 Yes, last class we were discussing Borel sigma algebra on 0 1 and we were building up towards building building up towards defining the Lebesgue measure or the uniform measure on 0 1 right. So, the Lebesgue measure it tries to formalize the notion of length right that is really all it is intuitively, but it is a it is a legitimate measure right. So, we took this collection f naught of subsets of 0 1 which includes the empty sets and sets of this form right by which I mean. So, this a 1 so a 1 open closed b 1 and this finite union of such sets where these guys are all disjoint right. We said that f naught we should we could easily prove that f naught is an algebra we could also prove that f naught is not a sigma algebra and we finally, said that the smallest sigma algebra containing f naught is in fact, the Borel sigma algebra right. So, these 3 facts are very important. So, we will use the use it very soon. So, what do we really want right. So, on sets of this form I want the measure of let us say let me call this set let us say f let us say have an f of set f of this kind I want ideally I want a probability measure which assigns probability of the set f is equal to what I want it to be b i minus sum over b i minus a i i equal to 1 through n right. I have to define a probability measure such that this is satisfied this is what I really want right I want the measure of intervals to be b i minus a i right. Now, the problem is this p has to be defined on a sigma algebra right I only have an algebra right. And on this algebra f naught I know what I want on this algebra right, but on the sigma algebra I do not know what it is right. Because the sigma algebra will contain more complicated sets the sigma algebra after all is the Borel sigma algebra right. And it has some complicated sets they are not necessarily nice sets of this form. So, I do not know what the measure on the sigma algebra is, but I know what I want on the algebra f naught right. So, that is the key step now. So, the key step is in going from the algebra sets of this form to the Borel sigma algebra right. So, this is accomplished by a very well known theorem in measure theory called Carre Theodoris extension theorem right, which I will state without proof we will just utilize the result rather than worry about its proof ok. Let f naught be an algebra on omega and let f is equal to sigma of f naught. Suppose p naught is a mapping from f naught to 0 f naught to 0 1 such that p naught of omega equal to 1 and p naught is countably additive on f naught. Then p naught can be uniquely extended to a probability measure p on omega f that is there exists a unique probability measure p on omega f such that p of a is equal to p naught of a for all a in f naught. This is the famous Carre Theodoris extension theorem. So, our aim will not be to prove this and our aim is to simply comprehend what it saying and apply it. So, the proof is the proof is a constructive proof which can be found in David Williams book for those who are interested, but it is not something that we will bother with. So, let us try to understand what this theorem says. So, you are given some sample space omega and you are given an algebra on omega not a sigma algebra just an algebra on omega and you are trying to. So, and what you are doing here is to define some kind of a pseudo measure on the algebra this p naught is see this is not defined on the sigma algebra it is defined on only the algebra f naught. And it has and it is a mapping from f naught to 0 1 it has the property that p naught of omega is 1 and p naught is countably additive on f naught. If these two conditions are satisfied then p naught can be uniquely extended to a probability measure on omega f and what is f? f is this guy sigma algebra generated by f naught f naught is only an algebra and you are making it into a sigma algebra. So, to speak you are taking the smallest sigma algebra containing f naught. So, you can extend this pseudo measure p naught to a legitimate measure p on omega f such that this p agrees with your p naught on the algebra correct is this clear. So, there is a few things I have to clarify here. So, this p naught is not a probability measure that is that should be very clear to you p naught is only if it is a pseudo measure defined only on f naught which is an algebra it satisfies p naught of omega is equal to 1. Because, you ultimately want the true probability measure p to satisfy p of omega equal to 1 and. So, this is this is important it should be countably additive on f naught p naught the pseudo measure that you have should be countably additive on f naught. Now, comes a problem. So, this f naught is what kind of an object algebra. Now, what is countably additive? If I have the p naught of a countable union of disjoint sets must be equal to sum of the p naught of those individual sets. Now, the problem is if you take p naught of let us say you take p naught of countable union of disjoint sets the countable union of those sets in f naught may not be in f naught why it is only an algebra f naught is only an algebra it is not a sigma algebra. So, then what do we mean by countably additive on f naught p naught may not even see p naught is defined on elements of f naught. But, if I take a countable union