Welcome to today's lecture on measure and integration. This is the eighth lecture on measure and integration. Today, we will be looking at a problem called the uniqueness of measures on algebras and sigma algebras. So, for this, we will need to define some term terminology. So, let us look at the uniqueness problem for topic for today's lecture. So, the problem is as follows. We are given c an algebra of subsets of a set x and S of c is the sigma algebra generated by c. So, c is an algebra and S of c is the sigma algebra generated by c. We have got two measures mu 1 and mu 2 defined on the sigma algebra generated by c such that mu 1 of a is equal to mu 2 of a for every a belonging to c. So, for all elements in c, mu 1 and mu 2 agree. The question is, can we conclude that mu 1 of e is equal to mu 2 of e for every element in the sigma algebra generated by c? So, this is a general uniqueness problem which plays a role later on when we extend measures to general settings. So, to answer this question, let us make some definitions. So, first of all, this is not true in general for all measures. We have to put some conditions on the measures. So, let us look at what is called a totally finite measure. A measure c is called totally finite if mu of every subset a is finite in that domain of mu. So, c is a collection of subsets and mu is a set function. So, we say mu is totally finite or sometimes we also say it is finite if mu of a is less than plus infinity for all a belonging to c. Then note that in case c is an algebra and mu is finitely additive, then mu is finite if and only if mu of x is finite. So, to look at that, let us assume c is an algebra and mu from c to 0 to infinity is finitely additive. So, suppose mu of the whole space is finite. Note the whole space x belongs to c because c is an algebra. Then we had seen earlier finite additivity of mu implies mu is monotone. We had seen this property earlier. So, we will not go into the details of this again. So, because mu is monotone, thus for every a contained in x, mu of a will be less than or equal to mu of x, which is finite. So, thus implies mu of a finite for every a subset of x and converse is obviously true. So, converse is true because x belongs to c and so mu of x is finite. So, whenever we are dealing with finitely additive set functions, saying mu is totally finite, it is enough to say that mu of x is finite and as a consequence, mu of every subset will be finite. So, this is a easy consequence of saying for a finitely additive set function on an algebra, mu of the whole space finite is same as saying mu of every subset a of x, a in the algebra of course is finite. Let us look at the property that let us take a set function mu and we say that it is sigma finite. So, we define what is totally finite and now we define what is called sigma finite. So, mu is said to be sigma finite if we can write the whole space as a union of sets x n and 1 to infinity such that these sets are pair wise disjoint. So, we want these sets to be pair wise disjoint and mu of each x n to be finite. So, each x n should be element in the domain of mu in the class c and mu of x n should be finite. So, essentially saying that for totally finite, we said mu of the whole space is finite and sigma finite means what that x can be cut up into pieces x 1, x 2, x n so on and mu of each x n is finite. So, this is what is called sigma finite set functions. So, what we are going to let us look at some examples of set functions. So, the length function lambda on the class of all intervals is sigma finite. So, that is easy to see. Why is the length function sigma finite? So, here is the length function lambda on the class of all intervals taking values in 0 to infinity. The whole space that is real line of course is the interval minus infinity to plus infinity and length of r, we know it is not finite, it is equal to plus infinity. But we can write r as a disjoint union of the intervals n to n plus 1, for example, n belonging to integers.  So, here is the real line. So, here is 0. So, take 0, 1, 1 close to, open to close 3 and so on and on this side minus 1 and so this is minus 2 and so on. So, we have cut up the real line. We have divided the real line into countable many disjoint pieces. Each one is an interval and length of n to n plus 1 for every n is equal to 1, which is less than of course infinity for every n. So, the whole real line is written as a countable disjoint union of intervals, each one having finite length. So, lambda on the class of intervals is sigma finite. Of course, it is not finite because the measure of the whole space, length of the whole space is equal to plus infinity. So, this is an example of a set function, which is sigma finite. We will give another example of a set function, which is totally finite. So, for that, let us look at the length function restricted to any finite interval. So, let us look at, say for example, let us look at the interval a to b and let us look at all subinterval, so let us look at i tilde a b to be all subintervals of a b and define of course the length function as before. So, a b to, so this will be a function from 0 to b minus a, length of i is equal to the usual definition of length i for i contained in a b and we know length function is finitely additive, it is countably additive and so on. So, on these intervals also it is countably additive and lambda of the interval a to b we know is equal to b minus a, which is finite. So, lambda, so one says lambda restricted to subintervals in a b is finite or totally finite for every a and b. So, this is the example of a measure, which is totally finite. Let us look at another example of a set function, which is not, so let us look at example of, let x be any set and for any subset a of x, let us define mu of a to be equal to plus infinity, if a is non-empty and mu of 0 to be equal to 0. Then obviously, this is the measure on p of x, this is a simple property that one can easily check, because mu of empty set is 0 is given and if a is any set, which is a countable, this is not union or not, so mu of the union will be equal to again plus infinity, which is equal to sigma mu of a i's, at least one of them has to be non-empty. So, I hope it is clear that this mu is countably additive, this is a measure on p x. So, if not, let us look at supposing a is equal to union of a i's, i equal to 1 to infinity. If a is equal to empty set, then a i is equal to empty set for every i. So, implying mu of a, which is 0 is same as sigma mu of a i's, i equal to 1 to infinity. The second possibility, if a is not empty and a is equal to union, so that implies there exists at least one i, i such that a i is not empty. Then, so let us say that such that i says that, so let us say that is a i not. There is at least one i, so let us say that is i not. Then, union a i not is not empty and implying that not empty. So, that is a is not empty anyway, so this is not required. Then mu of a is equal to plus infinity is equal to summation mu of a i's, because at least one term here is not empty, so this is equal to plus infinity. So, mu of a i not is equal to plus infinity, so that says this is also plus infinity, so they are same. So, this is a measure on the class of all subsets. So, mu of a plus infinity, if a is not empty and mu of empty set is equal