 Ok, so, ok, so last time we, I conclude the lecture by writing a statement of this proposition which was labelled characterisation by approximation, posimčenje, lepega, zeločenja seta. Zeločim, da se pričeš, ovo je taj statement. Zeločim, da je taj set, nekaj set, nekaj pričeš, zeločim taj posimčenje je zeločen. mezoljeno. I statement I says just E is measurable. 2I says that for any epsilon positive that exists, an open set, open such that O contains E and we have that m star of O minus E E is less than epsilon, then in analogy we have that three so for any epsilon positive there exist a closed set F, such that this time the inclusion is reversed, E contains F and the outer measure of E minus F is less than epsilon. And then somehow if we choose set in a larger collection we can do better than this. So we have that there exist a set G in the collection G delta, such that G contains E and such that the outer measure of G minus E, this time is zero. And again in analogy we have that there exist a set that they called, I call F, okay call it F again, but in the collection of sigma such that you have that F is contained in E and such that the outer measure of E minus F is equal to zero. And then if we further assume that, call it here, if in addition we assume that the outer measure of V is finite, we have also that all these statements are equivalent to statement 6, it's tell you that for any epsilon positive there exist a finite collection of, there exist a finite, so union call it union U of open intervals. Okay, such that we have that the symmetric difference between E and this union is small, okay, is arbitrary small. Okay, so last time I told you that we will proceed by step, there are three steps. I start by step A, which, okay, we prove a step A under the hypothesis that M star of V is finite and we are going to prove the following chain of implications, so you have that I, we will prove I implies 2, I, which implies, which coin implies 6, okay. Okay, so start, change it. Okay, so start by I implies 2, I, okay. Okay, so by a formal result, okay, so now we know that E is measurable, okay. By a formal result, okay, we know that for any epsilon positive, okay, there exist an open set, okay, an open set O, such that O contains E and in addition, we know that M star of O is less or equal to M star of V plus epsilon, okay. This comes from the definition of outer measure. Here we can get rid of the star because we know that it is measurable. Okay, so call it A. Okay, so now we use the fact since E is measurable, we have that the following equality holds and we use O, this open set O has a test set. So we know that M of O is equal to M star of O intersected E plus M star of O minus E. So which is equal, this is equal to M star of E because O contains E plus M star of O minus E, okay. Okay, so we combine these two and we get, and we use, so if we combine these two, we get that M star of E plus epsilon is larger or equal to M star of E plus M star of O minus E, okay. Now we use the fact that this measure, okay, in that case we can get rid of the star because we know that this measure is finite, so we can cancel out. So what we get is precisely what we want, less than epsilon, okay. Now we prove that 2 E implies 6, okay, so prove this step here. Okay, so we start by an open interval, an open set and we saw that since O is open, okay, we can, O, it is the countable union of this joint open intervals. So we saw this, I guess last time this joint open intervals, okay. Okay, so we set O equal union of IK, open intervals IK, okay. We have this joint. Okay, moreover we know, my hypothesis, okay, we know that M of O is less or equal than M star of E plus epsilon. And on the other end we know that M of O is equal to M star, okay, to M of the union of IK, which is equal, we prove this to the sum of the measure of the IK, okay. So this is the finite, this is the countable additivity property of M, okay. Okay, but we know that from this we know that M star of O, this also tells you that M of O is finite, okay, because M star of E is finite, so we are just adding epsilon. Okay, so this is finite, this means that also this sum is finite, so we look for the detail of this sum, which will go to zero, okay. Okay, so this is finite because of this. And so we have that, the limit as K tends to plus infinity of the sum of N, which goes from K to infinity of the measure of this IK, go to zero, okay, because this is a convergent series, so this means that the details of the mass vanish, okay. Okay, and now we choose K, we choose the index K in a suitable way. Okay, so let us choose K, K here, such that we have that the sum from K to plus infinity of the measure of IK is less than epsilon, okay. We can always do this. And in analogy we define, we define this set U as the union from N, which goes to one K minus one IK. So somehow this will be our candidate to satisfies the thesis, okay. This is a finite union of open intervals, okay. This joint open intervals. Okay, so we have that the measure of O minus U is equal to the sum for N, which goes from K to plus infinity of the measure of IK, which