 Welcome to lecture 7 on measure and integration. If you recall, in the previous lecture, we had started looking at the countable additive set functions on intervals and we proved some properties of such countable additive set functions. We will recall that the theorem that we were proving and then continue the proof. If time permits, we look at a characterization of countable additive set functions defined on algebras in the later part of the lecture. . Let us just recall what we were proving in the last lecture. We were trying to show that if mu is a finitely additive set function defined on the collection of all left open right closed intervals, which was denoted by i tilde, if such a finitely additive set function is given with a property that mu of any finite interval is finite. So, mu of left open right closed interval a, b is finite for every a and b. Then, we want you to characterize such countable additive properties of such functions and relate it to a class of functions on the real line. So, the claim of the theorem is that then there exists a monotonically increasing function f from r to r such that the value mu of the open left open right closed interval a, b is given by f b minus f of a for every a and b belonging to r. We want to show that given a finitely additive set function on the class of all left open right closed intervals, it must arise from a monotonically increasing right continuous function f with the relation that the value mu of a, b is given by the difference f b minus f of a. And in addition, if mu is here mu was only finitely additive, if we assume mu is countable additive, then f this function f can be selected to be right continuous. So, let us just recall how did we define this function. We looked at the function f defined by f at a point x is defined as the measure mu of the interval 0 to x if x is bigger than 0 and it is 0 if x is equal to 0 and is minus mu of x to 0 closed at 0 if x is less than 0. So, this was the definition of the function f and we proved the property that this function f indeed is monotonically increasing. And for that if you recall, we use the fact that f is the measure mu this mu is a finitely additive set function. So, in the next we if we assume that mu is countable additive, we wanted to show that this function f is right continuous at every point x in r. And we had started looking at the proof when x is bigger than or equal to 0. So, we had proved that for any point x bigger than or equal to 0, f is right continuous at the point x is equal to 0. So, let us today we will start with proving the other part remaining part of the proof namely if x is less than 0, then also f is right continuous at x. So, let us look at the proof. So, we want to show f is right continuous at a point x where x is less than 0. So, here is the point 0 and here is x. So, to show right continuity at the point x, let us take, so let x n belong to r. So, let us take points x n a sequence in r say that x n decreases to x. That means x n is converging, all the x n's are on the right side of x and is converging to x. So, because all the points x n's are on the right side of, so here may be x n, x 1, here may be x 2 and so on. So, after some stage x n has to cross over the point 0, the value 0. So, what we are saying is without loss of generality, assume that all the x n's are bigger than 0 for every n. Because x n is going to converge to x and x is less than 0, so at some stage it has to cross over. So, analyzing we can start analyzing the sequence from that point onwards or one writes this as without loss of generality, the proof is not changed if we assume x n is less than 0 for every n. So, here is the situation, here is the point x, here is the point 0 and here is the point x 1. .. So, now let us look at, so here is x 2 and so on. So, let us observe that the interval left open, right close 0 can be written as x 2, x 1 union x 1 to 0. So, I can write this as from this point to x 1 and from this point onwards to this one. And now this interval x 2, x 1, I am going to split further into a union of intervals. So, my claim is that this x 2, x 1 is same as x 1 to x 2, union x 2 to x 3, union x 3 to x 4 and so on. So, we are going to claim is that this is same as x n plus 1 comma x n, left open, right close union n, n equal to 0 to infinity union x 1, 0. So, the interval x 2, x 1, this part we are splitting it into left open, right close, left open, right close and so on. And because this is true, this is the equality because x n is decreasing to x. So, at any point here in between, if I take from any point in between x and x 1, then that stage has to be crossed over by some x n. So, that point will belong here. So, this the interval x 2, x 1 is a union of the intervals left open x n plus 1 to close x n, any equal to 0 to infinity. So, and also observe that these intervals are all disjoint. So, these are all disjoint intervals. So, I can write mu of using countable