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.  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 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.  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, 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.  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.


Hello and welcome to Dispatches from India Show by People's Dispatch. The main story across India this week continues to be the horrific COVID-19 wave which has had an impact on almost every region of the country. On Friday, India reported over 400,000 cases of COVID-19 in a span of 24 hours, the first country to do so. The country also reported over 3,500 deaths. Behind these numbers lie untold horrors, desperate struggles for oxygen, the lack of ICU beds and ventilators, overflowing crematoriums. Amid a systemic collapse, many have been left to fend for themselves to seek help from their fellow citizens. Stories in the media abound of people trying very hard to ensure care for their loved ones, often in vain. One of the most heart-rending aspects has been the struggle for oxygen. Despite repeated statements by those in power that there is no shortage of oxygen, social media and all communication channels are full of desperate pleas for that over the past two weeks. Despite having a year to prepare, the central and state governments were clearly not ready to meet this demand. And this has been one of the key factors in the collapse of the health system in various parts of the country. News Clicks Praveep Rakhaisa addresses key issues in oxygen supply in India. You know, there are two aspects to this issue. One is that why has the demand gone up so rapidly? And that I think is the key question. That this has caught the government completely unaware. They should have prepared for it. They should have had the ability to project what kind of requirements of oxygen they might get. They had done some projection that they knew that they could manage 100,000 patients at a given point of time, numbers of new cases coming up, that kind of surge that we saw last year, they could manage. But beyond that, they would have a problem. But the question is that did we prepare for this eventuality, that the numbers could go up from 100,000, 200, 300,000. Right now we are really looking like we are going to reach 400,000 at the speed at which we are going and probably even 500,000. And if that is so, what would happen to both the ability of the system, the hospitals to handle the patients and the ability for us to be able to supply them oxygen, which is the crucial input that you do require. And that's not something you can ramp up very quickly, but get medicines if you want, you know, plain loads of them can be flown from here and there. That's possible. The possibility exists, but not when it comes to oxygen. That's the key mistake that we seem to have made. The second part of it, if we look at the other part, the issue that is there, that there is a complete lackadaisical attitude towards even doing the minimum that was talked about earlier. It took them 10 months to order what are called PSA oxygen plants, which are smaller oxygen plants. They can be put very close to hospitals within the hospitals for that matter. And therefore the provide like you have for the electrical system, a backup supply, you get a backup supply, it's not a solution for hospitals, but you do get a backup supply that could be there. But the number of plants ordered were not very large. The amount that was being spent on this was again not very large. We're talking something like 30 crores or so. And if you take the 162 plants that the government ordered, it took them 10 months to order. And even after that, if we, it's taken five more months for them to install only 30 odd plants. And the tragedy is that at the current moment that we are seeing, it only meets about max 2.5% of the requirement, even if all the plants have been functional. So let's face it, this is not the, this is not a solution, but even that, a palliative arrangement, particularly for far flung areas where oxygen supply reaching liquid oxygen is difficult, this would have been a some kind of a stop gap arrangement. It wasn't the solution for cities like Delhi, Mumbai or Pune or any of the big cities because there the oxygen requirement is in hundreds of tons. So therefore to think that a small PSA plant or a set of PSA plants could do the job was certainly not going to happen. Meanwhile, the vaccination process continues, those at still a very slow rate. In fact, the daily vaccination rates have declined in the second fortnight of April as compared to the first half. At the same time, questions remain about supply, especially considering that in May, vaccinations will also begin for the 18 to 45 age group. This will see a huge rise in demand. Journalist Anandya Chakravarty talks about what India needs to do as the vaccination process continues. So I mentioned the seasonal flu, we all know that it comes sometimes in October November and then lasts through the winter months and then peaks around February March. There's no reason to right now believe that that is not going to be the case with COVID-19. We don't know whether there's a seasonality here, but the waves tend to show that there might be one. At least one should prepare for it, even if it doesn't happen. If it does happen, if it is something that comes seasonally, we have to remember that we'll not be able to vaccinate our people by October November when the next season, next wave could start. So for that we need to prepare right now because why can't we vaccinate? Because as I said in my last episode, we have about 94 crore adults in this country and each of them will need two doses and if we have to produce 188 crore doses of COVID vaccines, that can't be done before January, February or even March, April next year. So we have almost a year to go before we achieve 100% adult vaccination of COVID in India. And what we need to prepare for is just two or three things. We need to be ready with that. Number one is testing. We need to scale up testing dramatically because we know 80% of COVID cases are asymptomatic and it is these asymptomatic people who spread it. Why? Because they don't know. It's not as if it's their fault. And what happens is that they might wear masks outside, but when they come home, they take off the mask. Right? They have no symptoms. They don't feel unwell. They have no reason to test. So what do they do? They share meals. They sit in the same room. They chat. They hug their children. They are together with their old parents and they pass on the COVID virus, the coronavirus even though they themselves are asymptomatic. If they're tested regularly, those who are going out of the home, if they're tested regularly, tie up with companies, tie up with workplaces, tie up with various places, NGOs. If they can be given a test once a week, even once a week, then we'll be able to control many asymptomatic cases as in monitor those asymptomatic cases. And what will that do? As soon as you have an asymptomatic case, you need step two, which is isolation centers, decent isolation centers with decent beds, maybe good food. Important thing is sanitation, right? Good sanitation, enough toilets for people to use, nutritious food, where they can go, where they can go and sit. They're already infected, remember? So it's not as if they're going to get more infected by being around other COVID patients. So they stay there. If they get worse, then you essentially move them to hospitals or treat them. In some of these places, you can even set up broadband