is less than epsilon. Okay, then at this point we use the following inclusion. Maybe, which is, so we use this fact that E, symmetric difference with U is contained in the union of the symmetric difference between E and O, union, the symmetric difference between O and U, okay. I mean, this is, if you want to do it as an exercise, it's a boring one, but you can do this. You have that A, symmetric difference with B is contained in A, symmetric difference with C, union C, symmetric difference with B. Okay, otherwise believe me that it's true. Okay, so what we get? We get in this, that M star of E, symmetric difference U. We use this, it's less or equal than M star E, symmetric difference with O and plus M star O, symmetric difference with U, okay, which is equal, since O contains C, which is equal to M star of O minus E, plus M star of O minus U, which is less or equal. So this is less or equal than epsilon by hypothesis, and this is also equal than epsilon by construction. So at the end we end up with, this is less or equal than 2 epsilon. So this is arbitrarily small, okay. Okay, so this proves this implication. Okay, so now, okay, now we prove implication 6 implies 2i. So by hypothesis, so we have that there exists finite union of these joint intervals, for any epsilon positive, there exists such a union of i k, k equal 1, 2. Okay, let me call them k, and such that M star of E, symmetric difference U is less than epsilon. Okay, these are open intervals. Okay, again we use a former result. So we have that by a former result, we know that we have that there exists a set G open, such that E is contained in G, and such that M star of G is less or equal than M star of E plus epsilon. Okay, now at this point we define, we define O as the union between U and such G. So of course O is open. Okay, then the first thing that we observe is that O is open and O contains E. Then, no, no, this is, I mean, here, no, no, I mean, here we are proving another inclusion. Yeah, yeah, yeah, but it's the same, I mean, here is the thesis, and there is the hypothesis. This is, ah, you mean, I, and, yeah, sure. Is this what you mean? Yeah, yeah, yeah, yes, thank you. Okay, may I erase here? Okay, so, so we notice this. So, O minus E is equal to O, symmetric difference with E, because E is contained in O, and this is contained in O, symmetric difference U, union, U symmetric difference of E with E, and this is equal, for the same reason, to O minus U, union, again, U symmetric difference E. Okay, okay, from this we have, we deduce the following, which we have, okay, we want to estimate the outer measure of this, okay. We have that. Let me call, okay, M star of O minus E is less or equal M star of O minus U plus M star of E, symmetric difference, okay. No, sorry, U, okay, E symmetric difference with U or the converse, this is symmetrical. Okay. So, again, I repeat M star of O minus U plus epsilon, okay, because we are using the hypothesis, okay. This is less or equal. Okay, this is less or equal than M star of O minus M star of U plus epsilon. Okay, I can do this, because actually they are both measurable, so if you do some algebra, you will see that it's allowed to do this, okay. Because O and U are measurable. So, the outer measure of the difference set is just the difference between the outer measure. Okay. Okay, we call this inequality B. Okay, then we use this other fact. So, we want to estimate from above these outer measures, so we have that M star of O, okay, it's less or equal than M star of G plus M star of U minus G, okay. So, I take G one time, so then I just add this thing. Okay, this is less or equal than M star of E plus epsilon. And this comes from this. Okay, plus M star of U minus E, because E is contained in G. So, in principle, this set will be larger than this, bigger than this, okay. Okay, this is less or equal, and then I copy M star of V plus epsilon plus M star of the symmetric difference between E and U. Okay, now I can use the hypothesis here. So, this is less or equal than M star of V plus epsilon, okay, plus two epsilon, okay, because this is less than epsilon by hypothesis. Okay, then I combine, okay, this, call this C. I combine these two. So, by combining, we get the following M star of O minus E is less or equal than M star of V minus, okay, M star of U plus three epsilon, this dot, and then we have that since E, you can split E in such a way, so E is E intersection U union of E minus U. We have that by subaditivity, subaditivity in equality dot leads to, subaditivity dot leads to the following to the fact that M star of O minus E, okay, is less or equal than M star of E intersected U plus M star of E minus U minus M star of U plus three epsilon. And, okay, again, since M star of E minus U is less or equal than M star of E symmetric difference with U, which is less than epsilon, we have that M star of E intersected U minus M star of U is, if you want, less or equal than zero. Okay, which means that M star of O minus E is less or equal than epsilon plus three epsilon, which is equal to, okay, four epsilon, which means, which is, that is small, you can make it as