 Welcome back. So, we were discussing Borel sets, we bring to discuss Borel sets, Borel sigma algebra. So, we said that, so we were looking at the 0 1 interval, I took omega equal to 0 1. So, why I took open here and close here, we will do not worry about it so much. So, at this point, so I took this uncountable sample space omega equal to 0 1. And I, so one thing I said is that, if you consider all possible subsets of omega, it is too complicated, too large sigma algebra to assign even a uniform probability measure on. Remember that, it is an impossibly result. So, we started out saying that, we cannot do with such a large sigma algebra 2 power omega. So, we need to settle for a simpler smaller sigma algebra. So, we said that, the sets of interest to us are in fact, the open intervals. So, we wanted to keep intervals as being interest, sub intervals of 0 1 being interesting. So, we said C naught is the collection of all open intervals, intervals contained in omega. So, I took this collection and then I said, the sigma algebra generated by this open intervals is your Borel sigma algebra. In other words, the Borel sigma algebra is the smallest sigma algebra containing all open intervals. So, if there is any sigma algebra on omega, which contains the open intervals, the Borel sigma algebra is a sub sigma algebra of that. Is this clear? So, now and elements of, so this is denoted by scripted B, because it is a collection of subsets. We have to, so and elements of this B are called Borel sets. It is a simpler set. So, what is the Borel set? It is an element of the Borel sigma algebra and the Borel sigma algebra is the sigma algebra generated by open intervals on 0 1. Now, this Borel sigma algebra is, it turns out that it is not 2 power omega. It is a much smaller sigma algebra than the collection of all subsets of 0 1, which means there are subsets of 0 1, which are not Borel sets. In fact, what can be shown is, it has been proven that the set of all Borel sets, there are only as many Borel sets as there are real numbers or it has the same cardinality as that of the continuum. Whereas, 2 power omega has a strictly bigger cardinality. So, it is a fairly small sigma algebra, this Borel sigma algebra, but on the other hand what is nice about it is that it contains practically all sets, which are of interest to us. In particular, it contains intervals of course, and it contains also many bizarre sets, many interesting sets are included in the Borel sigma algebra. Although the Borel sigma algebra is a much smaller sigma algebra than 2 power omega, it will be very difficult for you to think of a set, which is not a Borel set. There exist such sets, there are lots of sets of subsets of 0 1, which are not Borel sets, but anything that you can normally, you and I can normally construct is likely to be a Borel set. So, it is non Borel sets are very bizarre, they are only of academic interest by enlarge. So, anything like this is one thing we will prove immediately, we will prove that the singleton set, the set only containing one real number in 0 1 is a Borel set. So, let us prove this proposition, let B be in omega, so B is in 0 1, then the singleton B is a Borel set, Borel measurable set or Borel set. So, how do you prove that some set is a Borel set, see the Borel sigma algebra is generated by open intervals in omega. So, the one thing you know for sure is that all open intervals are definitely Borel and what is. So, what is if you want to prove that some other set is some set, which is not an open interval is a Borel set, what do you want to prove, what do you have to prove? It should be either a it should be expressible in terms of either compliments or countable unions or countable intersections of the generating class, which is the open intervals. So, if I have to show that the singleton is the singleton set is not an open interval, it is just has that 1 point B. So, I am looking at this 0 1 interval and I am taking some point B, this is the singleton set containing that B alone is my set. So, I want to prove that is Borel and the way to do that is somehow express the singleton as a either a compliment or countable union or countable intersection of open intervals, which is the generating class. So, there is a standard way of doing that let me do this correctly. So, you can write the singleton B, you can verify this, you can write the singleton B as intersection N equal to 1 to infinity B minus 1 over N open interval B minus 1 over N B plus 1 over N. So, you buy that. So, I am saying that this set is equal to that set, which means I have to prove B is contained in this intersection and nothing else is contained in this intersection. So, in each of these open intervals B is certainly contained. So, the intersection certainly contains the element B that is very clear. So, if you want to prove that no other element let us say some other element H is not contained in this intersection. So, you will write H. So, you will if B is different from H you will at some point N will become. So, large that B plus 1 by 1 will exclude that H, if even if H and B are very close together at some point N will become large enough to exclude that H. And therefore, you can prove that no other element is contained in this intersection. So, there is one little thing I have. So, this you will buy I think right. So, are you ok with this intersection any questions on this this think about it is not. So, difficult you just have these are