facilities, give them systems by which they can work from isolation center, right? Now instead of working from home. So the most important thing that you do then is that before they start passing on the COVID virus to others, the coronavirus to others in the family, they've been isolated. They've been moved out of that third most important thing is increase oxygen supply. Make sure that you can easily certify industrial grade oxygen, which by the way, is pretty good quality oxygen into oxygen that can be used for medical oxygen and then transported because transporting is a big cost. Transporting oxygen. Put up oxygen plants in hospitals. Create oxygen bed hospitals. And what do I mean by that? I'm not using a technical term. It's just a standard. I mean, you understand what I'm saying, right? Get, take over a stadium. Put a thousand beds there. Put partitions. Put oxygen supply tanks. Put connections so that you can give at least low flow oxygen to these people. Set up a nice U nearby. One of the reasons why people are dying right now is because they're not getting oxygen. So as Faro Kodwadia, one of India's, Mumbai's most respected doctors have said that oxygen needs to be treated as a drug in this case. So oxygen is a drug. If oxygen levels have dropped significantly and oxygen has to be given, that is the first line. The second line is once a person is on oxygen in some cases, they need steroids if they have a lot of inflammation in the body. Then the third drug needed is some blood thinner to ensure there are no blood clotting inside the body. And these are just the three key drugs that doctors across the world say work, right? Nothing else actually has been proven to work. Many have been, I mean, WHO has said that some of them that are being recommended by doctors don't work. Remdesivit does virtually nothing except reduce fever by five, six days. It has no impact on mortality, but you'll see people running from pillar to post for Remdesivit, convalescent plasma treatment, controversial. No one has been able to prove that it works. And there's only some minor recommendation here and there by doctors. But again, we see appeals on social media for plasma. The key things are oxygen, some steroids and blood thinners and of course, paracetamol to bring fevers down. That is what we need. We need temporary oxygen beds, oxygen hospitals to be created, isolation centers to be created. And let me tell you, just this year, China created these things in five to 10 days. If they can do that in five to 10 days, yes, you'd say it's an authoritarian country. If an order comes, people will be too scared to not do it and they do it. Okay, we are a democracy, let's say. And therefore it'll take three, four times that amount of time, but we still have time. Then one, two months to build things right now, set them up right now for the future. And finally, we continue with our reporting on the impact of the pandemic on various sections to the people. Today, we go to the state of Tamil Nadu where the transgender community feels betrayed by the government's approach. The community is relatively well organized in the state as opposed to many other regions in India and is also covered by welfare measures. But they have not been receiving the benefits they should be getting. Here's a ground report. Pusklik met with activists of the Tamil Nadu-Arvanegal Association, a transgender union in the state. They talk about the state assembly elections, as well as how the pandemic-induced lockdown has impacted them. They also talk about the state government's policies towards the transgenders. I am an MP-Election candidate, I started out of Tamil Nadu as an MP-Election candidate. I took four-digit votes. But the thing is, now, we have a lot of work to do. Secondly, we have a lot of expenses. NASA has given us the judgment. Secondly, we have given a lot of government supreme court in the case of the third semester. We have not implemented any of the government we have now. If we go to the social welfare board and ask them, do we not have to pass the NGO in the government? We have not done that, we have not done that. They have been telling us about some of the reasons. So, when we go to work, when we go to study, when we go to everything, we have a struggle there. We have to struggle in the middle of every struggle, we have to do everything. So, even now, for the transgenders, if everyone says ADMK, ADMK in the government, then we do not talk about anything. But only for the female and male, everyone says, we do this, we do that. But, this politician does not talk about transgenders. So, many of these transgenders do not have a house, they do not have a job, they have education. So, even after studying education, many of them do not have a job. So, we all have to give them a job. I think it would be better if we give them a lot of salary like this. The transgenders and the people have considered it. They took it in 2009, I knew it. I have not taken it since then. In 2020, the corona period has come. So, they have not taken it, the government of Tamil Nadu. So, they do not know how many Tamil Nadu, Indian and transgenders are there. They do not know what to give them, what to do, what to do. They are Indian government. So, up till Chennai, there are only 1000-800 transgenders there. They are considering it. But, up till Chennai, there are 10,000 transgenders there. There are 4000 people who want to get an ID card. So, there are 1000 people there. Secondly, they are giving an OAT ID. OAT ID is for some transgenders and females. For some transgenders, they would have given a photo of the ladies photo. But, they would have given a male photo. The other gender photo is very less. So, there are a lot of murder cases. The government did not help us during the corona period. Because, we have received 1000 rupees per month. They would have given 1000 rupees in the past 3 months. Secondly, they have given an ID card. It is an ID card that no one can eat. The ID card may not have been given. They have given 10 kinds of ID cards. We have not received anything else. We have not received any ID cards or anything else. But, they have helped us a lot. Public people only have helped us more. They have given us a lot of clothes. They have given us a lot of clothes. They have given us a lot of ID cards. They have given us a lot of things. The ID card that they have given us is very good. But, the government did not allow us to eat the ID card. But, the government has not done anything for us in the corona period. They have told us that they have done a lot of things for transgenders. But, it is not true. The news is very clear. They are playing a game with transgenders. They are watching us with transgenders. They do not know what to do for transgenders. But, they have told us a lot of things. They have not done anything for the government. They have just told us that they have given us a job. When they give us a house... ...they say that it should last for about 20 to 25 years.  The government has already managed to do this problem as per the skin care order. So, it should last that for 20 to 25 years.  