small as you want. Okay, now we have to deal with step two, with step B, I call it B, I think, and so the idea here is that basically we prove the same implication, but we remove the hypothesis that M, that he has finite outer measure, okay? So, of course, we will use what we just prove, but we remove the hypothesis that M star of E is finite, okay? Because all these implications have been proved under the hypothesis that M star of E is finite, okay? Okay, so step B. Okay, so B, so remove the hypothesis M star of E finite and prove following. So we prove that I implies two I, implies, which implies I, okay? So it's close. Okay, so we start by this implication that I implies two I. So we start by a measurable set E, which in principle might also have infinite outer measure, okay? Okay, so consider E, a set, and, okay, so somehow I told you that we want to use the step A, so we want to use that this is valid for E with finite measure. So, in order to do this, we observe this very easy thing that E can be expressed as a countable, is a countable union or set of finite measure. Okay, so is a countable union of set of finite measure called an EK. So do you see how it can be expressed in such a way? How can you construct this set EK, which are measurable and have finite measure? Okay, recall that, of course, we are, E is a subset of R. It's very easy. I mean, for instance, you can see that E, if you consider this set EK taken as an intersection of, for instance, this is minus K and K, this is a way. I mean, there are infinite many ways. You can see that E is the union of this EK. Okay, they are measurable because we, of course, know that this is an integral to this. Okay, so we know that, maybe we can work with that set because we know that this implication is valid for this set EK because we just proved this. So we know that for any epsilon positive and for any K positive, okay, with K, of course, I denote an integer, a positive integer, okay? K integer. So there is a set okay open such that you have that. EK is contained in this okay and the outer measure of okay, in that implication I can remove the outer measure, I can speak about the measure is, okay, the measure of okay minus EK is less, now I need epsilon and then I use, I have to do this trick of multiplying for this factor, okay? Because this is the general term of a converging series then when I will sum up you will understand why I need to multiply epsilon for this 2 minus K, okay? But, okay. Okay, so let us now define okay, how would you define this set O that we want to provide? Yeah, the countable union of this set, okay? So let us define O as the countable union of this set, okay? Okay. And, okay. Okay, O is open, of course, is open and E is contained in O, okay? Okay, now we notice the following fact moreover, since O minus E is what? Is the union for K, which goes from 1 to infinity of OK minus EK. Okay, it is contained in principle. It is contained in principle. We have that by monotonicity and by the countable additivity property that M star of O minus E is less or equal than the sum for K, which goes from 1 to infinity of M star of, okay, M of K minus EK. Okay, which is equal, now we use this inequality, okay? It is equal K from 1 to infinity of epsilon to minus K, okay? And this is equal to epsilon, okay? This is just a trick to Okay, because this series, okay, of course you can put epsilon out, so this series this converts to 1. Which one? EK, I mean I is measurable, okay? E is measurable. So this is an interval, so we prove that the interval are measurable. So the intersection of measurable set is a measurement. Okay, so now we have to prove implication to I implies. Okay, this is quite easy, so so we set epsilon so we need a countable countable in many epsilon, so we just fix epsilon equal to 1 over M and by hypothesis we know that there exists no, now we prove I before you mean yeah, but now we are under the hypothesis that E is not measurable no, E is not finite, as not finite measure, okay? No, I don't think so. Probably I know what you mean, but we prove that lem under the hypothesis that E was measurable, maybe. Okay, let me prove the statement. Okay, then there exists OM open set such that E is contained in OM M star of OM minus E is less or equal than 1 over M. Okay, now we define okay, G has the intersection of OM and so we have that G by construction belongs to G delta and of course we also have that E is contained in G and then then we have that M star of G minus E is less or equal than M star of OM minus E, which is less or equal than 1 over M and since this holds for any M we conclude that M star of G minus E must be equal to 0. Okay, now for step B it reminds to prove that 4 implies I. Okay, so we define F as E okay, by hypothesis we have that M star of F is zero measure and since it has zero outer measure then we know that F is measurable, okay? We prove this at some point, okay? Okay Okay, so F is measurable, G is measurable because it's a border set and as we have that since E is equal to G minus F we have that E, okay, it's measurable as well. Okay, so this concludes the step B and now