identity property of the set function. I can write this is equal to summation n equal to 0 to infinity mu of x n plus 1 x n plus mu of x 1 to 0. So, here we have used the fact that mu countably additive implies this property is true. And now, this right hand side is a sequence of non-negative real numbers, possibly extended real numbers. So, I can write this as limit k going to infinity sigma n equal to 0 to k mu of x n plus 1 x n plus mu of x 1 to 0, close here 0. Now, we will write everything in terms of f. So, by definition mu of x to 0 is minus f of x is equal to limit k going to infinity summation n equal to 0 to k. And this is nothing but f of x n minus f of x n plus 1 plus f of x 1 to 0. So, that is in fact minus f of x 1. So, now, let us note what is this. So, this is limit k going to infinity. And what is this sum? This is starts with n equal to n equal to 0 will give you x 0 that is not. So, let us, so there was a mistake here. I should have written as union from n equal to 1, because it is 1 to 2, 1 to 2 and so on. So, that was the mistake here. So, this sum is from n equal to 1 to n equal to 1 to n equal to 1 to k. So, what is this sum? So, n equal to 1. So, that gives you f of x 1 minus f of x 2 plus f of x 2 minus f of x 3 and so on plus f of x n equal to k. So, that is x k minus f of x k plus 1. And now, so that is this part, this sum and minus f of x 1. So, now, we observe that in this x 1, x 2, this will cancel out and what is left is this is equal to f of x 1 minus f of x k plus 1 minus f of x 1. And now, in this equation, so this cancels with this. So, minus f of x, sorry, there is a limit outside, so limit of this k going to infinity. So, what we get is f of x is equal to, so this gives us that f of x is equal to limit k going to infinity of f of x k plus 1. And that proves the fact that f is right continuous at x in the case when x was less than 0. So, hence f is right continuous for every x. So, this proves the theorem that if mu on the class of all left open right closed intervals is a measure is countably additive with the property that mu of AB is equal to mu of AB is finite for every AB in R, then this implies there exists a function f, which is monotonically increasing, which is right continuous. So, monotonically increasing right continuous such that mu of AB is equal to f B minus f of A. So, this is says, so what we have shown is that to every monotony, every countably additive set function mu on left open right closed intervals, you can associate a monotonically increasing right continuous function. And we will next show, so this is proved, this completes the proof of the fact that to every monotonically increasing right continuous function, we can associate to every countably additive set function on the class of intervals, we can associate a monotonically increasing right continuous function with this property. In fact, the converse of this statement also holds, so what will be the converse of such a statement, the converse of such a statement would be that if you are given a monotonically increasing right continuous function f, then we can define a set function mu on left open right closed intervals in such a way that this equation is satisfied, this equation is satisfied. So, that will prove that the only way monotonically, only way we can construct countably additive set functions on the class of intervals is via monotonically increasing right continuous functions. So, the converse part of the theorem says the following, let f be a monotonically increasing function from r to r, so define mu f a set function from the class of all left open right closed intervals as follows for any two real numbers a and b, we want to define what is mu f of the closed left open right closed intervals a b. So, because this is the property has to be satisfied by f, so that itself gives us the defining property of the set function mu. So, mu f of the left open right closed interval is defined as the difference f b minus f of a for all real numbers a and b and now the question comes what happens if b is equal to plus infinity or a is equal to minus infinity or both of them. So, in that case we write this as for mu f of the infinite interval minus infinity to b, so it is open on the left side and close on the right side b, so it is left open right closed interval on the real line. So, what we do is we take the definition as f b minus f of minus x x going to infinity, so as x goes to infinity minus x will go to minus infinity. So, we are using we are defining it via limits, so look at the interval minus x to be left open, so that is the value of the mu of f and then take the limit of that as x goes to infinity. So, this is the definition of mu f of minus infinity to b and similarly if it is on the right side, so if a to infinity, so what we define as take the interval a to closed x, so then the value of that will be f of x minus f of a and now take the limit of that as x goes to