standard things you should get used to doing. So, the one little problem here is if I do let us say I take N equal to 1. So, this intersection this interval will be B minus 1 to B plus 1 and this may go outside omega. So, just to be certain I will intersect all these guys with. So, now these are all subsets of omega which are all. So, they are all intersections of. So, I am intersect omega is definitely borel because say the subset has to be in the sigma the sample space has to be in the sigma algebra. And these guys are all intersecting generating class open intervals with omega this. So, this whole thing is borel measurable right. And therefore, countable intersections of these borel measurable sets are in fact borel measure question. So, no no no. So, this intersection. So, I want to make this perfectly clear this is not a limiting operation I said this many many times. So, this is not to be understood as a limit of a finite union. This consists of all those elements which are contained in each one of these sets. You are not you are not limiting N to infinity you are not taking a finite intersection and limiting N it is not what you are doing. You are considering this sequence of sets you can call it B N or whatever you want. I am you are considering those elements which are common to each of these B N's. And the claim is that the only such element is little b all open intervals which are contained in 0 1 no no no. So, I am taking C naught. So, that is only for probability measure that is but you want to have closure under all countable unions or intersections. And I am taking the generating class to be all open intervals contained in 0 1. Now, all see C naught contains all those open intervals which are subsets of 0 1 that is it all of them. So, I am saying that see there are 2 cases right. So, you understood why I intersected with omega because if my N is very small like 1 or something this set may actually be bigger than omega. So, I just want to talk about subsets of omega. So, I am just intersecting with omega whenever my set is whenever this guy is going outside my sample space that is just a hack. But otherwise these guys are subsets of open intervals in 0 1 they are all Borel sets. So, if you intersect it is omega also you will get a Borel set right by definition omega is Borel measurable is not it. So, this so this intersection that is a Borel set. And I have a countable intersection of Borel sets must be a Borel set because it is a sigma algebra right. So, I have expressed my single term as a countable intersection of sets which are certainly Borel sets. And since the Borel sigma algebra is a sigma algebra I have to this must be a element of the sigma algebra. And therefore, the single term is Borel alright is this clear. So, next so the next question is this to. So, if from this you can so for example, you can next prove that. So, proposition you can prove that A B a set like that right where A and B are both contained in. So, this where this interval is contained in 0 1 is a F measurable set. So, this A B closed A B closed A open on that side are Borel sets. Why see I only said that see open interval A B is a Borel set by construction. But I can write. So, if you want to prove that that is a Borel set what would you do? You can simply. So, I have proven that the single term is Borel. So, I can write this as open interval A B union single term B. So, it is a union of two Borel sets is a Borel set. And similarly, this and this or if you want to do it from first principles you can still do it right. You can write A comma B let me do this correctly is equal to intersection n equal to 1 to infinity A comma B plus 1 by n open right. So, you can do that trick also if you want to do it from first principles these are all Borel sets because they are all open intervals. But the intersection what will happen is B will be included in all of them, but nothing beyond B will be included you can prove that again right. Of course, I have to intersect with omega right just to be sure that B plus 1 does not go outside omega right. So, this is a standard trick is this clear or of course, you can just take union of single term with open interval A B that is also fine right. So, you have so all these intervals are now Borel sets right. And anything which is a countable union of these any of these intervals or countable intersections of any of these intervals complements of any of these intervals they are all Borel sets. In fact, the Borel sigma algebra also contains some fairly complicated sets. So, we remember we encountered canter set right. So, the canter set. So, we defined it as the set of all subsets of 0 1 set of all numbers in 0 1 whose ternary expansion can be written without a 1 in it right only 0s and 2s right. So, another way to more geometric way of looking at it is to you start with this guy. So, you start with this 0 1 interval and then you will take the mid open third and remove it. In fact, you can show that all numbers here have start with the first trit the first ternary bit whatever is will be a 1 and these guys will have 0 this guys will have 2 right. So, this you knock off. So, you will be left with only 0 1 by 3 and actually you have to I mean canter set is defined on the closed interval 0 1, but it is not such a I mean it is not such a big deal. And then you will do the same thing again right you will knock off the middle third of those right knock those off and you keep that right. And