of sets in f naught it may not even be in f naught then what am I talking about as countable derivative. So, this should be understood as if it so happens that the countable union of sets in f naught is an f naught it may not be it may not always be right because f naught is only a algebra not a sigma algebra. So, if it so happens that if you take a countable union of disjoint sets in f naught if it so happens that the union is also an f naught then you must have countable derivative that is the sense in which this should be understood. So, if these two are satisfied then I can uniquely identify a measure p which agrees with p naught on my nice sets on my algebra. But, the thing about this p is defined for all elements of f not just on f naught there exists a unique extension not just there exists an extension, but there is a unique extension are there any questions on this statement all right will be. So, this is the theorem that helps us go from what we actually want. So, what have I done right. So, I want a probability measure to satisfy this and a. So, essentially I know what I want for subsets in f naught I know p naught of f should be this right. But, if I verify the conditions of caratiodary I can actually go to a legitimate measure on 0 1. So, the steps are as follows. So, clearly see f naught with the f naught defined here is already an algebra right. So, going back to that. So, we have f naught defined earlier is an algebra and sigma of f naught is borrel right that much we already have. And now define p naught from f naught to 0 1 such that p naught of null is equal to 0. See I have to define p naught for all elements of these f naught f naught contains the empty sets and sets of this form. So, for empty set I define it to be 0 as I must right. Because, I eventually want this p naught to become a measure right and p of f where f is a set of that form sorry p naught of f set of that form should be simply i equals 1 through n b i minus a i right. So, this f is I am not repeating this this f is a set like that. So, I have defined this pseudo measure on the algebra f naught now the only non trivial step which take some work. And I will not do here is that you can verify countable additivity of p naught on f naught what do I mean. So, if you take a collection of sets of this form collect. So, you take a countable collection of sets f i which are of this form and there this joint let us say. If it so happens that the countable union of several such sets is also a set of this form see it may not be for one thing right. Because, again as I showed an example of a set of this form where you take countable union you end up with something that is not like this right yesterday we proved that right. So, it may so happen that if you take countable union of several such sets it may not be a set of this form. But, if it is a set of this form you can in fact show that p naught is countable additive you follow what I am saying this is not this is not way trivial again right. This if you are interested this is not a very inspiring proof, but if you want to really look at the proof you can look at David Williams, but this can be done right. So, next so non trivial step is to verify countable additivity of p naught on f naught. So, you do that it actually it can be done there is a proof then what you have you can apply Cala Theodore's extension theorem right the conditions of the theorem are satisfied right. Then, so you that implies that this p naught can be uniquely extended to a legitimate probability measure on omega f, but what is your f now boron sigma algebra right which means. So, what did I just say so there exist a unique probability measure on omega comma b which agrees with the notion of length for intervals right, but it is a legitimate probability measure in the sense it is defined for all boreal sets is that clear. So, this Cala Theodore's implies by Cala Theodore's there exist a unique probability measure p on omega b which agrees with p naught on f naught. So, I have just indicated the program right I have not really proved all this, but I think for our purposes since we are not professional mathematicians I think we can afford to skip the proof of Cala Theodore's theorem and all right. So, we can just to understand how this Lebesgue measure is properly defined the program for doing it is what is relevant to us. So, what the statement means in plain English is that for all boreal sets on 0 1 I can define a probability measure a unique probability measure which corresponds to the notion of length after all what is p naught p naught is length right simply if you give it a interval like open a closed b gives you the length of it right. So, it is a measure unique measure that agrees with the notion of length, but it is a legitimate measure in the sense that if you give it something some complicated boreal set which is not necessarily just intervals it is still produce a measure it produce an answer right. You may not be able to a priori tell me what the length of something like the canter set is because it is because it is so broken up right yesterday we discussed that right. So, you keep removing these middle thirds you ended up with the scattering of points which is so bizarre that it contains no intervals