In the second part we are told to wait for the bride, children and their families to come closer. But, in the second part, they stop watching us... We have to do that to the government. Secondly, they are giving a safe house. They are giving a house in a rowdy place. When we go out of the house alone, there is a lot of robbery and murder going on. They are giving a house in a safe place. That should be very important. If we are given a safe place, it will be good. There is nothing to do with all the robbers. They are giving a safe house to the robbers. Recently, they have accepted robbers in Kerala. Before that, they accepted robbers in Tamil Nadu. Because of the DMK politics, Kalanchar and others were named as robbers. They were very helpful to us. After that, they named us as robbers. We didn't do anything. But now, the latest Kerala government has accepted robbers. When they accepted robbers, they are giving a lot of salary. They are giving scholarships for their studies. They are giving fees for books and government. They are giving 2 lakhs for the surgery. All of them are given by the government. In the lockdown, they gave a kit bag like Kerala government. They gave a kit bag, rice, oil, etc. They gave the provisions for the people. They gave a kit bag. It was good to be in the city. But it was not like this. They gave a kit bag, oil bag, rice bag, etc. They gave a kit bag, oil bag, rice bag, etc. It was like this. Tamil Nadu is treating the robbers very hopelessly. We have to do more and do more. We are fighting for each other. We are fighting for each other. But the government didn't do the robbers properly. We don't get the salary of any robbers. We don't have to come to Thirunangu for the robbers. Thirunangu does not have to come. If the robbers came from the government, they would get the salary we gave them. But we don't get this salary. What kind of salary we will get? How many robbers are there without a leave? They don't have a job opportunities. We will at least get the salary we had. Thirunangu can be good for it. We go to the welfare resident board and ask them what to do, what to do with the doctors. What do you do? The government has given you an officer and a salary. What do you do? What do you do? If you want to say something, you go and talk to the mayor. If you talk to the mayor, he doesn't answer. So, we are fighting with each other. Kerala government is doing very nice. Kerala government is doing very well. For the transgenders. During the lockdown period, everyone is going to the sexual part. Everyone is going to take a picture. Only 50% of the people are going to the sexual part. The rest of the people are studying. They are doing something new in some place. They are doing self-employment. They are not doing self-employment. They are going to the temple. Many people are going to work at home. There are such things. But during the lockdown period, everything was affected by the temple. Not only for the transgenders. Everyone was affected. The sexual workers were affected by that too. The debtors were affected by that too. But what the government should have done is, they should not go to such festivals. They should not go to festivals. They should not get money again. They should not get money again. So, the government should have a good help for the transgenders. They should have given 1000 rupees. They gave 1000 rupees for the month. They gave 1000 rupees for the three months. It is not enough. But the government should have done a lot. They did not do it. But the transgenders did not do anything. They should have given 1000 rupees for the fight. They would not have given even that. They would not have passed the G.O. This is what the government should have taken. They should have told the government to us. That's all we have time for. We will be back next week with more news from India. Until then, keep watching People's Dispatch. Watch our YouTube channel.


Mr. Speaker, as I just said, it was never my intention to have spoken on this bill. I mean, one would expect that it was straightforward. It was meant to have dealt with the COVID ramifications and one would have thought that persons would be wise enough not to open the door. But again, you see, Mr. Speaker, when one wants to enter the realms of political relevance and he or she seeks embrace by his own political party, he must show that he's metaphorically the people of the day. And he speaks of COVID mismanagement. I could not believe, Mr. Speaker, that this was the leader of the opposition there. I could not believe it. Have the audacity to talk about mismanagement? Speaks of COVID, the very man who took $7.3 million of taxpayers' money to give to his brother-in-law, gave to his brother-in-law, the people of this country are yet to fully recover the $7.3 million, have the audacity to talk about mismanagement? And then he speaks of crime and he has nothing. He never made a statement apart from one time when they killed a restaurant owner, a reputed one. You know? Well, I don't even know his shame. I don't even know how much melanin was in his skin. But that was the only time. Mr. Speaker, COVID did not hit St. Lucia alone. So once he hides behind the mask of COVID, COVID was a global pandemic. It did not reside in St. Lucia. Yet, he took our economy from the best in the OECS and delivered it to Village JPA at the worst. At the worst, contraction was almost everywhere. But we recorded the most contraction in the Caribbean and faith was in the world. That is the man's record. That is his record, Mr. Speaker.  The man wants to complain. But 11 days before the election, Mr. Speaker, he sought to enter into an agreement with Freistat for $70 million to be paid in one year at a daily interest rate of 1%. Daily interest rate. You get a bill for $228 million. Every day you don't pay $280 something thousand dollars. I don't even want to talk about that. I don't even want to talk about it, Mr. Speaker. It's a so point for me. It's a very so point. We own them all. The man sold it for $13.5 million. Turn around and rent it at a million a month. Are you hearing me, Mr. Speaker? You know, $13.5 million. And I'm not blaming the beneficiaries.  Because, Mr. Speaker, who does that? You own a facility. We had already expended over $50 million on the facility. You turn around, sold it for $13.5 million. And there it is, Mr. Speaker. He sold it outright for $13.5 million. In the agreement, we had to give the man back $12 out of the retrofit for us. So we are left with one and a half million dollars and then to pay almost a million dollars in rent for 16 and a half years. You know, who does that? And you today want to talk about management. Mrs. Speaker, honestly, I have a list and when I saw it, I went for it.  He stayed in a hotel at 2,000 pounds a night. 2,000 pounds a night. He cares about us. He cares about us. The vehicle he drove, which the prime minister inherited, I don't know why he did that. But again, being as humble as Philip J. Pierre is, he would always do. Mr. Speaker, only now this vehicle goes off. When he leaves his home in the morning until he gets back in the night, the vehicle runs all day. Even if he's not in there, he says he never knows at what time he will leave. So it has to be cool. And you talk about you care about people. You know, Mr. Speaker, some of those things, when you hear those things coming from the range developers, 32 million, local bee 32 million per man do 13 million, you know, Sanders collected money on our behalf. It knows how many they owed us and you give them a reprieve. They merely were the repository of our money. You turn around and ask them to keep it, you know, and we are suffering. We are suffering our health care, Mr. Speaker. This government will have to do something. We need to raise revenue, Mr. Minister of Finance. We need to raise revenue. You cut the vat rate by two and a half percent, depriving this government of about 270 million dollars over five years. And in addition to that, you squander everything that we had.  Mr. Speaker, let me, let me stop here. I'm waiting for the real I'm quem de la quem, the debates on the accession to the CCG. I thank you.