we prove the step C which is then we can conclude all the implication. So in step C we use B to prove that I implies 3I implies 5 and in turn implies I, okay? Okay, so start by this one okay, so we start by a measurable set, I mean here the trick is to to use of course step B and then to work in the complement set because, okay, so since we have that E is measurable by hypothesis then if we define G as the complement set is measurable as well, of course and okay, we know by step by step A okay, sorry in particular in implication I implies 2I okay, we know that for any epsilon for any epsilon positive that exist an open set O such that O contains G okay and the outer measure of O minus G is small, is less than epsilon okay, now it's quite natural to define F the set that we are looking for as the complement of O, okay? Okay, so let us define F as O complement which is closed so it is closed and moreover we have that F is contained in in E, okay? F is contained in E and moreover we have the following set of inclusion we have that E minus F is equal to E intersection F complement which is equal to E intersection O which is equal to O minus E complement equal O minus G and so you have that M star of E minus F okay, in this I can get rid of this because we are talking of measurable set is equal M star of most in this case we can just deal with true measure, okay? M star of O minus G which is less than epsilon and so we are done, okay? Okay, and then step three i's five okay, so we have that for any for any M for any integer M the resist corresponding FM closed set, okay, such that FM is contained in E and we know that M star of E minus FM is less or equal than 1 over M, okay? Okay, this time we define our the candidate F will be the union of this FM, okay? So let us define F as countable union of such FM this is in principle we don't know if it is closed but for sure it belongs to F sigma it's what we need and F is contained in E okay, moreover it's always the same argument M star of E minus F is less or equal than M star of E minus FM which is less or equal than 1 over M for any M and then we have M star of E minus F which is equal to zero okay, now the last implication of five implies okay, so we start by a measurable set let E no, no, no, sorry we want to prove that E is measurable so let E be a set and so belonging to the set of parts of R and F belongs to this collection of sigma and this F is such that F is contained in E and the outer measure of E minus F is zero, okay then we define let us define H as E minus F and again by the same argument that we used before with F that the outer measure of H is zero, which means that H is measurable so finally it reminds to observe that E, okay, since E is the union of H and and F we have that E must be measurable, okay because this is a Borel set so it's measurable and this we just prove that it's measurable okay, so this concludes the proof of all of all of the proposition, okay okay, so now we shall slightly change argument so we will first give we would like to give you some if I find some exercise so I prepare some exercise for you should be enough I think maybe let me fix okay if you see exercise 2 yeah, exercise 2 you, I ask you to prove this the limit the outer measure of A is equal to the limit of M star of A N first under the hypothesis that A, N are measurable and I will prove this now and then what you have to do is just to prove under, by removing the hypothesis that N are measurable, okay okay so let EN be an infinite increasing sequence is a sequence with EN for any N okay, then we have that measure of EN is equal to the limit plus infinity okay so somehow we may assume that the measure of EN are all finite because for any N otherwise I mean it's it's trivial, okay otherwise otherwise the star is trivial, okay so let's argue under this this hypothesis okay so E is the union so let E which is the union of all this set we can decompose E as the union of E1 E2 minus E1 and so we have EK minus EK minus 1 and so on okay, so this the set the set E1 and this kind of set are measurable first of all okay, and and this joint okay, so we have here E1 E1 this is E1 yeah maybe come closer okay, so we have that measure of E of this countable union is equal to the measure of E1 plus and goes from 1 to infinity of the measure of EN plus 1 minus EN okay, now we notice the following fact okay, that EN plus 1 is equal to EN union EN plus 1 minus EN which are disjoint disjoint and we have that the measure of EN plus 1 is equal to the measure of EN plus EN plus 1 minus EN maybe ah, here this one ah, yeah, yeah, sure okay, thank you okay, now okay, at this point we use the fact that this measure is finite okay, so since the measure of EN is finite we can deduce that the difference, the measure of EN plus 1 minus the measure of EN is equal to the measure of EN plus 1 minus EN okay, so so we continue here which is equal to the measure of some this difference here okay, here you have the longest so this is measure of E1 plus the limit as k goes to plus infinity of 1 to k measure of which is equal ah, no, here we start from 2 k plus 1 then