at all right, but it is a boreal set you will prove in your homework that is a boreal set right. So, even if you give it a complicated boreal set this p will produce an answer because it is defined on all boreal sets, but this p naught will not with the p naught does not operate on complicated sets like the canter set it only operates on these kind of sets is this clear probability measure p naught is not a probability measure because it is defined on an algebra right it is a pseudo I said it is a pseudo measure because it is only defined on the algebra, but we want it to become a legitimate probability measure on the boreal sigma algebra well no see countable additivity means see countable additivity of probability measure means if you take a countable union of disjoint sets you should have the sum of the measures on the right. So, the point I was making is p naught is not really a probability measure, but it should still satisfy countable additivity on the algebra, but of course because f naught is only an algebra countable union of disjoint sets may not even be in the algebra right. So, this statement will be interpreted as if at all the countable union is in your algebra of it. So, happens to be in your algebra then it should have countable additivity yes that is the that is the property of it, no I have not no the translation invariant part see translation invariant part earlier what I said this that there is no translation invariant measure on omega 2 power omega right. What I have said now is that I have a measure that uniquely satisfies. So, it satisfies the notion of length on intervals and it is a legitimate measure on the boreal sigma algebra right. What you can prove now is that this p is in fact translation invariant on the boreal sigma algebra not on 2 power omega p is only on boreal right. You cannot do p on 2 power omega right because it is to be there is an impossibility theorem. So, now this p has satisfies this notion of translation invariant and corresponds to a notion of length on boreal sets any other questions. So, now the task is. So, we know that there exist this measure right now the question is. So, but we only know the measure and explicit form for sets like that right. If I give you a set open a closed b you will tell me that the measure is b minus a right. But if I give you some other set you do not know you know that unique measure exists on boreal sigma algebra, but you do not know what it is. So, what we will try and do now is give you measures of compute measures of certain commonly encountered subsets. So, definition. So, let me actually not let us forget this p is called let me just right here this p Lebesgue measure 0 1. So, I strictly I should say 0 1 comma b right. So, this unique measure Karate theorem tells me that there is a unique measure on the boreal sigma algebra which corresponds to the notion of length all right this measure is called the Lebesgue measure. It is also as see because this interval is only the length of this interval is 1 it is also a probability measure on 0 1 why because p naught of omega itself is 1. So, if it so happens that your interval is 0 2 right then the length will be 2 right. So, in that case you can similarly define a Lebesgue measure, but it will not be a probability measure in order to make it a probability measure on 0 2 you have to scale things down by the length of the interval right length of your omega right. So, once you have defined uniform probability measure on 0 1 you can do it for any interval a b right you can define a uniform probability measure on any interval a b of the real line. So, it so turns out you can also do boreal sigma algebra and Lebesgue measure on the whole real line that is something you will get to very soon, but before that I want to explore this Lebesgue measure on 0 1 a little bit more. So, everybody with that so that is just a terminology this measure p unique measure p is called Lebesgue measure. So, now so we have singleton b all right. So, we know that a singleton is a boreal measurable set we proved it yesterday right all the day before I do not know right we prove express b as a countable intersection of open intervals and then proved that therefore, singletons are boreal measurable sets if it is a boreal measurable set or Lebesgue measure assigns a value to it right intuitively what you think should the value of a singleton should be 0 it should be 0 right. Because you are talking about length and you ideally want the 1 point to have 0 Lebesgue measure 0 length right that is true right, but you have to prove it right. So, you have to prove that p of b is 0 how do you prove that see remember I only know the Lebesgue measure for the empty set and sets of the form open a closed b or countable unions there of this is not a set of that form correct. So, what do I do I have to do some trick like that right. So, I have to do let see if this is right. So, I can do p of intersection n equals 1 through infinity open b minus 1 over n b plus 1 over n I should close it here right because I know only the measures of sets of that kind right, but I should also intersect with omega just. So, that it does not get out of my sample space intersection omega that is not a big deal or all the brackets in