infinity. So, for the infinite interval unbounded on the right side left open right closed, so a to infinity is defined as limit x going to infinity f of x minus f of a and if it is the whole real line, then we define mu of f of the whole real line to be limit x going to infinity of f of x minus f of minus x. So, look at the interval minus x to x and let both sides go to infinity, so this is the way we define mu of f and now note that this is a generalization of the length function. If f is the identity function namely f of x is equal to x that is a monotonically increasing function, then this is nothing but b minus a, so mu of a b is nothing but b minus a. So, this mu f is nothing but the length function when f is a monotonically increasing function and one can write down a proof of this on the lines of when we proved that the length function is countably additive. So, on the same lines one can write down the proof of the fact that this set function mu f is also countably additive. One can wonder where one will be using the fact that f is right continuous. So, the fact where we will be using the right continuity of this f to prove that it is monotonically increasing to prove that mu f is countably additive. So, this f for monotonically increasing we can define this mu f is finitely additive, but to prove countably additive we need f to be a right continuous function. So, if f is right continuous then one can write down a proof similar to that of the case of the length function. One uses the fact right continuity because one has to deal with the intervals which are left open and right closed. So, we suggest that if you are keen to know a proof of this you better write a proof yourself trying to see that the steps given for the proof of the length function is countably additive can be suitably modified to do this. So, we leave it as a exercise and if you feel it is too tough and exercise let us assume this and go ahead. So, mu f is a finitely additive sets function and using if f is right continuous one proves that mu f is also countably additive. So, this gives us this is this function mu f is called the set function induced by the increasing function f. So, this gives us a complete characterization of non-trivial countably additive set functions why non-trivial because we are looking at mu of the left open right closed interval a b to be finite in terms of functions which are monotonically increasing and right continuous. So, in some sense there is a correspondence between measures on the class of all intervals and monotonically increasing right continuous functions. In case that countably additive set function mu has the property that mu of the whole real line is finite then one can select this monotonically increasing function to be mu of minus infinity to x because then this is defined we do not have to restrict the fact that mu of a b is finite that will be true anyway because this is finite. So, a more canonical choice for the monotonically increasing right function monotonically increasing right continuous function is mu of minus infinity to x when mu of the whole space r is finite. So, in that case this function f is called the distribution function on r and this plays a role in the theory of probability where monotonically increasing right continuous functions are studied via what are called probability distributions. We will not go into that. So, we will just make a note of it in case we have finite condition that mu of r is finite we will take mu f of x to be in this function. So, this proves the we have characterized all countably additive set functions on the class of all intervals. The aim what we will do now next is the following we will study what are called set functions on general class of sets called algebras where. So, let us start with looking at a and algebra of subsets of a set x and mu a set function defined on this algebra taking non-negative real values. So, taking values in 0 to infinity then we want to show that the following holds that if mu is finitely additive and mu of the set b is finite for a set b in the algebra a then mu of the difference b minus a is equal to mu b minus mu of a whenever a is in the algebra and a is a subset of b. So, what we are saying is the following. Let us take two sets a and b belonging to the algebra a and we are given of course a is a subset of b and mu of b is finite. So, here is the set b and a is a part of it. So, this b so that is a. So, this part is a. So, we can write b as a union b minus a. So, this is the part b minus a and note that a and b minus a both are disjoint sets. So, b is written as a finite union. In fact, union of the two sets a and b minus a and they are pair ways disjoint and mu finitely additive implies that mu of b is equal to mu of a plus mu of b minus a and now let us note that this all are real numbers. Mu of b is a real number because it is finite, mu of a is a real number because a is a subset of