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A permutation matrix is an n by n identity matrix whose rows have been rearranged. In other words, it's a permutation of the rows of i. For example, let's find the permutation matrix peruse by switching the second and third rows of the 3 by 3 identity matrix. So we'll start with the 3 by 3 identity matrix, we'll switch the second and third rows, and we weren't told to do anything with the first row, so we'll leave it where it was. Now why is the permutation matrix called the permutation matrix? The obvious answer is that it's a permutation of the rows of the identity matrix. But if the only thing you learn in linear algebra is that every matrix corresponds to a linear transformation, you'll probably fail the class as there's a lot more you should be learning. But at least you'll have learned the most important idea. In the case of a permutation matrix, it corresponds to the permutation of the components of a vector. So we might take our permutation matrix and apply it to the vector A, B, C. So in order to do that, we'll need to rewrite the vector as a column vector, and then multiply, and we get... Notice that we produce the permutation matrix by switching the second and third rows of the identity matrix, and when we apply the permutation matrix to a vector, it switches the second and third components of the vector. And in general, the permutations of the rows of the identity matrix correspond to the permutations of the vector components. And this means we can also go backwards. Suppose we have a permutation to find the corresponding permutation matrix, permute the rows of the identity matrix in the same way. This is actually a little trickier than it sounds. So suppose we want to find a permutation matrix corresponding to the permutation A, B, C, D, E goes to D, B, A, E, C. It's important to keep in mind the permutation is the rearrangement. So while we could read this permutation as A becomes D, that's not the permutation. The permutation is that A shifts from the first position to the third. So we would move the first row of the identity matrix to the third row. And remember, if it's not written down, it didn't happen. We need to move R1 to R3. Also note that this is not a switching of rows one and three. Row three goes someplace else, which we'll have to determine. Next, the second element, B, stays the second element. Strictly speaking, we don't need to write this down as a row switch, but we should just so we know we've taken care of it. Remember, if it didn't happen, it should still be written down. So we'll indicate that R2 goes to R2. The third element, C, moves to the fifth position. So that's R3 becomes R5. The fourth element, D, moves to the first place. That's R4 goes to R1. And the fifth element, E, moves to the fourth place. The fifth row becomes the fourth row. And so applying these permutations, we get the permutation matrix. And remember, if you don't catch your mistakes, someone else will. This permutation matrix is supposed to produce this permutation, so let's verify that this produces the correct permutation by applying it to the vector. So we'll multiply the matrix by the column vector. And we find which is what we wanted to get. Given the matrix, the first question you should ask is, does it have an inverse? Since permutation matrices are permutations of the identity matrix, they have inverses. So how do we find them? So here's an important observation. Every row of a permutation matrix consists of zeros and a single one. So if u and v are the vectors corresponding to different rows, we know the dot product will be zero because the ones won't align. Meanwhile, the dot product of any row with itself will be one. Since the product of a matrix and its inverse should be at the identity matrix, which means that all of the entries should either be zero or one, this suggests that if we could multiply the rows of the permutation matrix by other rows of the permutation matrix, we might get the identity. But remember, the transpose of a matrix turns row vectors into column vectors. And this suggests that if p is a permutation matrix, its inverse is its transpose. Well, let's prove this. Suppose p is a permutation matrix where we'll indicate our row vectors as p1 through pn and p transpose will be, so let's find the product pp transpose, which will be. Now, since every row pI consists of all zeros with a one in one position, then the product of any row with the transpose of any row will either be one if the rows have the same row or zero if the rows are different. So the entry in the first row first position will be p1 times p1 transpose and that's got to be one because the indices are the same and the remaining entries will be zero because the indices will be different. And so that gives us the first row of our matrix. In the second row, all entries will be zero except for which will be the second row, second column entry. And in general, all entries in the ith row will be zero except for the entry in the ith column. The ith row, ith column entry will be one and our product will be the identity matrix. Now remember the inverse has to work on both sides so we need to prove that p transpose p is equal to I. To prove this as the case, we'll make it a homework problem. But once you've done that, we have the following result. If p is a permutation matrix, its inverse is its transpose.