to 0 is a measure and this obviously is not sigma finite, because there are no subsets anyway whose mu is finite. So, this is an example of a non-sigma finite measure. .. So, the theorem we want to prove today is the following, namely let us take see a semi algebra of subsets of a set x and S of c be the sigma algebra generated by c. Let mu 1 and mu 2 be to finitely additive set functions on S of c, such that mu 1 of e is equal to mu 2 of e. For all e belonging to c, then we want to show that mu 1 of a is equal to mu 2 of a. For all a belonging to first a of c, where a of c is the sigma algebra generated by c. So, we are saying as a first step we are going to prove that if two measures mu 1 and mu 2 defined on a semi algebra agree, then they also agree on the sigma on the algebra generated by that semi algebra. So, this is what we want to prove. So, let us see the proof of that. So, c semi algebra a of c that is the algebra generated by c and we are given mu 1 of e is equal to mu 2 of e for every e belonging to c to show mu 1 of a is equal to mu 2 of a for every a belonging to algebra generated by c. So, how do you prove it? So, let us start. Let us take a set a which belong to a of c, then that implies. Recall, we had shown a characterization of elements of the algebra generated by a semi algebra. So, we showed that if a is an element of the algebra generated by a semi algebra, then this a must look like a finite disjoint union of elements c i i belonging to n, where c i is belong to the semi algebra c. So, every element a in the algebra generated by a semi algebra we had shown must have a representation like this. But then mu 1 of a is equal to mu 1 of this finite union and we know mu 1 is finitely additive. So, that implies this must be equal to sigma i equal to 1 to n mu 1 of c i, but each mu 1 is equal to mu 2 on each element of c and c i are elements of c. So, that implies that this must be equal to 1 to n mu 2 of c i, but again by using mu 2 is finitely additive I can write this as mu of a, because a is a finite disjoint union of elements of this. So, mu 1 of a is equal to mu 2 of a whenever a belongs to a of c. So, this proves theorem that whenever two measures finite, whenever two finitely additive set functions mu 1 and mu 2 agree on a semi algebra, then they also agree on the algebra generated by it. . So, let us go to the next step of this uniqueness problem. So, that is saying that c be a semi algebra of subsets of s at x once again and s of c be the sigma algebra generated by c. So, this we have already proved. So, let mu 1 and mu 2 be sigma finite measures on s of c such that mu 1 of e is equal to mu 2 of, so this is a misprint, mu 1 of e should be equal to mu 2 of e for all e in c, then mu 1 of e is equal to mu 2 of a for all a belonging to s of c, whereas of c is the sigma algebra generated by it. So, let me state and we will divide the proof into steps of course. So, let us look at the statement of the theorem. Once again we were saying that let mu 1 and mu 2 be two measures which are sigma finite defined on the sigma algebra generated by a semi algebra c measures c semi algebra s of c, this s of c is equal to sigma algebra generated by c. Then mu 1 of a is equal to mu 2 of a for every a belonging to the semi algebra to show mu 1 of a is equal to mu 2 of a for every a in the sigma algebra generated by c. So, this is what we want to show. So, to show this the first step let us look at the first step. So, we may assume that c is an algebra. So, first of here we are given that c is a semi algebra. So, step 1 says we may assume that c is an algebra and that is because of the fact that we have just now shown that if mu 1 and mu 2 agree on the on a semi algebra then they also agree on the algebra generated by it. So, by the given hypothesis mu 1 of a is equal to mu 2 of a for every a belonging to the algebra generated by c. So, we already have mu 1 and mu 2 agree on the algebra generated by a of c and we want to show that this implies mu 1 of a is equal to mu 2 of a on s of c the sigma algebra generated by c. But note this is same as the sigma algebra generated by a of c that also we have shown that given a semi algebra you can directly generate the sigma algebra or you can generate the algebra first and then generate the sigma algebra both are same and just now we showed whenever two measures agree on a semi algebra they agree on the algebra generated by it. So, mu 1 and mu 2 agree on the semi algebra they show they agree on the algebra generated by it and we want to show that they agree on the sigma algebra generated by it and which is nothing but s of c. So, that proves the first step. So, as a first step in our proof we are saying that we can assume that the given class c on which mu 1 and mu 2 are defined is actually an algebra. So, that is the first simplification in the proof that without loss of generality we may assume that c is an algebra. So, next step let us look at next step says that we may assume that both mu 1 and mu 2 are totally finite we are given mu 1 and mu 2 are sigma finite. So, next step is that we may assume that we may assume mu 1 mu 2 are totally finite. So, what is the meaning of saying we may assume that if the statement. So, this is same as saying if the statement if the statement mu 1 of a equal to mu 2 of a for every a belonging to s of c is true when mu 1 mu 2 are totally finite then it will also be true when mu 1 mu 2 are sigma finite. So, that is the meaning of saying that we may assume that mu 1 and mu 2 are totally finite. So, let us see why that is the case. So, let us take a set A contained in. So, what we are given is mu 1 and mu 2 are sigma finite. So, mu 1 mu 2 sigma finite imply I can write x mu 1 is sigma finite. So, I can write x as union of x i's i equal to 1 to infinity where x i's belong to c and mu 1 of each x i is finite. Similarly, mu 2 is sigma finite. So, I can write x as union some j equal to 1 to infinity y j where y j's are subsets in c and mu 2 of each y j is finite. But then from both of these statements I can write x as. So, this implies we can write x as union over i 1 to infinity of x i's, but that I can decompose into union of y j's. So, x i intersection y j, I can write that. So, I can write this as a decomposition of x into subsets x i intersection y j. Now, what we have achieved is the following, mu 1 of each x i was finite, mu 2 of each y j was finite, but now this implies note that mu 1 of x i intersection y j is finite and mu 2 of x i intersection y j is also finite. So, now both mu 1 and mu 2 are finite on this piece. So, in the picture you can think of this as x.  