b. So, that is and mu of a will be less than or equal to mu of b that is finite. So, that is so these are all this is a equation in real numbers. Anyway that is not really important here, but note that all are non-negative quantities. So, that implies that mu of b is bigger than or equal to mu of a that is one thing that we observe that because this is non-negative. So, this also implies that mu of a is less than or equal to mu of b which is finite. So, that implies mu of a is finite. So, in this equation now I can say all are real numbers. So, I can manipulate this as equation in real numbers. So, this equation implies that if I take it on the other side. So, mu of b minus mu of a is equal to mu of b minus a. So, that is what we want you to prove and note here we have used the fact that mu of b is finite and hence mu of every subset of it is finite whenever that set is in the algebra. So, you can manipulate this as a equation only when they are real numbers if they are equal to plus infinity or minus plus infinity at any one of them then I cannot transpose them on the other side and write this equation. So, we have used the fact that mu is finitely additive and mu of b is finite that implies for every subset a of b which is in the algebra mu of a is also finite and mu of b minus a is equal to mu of b minus mu of a and in particular supposing I take b equal to a. So, this gives mu of empty set is equal to 0. So, in particular mu of empty set is 0 if mu is finitely additive and mu for at least one set b is finite. So, these are consequences of a set function being finitely additive. So, what we are trying to show is if a set function is finitely additive what are the possible consequences? We showed finite additivity implies monotone and if b is finite then I can interchange and write mu of b minus a to b equal to this. Finitely additive plus mu of at least one set is finite implies mu of phi is equal to 0 and let us. So, mu of monotone that we have already shown. So, let us look at the next property that is a very important thing characterization of countable additiveness of the set function. So, suppose mu of phi is equal to 0 then we want to claim that mu is countably additive if and only if mu is both finitely additive and countably sub additive. So, we want to characterize countable additive property of the set function defined on a algebra in terms of it being finitely additive and countably sub additive. So, let us prove these properties. So, let us start while one way. So, let us assume that mu is countably additive to show mu is finitely additive and countably sub additive. So, let us look at the first thing. To show it is finitely additive what we have to do is. So, let a be equal to a disjoint union a i i equal to 1 to n. So, whenever the union is disjoint set square by disjoints we will write it as by a square union symbol for cup with a square kind square instead of writing it as usual. So, where a i is belong to the algebra a. Now, I can also write it as union of a i i equal to 1 to infinity where a i is equal to empty set if i is bigger than n from n onwards. Let us put them as empty sets. Then a is a countable union of pair wise disjoint sets. So, implies by countable additive property that mu of a is equal to summation mu of a i is i equal to 1 to infinity. But that is same as sigma i equal to 1 to n mu of a i because for i bigger than or equal to n plus 1, the sets are empty and mu of the empty set is given to be 0. So, therefore, implies mu finitely additive. On the other side, let us try to prove that mu is countable is sub additive. So, let us take a set a in the algebra and let us say this is contained in union of a i is i equal to 1 to infinity. Now, let us observe the following namely this union a i i equal to 1 to infinity where a i is are in the algebra a. If you recall, we had shown that any countable union of sets in the algebra can be written as a countable union of disjoint sets in the algebra where again b i is are in the algebra, but this is a disjoint union. How did we do that? Let us just recall, we defined b 1 to be equal to a 1 and in general b n to be equal to a n minus union a i i equal to 1 to n minus 1 and so on. So, that is how we have defined those sets b i and note at every stage b 1 is a 1 a 1 in the algebra. So, b 1 in the algebra similarly b n is a n which is in the algebra finite union a i 1 to n minus 1 that is in the algebra and the difference of the two sets in the algebra is again in the algebra. So, each b n is a element of the algebra these are disjoint and they are union because union of b i and b up to b n the same as union up to a 1 a 2 a n and that is true for every n. So, this is equal to true. So, once that is done so using these two things now let us write that mu of a is a subset of this. So, this is mu of the union a i is i equal to 1 to infinity will be equal to summation mu of b i's i equal to 1 to infinity because this union a i's