place. So, this is the p of all this intersection that you will agree because b is we wrote b as intersection of all these guys I am just closing it now this side now because I only know the measure of these sets correct. Now, so first step now what can I do now see these guys let us call that set some b n or something. Now, b n's are what kind of sets. So, b 1 will be b minus 1 b plus 1 intersection omega and b 2 will be b minus half b plus half intersection omega then b 3 will be. So, what kind of sets are these nested decreasing they are Russian dolls remember Russian dolls. So, there they go down like that right. So, b n's are nested decreasing sets a sequence of nested decreasing sets. Now, if I have probability of an intersection of nested decreasing sets I can use what theorem continuity of probability measures to write this as limit n tending to infinity probability of b n right. This is by continuity of probabilities write down if you want on this arrow right and this is because of continuity of probabilities this is a non trivial step remember how difficult not difficult, but it was fairly involve to prove a continuity. So, this is an important step now what can you say about the probability of b n do you know the measure of b n b n see if it is after all if it is just b minus 1 b n to b plus 1 b n the measure is 2 by n except you may cut it off with omega at some point right. So, this is definitely a Borel set b n is a Borel set and it is measure is at most 2 by n it could be equal to 2 by n or something even smaller right because it may cut it off with a sample space right. So, this is so this is less than or equal to limit n tending to infinity 2 by n right because p of b n is less than or equal to 2 by n and that is 0 correct. So, this requires a proof right it is not. So, you only know the measures of sets of this form right. So, in order to work out what the measure of singleton is you have to use continuity of probabilities. So, without this step it is not correct at all right this clear to everyone see now you can figure out a lot of things. So, now if I ask you what the measure of open interval a b is what will you do. So, I will write I know the measure of open a closed b which I will write as union of open interval a b and union of the singleton, but I know the singleton has 0 measure right. So, from this I can conclude that all these intervals which are either open one side or close to one side close both sides they all have same measure b minus a right. So, which is very nice right. So, the length is does not depend on whether you are closing it on this side or that side right. So, you can easily prove that this p I am of course, referring to the Lebesgue measure throughout by this p. So, you can prove that yes. .Can we do with b maybe you can do with b right you can do with just b I guess right. I think if you suppose that nothing changes I think right I think it is still fine this is answer to the same right. So, you do not have to you can get it of that if you want when you have 1 by n instead of 2 by n right can show probability of open a b equals probability of closed a open b equal to probability of closed a b all equal to b minus a. So, these are all subsets of 0 1 let us say right. So, far because I know that the singleton has 0 probability I can I only know the probability of open a closed b, but I can infer the probability of these guys using the probability of singleton is 0. So, now, we have figured out the probability measure of singleton and all intervals all kinds of intervals right. What if I give you something more complicated not necessarily more complicated something like this probability of all rationals right. So, what I am talking about is probability of q intersection omega is 0 1 right. I am looking at the probability measure assigned to all the rational numbers in the set of all rational numbers in 0 1. Now, first of all all this is the set of all rational numbers borrel set on 0 1 why countable. So, it is a countable union of singletons singletons are borrel. So, rationals are borrel right. So, rationals have a measure right what is the lebesgue measure of the set of all rationals why singletons. Yeah. So, the probability. So, you have to write it properly right. So, this is the set of all rational numbers is a countable union of disjoint singletons right which are in fact your rational numbers probability of each singleton each rational number is 0. So, by countable additivity countable additivity of measures I will have this is 0 fine. So, this is. So, here you encounter a situation where see there are so many rationals in 0 1 right in any little interval of 0 1 you have infinitely many rationals. But, what I am saying intuitively what I am saying is if I am throwing a dart uniformly to land on 0 1 interval it has probability 0 precisely 0 of landing on a rational number. See the probability is not approximately 0 it is 0 it is actually exactly 0 right. So, here is the now here we come to the confusion that students usually have. So, the if you say that the probability of an event. So, in this case the event is getting a rational number right if you say the probability of an event is 0 it does not mean that the event will not occur that is not what we are saying right