 Užržeme,11 never.icede we introduced the concept of Outdoor measure. We define it, we denote it by m star. And we define it as a set function form the set of parts of r in to the extend the numbers, so 0 plus infinity. And it was defined that way. je infimum v tem zelo v poslednjih vse, tako in in oz. A unijonega taj poslednjih vse je oz. Prost vstavljaj je najboljih jaz, načo je vse poslednje. zelo, da... Intervali, zelo, da bi bilo intervali. Zelo, da počekajte, da vzelo, počekajte, vzelo, nekaj za intervali. OK. Tako za intervali. OK, zelo, da smo počekajte. Aja, zelo... Zelo, da je zelo, da počekajte, počekajte, da reč socialesk subpravitodne zelo. Zelo. OK. Posevamo to voltaj, koncentreramo da zelo sliža in zelo. because koncentralen u bl винjini od dve entries. Naredil je začešljenje. To je to, da je to druga. the other measure of the countable union of this set can be bounded from above by the sum of the outer measure of the set a n. The first thing that we observe is that if... to nekaj lahko bilo zelo izlično, ne? Če bomo bi nekaj, da je, da je, da je, da je, da je, da je, da je, da je, da je, da je izložite, danes je je zašločila, ne? Ne? Ne, ki je, ki je, ki je, ki je, ki je izložite, nekaj. Ne? in kvaliti. Z nekajh spremstv, si to... ki se na koncu nešto, na razljah m starov n svojih n, je finit. Povej, n professionals. Nared, svojih pzitiv, je... je to z z definicijama mstar. Zdaj je to seqen, kaj je izgledaj, izgledaj ime. Zdaj, kaj je nekaj, in nekaj, ko je deločno od 1 do infinity, deločno izgledaj. najгиžne. Musite začelnje. The sum of the length of i and i over the index i is less than m star of a n plus here we chose epsilon to the minus n. O moramo posrediti ta znik, da povisem svoje tega človek sem pogorsila. Ok, da so me pridaj... Ok, da so me pridaj na srečke in pomegrati. Prej, da smo vzivo, tudi z tako izgleda uskoda, pribefete, da je hrab herejdete! Ok, da so se vrdu. So we consider what we want to prove and start of the union of A n. Vz Ji se v preserved equation, then the sum of n and the sum over I. So we are dealing with positive times which is less, so at this point to we make use of 2 stars. je nekaj nekaj suma m stara a n plus epsilon 2 n. To je nekaj nekaj suma m stara a n plus epsilon. Now, as usual, since epsilon is any arbitrary, small number, we get what we want. So we get that the outer measure of the union of a n is less or equal than the sum over n of m star of n. So basically, we see that so yesterday we saw that we want not the countable sub-additivity property, but we want indeed for these joint set the countable additivity property. So this somehow is not enough. So the way to seem in equal here would be somehow to restrict the set in the domain of of m-style, kaj? Yes. Yeah, yeah, yeah, yeah. I didn't open intervals. Yeah, I mean, you mean fix a n, think at n, the index n fix, and you see it has, and also here, think at this n as a fixed index, and you run i from 1 to plus infinity, and this is an open covering made of opening intervals of a n, the one that you choose in the definition of m-style, kaj? This one, yeah. Probably it's, no, no, I'm sorry, this is an error, it's an index i here. Yeah, thank you. No, let me, no, no, no, sorry, sorry. No, no, no, no, no, no, no, no, no, n was okay, yeah, yeah, yeah, yeah, n was okay, so, no, no, no, no, no, sorry, no, n was okay. No, no, think it's this, no, you fix them, so you do it like this. You fix epsilon, you fix epsilon, and then you know that by definition of infimum, you can, somehow you can make this right hand, this part as small as you want, okay? So think of this, you fix for any a n, think of this, fix a n, and then instead of epsilon, you can fix epsilon to minus, if for you maybe it's more clear, as a reason, you have epsilon, then you know that by definition of outer measure, for any epsilon times 2 to the minus hand, you can get, you can find, you can find this open covering, such that this holds, okay? No, I mean, this is just a trick to have the fact that when you sum up and you get, you just obtain epsilon, no? This is the way. Okay, and immediate consequence is the following corollary, so consequence of the countable sub additive property is the following, so if you have that, if a is countable, the countable set, then you have that m star of a is zero, okay? Because yesterday we proved that the outer measure of the single set is zero, okay? Okay, and another corollary, which is really also easy, is that you can prove that the set, for instance, the set zero, one, that of course you already know is countable for other reason, for other argument, you can see the fact that the set zero, one is not countable also as a consequence of what we proved. This is another way to see it. Okay, because of course we know that m star of zero, one coincide with the length of zero, one, which is equal to one, so it's not zero, okay? Yesterday at the end of the lecture I introduced some terminology. I will briefly recall this definition. So we decided to denote with this italic g, the set of, I mean, let me call it collection of open set of R, with g delta, g delta. Okay, no. With g delta is the collection of set obtained as intersection of set, of open set of set obtained as intersection of set, set in g, so open set. Then with italic f we denote the collection of closed set. So and in analogy we denote with f sigma the collection of set obtained as countable intersection. It's important. It's countable intersection and obtained as countable union of set in half. Okay. So the aim of introducing this set is that