So, they are decompositions like this. So, this piece is nothing but x. So, this piece is y j. So, this is this piece here is x i intersection y j. So, the whole space is cut up into pieces. So, this is what this statement means, where each one of them is finite. Now, let us note, so here is observation. So, note, mu 1, mu 2 restricted to x i intersection y j are totally finite. So, what is the meaning of this statement? They are restricted. That means, if we look at the subsets, that is, for every a contained in x i intersection y j, a belonging to C mu of a is finite, mu 2 of a is finite. And for totally finite measures, we have already proved that we have already assumed that this statement is true and we are trying to show it for our sigma finite. Now, for any set a contained in x, mu 1 of a can be written as summation over i summation over j mu 1 of a intersection x i intersection y j. So, that is because a is equal to union over i, union over j, a intersection x i intersection y j. So, this is a countable disjoint union, mu 1 is a measure. So, this must be true. And now, note that this a intersection this is a set in the set x i intersection y j, where mu 1 is finite and then that we know that there the statement is true. So, here a is contained in x. Of course, a belonging to S of C. So, the statement is true. So, that means what? So, by the assumption that statement is true for finite measures, we conclude that this is same as a intersection mu 2 of a intersection x i intersection y j and once again that is equal to mu 2 of a. So, here we use. So, a basic idea is for any set, we can bring it to the finite pieces. There we know it is true and go back to the original piece. So, this is the proof of the second step that we may assume without loss of generality that our measures mu 1 and mu 2 both are finite. So, we have made two simplifications in our proof. The first one being we may assume that C is an algebra and second one that mu 1 and mu 2 are totally finite. So, what we want to prove now? So, we only are left with the case to prove that if C is an algebra mu 1 and mu 2 are totally finite, defined on the algebra C and if they agree on C then they will agree on the sigma algebra generated by C. So, that is the next step we want to show. So, for that let us write. So, to prove the final step let us write m to be the class of all those elements of S of C where mu 1 and mu 2 agree and what is the aim to prove? Our aim is to prove that this collection m is nothing but S of C. We are picking up subsets of S of C. So, m is a subclass of S of C. We want to prove that this is equal to S of C and that is proved as follows. First we will observe then m is a monotone class. We will prove that once we have proved m is a monotone class. We will also observe that we are given that mu 1 and mu 2 are equal on C. So, C is a subclass of m and C is an algebra and C is contained in m. m is a monotone class. So, that will mean what? That the monotone class generated by C must be inside m, but C is an algebra and the monotone class generated by an algebra is the sigma algebra generated by it and that also we have proved. So, that will prove as a step 4 that m is equal to S of C. Let us prove step 3 and then conclude from it step 4. So, step 3 we want to prove. We are given that mu 1, mu 2 totally finite. We are in this case C algebra mu 1 of A equal to mu 2 of A for every A belonging to the algebra C to show that mu 1 of A is equal to mu 2 of A for every A belonging to the sigma algebra generated by C. So, that is the question. So, we are saying the proof define m to be the class of all subsets belonging to S of C for which this property is true. So, that means mu 1 of E is equal to mu 2 of E. So, claim that this m is a monotone class. So, what is a monotone class? Recall a monotone class is a collection of, so a monotone class is a collection of subsets of a set X which is closed under increasing unions and decreasing intersections. So, these two properties have to be checked. So, let us check that. So, let us take, so let E n be a sequence in m such that E n is increasing. So, E n is inside E n plus 1 for every n to show union of E n's belong to m. Now, let us note that E n is an increasing sequence, E n increases to E which is union of E n's. E n's belong to m. So, keep that in mind. So, and we want to show E belongs to, so that is the set E, we want to show that E belongs to m. That means mu 1 of E is equal to mu 2 of E. So, now note, so let us note what is mu of E? How do we compute it? Let us observe that mu 1 of E is nothing but limit of mu 1 of E n's and why is that? That is because mu 1 is a measure, it is countable additive. So, we had proved that countable additivity of the set function implies whenever a sequence E n increases to a set E then mu of E must be limit of mu 1 of E n's. That was the characterization property for countable additivity. So, this go back and refer that was because of countable mu, countable additive. So, that is the property being used here or equivalent form of it. Now, each E n belongs to m. So, that implies that E 1 mu of 1 of E 1 is equal to mu 2 of E n. So, this is equal to limit n going to infinity mu 2 of E n. So, that is we are using mu 1 equal to mu 2 on mu 1 equal to mu 1 equal to mu 2 because sorry this is because E n belongs to m. And once again mu 2 is countable additive. So, there mu 1 was count and that implies that this is so this is mu 2 of E using the fact that mu 2 is countable additive. So, let us once again we have used lot of things from which we have proved earlier mu 1 is a measure E n is increasing to E. So, by countable additivity mu 1 of E must be equal to limit n going to infinity mu 1 of E n by countable additivity. Now, each E n belongs to m. E n is a sequence E n is a sequence in m. So, that means mu 1 of E n is equal to mu 2 of E n. So, this is equal to this. So, this is the second step equality. And now once again mu 2 is countable additive E n increases to E. So, by countable additivity this limit must be sorry this limit must be equal to mu 2 of E. So, it says mu 1 of E is equal to mu 2 of E. So, that implies that E belongs to m whenever. So, E belongs to m whenever E n is a sequence which is increasing to m. So, this is for increasing and the corresponding thing we have to prove when it is decreasing. And that is where we are going to use the fact that mu and mu 2 are totally finite. So, for the second case let E ns belong to m. E n include E n plus 1 for every n decreasing and E b equal to intersection of E ns and n equal to 1 to infinity. So, we want to show that E also belongs to m. So, for that once again mu 1 of E is equal to limit n going to infinity mu 1 of E n because mu 1 totally finite mu 1 of x finite and mu 1 countable additive. And that as above earlier is same as mu 2 of E n because each E n belongs to m and that is equal to mu 2 of E. Once again mu 2 is finite and E n is decreasing to E. So, that proves this also. So, this proves that m is a monotone class. So, the class m which was equal to all subsets E belonging to S of C such that mu 1 of E is equal to mu 2 of E is a monotone class. And we are given that mu 1 of A equal to mu 2 of A for every A belonging to C. So, what does that mean? An equivalent way of stating that is saying that the collection C is inside the collection m. So, that is what it means by very definition. So, m is a monotone class, C is inside it so that implies that the monotone class generated by C must be inside m. So, because recall what is monotone class generated by a collection of subsets of C? It is the smallest monotone class of subsets of x which includes C. During the smallest it must be inside it. But note, C algebra implies m of C is equal to S of C. So, this is an important theorem which we have proved that if you take an algebra and generate a monotone class out of it that is same as generating the sigma algebra out of it. So, this is same as saying that S of C is contained in m. But m is a collection of subsets of S of