then you keep doing this right again knock off the mid third of this mid third of this mid third of this and so on right. So, you are successively removing open intervals right and the canter set k is the set of points that remains right that is the equal in geometric way of looking at it right you can. So, this canter set contains it will not be empty ok. So, it is not like you remove everything at some point right it has for example you can show that 1 by 4 is a element of canter set it is not an empty set. In fact, we showed that the canter set is there is infinitely many point it has uncountably it is an uncountable infinity of points on 0 1 correct. Because, it is in bijection with 0 1 power infinity right good. So, this is. So, that is the canter set now this canter set is a fairly complicated set right it has a certain interesting property of a fractal self similar nature. So, no matter how much you zoom in to the canter set it will look the same it will have this structure that middle third will be missing right no matter how much you zoom in it will have this middle third missing and these 2 intervals and you zoom in further it will have the same structure right. So, this has a certain fractal structure. So, it is a very interesting and somewhat complicated set on the 0 1 interval right it does not have any simple structure as an interval or anything it is a very complicated scattering of points right. In fact, you can show that the canter set does not contain any interval because you will chop the middle third if it did right. So, this canter set is in fact, a Borel set you will prove that in your homework how will you prove that. So, you have to prove that either this canter set is can be expressed as a countable union or intersection of any of these intervals actually if you look at it is actually fairly simple and the complement of the canter set which is a set of all things that you remove right. That is a countable union of open intervals because the first stage I knock off that open interval, second stage I knock off 2 open intervals and so on right. So, the complement of the canter set if you canter set is k the complement of k will be a countable union of open intervals which you know to be Borel. So, k complement is Borel so k is Borel. So, that is how you argue these things. So, even fairly not just intervals it also contains certain fairly complicated subsets of the 0 1 interval canter set being 1 example right. In fact, in order to construct I said there are subsets of many subsets of 0 1 exist which are not Borel right. So, in order to construct as I said you have to work very hard it is very difficult for you to think of a set which is not Borel set, but they exist actually. So, if you want you can look at there are examples of. So, let me so extra reading so I will just mention a few resources. So, this is not so I am just putting it as extra reading just for your curiosity. So, this is by no means something that I will hold you accountable for. So, this extra reading so I will first put non Borel sets certain is the most famous examples in terms of a continued fraction there is a certain set of continued fractions you can actually look it up on Wikipedia just search Wikipedia for non Borel sets it has a certain example of a certain collection of continued fractions and the set you can show to be non Borel. And there is an example of a so just going a little bit more crazy there is a set called a Vitale set which is non measurable which I mean this is not really immediately relevant to us, but this is also very crazy set on which you can assign no measure to the set is an example of a non measurable set. So, this is a non Borel set this is a non measurable set. And finally, while talking about these crazy sets probably the one of the craziest of all is given by this paradox called Banak Tarski paradox. This Banak Tarski paradox is a very bizarre paradox where you take a sphere or a just a three sphere. And then what you can do is you can divide up the surface of the sphere into some let us say five parts in fact five is enough. And you can reassemble these five parts to form two such spheres of the same size it is possible it seems very bizarre. I mean you have a sphere of certain surface area and you break it up into let us say five parts. And those five parts you can reassemble back. So, as to get two spheres of the same size the reason all this is happening is because of this bizarre sets called non measurable sets. I have in the notes that I uploaded there is a link to this Banak Tarski this is a very bizarre phenomenon. So, this is all stuff that you will not be accountable for, but certainly will find interesting there is some really bizarre subsets of 0 1 or the surface of the sphere. And they are all they are not these nice sets they are all these non measurable sets to which you cannot assign any measure like length or area or probability or any such thing. So, if these sets were in fact measurable if you then you would can do bit of contradiction how can you have the same surface area doubling. I mean some people I mean think of I mean think of this being a gold ball you can generate two gold balls with it. So, that is the most you know an appealing way of proposing this paradox I guess it is all very bizarre because this pieces are very weird scattering of points which are non measurable. So, these things happen. So, that is for your own edification. So, but I hope