 Welcome to lecture 6 on measure and integration. If you recall, in the previous lecture, we had started looking at various properties of the length function. In today's lecture, we will continue looking at the properties of the length function and then we will try to characterize some other countably additive set functions on the class of all intervals in the real line. So, the first topic will continue is the length function and its properties and then countably additive set functions on algebras. Let us just recall what are the properties of the length function that we have already proved. So, length function was defined on the class of all intervals that is i and to every interval with n points left and point a and right and point b or need not be left and right. Normally, we will write the left end point a first and right end point b later. So, for interval with n points a and b, its length lambda of i we defined as the absolute value of b minus a. If a and b are real numbers and in case either of it is plus infinity or minus infinity, we will define the length to be equal to infinite. So, for all finite intervals, the length is a usual concept of the difference between the values of the end points. So, that is the absolute value of b minus a and plus infinity if the interval is infinite. So, we proved the properties that the length of the empty set and that is the interval is 0. Then, we proved the monotone property of the length function namely length of i is less than length of j if i is an interval which is inside the interval j and then we proved finite additive t property namely if a interval i can be written as a finite disjoint union of intervals j i i equal to 1 to n then the length of the interval i is same as summation of lengths of the individual intervals. So, if i is a finite disjoint union of intervals, the length of i is summation of length of j i's and then we looked at a slight extension of this property namely if i is a finite or an infinite interval actually we looked at that and it is contained in a union of intervals i i's that is i is covered by a finite union of intervals which need not be disjoint then length of i is less than or equal to summation length of the intervals i i's 1 to n the finite number of them. Then, so this was called for the finite i is covered by a finite union and now then we extended this property to the arbitrary countable union. So, if i is an interval which is covered by a countable union of intervals i i which need not be disjoint then we proved that length of i is less than or equal to summation length of length of the individual intervals and if you recall this property use what is called the Heimgorl property on real line. Let us continue our study. So, next thing we want to prove is the following that if i is a interval which is a any interval which is a finite interval say and it is a union of intervals i n's n equal to pair by disjoint intervals then the length of i is equal to sigma length of i n's and this property in fact we had proved it. So, let us prove it to once again. So, let us look at this property. So, if i is an interval which is written as a union of intervals i n's n equal to 1 to infinity i is finite. So, keep in mind we are keeping i as a finite interval and i n intersection i m is equal to empty then that implies length of i is equal to sigma length of i n's 1 to infinity. So, recall we have already shown that length of i is less than or equal to sigma length of i n's that is because of the property that just now proved i is covered by a union of intervals so length of i must be less than or equal to to prove the other way around. So, to show length of i is bigger than or equal to length of sigma i equal to 1 to infinity length of i i's this is what is to be shown. So, let us note that for any n i 1 up to i n these are the intervals which are contained in i and i is finite. So, let us say this is the interval with n points a and b that is i and i 1 is an interval which is inside a b. So, it has n points say a 1 b 1 i 2 has n points a 2 b 2 and i n has n points a n b n but these being finite numbers and they are disjoint. So, we can arrange the intervals like a 1 here b 1 here may be a 2 here b 2 here and so on and a n here and b n here. So, what we are saying is we can assume so without the loss of generality we can say that a is less than or equal to a 1 is less than b 1 less than or equal to a 2 less than less than a 2 less than strictly less than b 2 and less than or equal to and so on and less than or equal to b n less than or equal to b. And once that property is true we can so immediate that b minus a is bigger than or equal to b n minus a 1 which is bigger than or equal to now we can add and subtract consecutive terms. So, b n minus a n plus b n minus 1 minus a n minus 1 and so on plus b 1 minus a 1. So, add and subtract terms which are subtract a bigger term add a smaller term and so on. So, which is equal to sigma i equal to 1 to n length of i i's and this b minus a is length of i. So, what we have gotten is this is for true for every n so that implies length of i no sorry this is bigger than or equal to. So, this