they will provide us with a way to somehow to approximate the Lebesgue measurable set by means of somehow elementary set somehow. So now we will first give an example of this kind of approximation. G and g delta. G is just the collection of open set of R. We choose all the open set. G delta is the collection is the set obtained as countable intersection because in general the union of open set is open. So if you take the union you will not add anything more. So just to remind you you take the intersection of open set. So it will contain open set and in addition all the set that are obtained as countable intersection of open set. Which in general is not open. You can find. And in analogy here you choose you consider closed set. You know that intersection of closed set is a closed set. So here you consider the countable union to add something more. Okay. Now you consider given A any set in R. Given A belonging to the set of parts of R and consider any epsilon positive then there is you can construct an open set O. Set O. So it's open. This is any set. This is open. Such that A is contained in O. And in addition you have that m star of O is less than m star of A plus epsilon. Okay. So do you have any suggestion on how to construct this set O? This open set O? Yeah. But you have to just to construct it. So think at the definition of m star. Yeah. Exactly. So. So the m star of A is the summation of the length of all intervals of some cover. Yeah. Then such that this summation is infium. Okay. So from the definition of infium adding any epsilon we are then zero to this infium which is m star of A. We get another cover which is the union of open intervals such that okay. So all would be just the union of the interval. Okay. So. Okay. Just think of the definition of the infimum. Okay. So you have that for any epsilon by definition n star infimum infimum of something. Oh. You have that the resist accountable collection intervals in. Okay. So that you have that the sum of the length of I n is less than m star of A plus plus epsilon. Okay. Okay. So you take O as we said as the union of these intervals of these open intervals and we are sure that O is open. And then we observe that to prove this. You have that. M star of O now we use that the accountable subatitivity subatitivity property of the outer measure is less equal to m star of I n. Okay. If you want we know that this is less take me side notional interval which is less m star epsilon. Okay. Okay. And this is one thing. So this is a first maybe elementary way to approximate A. Another way is the following. So we claim that there is a set g in the collection g delta which was the collection obtained as intersection of open set. Okay. Such that A is contained in g and this time we have more sharp inequality in the sense that we can say that m star of A is precisely equal to m star of g. Okay. So to prove this we use this fact somehow. Okay. So we have that but by the first what we just prove we have that for any k positive there exist a set an open set, okay. Open. Such that okay contains A and we have that m star of okay is less than m star of A plus for instance 1 over k. Okay. So now we define so how would you define g this time? Even without doing a lot of you know we know that g delta are obtained as of the intersection of this okay, okay. So it would belong to g delta and we have that A would be contained in g because it's contained in any in any okay now since A. But we Yeah. I can say from k greater than 0 from k belong to perhaps m. Hm? From k greater than 0 from k greater belong to m. Natural number. I didn't get your okay. Okay. G delta is in the set of all interest count of count of intersections of opposites. So if we say 40 greater than 0 this intersection won't be counted I think we have I mean but countable yeah but you can include also the empty set and it comes finite. Yeah but I mean you are interested in k you cannot divide it for 0. Huh? Okay take k from k for any k put from 1 to infinity. I mean k I mean k So k belong to m. Hm? Yeah. I mean k belong to m. I mean k belong to m. I mean k belong to m. Hm? Yeah. Okay. Yeah yeah I didn't What I mean Yeah yeah I mean yeah usually with k you denote an integer a positive integer. An integer? Okay. I didn't get you. Okay so we have that a is contained in g since a is contained in any okay. And we have that m star of g is larger or equal than m star of by this of a. Okay since g contains a and on the other end we have that m star of g is equal by definition as m star of intersection of okay plus which is less than m star of a plus 1 over k for any k. So since this holds for any k you get that m star of g is less or equal than m star of a. And so you combine this last two and you get what we want, okay. Okay and raise. Okay what you can prove for the next exercise is that the other measure is that translation invariant. Okay okay now we start we begin to to introduce the notion of the bag measurable set. Okay so as already observed I mean the other measure is not what we want because it's not countable additive so we have somehow to reduce the domain of m star. So how to do this. Okay so I will give the following definition so we should say that a set e is measurable with measurable I mean the bag measurable. Okay here in the following I just refer to the bag measurable set is measurable set because but this is the correct definition okay bag measurable. If for any set a we have the following the composition we have that m star of a the other measure of a can be split as the sum of m star of a intersected