to at least one of the intervals i i's. Let us name any one of them which contains the point a to be i 1 and let us say the end points of that is a 1 b 1. So, the point a belongs to one of the intervals i i's because it is in the union. So, it will belong to one of them say i 1 and let us say the end points of i 1 or a 1 b 1. So, here is the end point a 1 here is the end point b 1. Now, the possibility is this b 1 is on the right side of b. So, one possibility is it is on the right side of b. So, either b 1 is bigger than or equal to b that means my picture looks like here is a 1, here is a, here is b and here is b 1. Then length of i which is equal to b minus a is less than or equal to b 1 minus a 1 which is equal to length of i 1 and which is obviously less than or equal to summation length of i i's equal to 1 to n. So, in the case b 1 is on the right side we are obviously through by this case. So, what is the other possibility case 2? So, suppose this is the picture. So, suppose this is the picture namely we have got a we have got b and here is a 1 and b 1 is on the not on the right side, but on the left side of b. So, let us take that as the picture. So, in that case the point b 1 belongs to that union. So, b 1 is in the interval a b. So, it will belong to that union. So, it belongs to b 1 belongs to i. So, it belongs to union. So, it will belong to one of the intervals in the i i's. So, let us call that as some interval i 2. So, b 1 belongs to i 2. So, that means a 2 must start here and b 2 either it will be somewhere here or it will be on the right side. So, and if it is on the right side of it that means what? So, let us say it is on the right side. So, here is b 2 instead of here. Let us say b 2 is here. Then the length of the interval i. So, length of the interval i which is equal to b minus a which is b minus a is less than or equal to b 2 minus a 1 b 2 minus a 1 which is less than or equal to b 2 minus a 2 plus b 1 minus a 1. So, b 2 minus a 1 is less than or equal to b 2 minus a 2 plus b 1. We are adding something bigger and then a 1. So, that is length of i. So, in that case length of i will be less than or equal to length of i 1 plus length of i 2. So, that is anyway less than or equal to summation length of i i's i equal to 1 to n. So, if you go on repeating this process. So, what does it mean? So, in the next stage what is the other possibility that b 1 is inside? So, that means here is here is a, here is b. If it is not outside, that must be inside. That means here is a 1, here is here was our b 1, here is a 2 and somewhere here is b 2. It is not on the right side, it is on the left side. So, once again b 2 belongs and then we can proceed in the same way. So, either at some stage will be through or eventually if not then we will have a 1 is less than equal to a, is less than or equal to a 2, less than or equal to b 1, less than or equal to b 2, less than or equal to so on, less than or equal to a n, less than or equal to b, less than or equal to b n. So, what we are saying is either will be through at some final stage or we can rearrange eventually after n stages the end points in that way. In that case again, lambda of i which is equal to b minus a, so here is a, here is b is less than or equal to same idea b n minus, less than or equal to b n minus a 1. So, go on adding and subtracting less than or equal to b n minus a n plus b n 1 minus 1 minus a n minus 1 and so on and plus b 1 minus a 1 and that is equal to sigma lambda of i j, j equal to 1 to n. So, it is just whenever we are on a finite stage the end points can be arranged nicely and we get the property namely the length function is having the property whenever a interval i is covered by a finite union of intervals and the length of i is less than or equal to summation length of i i's. Let us look at an extension of this property namely supposing i is a finite interval such that i is covered by a union of intervals i i's 1 to infinity. That means the interval i is covered by union of a countable union of intervals i i's then again the claim is length of i is less than or equal to summation length of i i's. So, let us prove this property and keep in mind here we are assuming our interval i is a finite interval. So, interval i is contained in union of intervals i i's i equal to 1 to infinity these are intervals i finite this implies length of i is less than or equal to summation length of i i's i equal to 1 to infinity this is what we want to prove. Obvious case note if any one of the terms on this side lambda of i i is infinite then we are through. So, that is note if i i is infinite for some i then what will happen length of i i will be equal to plus infinity which is bigger than or equal to length of i whatever it may be whether finite or infinite. So, that is so implies sigma length of i j j equal to 1 to infinity is also bigger than or equal to lambda because one of them is infinite so that case is obvious. So, let us assume not only i is finite all the intervals i i's are also finite and we want to check this property. So, what we want to check is the following i finite say n points say a b we can assume it is a closed interval because the length of i is not going to change each i j is finite with left end point a j right end point b j. We are not saying that we are assuming these i j's are open or closed or anything we are just naming the end points the left end point of i j. So, we are saying i looks like a and b and each i j is a j b j we are not saying that these end points are included and we are given that i which is a comma b is contained in union of i j j equal to 1 to infinity and if this was finite then we already know how to manipulate that that we have already done earlier in the previous case. So, the idea is from that infinite union bring it to a finite union and here is a closed bounded interval contained in a infinite union and we want to say this is going to be contained in a finite union. So, somewhere the compactness property of the interval a to b is going to be used. So, but for that we need the intervals to be open. So, let us make these intervals i j is open but of course the lengths will change. So, let us fix, let epsilon greater than 0 be fixed, select an open interval say call it as j j such that this j j includes our interval i j and does not change the length much. So, length of i length of this j j is equal to say length of i