As-Salam-Alaikum. Welcome to lecture number 37 of the course on statistics and probability. Students you will recall that in the last lecture I discussed with you some concepts of interval estimation including the concept of the determination of sample size. After that I began the discussion of the very important area of hypothesis testing. In this regard I discussed with you the formulation of the null and alternative hypotheses and the concept of type one and type two errors. In today's lecture I will continue with some of the very basic concepts of hypothesis testing after which we will apply this whole procedure to some examples. First let us revise a little bit about the two basic concepts that we discussed last time. The first thing is that the hypothesis in your mind about a certain phenomenon you express in a mathematical way. The hypothesis that you want to test is called null hypothesis. For example, if you might want to test that the average weight of the people in a certain community is so much, then you will write that in null hypothesis. The opposite of that is that the weight is not that much, that will be your alternative hypothesis. The second concept that was very interesting, I shared with you that there are four possible situations. The hypothesis, the null hypothesis is actually true and you decide that you would like to accept the hypothesis based on your sample data. That is actually false and the data you collected, you later decided that I would like to reject this hypothesis which I initially formed. So students, this is also a correct decision that it was wrong and you also rejected it. But you remember that there are two situations in which an error commit. If the null hypothesis is actually true, but you decide that you want to reject it, then you are committing type one error. On the contrary, if the null hypothesis is actually false, but you decide to accept it, that is called type two error. I also told you that the probability of committing type one error is denoted by alpha and the probability of committing type two error is denoted by beta. Alpha is also called level of significance. These are the two things. Let us move on. The third very important concept is that of the test statistic. And as you now see on the screen, a statistic that is a function of the sample data not containing any parameters which provides a basis for testing a null hypothesis. This is called a test statistic. Let us try to understand this. Students, in the very beginning, in the first lecture I shared with you that the statistic means any quantity such as the mean, median, mode, standard deviation that you compute from a sample. And here we are saying that we will determine that statistic which enables us to test our hypothesis. So, we call it the test statistic. Now, in different situations, we will have different formulae or different statistics that we will be using. That is the testing about the mean and the statistic that we will be using is Z. The standardized version of X bar. This is the concept of the critical region in comparison with what is called the acceptance region. As you now see on the slide, all possible values which a test statistic may assume, they can be divided into two mutually exclusive groups. One group consisting of values which appear to be consistent with the null hypothesis. In other words, values which appear to support the null hypothesis. And the other group having values which lead to the rejection of the null hypothesis. Students, you may remember that last time when I discussed the hypothesis testing, I said that intuitively you will agree that for example, if you have hypothesized that the mean height of this particular population of women is 64 inches, that is 5 feet and 4 inches. After that, you have drawn an example from the same population. And you have to answer that the answer is 65 inches. So, intuitively you will feel that my sample result is X bar that is supporting my hypothesis that mu is equal to 64 inches. Why? Because 65 inches or 64 inches gap that is small. Now, the critical region or acceptance region is that group of X bar values which seem to support our hypothesis, they will constitute what we call the acceptance region. I mean, if my X bar falls in that region or one of those values, then I will accept the null hypothesis. This is because that group of X bar values which are so far away from my hypothesized mu, that I have to say to myself that this is so different from what I hypothesized that this means that there is something wrong with my hypothesis. So, I cannot accept my hypothesis on the basis of this real data, this real evidence that I have obtained on sample basis. Then, students will be calling them the critical region, that group of X bar values constitutes the critical region. And here, another very important point is that the critical region is not going to be determined arbitrarily. It has a very, very vital connection with the level of significance or as we go along, you will be learning that connection in a proper way. At the moment, I would like you to have a look at the slide which gives you at least an introductory idea regarding the acceptance region and the critical region. If your sampling distribution is that of X bar, which means that you are taking all possible X bar values along the X axis. And then, if you standardize your X bar and convert it to Z, then of course, the mean of the sampling distribution is zero. So, as you are seeing, critical values are both on the right tail or on the left tail. And these are those values beyond which you have the critical region. The portion which is between the two critical values is what we call the acceptance region. So, we have two different parts. One is the left tail and the other is the right tail. The portion which is between the two critical values is what we call the acceptance region. The point which I just said earlier is that the values of X bar, which are close to our hypothesized mu, they will be constituting the acceptance region. The point which is with this diagram tells students that we always begin by assuming that H naught is true. And if H naught is true and if mu is equal to 64 inches, for example, then of course, the mean of the sampling distribution of X bar will be 64. The reason is that we have already learnt that mu X bar is equal to mu, that is equal to the population mean. And if we are saying that the population mean is 64 inches, then the mean of our sampling distribution of X bar is also 64. And when we convert our X bar to Z, then of course, 64 is replaced by 0. So, that is the mean of our sampling distribution of X bar is also 64. So, that is the hypothesized value of Z. And on the right side and on the left side, the acceptance region you are getting is until you reach the critical values, which are quite far from the central hypothesized value. So, if our Z is far from the values beyond the value of Z, then we say that it is so far away from the hypothesized value that we cannot accept our hypothesis. Students, you will say that this is a very long discussion, but believe me, this is the crux of the matter. And if you understand the basic philosophy of this concept, you will be really well off. Because, you will see that T distribution, chi-square distribution, F distribution, when you are conducting hypothesis testing, you will be doing the same thing. So, at this time, the more you work on it, the more time you take to understand the basic philosophy, it is really worth it. So, till here, we have reached a sampling distribution of X bar and its mean that is equal to mu as we have read earlier. And now, if our hypothesis is that mu is equal to a certain number, then we are saying that our distribution is centralized on that number. And if our X bar, which I will take in a real sample and my X bar is close to the central value, then it is in the acceptance region. And I will accept that my X bar is so close to the hypothesized value mu that I should accept H naught. If it is far away, in the tail, to the left side or the right side, then we will get some doubts. But now, let us think about the next thing. I had told you that the level of significance of this thing is related to the critical region tails, or the acceptance region inside, how it will be established. The thing is that the one critical value we marked on this side and the other one on this side, how do we know where to mark that point? Should it be closer to mu or should it be further away? You cannot take these decisions objectively.  