as usual you end up with the telescopic series okay, so this concludes the proof okay okay, so so in your exercise so in the exercise 2 so the idea is I can give you a way to solve it the idea is that you so you start by any set okay, you try to approximate in a clever way this set by measurable set in specific by Borel set okay then you use the fact that this Borel set which would be the union okay, I cannot say much but okay, are constructed in a clever way satisfies this property, this limit that we just proved because they are measurable and then you see that the property is preserved also for the nonmejorable set okay just a matter of understanding how to define this set this measurable set which approximate your arbitrary set okay, and then now so far we just saw example practical example of measurable set because we dealt with interval which are measurable with open and closed set which are measurable, Borel set are measurable so one might ask but there exist a nonmejorable set it's indeed true that the sigma algebra of measurable set as I claimed some lesson ago is strictly contained in the set of parts of R is it true can we provide nonmejorable set which is here but not here so the answer is yes and the construction require the use of the axiom of the choice I mean it's quite tricky the construction but okay we will see so we start by defining by giving the following definition so the title is non measurable okay okay, this is very probably we already did this so we start by a set E in R any set we fix an element Y of R and we denote the translation of E as the set of element RZ which looks like Z Z are equal to X plus Y with X in E these are just translated set now we prove no I don't prove I just give you as an exercise so this lem this is a lem you can prove by yourself so we have that let E in R this time we require E measurable okay then for any for any Y in R we have that the translation E plus Y is measurable as well and moreover the two measure coincide of E is equal to the measure of E plus plus Y, okay, you can prove this mean by yourself but you have just to somehow to apply the fact that the outer measure is translation invariant and okay, now we define a new operation between between element of R and I call this operation some module 1 some, okay okay, so we consider this interval which is half closed and half open, so zero is included 1 no and we define on such an interval this operation some module 1 some module 1 of X and Y of X and Y and Y as follows so we have that for any X and Y in this interval okay, we denote some module 1 so we use this symbol to denote this operation so a plus with the circle at the top of the above the plus is defined in such a way it's just the sum the ordinary sum X plus Y if X plus Y remains is strictly less than 1 so it remains in this interval otherwise you shift of 1 so we have X plus X plus Y minus 1 if on the contrary X plus Y is larger or equal than 1 here you have a strict minus then less okay so observe that the supremum the supremum of X plus Y is 2 which is never achieved so when you subtract 1 you always remain in here okay no no it's never greater than 2 because you took X and Y here okay so what can we say about this operation okay just this sum of the one is a commutative and is a mutative okay probably you can check by yourself and associative operation that takes pairs of number in this interval zero in two numbers in the same in the same interval okay so in analogy with the translation of the classical translation of a set we define the translation of a set by means of this operation so the definition is the following so let E be any subset of course we have to restrict ourselves in this interval subset of zero one then we define the just some terminology that translates model one of E to be E plus Y okay Y is still an element of this interval okay so Z of course Z will be X some model one of Y with X in E so this is completely analogous to the classical translation okay now what we want to prove is that the Lebesgue measure is also invariant under the translation model one okay so let E zero one be a measurable set okay okay then for each Y in in this interval the set E translation model one with Y is measurable and moreover we have that the two measure coincide measure of E is equal to the measure of translation model one okay okay so to check that for instance this is measurable and this relation hold we just we rely on the definition of measurable set okay so we need two set to check the definition which are so we will so we consider this two set we define this two set so E one which is E intersected zero one minus Y and E2 which is the complement is E intersected one minus Y one so just to visualize they are put in that way here we have zero, here we have one so pick Y here for instance actually one minus Y here so E1 will be be contained here and E2 will be contained okay this is just to say that so when you pick element here the sum model one is just classical sum when you pick element here in E2 then you have to shift of one it's just to so we have that E1 and E2 measurable disjoint and