e plus m star of a minus e. Okay so this is just the definition. So just let me recall you that you can express a minus e as a intersected e complement. Okay. So in that way you can have two intersection on both that. Okay so just let me remark that only one I mean you can see this equality as two inequality. So one inequality is always true because you serve we observe that we have that a is contained in a intersected e union a intersected a minus e. So basically by the subatitivity property this side always hold. Okay. So this is always true. So somehow you can say that this is equivalent this property star star would be you can say that star is equivalent to prove the reverse of this okay to prove that m star of a is less is a larger equal than m star of a that e plus m star of a minus e. Okay. So basically in the following we just prove this when we want to show that e is measurable. Okay. Okay another remark is the following you have that the definition of measurable set of measureability is symmetric in e and e complement. Okay. Because you have that a as I told you a intersected e complement is a minus e is equal e complement is a intersected. So if e is measurable then also e complement would be measurable. Okay. Okay. Then another easy so just to show some the first example of measurable set we have the trivial case like an example. Okay. You have that the empty set is measurable. Okay. Because take any set in the set of parts of r then you have that m star of r is really equal to m star of the empty set intersected a plus m star of r minus empty set. Okay. This is just okay. This is zero and this is precisely equal to then another example another easy example is that the full real line is measurable. This is also very easy. So just to be here m star is equal to m star r intersected a plus m star of a minus r this is the empty set and then we try to prove step by step a kind of finite additivity property for the measurable set. Okay. So if we have that if you consider e1 and e2 to measurable set then we have that also their union their union is measurable as well. Okay. If you want to see it we choose a test set a so let a be any set in r and for first we use the fact that e2 is measurable. Okay. That be any set and since e2 is measurable then we have that. Okay. Then we have that m star of okay. We have the following. Here we consider as best set a minus e1. Okay. It's any set. It doesn't need to be measurable. We just say that this is any set and we use the fact that e2 is measurable. So we have that. This is equal to m star of a intersection no, sorry a minus e1 intersection e2 plus m star of a intersected e1 complement minus e2. Okay. Okay. I call it one. Okay. What we want to prove so that our thesis is the following. Okay. We want to prove the following. We want to prove that as I told you one side equal is enough that m star of a is larger or equal than m star of a intersected the union of e1 union e2 plus m star of a minus no, m star of a minus this is our our task. Okay. So we observe that two call this is a set equality of that a intersected union of e1 e2 is equal to a intersected e1 union a intersected e2 e1 as from two okay. We got that call it three m star of a intersected union is less or equal by the subjetivity property of m star of a intersected e1 plus m star. Okay. Now we start from from here. Okay. We'll just keep inequality one. So we have that m star of a intersected e1 union of a minus e1 e2 is less or equal this is by three intersected e1 plus m star intersected e2 e1 plus m star. Now we use one. Okay. We use one in the sense that so we replace this one and this is the same and we can say that this is equal to m star of a intersected e1 plus m star of a minus e1. Okay. But so far we did not use that e1 is measurable and we use it here is less or equal than m star because e1 is minus equals the proof. So basically with all these few properties that we prove for measurable set what we what we can say about the collection of measurable set in term of algebra or sigma algebra. We still don't know if it is a sigma algebra because we just proved that the finite union is is a closed operation but we can say that at least is an algebra. So as a corollary we can say that the family and simultaneously we also introduce this notation which is italic m the family m of measurable set is an algebra. Okay. So why? Because okay. We saw that for instance the empty set belongs is measurable. Then we saw that the definition of measureability is symmetric in e and e complement. So we have that if e belongs to m if e is measurable then we can also deduce that the complement is measurable as well. And finally with the latter we proved that if one and the two are measurable set then also the union is is measurable. Okay. So but we want more but we will prove that. Also we have more property. And now we prove something which is finite and utility property of this land. Okay. So again let a be any set and we have that e1 e2 e n is finite. So now we proceed by step. So we start by with a finite sequence of measurable set of this joint. It must be also this joint. This joint. Okay, measurable sets. Then we have the following decomposition and we have that m star of a intersected the union of e i where i goes from 1 to n is equal to what? Is equal to the sum for i which goes from 1 to n of m star of a intersected e i. Okay. We just as a remark. Okay. This fact is slightly more general than finite additivity property for m star. Okay. Can you see why? So finite additivity property would be take a reset