What is the relationship between the level of significance of critical values? Suppose that our null hypothesis is that mu is equal to 45 and the alternative hypothesis is that mu is unequal to 45. As I said, it is that mu is unequal to 45. As I said, it is that mu is unequal to 45. That either mu is less than 45 or mu is greater than 45. Our alternative hypothesis is that mu is not equal to 45 or it is less or it is more. If this is the situation and if the level of significance is equal to 5 percent, then students, what you have to do is that you divide your level of significance by 2. So, 5 divided by 2 gives you 2.5 and this is exactly the area 2.5 percent that you would like to have on the right tail as well as on the left tail of the sampling distribution of X bar or of its standardized version Z. So, as you now see on the screen, if you do keep 2.5 percent area on the right tail and 2.5 percent on the left tail, then according to the area table of the standard normal distribution, Z comes out to be 1.96 and minus 1.96. So, this is our critical region, that is, if our Z value is 1.96 or minus 1.96, then we will say that it is so far away from 0 that we should reject our hypothesis because if our hypothesis is true, then why is it that our Z has come out to be so far from what we hypothesized? This is the method of establishing the critical values on your sampling distribution. I will repeat it one more time. If your alternative hypothesis is that mu is unequal to 45 or whatever number we have, if we say that mu is not equal to that number, then this means that we are conducting a two-tailed test. By a two-tailed test, I mean that half of my critical region lies on the right tail and half of my critical region lies on the left tail or critical values established. And then you consult the area table of the standard normal distribution and you find the Z value corresponding to that area. If your alpha is equal to 1 percent, then of course, alpha by 2 will be equal to half a percent and if you put half a percent on the right tail and half a percent on the left tail, then of course, your Z values will be 2.58 and minus 2.58. Now, what is the one-tailed test? Students, when your entire critical region is going to lie on only one tail, either it will lie only on the right tail or it will lie only on the left tail. And how will this be determined? Look, first of all, our hypothesis is that in this particular country, these women, their mean height, that is, at the most 64 inches, that is, it cannot be more than this. This is our hypothesis, it is in our mind. In this case, you can form null hypothesis in such a way that the minimum mu is less than or equal to 62 and your alternative hypothesis will be that mu is greater than 62. If you form null null hypothesis, then it will be in such a way that the mean height is more than 62 inches. If you hypothesize students in such a way, then the alternative hypothesis is greater than or equal to 62. Remember that if your entire critical region will lie on the right tail of your sampling distribution. Then, of course, your entire critical region would have been on the left tail of the distribution. If your entire critical region lies on the right tail of your sampling distribution, then, of course, your entire critical region would have been on the left tail of the distribution. It is the opposite situation. The mean weight of the adult males of this particular country is at least 155 pounds. It is not that much. The mean weight is less than that. This situation is less than 155. It is not that much. It is a come here. The mean weight is less than that. In this situation, whenever you have this situation, your entire critical region will lie on the left tail of your sampling distribution. Students, you will say that it is complicated. I told you last time that this is a concept in which a lot of small concepts are involved. After collecting them all, you will have a proper idea in your mind. But it will take a little bit of time and all I can advise you to do is to be patient and to persevere. We are wanting everything to happen very fast. But students, there is no short cut to true knowledge. It is so easy to pick up formulae which are sitting in the text room. It is so easy to pick up the text books and other books waiting to be used. You pick it up and you apply it. You feed it to the computer. There you go. But what is the point if you do not have the concept? Just relax. Try to get involved and enjoy this very interesting concept. Students, even so far, we have reached that if the level of significance is 5% or 2-tailed test in which half of the critical region lies on the left tail and half of the critical region lies on the right tail, then you should half the level of significance. If it is 5%, then half of it will be 2.5%. So, 2.5% area on this side, 2.5% on this side and according to the Z values that have come, minus 1.96 and plus 1.96, they are the critical values. If my Z is 2.01, then it means that it lies beyond 1.96, that is, it lies in the tail. It lies beyond my mu and therefore I will reject the null hypothesis. But till now, one thing has not been clarified. That the level of significance that we had, why did we do this by half the level of significance? We have not come to understand that the level of significance is in critical values. After all, the probability of committing type 1 error and type 1 error is that if H naught, our null hypothesis is actually true and we reject it, then we are committing this error. Now, the crux of the matter is that in any such hypothesis testing procedure, we always begin by assuming that H naught is true. If we have kept 62, then in the beginning, our assumption is that it is 62 and according to it, our sampling distribution, its mean value is 62 and after that, its standardized version, when we take it, then of course we will get zero there. What I said is vitally important that we begin by assuming that H naught is really true. Now, if it is true, then I should never reject it, right? That is, if my X bar is really 62, then I should accept it. If my X bar is 74, then I should accept 62 because it is true. But students, if it is 74, then it is far away from it and it is in the tail. And if it is so far away, then we will say that no, we have to reject it. Because it is not supporting me in this matter, it is so far away from it. So, if it is really true and I am starting to reject it, then I am committing type 1 error, right? Now, if it is so far away, then I will reject it. If it falls in that little region, whose area is 2.5 percent, or it falls in that little region on the other side, whose area is 2.5 percent, then what is the area? Probability. In other words, if my X bar is that X bar, whose probability is 2.5 percent plus 2.5 percent, 5 percent, if my X bar is that one, whose probability is 5 percent, whose probability is that I have to reject this mu. After doing this, I have to commit type 1 error. So, students, exactly