is same as union b i's and union of b i's there is a disjoint union. So, by countable additive property mu of the union is equal to this sum and now note b i is as each b n is a subset of a n. So, this is less than or equal to by finite entity property monotone property this is less than mu of a n a i i equal to 1 to n. So, what we have shown is that sigma. So, what we have shown is the following namely that mu of union a i i equal to 1 to infinity is less than or equal to sigma i equal to 1 to infinity mu of a i. Now, we just want to conclude that in fact mu of a is less than or equal to this quantity. Now, let us observe a is a subset of union a i. So, this implies I can write a is equal to union of a intersection a i i equal to 1 to infinity. I can just intersect and then this is an equality. So, that means mu of a is equal to mu of union i equal to 1 to infinity a intersection a i and this is less than or equal to because this is this union is a subset of the union. So, this is less than mu of union i equal to 1 to infinity of a i's because each one is a subset of this. So, this union is a subset of this and now from here this is less than or equal to mu of summation i equal to 1 to infinity of mu of a i. So, we have shown that whenever a is an element in the algebra is a subset of union of a i's i equal to 1 to infinity then mu of a is less than or equal to summation mu of a i's. So, that proves that mu is countably sub-additive. So, we have shown if mu is countably additive then this implies mu is finitely additive and also mu is countably sub-additive. So, that completes one part of the proof. Let us prove the other way around implication namely. So, we want to show assume mu is finitely additive and mu is countably sub-additive. To show mu is countably additive. So, let us show the proof. So, how to prove countably additive? You have to show let a belong to algebra and a be equal to this joint union a i's equal to 1 to infinity a i's belonging to algebra and we have to show mu of a is summation mu of a i's. Now, by countable additive countable sub-additive property which is given to us this implies countable sub-additive implies that mu of a is at least less than or equal to sigma i equal to 1 to infinity mu of a i's. So, countable sub-additivity implies this fact that this is less than or equal to this. So, we have to prove only the other way around. So, to show that mu of a is also greater than or equal to sigma i equal to 1 to infinity mu of a i. So, this is what we have to show and note. So, here is a small observation to show this enough to show it is enough to show that mu of a is bigger than or equal to sigma i equal to 1 to n mu of a i for every n. So, if we can show for every n mu of a is bigger than or equal to this, then it also will be true for i equal to 1 to infinity because this is nothing but limit of this partial sums. So, this is enough to show. So, we have to only show that mu of a is bigger than or equal to sigma mu of a i's i equal to 1 to n. To show that, let us observe. So, note that a equal to union a i i equal to 1 to infinity implies for every n, the union a i i equal to 1 to n is a subset of a for every n. We are in the algebra. So, this set is in the algebra. This is in the algebra. Mu finitely additive implies mu monotone and hence implies that mu of union a i i equal to 1 to n will be less than or equal to mu of a for every n. But again by finite additivity, this is nothing but sigma i equal to 1 to n mu of a i is less than or equal to mu of a for every n. This is happening for every n. So, this implies we can let n go to infinity. So, i equal to 1 to infinity mu of a i is less than or equal to mu of a. So, that proves the other way around inequality also of the required thing. So, this proves this. So, that proves that mu is countably additive. So, what we have proved is the following namely, we have given a characterization of countable additive property of set functions which are finitely additive. So, if mu of empty set is equal to 0, then mu is countably additive if and only if. So, note here if and only if we have proved both ways. So, if and only if mu is both finitely additive and countably sub additive. So, this is a characterization of countable additiveness of set functions. But of course, the domain of the set function should be an algebra that is important. So, this is a very useful criterion for countable additivity. We will prove another characterization of countable additivity of set functions in terms of limits increasing and decreasing limits. So, that is given in the and we will state the next theorem. But again that theorem is again about set functions defined on algebras. So, the theorem says the following. Let a be an algebra of subsets of a set x and mu be finitely additive and with the property of course, mu of empty set is equal to 0. Then, we want to prove that mu is countably additive if and only if once again it is a