and you would add that. Okay. This is the finite additivity property for m star. So why this contain this? Take a equal to r. Yeah? Yeah. No, no, no. Because if you take a equal to r this is this becomes a special case of this. Okay. So this. Yeah. Because take a this is just trivial but better to show. So this is finite additivity property. No, no, I have to prove this. No, no. This. I didn't get. I will show you. No, it's necessary that I will show you. Yeah, no, probably, I mean I understand probably what you mean. In this case we can get rid of star. I mean, but we still have to be produced now. Let it. We will define the Lebesgue measure as the outer measure for some special set. But this will come later. So for now take it in this way, okay? That comes. Yes, we will show later. Okay. Okay. We prove the lemma by induction. Okay. Maybe I mean finite. Okay. Just to be property. I don't want to of m star over measurable set. Let's put in that way to avoid confusion, okay? Because otherwise for any set doesn't hold, okay? You have to this holds if i are measurable, okay? Okay. We prove the lemma by induction. Of course, over the index n. Okay, for any one, of course is true property pn pn is true. This is just a triangle. Okay, so we assume that is true for the index n minus one. n minus one, so see that. So for any equal one is true because you just a intersected in one is equal to the outer measure of a intersected in one equal to the outer measure is total logic. Okay. So we assume that pn is true for the index n minus one. And of course we want to prove that it is true for also for the index n. Okay. Okay. Okay, so we consider so we know by hypothesis that pn is true for any two n of this joint. Okay. So pn we know that then we can we can consider the following set equality. So we have the following. Okay, we can write think that a intersection, the union of ei with the index i goes from one intersected in this is equal to a intersected in. So basically this intersection is redundant, okay. And you have also that a intersection. Okay, the union of one from one to n of ei intersection nc is equal to intersection of the union of ei i1 to n minus one. Okay. Here you have n minus one. Okay, now you use if you with a draw you immediately get these two. Here you have n and n. Okay. They are disjoint. Okay. Okay. Now we use the measurability of the last set of en. So we have that. Why that? So this is probably partial the answer to your questions. I don't know who ask where we use. You ask where we use. Okay. I don't measure up. Vlity en. Okay. We have that. So basically to prove this you use two facts. The measurability of en and this induction hypothesis. So you combine these two facts and these two equality. Okay. So we have that. M star intersected the union of ei i1 to en. Okay. And you use this as a test set. Okay. Is equal to m star of a intersected the union of 1 to intersected en plus m star of this minus en which is equal to intersection with en complement. Okay. A intersected the union of ei i1 to en intersected en complement. Okay. Okay. At this stage we can use this. Okay. This is equal to m star of a intersected en because of this plus m star of a intersected 1 n minus 1 ei because of this. Okay. Okay. And here in this passage we use the fact that the artist join basically. And now we use the fact that p n minus 1 is true because we assume that is true for n minus 1. Okay. Let me write here. Okay. So finally we end up with the following. So we have m star intersected en plus the sum n minus 1. Okay. Which is equal to the sum which goes from 1 to n of m star of a intersected. And this prove this leads to the thesis. Okay. Okay. The next step will be to prove that indeed the collection of measurable set is a sigma algebra. Not only an algebra but a sigma algebra. The collection measurable set is a sigma algebra. Okay. Just to... Okay. We already proved this. This we know. So what remains to prove is the fact that the countable union of measurable set is still a measurable set. So it remains if you have a collection of measurable set then the union of them provides you with a measurable set as well. Okay. Before... Okay. Before proving this theorem we shall need a kind of auxiliary lemma which basically tells you that instead of dealing with any set when you have to to consider this union you can express these unions in terms of a union of this joint set. Okay. Which would be easier to handle. So the lemma is the following. Okay. So let that EI be a collection sequence. Then we will prove that there exist another sequence. A sequence of set we call them VI. Okay. Still measurable but why they are almost useless. Measurable. Which are VI are these are my property they are with this joint. Okay. And such that you have that VI and VI is contained in PI. PI and moreover the union of VI gives you the union of EI. So you have these three factors baby. So how would you define this VI? So A1 B1 is equal to E1 because you have to start from something and then B2 yeah so you have this B1 is equal to E1 there you have that B2 is equal to E2 and then you don't need the U1 anymore. Minus E1 okay. Because this is already contained. Okay. And then you go on so you have because would be B3 would be A3 minus the union of A1 A2 so V N would be just AN minus A1 AN minus one. Okay. Okay. In such a way what we immediately can say that this BN are measurable because we are doing just union and this minus can be seen as a complement and so we stay in the in within this collection. Okay. Okay. Another fact that comes