this is what I wanted, that the probability of committing this error should be 5 percent. My X bar, if any of them came into a region, then it happened that since I have fallen in that region, here or there, I will reject this node, I will commit this error. But since the probability of my X bar falling in those regions is only 5 percent, in other words, the probability of committing this error is 5 percent, and this is exactly what I wanted. So, students, do you see? Everything falls in place. Jo mai chahti thi, an usi ke mutabiq, I determine these critical values. Isn't that interesting, exciting and fascinating? And one marvels at the intellect of the mathematicians who thought of this method of hypothesis testing. Ab dekhye kis kadar interesting hai, ke everything falls in place in a mathematical way. All right, I think, bahot achee aur lambi discussion hum kar chuke. Achee a buri, itto faisla aap karenge. But let us now apply this concept to an example. As you now see on the screen, a steel company manufactures and assembles desks and other office equipment at several plants in a particular country. The weekly production of the desks of model A at plant number one has a mean of 200 and a standard deviation of 16. Recently due to market expansion, new production methods have been introduced and new employees hired. The vice president of manufacturing would like to investigate whether there has been a change in the weekly production of the desks of model A. To put it another way, is the mean number of desks produced at plant number one different from 200 and this test is to be conducted at the 5% level of significance. The mean number of desks produced last year, 50 weeks because the plant was shut down two weeks for holidays. This mean number for last year is 203.5. On the basis of the above result, should the vice president conclude that there has been a change in the weekly production of the desks of model A students. Let us try to understand this interesting example. The variable of interest that is the number of desks produced per week and as you have seen, according to the record of the first years, on the average 200 desks were being produced at plant number one. These desks which are of a particular model, which we are calling model A. So the historical record is that 200 are produced on the average per week and standard deviation is 16. This means that every week exactly 200 are not produced. In some week there is less production and in some week there is more. This is the old story. One year or one and a half years ago, this new production method was hired by some new employees. Now the vice president wants to see that his aim was to increase the production. So he wants to test whether it has changed as much as it was before. It is also possible that it has become less. So to cut the long story short, what we are wanting to test is that the mean production is 200 desks per week and the alternative is that it is not. And as you now see on the screen, when we write it in a formal way, we say H0 mu is equal to 200 and H1 that mu is not equal to 200. Next we would like to specify the level of significance that is the probability of committing type one error. And usually in many situations we take the level of significance as 5% and this is the value that we are going to set in this particular example as well. The next step is the test statistic, i.e. that formula, that statistic on which we conduct our test. Students, in this problem the sample size is large. As you remember they have the data for 50 weeks and 50 is a large number, large enough for me to be able to use the normal distribution. Therefore our test statistic is z is equal to x bar minus mu over sigma over square root of n. Now you will recall that when I discussed with you the sampling distribution of x bar, I can wait to you that mu x bar is equal to mu and sigma x bar is equal to sigma over square root of n in those situations where the FPC is not required. So, when we substitute these values in the basic formula that z is equal to x bar minus mu x bar over sigma x bar, then we obtain this formula that you see on the screen. This formula is our test statistic on which we will test our hypothesis. The fourth step is to compute z for this particular problem. Now as you know the standard deviation is available to us and it is equal to 16 and x bar for the 50 weeks that were considered came out to be 203.5. Hence when we substitute these in the formula for z, we obtain 203.5 minus mu over 16 over square root of 50. This is the value about which we are trying to conduct this test. So, what should I do? What value of mu should I substitute in the formula? As I indicated a short while ago, students, we always begin by assuming that H naught is true and if H naught is true, then of course mu is equal to 200. Substituting this in the formula, our z value comes out to be 1.55. The fifth and very important step is the determination of the critical region. We have to take two things into account. Number one, what is our level of significance? As I indicated, we are taking it to be 0.05 and number two, is this a one-tailed test or is it a two-tailed test? Since our alternative hypothesis says that mu is unequal to 200, it means that mu could be less than 200 or greater than 200 or 2k less than sign or greater than sign. Dono involve hoge is liye, it is a two-tailed test. Half of our critical region is going to lie on the left tail and half of it on the right. Now, combining these two points, we obtain the critical values as plus and minus 1.96. The reason being that if we take 2.5 percent area, which is the half of course of 5 percent on the right side and also 2.5 percent on the left side, then our area table of the standard normal curve gives us z is equal to plus and minus 1.96. The last step is to draw a conclusion, should we accept H naught or should we reject H naught? Now, a short while ago, we found that our z value corresponding to our x bar came out to be 1.55 and as you can see on the screen, 1.55 is less than 1.96 and it is falling very much in the acceptance region and therefore, we do not reject H naught, rather we would accept this hypothesis. Students, you have seen that we have followed a methodically and we have arrived at the conclusion that we will not be rejecting but accepting the null hypothesis. Orgoya means that our result was that x bar for the last year is 203.5 i.e. the mean number of desks produced per week is 203.5. So, this difference between 203.5 and 200, which was the past record, we can say that this is due to chance. This difference is insignificant. This is a very important term in hypothesis testing and this is a technical term. Insignificant means that the difference between my x bar and my hypothesized mu i.e. 203.5 and 200. This difference is not large enough to signify that I should reject H naught. In other words, it is insignificant. If there was a situation where our data for last year was computed on the basis of x bar, supposing it came out to be 215.7 or against the value of z value, if it came out to be greater than 1.96, then obviously it would have fallen in the right tail of our distribution and in the critical region, then we would be rejecting H naught. If situation may, we would have said that our result, this value that we got from the sample, this is significant. The difference between this x bar value and my hypothesized mu, this difference is so large that it signifies that I should reject H naught. This is a technical term and you must keep this kind of an interpretation in mind. Let us now apply this hypothesis testing procedure to another interesting example. As you now see on the slide, a random sample of 100 workers with children in daycare show a mean daycare cost of