it just says the probability is 0 that is it. See the null set has probability 0, but it is not true that all sets that have probability 0 have are in fact empty not at all true right. So, in this case there are infinitely many sets. So, this set is infinite right there are countably. So, there is a countable infinity of rationals in 0 1, but still the probability measure assigned by this uniform measure is 0 right. So, which means that if you generate a random number in 0 1 uniformly at random according to this measure the probability of it being a rational number is 0 it is not approximately 0 it is in fact 0 right. So, take some time to digest this it does not mean that you will not get a rational number just means that the probability of getting a rational number is 0 right is this clear. So, this is some this is a standard confusion the minds of students to begin with because see in this kind of uncountable sample spaces these kind of seemingly bizarre situations occur right when in countable discrete probability is all very nice right there is nothing to worry about right things like this will never happen right. So, this is clear any questions on this. So, the probability of any countable set on 0 1 will in fact be it is not just rational right you take any countable set you want on 0 1 let us say algebraic numbers computable numbers any countable subset of 0 1 will have 0 probability right. Yes, so which is what I am saying that is that that intuitive understanding is wrong right. So, if you say that the probability of an event is 0 it does not mean that that event is null set it does not mean that the event will not occur. So, if you throw a dart at 0 1 it can land at 0.5 because 0.5 is a element of your sample space right every element of sample space is a possible outcome. So, it could so happen that your your dart lands on 0.5 or 0.3 whatever you want, but the probability of that happening is 0 probability of a running on a rational is 0 right. So, which is why so the this is a misconception that exists in the minds of many students because you are used to this discrete probability spaces right. So, you can have infinite sets assigned probability 0. In fact even more bizarre you know that the canter set is uncountably infinite it is a much bigger set than the set of rational right. It is a scattering of points which is this bizarre scattering of points you know it is a Borel set. So, it must have some probability measure it is an uncountable set right it is a much bigger set than rational right. So, it is an uncountable infinity of points which still has 0 measure. So, you will prove that in your homework because what happens is you keep removing these open intervals the middle third the measure of all the intervals that you remove in fact add up to 1. So, what remains is 0 measure set, but it is an uncountably infinite set. So, it is not as the uncountably infinite sets of 0 1 will necessarily have positive probability measure even that is not true right. So, you will prove that the canter set has 0 measure try it as a homework it will be in your homework in any case. So, for probability measures events here are in fact Borel sets it is just that the event has 0 probability right that is all there is to it. It does not mean that the event will not occur that is that connection you should not make sorry no I mean. So, I am assigning probabilities to subset of omega right I have not given any frequentist interpretation or any such thing I am simply putting measures on sets right. If you are imagining it to be this and that it is I mean that is not I mean it may not be right. So, I am saying that subsets all Borel sets will have a Lebesgue measure if you take sets like rationals they have probability measure precisely 0 not approximately 0 it is in fact equal to 0 there is no question about this. It does not mean that rationals cannot occur in 0 on it very well can it is probability is 0 that is all. So, these kind of events. So, when an event has probability 0 you say that in English. So, the terminology used is it almost surely will not occur it is again it is a terminology and if an event has probability 1 you say it almost surely will occur. So, if I say an event is almost surely going to occur that does not mean the event is omega no right. For example, if I take probability of irrationals in 0 1 it should be what 1 because this is 0 right. So, if I if you throw a dart uniformly at random on 0 1 almost surely the event of an irrationals will occur it almost surely an event of a rational will not occur right that is the terminology used. But, that does not mean that this set is empty or that this set is omega it is not right. No cannot all you know is that countable sets under countable sets have to have 0 measure, but it is not true that uncountable set necessarily have non 0 measure that is not true countable sets definitely have 0 measure under this uniform measure right. Uncountable sets may have 0 measure may have positive measure right. So, an example of a. So, this set sets like the intervals have non 0 measure, but a canter set is an uncountable set which has precisely 0 measure. So, if you throw a dart on 0 1 uniformly at random the probability of it landing on a canter point is precisely 