you understand these Borel sets. So, what did we say Borel sets they include all intervals and countable unions countable intersections and complements of intervals. They also include certain complicated sets like the canter set and it also I mean any pretty much any set you can reasonably think of without working too hard is very likely to be a Borel set that is. So, it contains sets all sets of practical interest to us. So, if you want to construct a non Borel set you have to work very hard to construct one is this clear. So, this is the Borel sigma algebra which is a very small manageable sigma algebra on the 0 1 interval, but which nevertheless contains most sets of practical interest. So, are measurable space from now on will be what omega comma B I will only take. So, omega is this guy and I will take. So, measurable space that I will take from now on is omega B 0 1 when the context is clear I will not write B 0 1 when the context is not clear I will write what Borel sigma algebra. So, for we started off. So, we come up with a measurable space omega and sigma algebra on the omega which is a smaller sigma algebra than 2 power omega, but which nevertheless contains lots of interesting sets like my intervals and countable unions and countable intersections there of. So, now aim. So, now my aim. So, what is the initial aim we started out with I want to assign a uniform probability measure on 0 1 I said I could not do it on 2 power omega. So, I settle for a smaller sigma algebra I just kept the intervals and generated a sigma algebra with those intervals and that I ended up with B as my sigma algebra. Now, the question is can I assign a uniform measure on this Borel sigma algebra on only on Borel sets. So, essentially if the answer is yes you can do it now it is a smaller sigma algebra and I can assign a uniform probability measure on omega B. So, I will assign probability measures to only Borel sets. So, there are many subsets of 0 1 like all these crazy servitale and non Borel sets right those guess I will not assign any probability measure 2 I will only worry about Borel sets. So, under the measure that I am uniform measure I am going to put only Borel sets will have a concept of a length or any probability measure on it for that matter right non Borel sets you cannot ask me what probability it is or what length it is it is meaningless to speak of this thing just like it is meaningless to speak of the area of these guys right. So, now I have to start telling you how to put this uniform measure on omega comma B are there any questions at this point. So, I want the following rate I want to put a uniform probability measure on omega comma Borel right. So, my omega is 0 1 I have the Borel sigma algebra on it I want to put a uniform measure on uniform probability measure on this space this measurable space right. Now, the question arises what kind of. So, I want. So, if I give you a nice Borel set let us say an interval from A B say let us say open A closed B right I know what measure I want for those sets I want it to be B minus A right the length of it right that is really what we started out saying right I want the property that the measure of the interval A B should be something proportional to B minus A and I also said it should be translation invariant right. So, I will start of saying that for the generating class or for all these nice intervals I want the measure to be the length of the proportional to the length of the interval. In this case it you can take it to be equal to the length of the interval because I mean the length itself is 1 and you will automatically get a probability measure right. So, I want to say that for these generating classes my measure should be proportional to the length, but that is not enough right I have to specify the probability measure for all Borel sets which can be fairly crazy right I mean for if you have something crazy like canter set right how do you it is not a it is not like a collection of intervals right it is a very weird scattering of points, but the measure I assigned should hold for all Borel sets, but I can only specify a measure for these nice intervals right. So, I have to be able to extend this measure right this measure I assigned to these intervals to all these crazy Borel sets also right. So, this is what our program is now. So, I will do the following. So, I want to take so I want to start with a very simple collection of subsets. So, let us again take omega equal to 0 1 as usual and I take f naught is the collection of subsets of omega which are finite unions of disjoint intervals of the form a B plus the empty set. So, I am saying so I am taking a collection of subsets of 0 1 which contains the following subsets it contains the null set. So, put the null set in so you are collecting subsets of 0 1 put the null set in first and then you put all subsets which are of the form open a closed B or finite union finite unions there of not countable just finite unions there of. So, a typical element of f naught will be of the form a b 1 closed union a 2 b 2 closed union dot some a n b n right where a 1 is less than b 1 less than or a 1 is 0 less than or equal to a 1 less than b 1 less than or equal to a 2 less than b 2 dot dot dot less than or equal to a n less than b n. So, f naught consists of sets which are finite unions of these semi open semi closed intervals it also contains null set. So, now what I am going to do is I am going to say that if