rupees 2650 and a standard deviation of rupees 500. Verify the department's claim that the mean exceeds rupees 2500 at the 5% level of significance with this information. Students, now there is a very important point in this. What we are trying to test is that the department whose workers are talking about, the department's claim is that the mean daycare cost for the children of those workers, it exceeds rupees 2500. Look, here the thing is not happening that the null hypothesis will be that it is 2500 against the alternative, that it is not 2500. If we say this, then we say that it may be less than 2500 or maybe more than 2500 but it is not 2500. This is not happening here. We are saying that the department's claim is that it exceeds 2500. And that is why they are recommending the government to increase their salary because this expense is even more than 2500 rupees. In this situation, we will not be applying a two-tail test but we will apply a one-tail test. And here, note that you will always keep that hypothesis in the null. The one which contains the equal sign. What I just said does not contain the equal sign. If I say mu, the mean daycare cost is greater than 2500, then it does not contain the equal sign. So, in this question, we will place this hypothesis alternatively and we will say that h1 is that mu is greater than 2500. The opposite of this is that mu is less than or equal to 2500. That can be placed in the null hypothesis. Why? Because when I say less than or equal to, I mean that it is not greater than 2500. It is either less than 2500 or at the most equal to 2500. So, the word equal means that mathematically speaking, equal sign just can that involve hogya, that is the one that we will have to place in the null hypothesis. So, as you now see on the screen, the null hypothesis is that mu is less than or equal to 2500 whereas the alternative is that mu is greater than 2500. The next step is to specify the level of significance and as usual we may set it to be 5 percent. The third step is the determination of the test statistic, that formula which will enable us to accept or reject our hypothesis and since the sample size 100 is large, we can once again very happily apply the normal distribution and our test statistic is z is equal to x bar minus mu over sigma over square root of n. But students, a very important point to note is that in this particular problem, sigma the standard deviation of the population is unknown. Appahenge ke we had the information that the standard deviation is equal to 500 but students that was the standard deviation for the sample of size 100. Islie that is not sigma but it is s. But as we studied in the previous lectures, when we do not have sigma available to us, we replace it by its estimate and therefore we can use s instead of sigma and hence our test statistic can be written as z is equal to x bar minus mu over s over square root of n. Now, the next step is to compute the value of z substituting x bar equal to 2650, s is equal to 500 and n is equal to 100. We obtain z is equal to 2650 minus mu over 500 over square root of 100. As indicated earlier, we always begin by assuming that H naught is true and if we look at our null hypothesis in this particular problem students, it is mu less than or equal to 2500. Ab isme jo equal sign hain agar hum usko isthemaal karin to fer hum is formulae may mu ki jaga pe 2500 substitute kar sakte hain and if we do that then z comes out to be equal to 3. The fifth step is the determination of the critical region. As in the last example, there are two very important things which we have to keep in mind in order to determine the critical region. Number one, the level of significance which in this particular problem is once again 5 percent. And number two, whether it is a two-tailed test or is it a one-tailed test? I hope that you realize that in this problem it is a one-tailed test because as I explained earlier, if our alternative hypothesis says mu is greater than something or even if it says mu is less than something, then of course we are dealing with a one-tailed test. To k iss problem mein, hamara alternative hypothesis ye hai, k mu is greater than 2500. Therefore, the entire critical region will lie on the right tail of our sampling distribution. Ab agar ham area table consult karein, we find that the z value is 1.645. The reason being that the area between z is equal to 0 and z is equal to 1.645 is equal to 45 percent. Or jo area 1.645 ke right side par hai, that is equal to 5 percent. Chunke hame poora ka poora 5 percent right tail palena hai, iss liye z is equal to 1.645. Now, the next step is to draw a conclusion. Our z value has come out to be equal to 3 and hence it is definitely to the right of z equal to 1.645 and therefore it is falling very much in the critical region. As such, we reject the null hypothesis and we accept the alternative hypothesis. Students aapku yaad hai na, ke iss problem mein jo hamara alternative hypothesis tha, that was what the department was claiming. Unka claim yehi tha na, ke jo mean daycare cost hai, unke workers ke bachon ke silsle mein, that is more than 2500. To abhi hamne jo reject kia hai, that was the null hypothesis which was that mu is less than or equal to 2500. Usko to hamne reject kar diya. Iska matlab hai ke jo department ka claim hai, that is supported by this sample evidence. Alright, aapke aapku ek aur bahut important or interesting point aapke sa chere karna chaati hu. Dekhi, dono problems jo hamne abhi kia, usme hamne level of significance liya 5%. Agar ham level of significance kum karne, if we make it only 1% and even then our null hypothesis is rejected, then we say that our sample statistic is highly significant. Iss problem mein, as you now see on the screen, if we set alpha equal to 0.01, then because it is to be a right-tailed test, because our alternative hypothesis has the greater than sign, then our Z value comes out to be 2.33. Iski waja ye hai ke when we look at the area table of the standard normal distribution, we find that the area from Z is equal to 0 to Z is equal to 2.33 is equal to 49%. Which means that the area beyond 2.33 is equal to 1%. Aap iss wak thamne chuke level of significance ko 1% set kia waha hai, isliye 2.33 is the critical value. Now, since our Z is equal to 3, therefore it is lying to the right of 2.33. In other words, our statistic is highly significant. Students, yeh jo baat main ab kahi IAS pe zara gaur krte. Dhe khe, pehle humara level of significance tha 5%. Yani, the risk of committing type 1 error was 5%. Aur itna risk hum allow kar rehte. Aur iss ke tahit humne kaha keji hum reject krtehe apne h0 ko. And so we said that our statistic is significant. Aap humne yeh jo risk hai na iss ko kum kar diya. Hum katein ke hum 5% kyu hum allow kar rehte. We would rather like to have a smaller risk of committing type 1 error. To humne usko sref 1% ab hum allow kar rehte. Aur humne dekha ke even 1% jab humne kar diya, even then our statistic has fallen in the critical region. Aur sref 1.645 hi ke beyond nahi tha. Bal ke 2.33 ke bhi beyond hai. To assi situation mein we say that our statistic is highly significant. Yani jab kabhi hum yeh term use karengi. To gaya hum yeh kahre hain. That even with as small a risk of committing type 1 error as 1% we are saying that we should reject h0. Students in today's lecture we have discussed in detail some of the very very basic fundamental and important concepts of hypothesis testing. And we have also applied them to a few examples so that we have had examples of one tail test as well as a two tail test. I would like to encourage you to attempt quite a few questions on hypothesis testing and in the next lecture we will apply this concept to testing about proportions. Best of luck and Allah Hafiz.