0 although the canter set is uncountably infinite set any other questions yes. So, the probability of irrationals in 0 1 is 1 minus the probability of rational because p of a is 1 minus p of a complement we that is the first probability property we proved. So, probability of rational is 0 because it is a countable union of singletons. So, this must be 1 right. So, I want to just indicate how this. So, if this is this the Lebesgue measure on 0 1. So, I want to quickly build this Lebesgue measure on R the theory is very similar, but it is not a probability measure any more right it is just length right. So, if you have. So, if you want to talk about Lebesgue measure on R. So, you should first build your sigma algebra right you should have to build your sigma algebra on subsets of R right all subsets of R. Now, how do you do it again you start of saying that I only certain sets are of interest to me right. So, one standard way of doing it is you take c naught as the collection of all open intervals on open intervals which are in R. And you take you generate the sigma algebra of those open intervals and you call that the Borel sigma algebra. Actually there are multiple ways of doing it at least 3 equivalent ways of doing it. So, you can let c naught be the collection of all open intervals and then B of R the Borel sigma algebra on R is defined to be sigma of c naught just like in the previous case right you just took the sigma algebra generated by open intervals doing the same thing except this B is on with the Borel sigma algebra on the entire real line. Now, see on see in the 0 1 interval we proved that you can generate the sigma algebra Borel sigma algebra using either open intervals or these f naught kind of sets or even closed intervals right they all lead to the same sigma algebra. So, similarly on R there are multiple ways of generating the same sigma algebra B. Another way is to define it let D be the collection of semi infinite intervals that is D is the set of all semi infinite intervals such that X is in R right. So, these are intervals like that minus infinity to right again you can what you can show is that the sigma algebra generated by D is in fact Borel sigma algebra. This is another way of generating the same sigma algebra right you take the semi infinite intervals and you create the sigma algebra generated by that that is that is one sigma algebra. See the problem is now you have defined the same thing B R in two different ways right whenever you define the same thing in two different ways what you have to do you have to prove that they are equivalent definitions right. So, you have to prove that the sigma algebra and the sigma algebra are one and the same right. So, in that sense they are not both definitions one of them is a definition and another must be proved as a result right or you have to say both are definitions and they are both equivalent right. So, this is one this is. So, Borel sets on R are either generated by open intervals on R or semi infinite in closed intervals on R right. In fact yet another way to define it is to say that a set on. So, in terms this is in terms of sigma algebra another way of defining Borel sets on R is to say that if you take a subset of R that set is a Borel set on R if intersections with each of these n n plus 1 intervals is a Borel set on n n plus 1. So, we know Borel sets on each of these length one intervals we already defined it. So, if these intersections taken with all these n n plus 1 intervals are all Borel sets you say that the set on R is a Borel set. So, you can say that is you can show that is also the same sigma algebra right. Now, again. So, now if you once you define this Borel sigma algebra on the real line. Now, you have to go ahead and define Lebesgue measure right. How would you do it? You take same story right you take f naught which are sets of the form A B A 1 B 1 union 8. So, on same story right and then you verify Karate Dory holds right and then say conclude that there is a unique measure that corresponds to length right. So, repeat the same story. So, that will lead to a measure on R B R it just repeat of the same story will give you a measure that is that is called a Lebesgue measure on R. So, that measure the Lebesgue measure on R is usually denoted by lambda. So, we will say that. So, I would have not really gone over this in great detail just because it is the same story repeated again except the Borel sigma algebra is defined in one of many equivalent ways right. So, what you will have is. So, lambda is the this is the Lebesgue measure on R B R in other words R B R lambda will be a measure space. This guy corresponds to length on R is this a finite measure? No right because you are associating length right and lambda of R will be plus infinity remember measures can be infinite right. So, the Lebesgue measure is not a probability measure it simply a ordinary measure it is just a measure it is not a probability measure. In fact, it is a infinite measure it just corresponds to the notion of length on R and the machinery to build it up is exactly the same once you have defined the Borel sigma algebra on all of the real line. Then you apply Karate Dory and the same story follows. So, I think it is time for me to stop. Thanks.