C. So, that is inside S of C. So, that is same as saying that m is equal to S of C, the sigma algebra generated by C. That means what? For all elements in S of C, mu 1 is equal to mu 2 of A. So, that proves the theorem, the uniqueness theorem namely f 2. So, we have finally proved in these four steps the theorem that if mu 1 and mu 2 are two measures defined on a semi-algebra of subsets of a set X and mu 1 and mu 2 are both sigma finite then and they agree on the semi-algebra then they also agree on the sigma algebra generated by C. So, this is an important theorem which we are going to use quite often. So, with this we come to an end of part, a part of our course. So, this is probably the right stage to revise what all we have done till now. So, let us revise what we have done till now. So, we started with looking at collections of subsets of a set X. We defined what is a semi-algebra. So, what was a semi-algebra? Semi-algebra was a collection of subsets of a set X with the properties. The whole space belongs to it, the empty set belongs to it. It is closed under intersections and the complement of a set is inside, not inside, not necessarily inside it, but can be represented. So, semi-algebra C, semi-algebra meant that empty set, the whole space belongs to it, one property. The second one A and B belonging to C should imply A intersection B belong to C and the third property that A belonging to C implies A complement can be written as a finite disjoint union of elements of C. So, N for some c i is belonging to C. So, that is a semi-algebra. Then we defined what is called an algebra. So, a collection C is called an algebra. The first property as it is empty set, the whole space belongs to it. A and B belonging to C should imply it is closed under intersections that also belongs to C. Now, we have something stronger, instead of just saying that A belongs to C, this complement is representable. Actually, we want that this complement also belongs to C. So, this is something stronger. So, we said this is a stronger property. So, semi-algebra implies semi-algebra and the converse need not be true, that we had said. Then we defined what is called a sigma algebra. A collection C is a sigma algebra, if of course it is an algebra first of all. So, it is phi x belong to C. It is closed under intersections A and B belong to C, if A and B belong to C. But this is not enough. So, actually we want a countable, this should be true for any countable collection. So, let us write whenever A i's belong to C that should imply that intersection A i's belong to C and because it is going to be closed under complement. So, this is the property 1, this is second and the third property is that whenever A belongs to C, it should imply a complement belong to C and that automatically implies that C is also closed under. So, this property of countable intersections can be equivalently stated as because of the complements that A i's belong to C imply union A i's also belong to C. So, a sigma algebra is a collection which is closed under countable unions and complements and of course empty set under the whole space belong to it. Then we define what is called a monotone class. So, what was a monotone class? A m was called a monotone class, if whenever a sequence E n belong to m, E n's increasing, E is equal to union E n's should imply that E also belongs to m. So, this is one property and second property we want that whenever E n's belong to m, E n's are decreasing and E is equal to intersection of E n's should imply that E also belongs to m. So, a monotone class is the collection of subsets of a set X with the property it is closed under increasing unions and decreasing intersections. Of course, sigma algebra implies monotone class, the converse is not always true. Then we looked at, so this was the first basic concepts of properties of collection of subsets of a set X we looked at and then we looked at what are called the algebra generated by a collection of subsets or the sigma algebra generated by a collection of subsets or the monotone class generated by a collection of subsets of a set X. In all these cases, basically given a collection in C, so let us just recall what was the meaning of saying generation. So, C any collection of subsets of a set X, so algebra generated by C generated by C is the smallest one. So, it was the intersection of all the algebras which include C and we showed such a thing exists. Similarly, the sigma algebra generated by C, we said it is nothing but look at all sigma algebras of subsets of X which include C and take their intersection, so that is a sigma algebra, so that is called the sigma algebra. So, another way of saying is the algebra generated by C is the smallest algebra of subsets of a set X which includes C. Similarly, S of C is the smallest sigma algebra of subsets of C which includes C and similarly, we have monotone class generated by C. It is the smallest monotone class of subsets of X which include C and we showed by these properties that such a object always exists. Then, we proved a very important theorem namely the monotone class generated by C is equal to the sigma algebra generated by C if C is an algebra. So, this was an important theorem that we had proved. So, these concepts were basically about collection of subsets of a set X. Then, we looked at functions defined on such collection of subsets of this set X and we called them as set functions. So, set functions are functions defined on a collection of subsets of a set X and the important class of set functions was the length function and we showed that the length function had important properties namely the length function which is defined on the class of all intervals in the real line was shown to be a countably additive set function which was also invariant under translations. So, that was an important property. Then finally, we had proved some equivalent conditions for countable additivity and these conditions are very useful. So, let us just recall these equivalent conditions. We have used one of them today also. So, one of the important conditions for example, if mu is defined on a algebra. So, this is a algebra and it is finitely additive. Then, we said that mu is countably additive if and only if mu is countably sub-additive of course plus. So, let us write plus mu finitely additive. So, let us remove this condition. Let us write mu to be an algebra and mu of empty set equal to 0. So, let us put. So, mu is a set function defined on an algebra and mu of empty set is 0. Then, we proved that saying that mu is countably additive is equal into saying that mu is finitely additive and countably sub-additive and this is quite useful in proving the countable additivity of set functions. So, this was one and second we proved that mu finitely additive we assume that then mu countably additive if and only if whenever e n's decrease of course, under the condition a is algebra to e should imply mu of e n's is equal to the decrease. So, let us write decrease to e then limit n going to infinity mu of e n is equal to mu of e. This provided we had to put an extra condition mu of x is finite. So, mu of x is finite e n's decrease to e and implies that limit of e n's is equal to mu of e. So, this condition is equivalent to saying mu is countably additive when we have this. If