is bigger than or equal to sigma i equal to 1 to infinity length of i i's i equal to 1 to infinity. So, that proves the other way round inequality also. So, hence what we have shown is that the length function has the property whenever a finite interval is written as whenever a finite interval is written as a countable union of disjoint intervals then the length of the interval i is equal to summation of the lengths of the individual intervals. We would like to extend this property to not only to finite interval in fact to any interval. So, for that we need a result namely suppose i is a any interval then we want to claim that the length of i is equal to summation length of i intersection the interval n to n plus 1. So, this is the property we would like to prove and in fact one can have here the interval which is left upon n and right close n plus 1 because n point is not going to matter. So, we want to prove that the length of an interval i is same as length of its species which lie inside the intervals n to n plus 1. So, to prove this property let us observe the following. So, let us observe this is a real line. So, we can write it as the intervals. So, here it is say 0, 1, 2 and so on and here is on the other side. So, here is minus 1, minus 2 and so on. So, let us take an interval i. So, if possibly if i is finite, if it is a finite interval then obviously it will lie between two some bounds. So, in finite then there exists some n and m such that i is inside n to m. So, there will be some. So, here is some n and here is some m so that i is inside this. Now, let us look at the pieces. So, let us inside let us look at the pieces of. So, this is n plus i and this is n plus i plus 1. So, intersection with this i. So, what we are saying is this i can be written as union of n to n plus i i equal to. So, 1 to up to up to n plus i equal to m so that i equal to n plus i equal to m minus n. Now, let us observe that these pieces these are disjoint union. This union is a disjoint union and a finite number of them. So, this will imply length of i is equal to summation i equal to 1 to m minus n length of n n plus i. Now, this interval i does not intersect with any other interval out which is bigger than m and which is less than n. So, all those intervals this is in the intersection with i is empty. So, what I can write is this is same as sigma of i equal to 1 to m minus n. This is the intersection of should have written the intersection with the interval i. So, the pieces because the interval may start somewhere here. So, let us write this is intersection with the interval i. Let me write this again. So, this is intersection with i. So, let us write this again. So, the i is written. So, i can be written as a union n to n plus from this is also not wrong. Let me just write n to i can be written as union n plus i to n plus i plus 1 intersection i i equal to it starts with n. So, 0 and goes up to when n plus i plus 1 is equal to so that is m minus. So, we want n plus this is equal to m. So, m minus n minus 1. So, that implies length of i because this is a finite disjoint union. So, this is equal to summation i equal to 0 to m minus n minus 1 length of n plus i to n plus i plus 1 intersection i. Now, for other parts they become 0. So, I can write as sigma over i belonging to integers length of n plus i n plus i plus 1 intersection i over all i integers all integers i because the intersection with the other intervals is going to be empty and that is going to be 0. So, this proves that whenever i is finite we are through. So, i finite case is. Now, let us prove the same thing when i is infinite. Let us take i infinite then I can write i is equal to union of same thing n to n plus 1 intersection i i belonging to integers. This is because keeping in mind that the real line is equal to union n to n plus 1 n belonging to integers. So, interval i is this intersection this and now because i is infinite let us say it looks like say something like a to plus infinity. So, in that case this intersection of. So, i infinite implies that n to n plus 1 intersection i is equal to n to n plus 1 for infinite ends. In fact, some stage onwards. So, in fact if n is say bigger than or equal to a then that is the interval. So, here is a and here is n then n to n plus 1 and so on. They are all going to be non-empty intersections with intersection being equal to that n plus 1. So, this implies that sigma length of n to n plus 1 intersection i is going to be equal to plus infinity over all n belonging to z and that is same as length of i because i is a infinite interval. So, that proves the property namely. So, this proves the property namely for any interval i for any interval i the length of the interval can be written as the length of its species, length of the species i intersection n to n plus 1. So, this is an important property. So, it says length of any interval is a summation of length of its species. This property and note that length of each one of this piece being a finite interval is a finite number. So, we have this says that any interval can be written as a countable disjoint union of intervals each having finite length. So, this