I am if you give me an element of f naught right some set like that I am going to say it is probability measure I want it to be equal to b 1 minus a 1 plus b 2 minus a 2 plus so on to b n minus a n right. But I want want to make sure that that measure also extends to all boreal sets right also to countable union this is only a finite union remember right. So, that is our program that is what we have to do. So, now, so the extension of this measure right I know what the measure of these guys should be right b 1 minus a 1 plus b 2 minus a 2 plus so on. But I want to extend that measure to the entire boreal sigma algebra all boreal sets. So, the way you do that is by a very fundamental theorem in measure theory known as Karate Odery's theorem Karate Odery's extension theorem Karate Odery's a British mathematician I will state the theorem without proving it. So, that is the main technical tool that helps us extend are what measure we want on sets of this form to all boreal sets. So, before that I want to get these certain properties of f naught I want to finish. So, I want to say that f naught is first of all an algebra. In fact, the moment I wrote down some f naught you probably guess it is an algebra and I mean because that is the notation I have been using all along right. I use f naught for algebra and f for a sigma algebra right I did not tell you that at this point right I will just say f naught consist of a certain kind of sets. But you can prove that f naught is an algebra how do you prove that f naught is an algebra it contains phi by definition it has null set then you have to prove complements are also sense a f naught. So, if you have some set like. So, open closed open closed right this is 0 1 and I have some set like open closed open closed then the complements will also be of that form right this guy the complements is also open closed open closed and so on right is omega in f naught the sample space is it an element of f naught it is right because the which is now you see why I took open closed here that is also an element of f naught right. So, all these simple properties hold and how about countable unions sorry finite unions f naught should be closed in the finite unions which you can show right it is a. So, it is a fairly simple homework problem I leave it you as a homework problem. So, this I leave you as a homework I just spoke out the proof you just have to write it out. So, closure under complementation and finite unions now this f naught is not a sigma algebra. So, this collection f naught that I have defined is an algebra, but not a sigma algebra. So, here is another example of a collection of subsets of 0 1 which is an algebra, but not a sigma algebra you already encountered such an example in your homework I think of some collection which is an algebra, but not a sigma algebra right. So, this is another example is f naught this collection is not a sigma algebra although it is an algebra how do you prove that. So, in order to prove something is not a sigma algebra you have to see the complementation will I mean that will not going to help you because I mean it is an algebra already right. So, you the only way it cannot be a sigma algebra is countable unions are not closed right. So, you can give a very simple example if you take let us say you take sets of the form let us say you take a n sets of the form 0 comma n over n plus 1 that. So, that will do I think you do that n equals 1 2 dot dot. So, a 1 will be 0 half right a 2 will be 0 2 over 3 closed and. So, this is the sequence of sets I have are these guys in f naught. So, a n's are in f naught for all n agreed why because they are of that form know they are of this form. Now, if you take countable union countable union n equals 1 through infinity 0 comma n over n plus 1 what you get 0. So, I am taking countable union not countable intersection then I am taking countable union of 0. So, I will the first set will be 0 comma half second set will be 0 comma 2 by 3 and so on they are getting bigger and bigger right. So, I mean so the point is this right. So, this union contains now you have to look at. So, you are looking at the set of all points contained in at least one of the a n's right correct. So, you can show in fact this is equal to open interval 0 1 this is open you can show that these two sets are equal. So, the open interval 0 1 includes all points in 0 1, but does not include 1. So, you can show that this none of these a n's include 1, but for any number 1 minus epsilon 1 minus a little bit will be in one of these a n's. So, this you think about this is fairly easy I mean just try and prove it vigorously. So, you get this union is equal to open interval 0 1 is open interval 0 1 in f naught no it does not it is not in f naught, because you only have open closed on this. So, this is a not an element of f naught correct. So, here is a concrete example that my f naught is not an sigma algebra although it is an algebra. So, proof of two finally, I want to state the third property that I am going to look at the sigma algebra generated by this collection f naught. So, f naught is sets of these forms collection of sets of these forms right. So, I am going to generate the sigma algebra. So, I am going to consider the sigma algebra generated by the f naught. So, I told you f naught is not a sigma algebra right. So, I have to make it into a sigma algebra right make it into by which I mean consider the smallest sigma algebra that