characterization if and only if the following property holds and the property says for any element a in the algebra a, we should have mu of a is limit of mu of a n's. What are a n's? Whenever a n is a sequence of sets in the algebra which is decreasing. So, a n is a subset of a n plus 1 for every n. The sequence a n should be decreasing and a n's should be increasing a n is a subset of a n plus 1. That means a n's are increasing sequence of sets in the algebra and a is the union of all these sets a n. So, this is a characterization of countable additiveness of the set function mu provided one can prove the following. For any set a and for any sequence a n of sets in the algebra which is increasing and a is the union we should have mu of a equal to limit mu of a n's. So, let us prove this property once again. So, to prove this what we have to show first? Let us prove. So, let mu be countable additive. So, let mu be countable additive. To show, we have to show the following. Let take a set a belonging to the algebra. Take a sequence a n's belonging to the algebra such that a n is subset of a n plus 1 and a is equal to union of a n's. We should show that mu of a is equal to limit n going to infinity mu of a n's. So, that is what is to be shown. .. So, let us now let us observe a is union of a n's and we are given something about countable additive t. So, the obvious thing is try to write this union as a countable disjoint union. So, we do that. So, prove let b n be defined as a n minus union a i i equal to 1 to n minus 1 for every n bigger than or equal to 1. Then as observed earlier each b n belongs to the algebra, b n's are disjoint and a which is union of a n's is also equal to union of b n's. Of course, this is disjoint. So, let me write it equal to this. So, implies by countable additive property mu of a is equal to mu of this union b n's and that is equal to by countable additive property that is summation n equal to 1 to infinity mu of b n's. So, that is by countable additive property. Now, but we do not want b n's. We want something in terms of a n's. So, here is an observation. This summation I can write as limit k going to infinity of the partial sums. So, n equal to 1 to k of mu of b n's, but b n's are disjoint. So, this is same as limit k going to infinity of mu of union b n n equal to 1 to k, because b n's are disjoint by finite additive property. This must be true. So, note once again mu is given to be countable additive and hence it is finite additive and by finite additive property this is true. Now, the observation is that the union of b n's n equal to 1 to k is same as the union of a's. So, this is same as k going to infinity mu of union a n union of a n equal to 1 to k. But note we are not use anywhere the fact that a n's are increasing. So, since a n's are increasing what is this union? This union is precisely mu of the largest set that is a k. So, that is mu of a k. So, what we are shown is that mu of a. So, we have shown mu of a is limit of mu of a k's going to infinity. Whenever a n is a sequence which is increasing whenever a n's is increasing and a is equal to union. So, we have proved one way. Countable additive implies the required property. Let us look at the converse. So, conversely. So, let us assume mu has the given property and what is the given property? Given property says whenever a set a is written as union of a n's a n's are increasing then mu of a is equal to mu of the union. So, we want to prove to show mu is countably additive. That is, let us take a set a which is disjoint union of sets a n, n equal to 1 to infinity where a and all a n's are in the algebra. We have to show that mu of a is equal to summation mu of a n's. But this, I can write it as union over k 1 to infinity union a n, n equal to 1 to k. So, instead of taking n equal to 1 to infinity, take union of sets a 1, a 2, a k and then take the union over k. Both will be same. But the advantage of this way is now if we call this as b k, then b k is a set in the algebra because it is a finite union of sets in the algebra. b k is increasing because we are taking union of more and more sets and their union is equal to a. By the given property, hypothesis mu of a is equal to limit k going to infinity mu of b k. Now, let us go back to represent b k in terms of a's. So, that is limit k going to infinity mu of union a n, n equal to 1 to k. Now, we use the fact that mu is finitely additive. So, this is limit k going to infinity summation n equal to 1 to k of mu of a n's and which is same as sigma 1 to infinity of mu of a n's. So, that says whenever a is a discount union of, countable discount union of sets in the algebra, mu of a is mu of sigma mu of a n's and that is countable additive property of set function. So, we have proved the theorem completely. Namely, if a is an algebra of subsets of a set x and mu is finitely additive with that property, then mu is countable additive if and only if mu has the property