we do not put this condition then this may not be true, but then I can also one can equivalently prove another thing that if e n's increase to e then that should imply mu of e is equal to limit n going to infinity mu of e n's. So, this was the property of saying that when is something countably additive and finally today we proved the uniqueness theorem saying that if mu 1 and mu 2 are two finite countably additive set functions defined on a semi algebra. So, this is a semi algebra mu 1 and mu 2 are sigma finite measures and mu 1 of a is equal to mu 2 of a for every a in the semi algebra. Then this implies mu 1 of a is equal to mu 2 of a for every a belonging to the sigma algebra generated by c. This is mu 1 and mu 2 should already be defined. So, we should say they are already defined in S of c, c is a semi algebra. So, c is a semi algebra. Let me state it once again c is a semi algebra mu 1 and mu 2 are defined on the sigma algebra generated by c. Both mu 1 and mu 2 are sigma finite and mu 1 and mu 2 agree on the semi algebra. Then they agree on the sigma algebra also. So, they agree on the whole domain. So, if they agree on the part of the domain which is a semi algebra, then they agree on the whole of the sigma algebra also. That is a uniqueness result and that we proved under the condition that mu 1 and mu 2 are sigma finite measures. We will see this how that is used in extension theory in the next few lectures when we come to them. So, that is what we stop today here. So, in the next lecture, we will start a new topic called extension theory. So, we would like to extend a set function defined on a class to a bigger class. For example, on the real line, we have the notion of length defined on a collection of intervals on the collection of all intervals. We would like to define the notion of length for any set. So, that is the aim of that is a motivating thing for extension theory. So, we will use that and prove the theorems in the next lectures. Thank you.


Hey guys, welcome back to our channel. My name is Andrea with Food and Mentory Adventures in Food and I am here today with an Aldi haul. I spent $112 in some change, so let me jump right in. I'll start back here with the produce. I got an Asian chopped salad kit and this will be for my lunch during the week. My husband doesn't really care for this, but I like it a lot. They were really low on their produce. They said that there's a supply issue. There's a side up. Normally I like to get their Caesar salad, but they did not have any Caesar salad at all. I also just got some Italian salad. Normally I prefer their butter lettuce, but they did not have any butter lettuce. I grabbed some avocados for my husband, a bag of the minis, and then he needed one that was ripe for our dinner tonight. I grabbed a couple of packages of pepper jack cheese for recipes. I already have like a bunch of cheddar and Parmesan in the fridge. These cherries looked so good. So I just grabbed a package. I think there were 4.25 or 4.99 a pound. I can't remember. I have wanted to try this marble cinnamon breakfast bread for a while. It is really heavy, smells really good, smells just like cinnamon. Let me flip it over on the back so you can see the ingredients a little bit. So here they are. I plan to just butter and toast this bread. I needed some rice and I really like their jasmine rice. I needed these slider rolls for a recipe. I prefer these or store brand to the regular name brand Hawaiian rolls because the Hawaiian rolls, the name brand, they have like soy flour and stuff in them and these don't. So I prefer these and then I grabbed some herb seasoned stuffing mix. I like to have this. This doesn't expire until next year. These were on clearance for $1.25 which is a great deal. I grabbed some eggs. I didn't need them but they were really good price. I think these were $1.12 or $1.18 for a dozen eggs which is a great deal. I occasionally have recipes that call for a family size of the cream of chicken soup and they had it at Aldi. So I grabbed two cans of cream of chicken soup. I grabbed some sweet potatoes just to eat as a baked sweet potato or I'm not sure what I'm going to do with it but I really like sweet potatoes. These bars I have had the hardest time finding them. I really like them a lot.  Most of the stuff I can identify and I just like to have these on hand. If you see them grab them because they are really hard to find. Another thing that's been hard to find are these fiber bars that my toddler likes. So I grabbed three boxes. Let's see what else. Let me go back here. I grabbed their yogurt. I think lemon flavor is a seasonal flavor. I bought this before and I really liked it so I grabbed it again. I needed some heavy whipping cream and half and half and I also needed some milk. Now their organic milk is about a dollar cheaper than what you'd find in the grocery store. So these were 303 each which is still cheaper than a gallon of organic milk. Grab some of this lemonade for my husband. It was on sale. I think it was like a dollar 30. I grabbed some jumbo flaky biscuits and then some cinnamon rolls. I can't remember if I've ever had their cinnamon rolls but wanted to give them a try. I needed some more clubhouse or club type crackers. So grabbed some of them and then grabbed some of these for my son. Let's see. I've wanted to try this salsa for a while. Chimichurri. You get the roasted tomatoes, parsley, garlic, oregano and red jalapeno peppers. I've never even heard of red jalapeno peppers. This is one that I used to buy all the time and I really like it a lot. Restaurant style. These little mini graham cracker crust were on sale for 99 cents each and I like to make a lot of cheesecake type desserts and this is perfect for portion size. And then last year I missed these fruit bowls on sale and you get, let me see, it drains the juice out and it keeps your fruit fresher longer so that you can either buy like a large bowl and I don't remember the size or the two pack. So these will do better or fit better in my fridge. So grabbed a pack of those and I think they were $9.99 each. Grabbed some pull ups for my toddler. I've heard a lot of great things about this, about these big dipper corn chips, Mexican style street corn and then another favorite of mine I haven't bought it in a while are these 100 calorie snack packs chocolate chip and they also have like an Oreo type flavor. So this is everything that I picked up on my Aldi haul. I hope you all enjoyed it. If you did please make sure to give us a thumbs up and if you haven't subscribed to our channel we'd certainly love to have you. Thanks so much for watching and we'll see you all next time.