immediately is that BN by construction is contained in AN Okay. For any N. And okay what remains to prove is that for instance that they are paired with this joint. Okay. Okay. Let us prove any N and then with N different from N we have BN intersected BN is equal to BN distract. Okay. For for the computation for instance we can assume that the index N is less than the index N. Okay. For the computation we assume Okay. So we express BN has I N minus we express by definition BN is A N minus the union of A I all right which goes from one to N minus one so in particular this is contained in A N. Okay. What about BN Okay. BN is A N minus Okay. I perfectly is longer but maybe you have minus A I union union A N minus one. And this is contained in A M compliment. Okay. Because okay basically because you have union at some point you have some A M and you have union and you have N. Okay. Because M is less Okay. So this tells you that the art is joined. Finally we have to prove that if we we take the union we don't lose nothing anything. Okay. So we prove to prove that that the union of this B I give you exactly the union of B I. Okay. Okay. One side of course is trivial because so you have that you know that B I are contained in A I so this union would be contained in this union so we have that since B I are contained in A I you can immediately say that the union of B I is contained in the union of this. Okay. The reverse require a bit of some argument. Okay. So we want to prove the reverse or namely that the union of B I are contained in the union of B I we start by we pick an element in this union and we want to prove that it belongs also to the other union. Okay. Okay. So let X be an element of the union of A I which goes from 1 to infinity. Okay. In particular we observe that there must be an index N must exist there will be an index such that X belongs to A N. Okay. So that X belongs to A N and simultaneously X does not belong to the previous A J. You can any J less than N. Okay. So you take the smallest index it must exist. Okay. And in particular we have that with this choice we know that X must belong to what? X will belongs to the set B N because I recall that B N is made of A N because this union of I N minus Y but we know that X doesn't belong to those so it must be here. And so we have that in particular X belongs to the union of the I. Okay. Okay. So we prove what we all we had to prove and now we are ready that indeed the measurable set the collection of measurable set is a sigma algebra. Okay. Okay. You already so what remains to prove is the countable the fact that the countable union of measurable set is still countable. Okay. Okay. So consider a sequence E N measurable set. Okay. Okay. Then okay. Then we consider this union this countable union and okay. But the lem that we just prove we can just without loss of generalize we can assume that E N has the property that we just so they are paired with this joint of course measurable and the union the set that we introduced before. Okay. But just let me stress that must be pairwise this joint. So by lemma by previous lemma. Okay. So we can always make this assumption. Okay. Then define this set F N as the union of E I for I which goes from one to the index N. Okay. Of course F N is measurable because we know that if we do finite operation we still remains in the in the algebra. And okay. F N is measurable as we get what? We have that we have that for any test set A set of parts of R we have that the usual we have the following. Okay. We have that M star of A can be split that has M star of A intersection F N plus M star of A. Oh, let me use this this way to express the defense. Okay. Okay. Okay. Let me now observe that F N complement contains E complement. Okay. E is the union of all the set and this is just the union. Okay. So this is by this fact and the monotonicity of M star what we get. We get that this is larger equivalent M star of A intersected F N plus M star of A intersected E complement and call it star. Okay. Now we use one of the previous lemma by previous the one that I show you that leads to the to the finite additivity property so this lemma but we have that since E I are disjoint measurable we have that M star of A intersected F N which is nothing but the union of the first N E I is equal to the sum of M star the finite sum 1 to N star of A intersected E I Okay. Okay. Now we insert this in star this part here so we get so by the above equality if we combine these two what we get we get that M star of A M star of A is larger than the sum of I 1 to N of M star of A intersected E I plus M star of A intersected E complement Okay. And call it 2 star Okay. Now what we can observe? So we show we see that the left hand side of this is independent of N left hand side Okay. So we can deduce that M star if we let N goes to infinity we can say that if plus infinity we have that M star of A is larger or equal than the infinite sum from 1 to infinity of M star of A intersected E I plus the same term of A intersected E complement Okay. So finally Grover if you want noticing we can write intersected E E I we have that the sum of 1 to infinity of M star Now we use the countable so the activity property and we have that this is M star of the union of A intersected E and this is equal to M star of A intersected E Okay. Finally for what we get we get that M star of A is larger than M star of A of E plus M star of A minus E Okay. Which is this tells you that E is indeed measurable Okay. So this concludes the proof Okay. I think for today we can we can stop here.