that mu of a is a limit of mu of a n's whenever a n is increasing and a n is equal to union of the sets. So, this is characterizing countable additivity in terms of limits of increasing sequence of sets and this property one says that mu is continuous from below at the point a. So, countable additivity for finitely additive set function is same as saying they are continuous from below at the point a from below because a is union of these sets. So, from below there is a corresponding result for sequences which are decreasing. So, let us state that result and prove it also that if a is an algebra of subsets of a set x and mu is finitely additive. So, that conditions are same as it plus we want additional condition that mu of the whole space is finite. So, this is the additional condition put to state the result namely mu of the whole space is finite. So, it says mu is countable additive if and only if the following holds namely for any set a in a whenever mu of a is equal to limit n going to infinity mu of a n and whenever a n's are decreasing. So, a n plus 1 is subset of a n and a is the intersection it says. So, countable additivity is equal to saying for every set a in the algebra if a is intersection of a decreasing sequence of sets a n's then mu of a must be equal to limit of a n's and the proof of this uses the earlier theorem. So, let us assume mu is countably additive and a n's decrease to a all in the algebra a. We want to show that mu of a is equal to mu of a n's to show mu of a is limit n going to infinity mu of a n. Now, we know something about increasing sequences. So, from decreasing we want to manufacture an increasing sequence and that is done via complements. So, define b n to be equal to x minus a n for every n. Then b n each b n belongs to the algebra a, b n is decreasing because a n's are increasing as a n's are decreasing and where do they decrease? So, b n's increase to x minus a because a n's are decreasing to a. So, by the earlier theorem we have. So, countable additivity implies whenever a sequence is increasing mu of x minus a must be equal to limit n going to infinity mu of x minus b n. Now, we use the fact that mu of x is finite. So, this is same as mu of x minus mu of a and this thing is equal to mu of x minus mu of b n and this is possible only because we have the fact that mu of the whole space is finite. So, everything is a finite quantity and we have already shown mu of the difference is equal to difference of mu's provided the things are finite. So, this is equal to limit of this. Now, x cancels negative sign limit. So, mu of a is equal to limit mu of a n's n going to infinity. So, this is countable additivity implies this. The other way round property. So, let us assume this has that property that whenever a n's increase. So, mu has required given property to show mu is countable additive. So, let us take a equal to a disjoint union a i's then this is what is x minus a that is intersection of i equal to 1 to infinity x minus. So, let us write this as union of i n equal to 1 to infinity union of a i's i equal to 1 to n. So, these are the sets. So, then it is x minus union a i i equal to 1 to n. So, that is equal to intersection i equal to 1 to n and this is nothing but x minus. So, let us call this as a set as b n. So, let us call this as b n. Now, note b n's are decreasing and they are in the algebra because the a n's are union n. This will be increasing. So, this will be decreasing. So, mu of x minus a by the given hypothesis is limit n going to infinity mu of b n's. What is mu of b n? Mu of b n is x minus this. So, that is equal to mu of x minus. So, limit n going to infinity mu of the union that is disjoint. So, summation i equal to 1 to n mu of a n's and this thing is equal to mu of x minus mu of a because everything is finite. So, this cancels with this. So, mu of a is limit of this which is equal to sigma 1 to infinity mu of a n's. So, that proves countable additivity. So, we have proved that when mu is countable additivity, if and only if for a decreasing sequence of sets a n, a equal to this intersection mu of a is the limit under the condition mu of x is finite. So, this is important and this condition cannot be removed that is the next. So, this kind of thing is called continuity from above and here is a remark that the condition mu of x is finite is necessary in the second part and cannot be removed. So, that will request you to construct an example. You can construct a very easily an example on the real line with length function as the set function and here is an exercise for you to do that is finitely such that mu of is finite. So, last part we said a n decreasing to a that you can actually reduce a bit says whenever a n's are decreasing to empty set that is also equivalent to saying that mu is countable additivity. So, these two parts will like you to explore and understand and answer these questions. So, thank you. Let us stop today. Thanks.