is an important property which is going to be called as sigma finiteness property of the real numbers of the length function. So, let us we are going to use this property to prove what is called countable additive property of the length function and that says that the length function if a interval i is written as a countable disjoint union of intervals i n's then the length of the interval i is equal to summation length of i n's. So, to prove this property so let us start looking at the proof of this property. So, to prove this property let us write so i is a interval which is written as a union of i n's n equal to 1 to infinity, where the intervals i n intersection i m is equal to m t to show that length of i is equal to summation length of i n's n equal to 1 to infinity. So, let us look at a proof of this case 1. So, assume so let i be finite actually finite or infinite is not important. So, let us let us take general case itself. So, note i length of i is equal to summation length of i intersection n to n plus 1 and this is because of the property that we have just now proved. And now also note i intersection n to n plus 1 can be written as this is a finite interval and i is equal to so this is n to n plus 1 intersection this interval i is a countable discharge union. So, it is a union of i j j equal to 1 to infinity. So, we can write this as union j equal to 1 to infinity of i j intersection n to n plus 1. And now this is a equality for finite intervals only because i intersection n plus n to n plus 1 is a finite interval which is written as a because i j's are disjoint. So, these intervals are disjoint and they are finite. So, thus this implies by the additive property for finite intervals which are disjoint that lambda of i intersection n to n plus 1 is equal to summation j equal to 1 to infinity lambda of i j intersection n to n plus 1. So, here we are using the fact that whenever a interval i is a finite interval which is a countable discharge union of intervals 1 to infinity then the length of i is equal to summation of length of this. Now, look at this equation here and look at this equation here. So, length of i is equal to summation n over integers length of i intersection n to n plus 1 and that is computed to be equal to this. So, combining these two we get the property that length of i is equal to summation n belonging to z of length of i intersection n to n plus 1 and that property we are going to put it here. So, summation j equal to 1 to infinity lambda of i j intersection n to n plus 1. And now keep in mind that all these are double summation of the series and all of them are non-negative. So, I can interchange the order of integration. So, I can write this as summation over j equal to 1 to infinity summation over n belonging to integers of length i j intersection n to n plus 1. And now once again I use the fact that the interval i j length of i j can be written as length of i j intersected with n to n plus 1 summation over n belonging to z. Just now we have proved that fact the sigma finiteness of the length function any interval length i is summation length of its species inside n to n plus 1. So, this is here. So, that gives me the fact that. So, this gives me length of i is equal to summation j equal to 1 to infinity and this is length of i j. So, that proves the countable additive property of the length function namely if a interval i is written as a countable disjoint union of intervals. So, this proves the countable additive property that if a interval i is written as a countable disjoint union of intervals i n then length of i is equal to summation length of i n. Whether i interval i is finite or infinite that does not matter. So, the length function is countably additive. That is what we have proved as an important property of the length function. Let us extend this property to coverings which are not necessarily disjoint. So, that is called countable sub additivity. So, that is says that if a interval i is such that i is contained in union of intervals i n n equal to 1 to infinity which may not be disjoint then obviously we should expect that the length of i is less than or equal to summation length of i n and its proof is very much similar to the earlier case. So, let us just go through the proof again so that we understand how sigma fineness of the length function is used. So, i is a interval which is contained in union of i n n equal to 1 to infinity. These intervals i n may not be disjoint. Now, what we do is look at length of i. I can write this is equal to sigma n belonging to z length of i intersection n to n plus 1 that is sigma fineness of the length function. And now the interval i intersection n to n plus 1 is inside because this i I can write as union over i n. So, this is equal to, so let me write, this is less than or equal to sigma length over n. So, this thing is less than or equal to sigma length of i j intersection n n plus 1 j equal to 1 to infinity. So, here what we have used is the fact that interval i to this intersection, this interval is covered by the union of these intervals. So, and this is a finite interval. So, length of this must be less than