reunion is E and you have that E1 union E2 is equal to E okay that as I told you E1 translation model one of with Y is equal to E1 plus Y so that the classical translation and so this give you that by this you know that this is measurable and that the measure of this set is equal to the measure of of E1 okay then so what about E2 yeah you are right okay what about E2 okay E2 is now E2 over one with Y is E2 plus Y minus one okay because if you have that if X is in E2 then X plus Y is larger than or equal to no one okay okay then you have that E2 again by this is measurable and the measure of E2 is the measure of the measure of model one is equal to the measure of E2 okay okay now we prove this okay now we have that E plus translation model one with Y is equal to E1 translation model Y with Y with E2 and the set R these two set are measurable and this joint this is because you can quickly see that they are this joint because okay because if you pick because if not you have a Z in the intersection that exist a Z which is equal to X1 plus Y and that is also equal to X2 plus Y minus 1 with X1 in E1 and X2 in E2 so if you combine these two at the end what you get you get that X1 is equal to X2 minus 1 X2 minus X1 should be less than 1 because X1 is larger than 0 and X2 is less than 1 and so you should end up with this so you obtain a contradiction okay just to say that these two are you can express this set as the union of two this joint measurable set okay so this fact you can immediately say that the translation model 1 of V is measurable because it's the union of two measurable set and then for what we prove we have that this whole this is less E2 so we prove that these two measure are equal to the measure of E1 plus the measure of E2 and this give you the measure of E okay so we this conclude the proof so I will start so all this machinery it's needed to to construct this no measurable set okay it's quite tough but we need it so I recall you the axiom of choice with Zermelo's axiom so you consider a family of arbitrary this joint indexed by a set A okay so you can just to for short you can see set E alpha with alpha in this set of index A okay then what tells you this axiom tells you that then there exist a set consisting of exactly one element from each E alpha okay so somehow there is a way to choose just exactly one and only one element E alpha in any set this is probably already no this axiom already no okay so how we will use this so first of all we need to introduce a class of equivalence within the interval 0, 1 okay so let us introduce on the set on the interval 0, 1 following equivalent relation so x is in relation with y is equivalent so x you can read x is equivalent to y if and only if they differ by a rational number so x minus y rational number so just recall you some consequence of this so the equivalence equivalence relation so is a partition if you want is a partition of 0, 1 into equivalence classes okay so any two elements of the same class differs by a rational number of course number okay then on the other end any two elements of different classes classes differ by any rational number okay this is trivial but so how we use this equivalence relation okay we use it in that way okay so we denote by k this set the quotient set of 0, 1 by this equivalence relation so 0, 1 is the quotient set so k so the quotient the quotient set 0, 1 by okay now it comes the axiom of this choice so okay by the axiom of this choice we can define this law the axiom of the choice we define this that exist this function ff so where f is defined over k so the domain of f is k and the codomain is our interval 0, 1 okay so such that for any you can choose in any class of equivalence so in any element of k so in one element belonging to the class belonging to the class of equivalence okay so now I wrote you so this function such that it maps any class of equivalence so which is an element of k any class of equivalence into equivalence let call it c in k so this c stays for a class of equivalence to an element little c in 0, 1 okay so that is so you have that c the class of equivalence is this maps of course this c belongs to to the class of equivalence and so the criteria with which you associate this big c to this little c is given by the axiom of the choice the axiom of the choice ensure that this is possible to provide a law with these properties so basically this big c makes the part the role of the set e alpha of the zermelo axiom okay okay so now we define okay so we define this law and we define t as the a set of image t is equal to f of k so t is the range of f okay we have that t of course is contained in principle in 0, 1 and so now we don't don't have time but tomorrow we will see that actually t is our no measurable set okay tomorrow we will prove the following proposition or rather a theorem is is not the back measurable, okay so to prove this we will take advantage of the sum model 1 that we just introduced so this is why we need to introduce all this machinery but now it's quite long and tomorrow we will prove this okay for now I think we can stop we can stop here