i plus epsilon. So, slightly increase. So, what we are saying in this picture take an interval from here to here the open interval from here to here call that as j of j. So, each i j i j which was from a j to here b j is enclosed in an open interval slightly bigger but the length. So, this is the length portion that you add that at the most is equal to epsilon. So, now what happens is the following. So, a a b is contained in union of i j's and each i j is contained in the union of j j's and each j j is an open interval. So, we have taken an open interval. So, we have got an open cover of the close bounded interval a b. So, implies by Heine Borel property of the real line which says whenever a close bounded interval is covered by an collection of open intervals then implies there exists some n such that finite number of them will cover it. So, a b will be contained in union of j equal to 1 to n j j's. Finite number of them will cover it and now implies by our earlier case that length of i. So, this was my interval i. So, length of i is less than or equal to sigma length of j j's 1 to n. Now, each one of them is less than or equal to sigma j equal to 1 to n length of i j plus epsilon. Now, we want to separate it out this summation and let it go to infinity and equal to 1 to infinity. But, the problem will come because of this summation epsilon added n times that summation will tend to become very very large. We do not want to happen that. So, what we go is we revise our construction. So, where we select it for a given epsilon fix, select an open interval j j such that this is so instead of epsilon for the interval i j, let us divide it by 2 to the power j. So, instead of having this extra length to be equal to same length as epsilon for every interval i j, for i j we want this extra length to be equal to epsilon by 2 to the power j. So, once we do that, we are in a better shape because now this estimate will be 2 to the power j. So, that means it is less than or equal to summation j equal to, I can put it 1 to infinity because it is less than or equal to lambda of i j plus summation epsilon by 2 to the power j j equal to 1 to infinity. Now, this series is convergent because it is a geometric series with common ratio 1 by 2 which is less than 1. So, this term is equal to epsilon. So, what we are saying is length of i is less than or equal to summation length of i j plus n number epsilon, but epsilon was arbitrary. So, let epsilon go to 0. So, we will get length of i is less than or equal to summation length of i j's. So, what we are saying is that the countable property that we looked at namely length of i is less than or equal to summation length of i i's whenever a interval i which is finite is covered by any countable union, then the length of i is less than or equal to length of i i's. So, this is we have extended that earlier property from whenever a finite covering is there, we have extended it to a countable infinite covering, but only for finite intervals. But we would like to extend this to even arbitrary intervals which are not necessarily finite. So, for that we will have to do a little bit of more work. So, let us look at the next property which says the following. Let i be a finite interval such that i is equal to union 1 to infinity i n where i n's are pairwise disjoint. Then at least we can conclude that the length of i is equal to summation length of i n's. So, whenever a finite interval is a countable union of pairwise disjoint intervals, then the length of i is equal to summation length of i n's. So, let us prove this property. So, what we have got i is equal to union of i j's j equal to 1 to infinity i j's are pairwise disjoint i finite implies length of i is equal to summation length of i j's. So, let us observe note, we have already proved just now that if interval is written as this a union of countable disjoint union, note length of i we have just now shown is less than or equal to length of i j's added up j equal to 1 to infinity, call it 1. So, length of i is less than or equal to this is just now we proved for finite intervals. So, only to show only to show that length of i is bigger than or equal to summation j equal to 1 to infinity length of i j. So, only this is to be shown and for that, so here is the interval a to b, i is finite. So, here is the interval finite interval i. Now, look at i 1 is a subset of i. So, it should be somewhere inside. So, somewhere is a 1, somewhere is b 1. Similarly, i 2 is also inside i. So, somewhere it has to be, either it has to be a 2 here, b 2 here or it could be here somewhere and so on. So, for every n, let us look at, let us consider the end points, the end points, a n, b n of i n's. We can arrange them, we can arrange, there are only finitely many of them such that here is a, here is a 1, here is b 1, here is a 2, here is b 2 and so on, here is a n and here is b n and here is b. That means, we can arrange them in such a way that a is less than or equal to a 1, less than or equal to b 1, which is less than or equal to a 2, less than or equal to b 2 and so on, less than or equal to a n, less than or equal to b n, which is less than or equal to b. Once that is done, so this implies by a simple algebra that means, length of i, which is equal to b minus a. Now, this is b, this is a, I am going to make it shorter b n and a 1. So, this is bigger than or equal to b n minus a 1, which is bigger than or equal to b n minus a n plus b n minus 1, the next one here minus a n minus 1 and so on plus b 1 minus a 1. So, this put together is nothing but, which is equal to sigma i equal to 1 to n length of i i. So, what we are saying is for every n, the end points of the intervals i 1 up to i n can be rearranged in this fashion. Hence, by looking at the ordering of this length of i is bigger than and this happens for every n. So, that implies length of i is bigger than or equal to sigma i equal to 1 to infinity, because it is happening for every n, I can let it go to infinity length of i. So, the other way round inequality is also proved. So, that means, we have proved that whenever i is a finite interval, which is written as a countable union of pairwise disjoint intervals, the length of i is equal to sigma length of i n. So, with that, we prove an important property of the length function for finite intervals. We will continue our study of the length function in the next lecture. Thank you.