or equal to length of this. Now, once again non-negative series of non-negative numbers I can interchange. So, this is equal to sigma j equal to 1 to infinity sigma n integers length of i j intersection n to n plus 1. And that once again, this once again is nothing but the length of the interval i j by the fact that just now sigma fineness of the length function. So, length of i is less than or equal to length of sigma length of i j's. So, this property is called countable sub-atiti property of the length function. Here is a very important property of the length function, which is called translation invariance. It says if I take a interval i and translate it by some number x, then the length of it does not change. So, it says length of i is equal to length of i plus x. When I take a interval i and translate, so this is a translated set, just shift it, push it by a distance x. So, i plus x is all y plus x, y belong to y. And this property is obvious, because if say for example, if i has got n points a and b, so this property of translation invariance is quite obvious, because of the fact that if i has got, let us say it is a interval with left end point a and right end point b, then i plus x is the interval with the left end point a plus x and right end point b plus x. So, length of i plus x is same as b plus x minus a plus x, which is equal to b minus a, which is equal to length of i. So, length of i is same as length of i plus x. That is for finite and the same proof very continues for infinite, because for example, if i is equal to say a to infinity, then what is i plus x? i plus x is a plus x to plus infinity and in either case, length of i is equal to plus infinity, which is same as length of i plus x. So, basically observing that if i is an infinite interval, its translation remains an infinite interval, so the values of both are equal to plus infinity. So, this is what is called the translation invariant property of the length function. Finally, let us prove what is called the finite additive property of the length function. We have used finite additive property of the length function for finite intervals and we proved countable additivity property for the length function. I just want to exhibit that the countable additivity implies finite additivity, when we have the fact that the length of the empty set is equal to 0. So, basically what we are going to say is, if i is an interval, which is union of ij, j equal to 1 to n, where ij intersection ik is equal to empty, then I can also write it as union of j equal to 1 to infinity ij, where I can define ij to be equal to empty set, if j is bigger than n plus 1, from n plus 1 to 1 put them everything equal to 0. So, then i is a countable disjoint union of intervals, so length of i must be equal to summation length of ij by countable additivity property and that is same as summation j equal to 1 to n length of ij, because from n plus onwards they are empty and so the length is equal to 0. So, countable additivity implies finite additivity whenever length of the empty set I can put it equal to 0. So, let us just recapitulate the various properties of the length function that we have proved, namely the length function is a set function defined on the class of all intervals in the real line with the properties that it is countable additivity, countable additivity, finitely sub additive, finitely sub additive and translation invariant. The important property is that it is countably additive. So, in view of this the next question that arises is the following, so is a measure length function is a measure which is translation invariant, because it is countable additive and length of the empty set is equal to 0. In view of this also observe the length of the singleton is equal to 0. This is also a property of the length function, because the singleton set can be written as an open interval with the or a closed interval with the same end points and so length will be equal to x minus x which is equal to 0. It is finitely additive and countable additivity, finitely sub additive, so that we observed. So here is the question that are there other countably additive set functions on the class of intervals? So, we will like to know length function which we just now proved is one such function which is countably additive set function on the class of all intervals. So are there other countable additive set functions on intervals? So, to answer this question let us make a notation. So, we will denote by i upper tilde, so there is a wave kind of a sign. So, this symbol is called calligraphy i, so the collection of all left open right closed intervals will be denoted by this symbol. This is calligraphy i with the upper tilde, so this is a collection of all left open right closed intervals. So, intervals whose left end point is not included, but right end point is included and keep in mind if it is infinite there is no right end point on the real line. So, this is the collection of all left open right closed intervals. So, what we are going to prove is the following, suppose we have got a set function mu on the class of all left open right closed intervals such that let us