contains all elements of f naught which is what this is sigma f naught. Any guesses on what this will be no see f naught is empty set and sets of this form right. I told you with an algebra it is not a sigma algebra, but in probability theory I want to work with sigma algebra. So, I am considering the smallest sigma algebra generated by f naught what it turns out is that this sigma algebra is no different from the Borel sigma algebra it turns out this is equal to b 0 1 this you can prove. So, what am I saying in effect what am I saying see the Borel sigma algebra initially was defined as the sigma algebra generated by open intervals. I am saying that even if you generate a sigma algebra by the semi open semi closed type of intervals it is still be the same Borel sigma algebra. Meaning that the sigma of f naught is same as sigma of c naught which is the open intervals. So, I have not proven it yet right. So, do I have the time to prove it I have 5 minutes let me try and prove this. So, how do you prove that. So, how do you prove something like this how do you prove that the sigma algebra generated by these class of sets is no different from the sigma algebra generated by open intervals. See first of all you should recognize that this is an equality of 2 what kind of objects collections right. So, this is a collection of some subsets this is some other collection of subsets of 0 1 and I want to prove that this is equal to that which means I should prove. So, this is contained in that and that is contained in this right that is what I need to prove. Now, which is the easier part. So, this let me just write this this guy is nothing but sigma of c naught by definition c naught is open intervals. So, this is you already know that. So, how do I prove that. So, I want to prove this is contained in Borel and each Borel set is contained in sigma f naught right. So, let me try and do that proof of 3 to show sigma f naught is contained in b 0 1 it is enough to show f naught is contained in b right why is this. So, this is the smallest sigma algebra containing f naught. So, I am saying that if the Borel sigma algebra is a sigma algebra containing f naught then it must follow that sigma f naught is contained in b by the theorem right with the theorem is generated about sigma c j j generate a sigma algebra. So, you buy this. So, if I prove this I prove this why because here is a sigma b is a sigma algebra containing f naught. So, sigma f naught must be a sub sigma algebra of b how do you prove that f naught is contained in b. So, in order to prove this I should prove that every set in f naught is a Borel set right and what is the typical element of f naught it can either be an empty set which is definitely Borel or it must be a set of this form right, but I know that half open half closed intervals are Borel. So, finite unions are Borel right. So, this is clear. So, this is easy right this you this is fairly easy I just argued it out prove that element. So, empty set is in b and also any element of f naught of this form you prove an already that they are Borel sets right. So, this follows if this is true then this is true. So, one direction is over the one direction is very easy right. Now, I have to prove which way I have to prove that next to show which direction B is contained in sigma f naught to next to show we proceed as follows. So, if you take any open interval a b which is a subset of omega open interval a b can be written as union let us see union n equals 1 through infinity a minus 1 by n is that correct just like this know I mean something I am writing something similar to this right. So, you agree with this equation verify this. So, now what have I shown any open interval is a countable union of elements of f naught right. So, these guys are all elements of f naught correct. So, countable union of elements of f naught will be a element of sigma f naught certainly right. So, what is this prove this implies every element of c naught is an element of sigma f naught correct. So, I am doing this proof I mean just as a practice for you I mean I think you should these you will encounter several such proofs. So, your so this implies c naught my collection of open intervals we certainly contained in sigma f naught. So, why is that true for any open interval a b I have written it as a countable union of elements of f naught. So, this countable union must be an element of sigma f naught fine everybody with me. So, then this follows correct now what this one step remains. So, I have identified a sigma algebra that contains c naught correct. So, therefore, sigma of c naught must be contained in sigma of f naught what is sigma of c naught borel right this must be borel. So, I proved it both ways right. So, I proved that the sigma algebra generated by sets of the form f naught is nothing, but the borel sigma algebra. In fact, in the homework you will prove that if you were to generate sigma algebra using closed intervals you will still get borel sigma algebra. So, you can generate borel sigma using open intervals or closed intervals it does not matter this everybody with me. So, this lemma is important. So, lemma has 3 component that says that f naught is an algebra not a sigma algebra and sigma of f naught is a in fact, the borel sigma algebra. So, we will use this lemma very heavily to define the uniform probability measure or the Lebesgue measure on the 0 1 interval which will be topic of tomorrow's lecture. Thanks.