And now, Auto-Lite and its 60,000 dealers and service stations present... Suspense! Auto-Lite brings you William Bendix in break-up. A suspense play produced and directed by Anton M. Lieder. Suspense, Radio's outstanding theater of thrills, is presented for your enjoyment by Auto-Lite and its 60,000 dealers and service stations. And friends, today is the very first birthday of Auto-Lite resistor spark plugs. Yep, and a more popular one-year-old you never saw. That's why car owners everywhere are switching to Auto-Lite resistor spark plugs and no wonder. For when you replace your narrow gap plugs with wide gap Auto-Lite resistor spark plugs, your car will idle smoother. You'll have better luck with lean gas mixtures, actually save gas dollars. What's more, wide gap Auto-Lite resistor spark plugs cut down spark plug interference with radio and television reception. So friends, don't debate, delay, defer or defray, but drive down tomorrow to your nearest Auto-Lite dealer and switch to a set of sensational Auto-Lite resistor spark plugs. You'll be glad you did. And now, here is William Bendix in a tale well calculated to keep you in... Suspense! Let him alone, Chick. Max, tell Chick to let him alone. All right, Chick. Just use him for a foot rest. It ain't often a fella gets a chance to wipe his feet on a cop. Where we taking him, Max? Oh, to the end of the line, Marty. Then we give him a transfer. A steel jacket transfer.  Chick gets the picture. When, Max? When we get over the bridge. Yeah, we'll make Kevlahan a good cop. And then we'll dump him. Chick, I said enough's enough. Make it stick this time, will you, Max? You're awful sensitive for an ex-cop. You with us, Marty? What do you think? I think maybe you're remembering that before they bounced you off the force, you and Kevlahan once teamed together on the homicide squad. Eh, once had long red curls, too. No, Max, what I'm remembering is the raw deal that Captain Brandt gave me. And the little murder rap Kevlahan's been just sweating to pin on me. Marty, tell you what. Yeah? You're all the time asking for a chance to climb up in the organization. I'm going to give it to you now. I'll let you combine business with pleasure as a personal favor. I'm listening. I'm going to let you put the quiet on Kevlahan all by yourself. How's that? I didn't hear you, Marty. I said, how's that? How's that, he says. How's that? I leave it to you, Captain Brandt. You dealt the cards for this hand. How is it? Me, Marty Connors, turning cop killer. Are you satisfied, Brandt? We're on the Queensboro Bridge now. Me and Kevlahan. Remember your tough team of Connors and Kevlahan? Only this time he ain't my partner, Brandt. He's my gun meat. I guess I should have expected anything. Anything at all once I started to roll on my own. No holds barred since the breakup. Just a lot of dirty infighting. From the day Brandt put the skids under me in front of those newspaper vultures. Morning, Chief. Hi, Kev. Hi. What are those reporters doing outside? Hand me a shield, Connors. What? I said, hand me a shield. You're suspended on decision of the board. But... You hurt me. You're going up on charges. But Brandt, give me a break. I'll give him a break, Captain. Break? Tell me, Connors, where were you last night? Well, I... I was... I had a date. At a bar, huh? With a dame you never saw before? Well, we had a few drinks and... And what happened after that? Well, I... You see... I'll tell you. You're the only one who doesn't seem to know because everybody else read it in this morning's paper. Strange dame got you dead drunk at a bar, rode you for your gun when you passed out, tried to stick up, shot the joint, the splinters, and was collared still carrying your gun. Now, you know? Do you still expect a break? Sure, I expect a break. In the first place, I was off duty at the time. And in the second place, I've been on the force 11 years. And third place, if it was 111, Connors, I'd still bounce you and bounce you hard.  Here! Edit to your collection, Captain Brandt. I'm sorry about the verdict, Marty, but don't blame Brandt. Oh, sure. Don't blame Brandt. Blame me. I'm the first cop ever gets drunk, huh? Brandt, don't drink. You don't, huh?  It's a laugh. Oh, stop it, Marty. What are you going to do? Going out to Taiwan on. What did you think I was going to do? What are you trying to prove? Oh, Marty, look, you didn't get jammed up because of a drink now, and then it was a dame. With you, it's always a dame.  You let him make a sucker out. Lay off, kid. It's the truth, Marty. I said, lay off! Lay off! Okay, then. Marty, let's not...  How, how do you fix for money? I got enough. And pretty soon, maybe I'll have a lot more. Big bundles of it. I've already got a call from Doc Williamson and Max Shale. What? Nibble on that one, Kiv. Marty, are you crazy? Steer clear of them. Why? Show me why. You know as well as I do. If you don't, well, just ask any of the boys on the Narcotic Squad. Here. Oh, Marty, stop. Shorter. I can see where your head at Justice Plain is. Listen to me, Marty. No, no, listen to me. You tangle with Williamson and Shale and no telling what can happen. One of us might one day have to go after you. It might be me. You ain't on Narcotics. Narcotics and homicide don't mix, huh? Mercy me. It gives his Sunday school. After the deal I got... Look, if you'll only mark time for a while and not lose your head, the newspaper blast will die down and I'm sure the Commissioner will give you another... Nuts to the Commissioner and Nuts to Brent. This to him, look, all five fingers. Now to you, Kivland. And you stay away from here. Stay away. I mean it. I'm going to go line up that job right now. So you're Marty Connors, huh? I can use you. I can be used? It pays a hundred and a half a week. I need a guard. You know the set up Max and me are in? We got a nice club, nice customers, a lot of them rich. It attracts cheap guns. And they figure maybe they can knock the place over and take all the door on the gambling tables. Now, you have a reputation for being hard. That might stop the trouble even before it begins. That would be part of your job. And the other part? I carry a lot of money and one or two other things. I want personal protection. Well? It's a contract. But is that all I do? Don't I get in on the big stuff? Not the gambling, the big stuff. I don't know what you're talking about. The dust, Doc, the dust, that's what pays off. I still don't know what you're talking about. But when I think you're worth more, I'll see that you get a chance to earn it. Until then, don't let your nose get dirty. You know what I mean? I'm a too hanky man. But who's the long-legged redhead at the piano? Do I get a chance to protect her too? I handle that myself. Leave it strictly alone. Understand? Sure, sure. And I'll blame you. Not the least little bit. So I went to work as a stalker for Doc Williamson and Max Shale. I'd call for Doc every night at one, take him to the club and then walk around the tables. And that was a bottom of the heap for weeks. I wasn't making any headwind in that direction. But I wasn't another. Rita, I learned her name was Rita. Sometimes she'd be there when I called for Doc. But I might have been collecting the laundry for all the attention she paid. Until one time I rang Doc's doorbell and she answered it. Hello, big fella. It's how Doc, I'm here. He's not in. When will he be back? Not soon. Maybe. Won't I do? Didn't he leave word? He didn't leave anything. Except me. I'll wait for him in the lobby. What's the matter, Marty? Don't you like me? I like Doc's watch and chain, too, but I don't heist it. Why not, you big enough? Maybe you take better care of it than he does. Maybe he treats it bad, leaves it around lonesome, and it's that mind case. Don't you need a watch, Marty? You can't tell. Might come a day when I'll need the right time, bad. I thought about that watch a lot the next few days. The next time I rang Doc's doorbell and she answered it and