say it is finitely additive, this mu is given to be finitely additive and also given the fact that mu for a finite interval is finite for every a and b. Then we want to prove that this can be characterized by the existence of a monotonically increasing function f from r to r such that mu of the interval left open right closed interval a, b is given by f of b minus f of a for every a belonging to r. So, what we want to show is that if mu is given to be a finitely additive set function on the class of all left open right closed intervals and mu with the property that it is mu of a finite interval is finite then we want to show that this must be given by a monotonically increasing function f with the relation that mu of a, b is nothing but f of b minus f of a and keep in mind this looks like if lambda is the length function if mu is the length function then the obvious choice for f is the identity function y equal to x. So, then it will be equal to b minus a. So, it is in some sense we are generalizing the length function that means if mu is any finitely additive set function then it must be given by this. So, to prove this let us observe that mu of a, b is given by f of b minus f of a. So, that itself tells us what should be the definition of the function f for example, if I fix here a point a if a is fixed that means f of a is fixed then I can calculate f of b as equal to mu of a, b minus plus f of a. So, this relation itself gives me a hint how should I define mu of the function f. So, let us fix n a and the most convenient point is to fix a to be the origin. So, and we will also show later on that if mu is countable additive then this function can be chosen to be also not only monotonically increasing, but the right continuous function. So, let us define our function f from the real line. So, f at any point x in the real line is defined as mu of the interval open interval 0 close at x. So, left open right close interval 0 x size or mu of that if x is bigger than 0 and it is defined as 0 if x is equal to 0 because that will mean mu of the empty set is equal to 0 it is countable additive and mu of f of x to be equal to minus of mu of x of 0 if x is less than 0 because f x is less than 0 then this point x is going to be on the left side of 0. So, left open right close interval this. So, with this definition of mu we want to claim that this function has the required properties namely first property that the first property namely. So, let us check these properties of this function. So, first this satisfies the required equation. So, we want to check that for a interval a b mu of a to b is equal to f b minus f of a. So, to check that let us take this is the point 0. So, if a and b are both finite numbers real numbers. So, let us say this is the interval a to b then mu of then f of b minus f of a is equal to mu of 0 b minus mu of 0 to a and this let us observe that mu is given to be finitely additive. So, this is same as mu of a to b because I can write 0 to b as union of 0 to a and union of 0 to b this joint interval. So, this joint p c. So, using that fact this is just mu of a b. So, that proves it and similarly if it is infinite. So, suppose the interval i is a to plus infinity then I can write it as then I can write this as equal to this is equal to mu of is finite. Now, let us observe one thing that we have not if the interval is i is infinite then what is mu of i equal to we have not defined what is the relation between. So, between this and the function. So, keep in mind we have defined f of x is equal to mu of 0 to f x is equal to finite and this is equal to if this is finite. So, we want to check that this satisfies the required property namely if i is equal to a to plus infinity then I want to check that mu of i is mu of i is equal to f of for a oh sorry this is only for finite intervals. I am sorry we want you to check that only for finite intervals this property is true. So, whenever a interval i is finite interval then we know this is finite and this property is true. So, now let us check the next property namely that f is monotonically increasing. So, let us check the property that this property is. So, let us take two points. So, let us take the case here is 0 here is x say here is y. So, we have got x less than y we want to check f of y. So, we want to calculate f of y. So, what is f of y by definition is mu of 0 to y and that I can write as mu of 0 to x using finite identity property I can write 0 to x union of x to y mu of that and that by finite identity property is mu of 0 to x plus mu of x to y and now this is equal to f of x by definition. So, this is equal to f of x plus mu of x to y and this is some non-negative quantity. So, I can write this is bigger than or equal to f of x. So, if x is less than y then f of x is less than f of y. So, that proves it is monotonically increasing in the case when both x and y on the right side of it and same proof will work if they are both are on the left side of 0. So, here is y and here is x. So, we have got y less than or equal to x. So, we want to look at what is f of y. So, that is equal to minus mu of y to 0 by definition. So, that is equal to minus y to 0. So, this I can write it as mu of y to 0 union 0 to y to x union x to 0 and that again by identity property is minus mu of y to x minus mu of x to 0 and this is equal to minus this is f of x. So, I have got this is plus f of x because f of x is defined as minus of this and this is a negative quantity. That means f of y is less than or equal to f of x. So, once again that property is true and the third case when the third case being let us take it is 0 here and y here and x here. So, in that case what is f of y? Let us look at y to x that is sorry what is mu of y to x from this I can write it as is equal to f of y minus f of x by definition and this is bigger than or equal to 0. So, implies f of this is f of sorry this is not f of this is f of x minus f of y. So, that means f of x y is less than or equal to f of x. So, once again in all possible cases we have checked that f as defined above is a monotonically increasing function. So, we want to check now that if mu is countably additive then this implies f is right continuous from the right. Finite additivity property gave us that f is monotonically increasing and we are claiming that if mu is countably additive then f must be right continuous. So, what is right continuity? So, let us take a point x let x belong to R and let us say take a sequence x n say that x n decreases to the point x to show that f of x n converges to f of x. So, that is what we have to show that f of x n converges to f of x. So, let us try to look at a picture. So, this is 0 and let us look at the case when x is bigger than 0. So, this is the case when x is bigger than or even equal to 0 we can take it. So, and here is the sequence x n x n decreasing to x that means here is x n here is x n plus 1 and so on and that is converging to decreasing to x. So, let us observe in this case. So, note look at the interval which is 0 to x n. So, this interval 0 to x n I can split it as interval 0 to x left open right close 0 to x and then x to union of x to x n union of this x to x n. Now, I want to split this portion also. So, the interval x to x n plus 1 x n this interval is same as let us start. So, this part so that is x n plus 1 comma x n the next part that will be x n plus 2 comma x n plus 1 and so on. So, can I claim? So, I want to claim that this is union of x n plus k to x n k equal to 1 to infinity. So, and that is because if I take any point between x to x n. So, take any point here this x n converges so it is going to cross over this point any point inside the interval x to x n. So, that means that it is going to fall inside one of these intervals and all these intervals are subsets of it. So, it is quite easy to check that x to x n this interval is a union of intervals x n plus k to x n k equal to 1 to infinity and here we are using the fact that x n decreases to x. So, this fact is being used here. So, now realize that x to x n is a countable disjoint union of these intervals and mu is given to be countably additive. So, what we have is the following property. So, that says mu of x to x n is equal to summation k equal to 1 to infinity of mu x n plus k comma x n. So, now let us write this in terms of f. So, that means f of x n minus f of x is equal to summation k equal to 1 to infinity of f of x n minus f of x n plus k. Now, this is a series of non-negative terms. So, I can write as limit of the partial sums. So, let us write limit of m going to infinity of summation k equal to 1 to m f of x n minus f of x n plus k and this is equal to limit m going to infinity. Now, this is a partial sum. So, what does that mean? So, this is k equal to 1 n to n plus k. So, n to n plus 1, n to 2. So, this is a sum where terms will cancel out. So, let me just write it. So, this is nothing but f of x n minus f of x n plus 1. The next term will be plus f of x n plus 2 minus f of x n plus 3 and so on. So, it will be going up to m. So, k equal to m. So, f of minus f of n plus m. So, that means what? So, note that n plus 1. So, we are starting with k equal to 1, n to n plus 1. So, n plus 2, k plus 1. So, what are the terms which are cancelling? So, this says that n plus 1, n plus 2, n to 2. So, n plus, sorry, it should be n plus 1 minus n plus 2. So, when the next term comes, n plus 1, n plus 2 and so on. So, these terms cancel. So, what will be left with equal to limit m going to infinity of f of x n minus f of x, sorry, x of n plus m. So, that means what we get is the following and this is independent of m. So, what we get is left hand side was f of x n minus f of x is equal to f of x n minus limit m going to infinity of f of x n plus m and this cancel out, negative sign cancels out, we get limit m going to infinity f of x n plus m is equal to f of x. So, that is same as saying that f is continuous from the right at x and that we approved when x is bigger than or equal to 0. So, when x is bigger than or equal to 0, this is right continuous. The other case when x is negative and still whenever x n converges to 0 or x n decreases to 0, we will show that f of x n converges to f of x showing that f is right continuous at x when x is negative also. We will do this in the next lecture. Thank you.