said... Hello, big fella. Tell Doc I'm here. He's not in. You lovely, lousy, long-legged redhead. Marty, Marty. Marty, Doc will be back in a minute. Where can we meet? I gotta see you. My place tomorrow? Yeah, your place tomorrow. In the next day. In the next. I'm afraid of him. I'm afraid of him. Right, Doc, he's just a yellow slop. I think he knows about you and me. We've been seeing each other the last few weeks. I'm scared. Marty, he beats me. Once he threatened to kill me. Well, the next time he tries to flash this, you'll see how quick he'll fall here. Take it, it's only a little twenty-five automatic. Put it in your purse. Go on, take it. Do as I say. Hold it in reserve. He raises a finger till you draw fast and watch him retreat. You know, I feel better already. Sure, that's why they call you the old equalizer. Was it about me he threatened to kill you? No. Who, another big fella? No, no, nothing like that. He once found me going through his bureau. What's the secret? He wears a girdle? All right then, don't tell me. Have another drink. Marty, you know what Doc and Max do for their really big money? Sure, they peddle dreams. That's no secret, not really why. You know the setup? Yeah, I'm just an arm, a yard and a half a week. Well, I know the setup. I'm the only one who does aside from Doc and maybe Max. Yeah, but go on. Max does most of the work. He's distributing the stuff.  You know why? Because before he can distribute the stuff... He gotta have it to distribute. Doc's the buyer. He has the contacts all over the world. And guess where he keeps what he knows? The address is cold shipping dates. Well, how would I know? In a little green notebook in a chamois bag, hung around his neck. That's why I hired you, Marty. Oh, Marty. Marty, do you know what it could mean to us if we had that notebook? Yeah. He wears it on him, huh? Around his neck. Rita, Rita, listen hard. This evening before I come to pick him up, snuggle up to him. See if he can feel that chamois bag. If he's wearing it, give me the nod when I call from at one. Do you follow me? Oh yeah, Marty, I follow you. Maybe I'm even ahead of you. I went up there at one. I rang and rang, but no one answered. When I tried to nod, the door swung open. Rita wasn't there. But Doc was. He would never be deader.  The room still smelled of cordite. A twenty-five shell lay near his feet. His body was still warm. I know because I felt it when I reached for the chamois bag. Which wasn't there. For suspense, auto light is bringing you William Bendix in radio's outstanding theater of thrills. Suspense. Did you invite Mary and me to the studio just to hear you talk about auto light resistors? Talk about them. Why, by conniving Cornelius, we're going to celebrate them. Because half you have the honor of being present at the first birthday party of those super sensational auto light resistor spark plugs. Gee, I don't know. Have you got a birthday cake? Have I got a cake? Well, I've got a cake studded with auto light resistor spark plugs instead of candles. Hey, pull the card off that nearest plug. Happen, read what it says. Well, let's see. It says, replace your narrow gap plugs with me if smoother idling you would see. If you want more miles to cost less money, cuddle up and be my honey. Assigned ALRSP, guess who? ALRSP, auto light resistor spark plugs, of course. Try another one, Happen. Many happy returns of the day, old light resistor spark plugs say, PS, to cut down interference on your radio and television, see that your car we is in. That's a rhyme? Well, now you wouldn't accuse me, would you, Happen? Well, anyhow, Happen, now that we've had our fun. Let's get back to suspense. And now, auto light brings back to our Hollywood sound stage, William Bendix as Marty Connors in Breakup. A tale well calculated to keep you in suspense. I got out of Doc's apartment fast. Left it just as I found it. Him dead on the floor with a shell beside him. I had to find Rita, poor kid. She'd really be in a mess now. Everybody knew she was Doc's girl. I had to locate her in the chamois bag. I took a cab to her flat and let myself in. No, Rita, that meant I would have to go hunting, but not without a gun, not now. I went home to pick up my police positive. It was in my bureau drawer. And snuggling next to it among the socks was Rita's little 25. With one chamber empty and a smell sharp as incense. So I was being suckered again. Oh, well, give your old son of a gun. Come in, come on in. Can I fix you a drink, give? I ain't here as social, Marty. No? No. Remember I told you I might be sent after you one day? Well, I'm sent. What are you talking about, give? The elevator boy found Doc dead a few minutes ago. So? So you're the only one he took down from that floor since 10 o'clock. Well, give, you know what to expect. From a sleepy elevator, kid. Why, uh, ain't you with Doc tonight? Well, it's just my night off. Give, you don't really think I've turned killer. Why not? Oh, by the way, I know about you and that redhead. Come on, let's go, Marty. Brant, what's to talk to you? Brant, huh? I should jump at that. I don't hop bells for Brant now, give. You know that. As far as I'm concerned, you and Brant...  You don't make... You don't make me. If Brant wants to talk to me, he comes here with a warrant yet. Then, I'll have to take you... If you want it, copper! You drawn on... Stay where you are, give. You ain't bringing me in. You big ape. You're trying to nerve yourself up to it? Oh, Marty, Marty... Back, give! A guy would think you meant it. I do mean it. I swear I'll let you out. Not you, Marty. Not a next... I ought to keep your nose out of it for a while, give. Sixth floor, sir. Huh? Oh, okay. Mr. Shale's apartment is just around there. Thanks. I've been so worried about you. I've been sitting here darned by inches, Marty. I didn't know...  I swear it. I swear it by all that's holy to me. Such as what, a dollar bill, maybe? Marty, listen to me. When I got to the doctor, he was waiting for me. He knew all about us. He began to hit me. I did what you said, Marty. I took out your gun. My finger must have slipped. Oh, yeah? What's your frame, Marty? I love you. Lots of men would like me to say that to them, but I love you. All right, stop it, stop it, stop it. I didn't mean to hurt you. After I lost my head, I went to your place, but you weren't there. I put the gun in the drawer. I thought it would be safe. And I waited for you, but you didn't come. So I came here. I thought maybe Max would have saved me. It's all right, baby. It's going to be all right. Now, shh, quiet. Let me have the chamois bag. Rita, the chamois bag. Let me have it. I, I haven't got it, Marty. I didn't even look for it. I couldn't have touched him. You're not lying to me now. It wasn't on him when I found him. He was clean. Rita, shh, shh, shh. Who would that be? Max, maybe? Max would have come right in. Who is it? A elevator boy, sir. What do you want? The lady, Miss Carter, lost the gloves in the elevator. You were wearing gloves? I don't remember. All right, boy, give them to me. Hold it, Marty. That's enough. Now you're covered. Okay, son. Thanks. Get back to your elevator. Yes, sir. Back in, Marty. Go on. Both of you, get back in. Well, you've lost a lot of savvy since you left the squad, Marty.  Huh. Marty Connors and Doc's baby together. Well, the DA won't have much of a job with this one. No, I don't know anything about it. Easy, baby. Easy.  A copper gun, Ratsey. Who'd have figured it? Get away from that phone, Marty. No, sucker, Kiv. We're getting a lawyer. You've got things to ask. Wait for him. You move again without I say moving. I'll plug you. You wouldn't, Kiv. Try me.  And I'm tricking you like one. The palsy stuff is through. You'll see Brandt first you and the girl, and then you can call a lawyer. Let me at that phone. Drop it, Copper. You're covered. Hand him, Chick. Make sure he's clean. Okay, Marty. Relax. This'll get you nothing but tears and moans, Shale. Is he clean, Chick? Okay, cold cocking. All right now. What's it all about? Well, when I call for Doc tonight... I'll get to you in a minute, Marty. Talk, Rita. Doc's dead. Good. Who did it? I did. He showed fight when I told him to hand over the notebook. Rita, you...  Go ahead. That's all. Except, uh, here's the notebook. Ah, nice going, baby. Got it at last. Rita. You've been stringing along with him? Sure she has. If it's any of your business, which it ain't. You lousy redhead. What's he so heated about? Well, I used his gun and planted it on him. You gave her your gun for the job? That guy's getting to be a habit with you. Framed spotting the frame and then talked out of the spot by a lousy redhead. Forget it, Marty. The frame is out. I need an arm like you, now that Doc is dormant. All sorts of guys will get ideas about moving into the business. Look, here's the pitch.  Uh, because he asked for it. He was afraid of something. Didn't say what. That's your story. Stick to it. I'll set up an alibi for you for the rest of it. Leave it to me. Okay? Okay. You lousy redhead. What does this copper know? Too much. We dump him right now.  Chick, take care of it. Eh, baby. You'll leave in town in a hurry before they grab you. Start packing. All right, Maxie. Now give me a hand with the cop, Marty. We'll take him off the back way. I said give me a hand, Marty. Going to play it smart or not? Yeah. Yeah, sure, Max. I'm going to play it smart. I'm still waiting for your answer, Marty. I've been waiting all the way across the bridge.  How's Kevlin, Chick? Nice and dark, ain't it, Marty? Eh, this'll do. Pull up here at the lights. I guess we're ready.  Chick, roll out the cop. Wait a minute, Max. Kevlin is... Kevlin goes. You don't pull that stuff, Marty. It's too late. Not Kevlin, Max. Chick, roll him out. Max?  We're two guns to your one. You're with me now. I'll help you go with Kevlin. Not Kevlin, Max. Chick, get him! Kevlin, you all right? Okay, Marty. Didn't touch me. You? I'm okay. Let me get Max's overcoat off. I'm cold. You don't need it no more anyway. Let's dump them both in the backseat. Okay. That does it. You want to take the wheel, Kev? Yeah, sure, sure. This is almost like old times, Marty. You and me together. Yeah. But... But I'm still going to have to take you in to see Brandt. It's a good trick if you can swing it. Huh? I told you, if Brandt wants to see me, let him come. Marty! Marty! How's he going to be, Doc? A matter of minutes. If Captain Brandt wants to talk to him at all... I've phoned him. He should be here now. Kev. Right here, Marty. Right here. You're going to be okay, Charles. Yeah, Kev. I heard the doc. The top coat I was wearing. The one I took off Max. It's right here, Marty. Yeah. Here it is. There's a little notebook in one of the pockets. I'll cut it so I can see it right away. Yeah, that one. Holy cow. Marty, you know what this is? What's in it? You see, Kev. I told you Brandt would have to come to me if... if he wanted to see... Marty. Marty! He can't hear you. He gave me this, Captain. You know what it is? It would have put him right back on the force. He was never off it. What? That's right. It was a plan. Right from the time he let the dame shoot up the saloon with his gun.  Don't look at me like that, Kev. It was a good idea and it worked. And I ain't ashamed of having thought of it. It was the only way to get the goods on these dope runners. Yeah, but that... It's for Marty. Well, he was a good cop. And after all, this is what he was paid for. Thank you, William Bendix, for a splendid performance. Mr. Bendix will return in just a moment. I don't know. Have you got any more birthday banter about auto light resistor spark plugs? No banter half. Just facts. For the fact is, those wide gap auto light resistor spark plugs are wonderful. They're sensational. They're ignition engineered. And there's a type for every make of car. Why, by Cornelius, everyone ought to switch to a set of auto light resistor spark plugs right away. And now, Hap, here's one more rousing spark plug rhyme. Let's read it together. Okay. You can tell the difference in your car with auto light resistor spark plugs, Thar. Well, Hap, you certainly ought to remember this auto light resistor spark plug birthday. And friends, here's something for everyone to remember. Auto light means spark plugs. Ignition engineered resistor spark plugs. Auto light means batteries. Stay full batteries. Auto light means ignition systems. The lifeline of your car. And now, here again, is William Bendix. Thank you. It's always a pleasure to appear on suspense. This is probably every actor's favorite program, and I know that... Help! Help! Let me out! There's someone in that closet. Well, Jack, Benny, what are you doing in there? Oh, hello, Bill. I am glad somebody let me out. Well, I don't get it. What's the idea, Jack? Well, you see, Bill, I'm moving my program to CBS Sunday nights, and I thought I'd drop in and look over the studio, take one out. And then, all of a sudden, all that shooting started, and I hid in the closet. Oh, well, I'm sure glad you didn't stop any of those bullets, Jack. We all want to hear you next Sunday night on CBS. Your program has always been on my must list, along with suspense. Well, thanks, Bill. By the way, what about next week's suspense? Next week, Radio's Outstanding Theatre of Thrills presents Gene Kelly and Ethel Barrymore in To Find Help. Another gripping study in... Bill, wait a minute. Wait a minute. Wait a minute. Well, let me say it, will you? You know, that one word. Let me say it. Oh, well, sure, Jack. Next week, hear Gene Kelly and Ethel Barrymore in To Find Help. Another gripping study in... Suspense. Who said I can't do drama? William Bendix appears through the courtesy of his own radio show, The Life of Riley, sponsored by Prell, and will soon be seen in the Paramount picture, The Streets of Laredo. Wally Mayer was heard as Kivlehem. Tonight's suspense play was adapted for radio by Joseph Ruskall and Walter Newman, from the original story by Thomas Walsh. Music was composed by Lucian Morawek and conducted by Lud Bluskin. The entire production was under the direction of Anton M. Lieder. In the coming weeks, suspense will present such stars as Danny K., Robert Montgomery, Dana Andrews, and many others. Make it a point to listen each Thursday to Suspense, Radio's Outstanding Theatre of Thrills. And next Thursday, same time hear Gene Kelly and Ethel Barrymore in To Find Help. This is the AutoLight Suspense Show. Take advantage of high prices for steel scrap. Turn in yours to your local scrap dealer. Good night. Happy New Year from AutoLight. This is CBS, the Columbia Broadcasting System.