 Ok. So, jaz recolju, sve definitije, ki smo zelo posledali jasnjev. Zelo smo posledali definitije delobnjih, zelojte vse, jaz jaz jaz iz je tudi z vrst, in zelojte, ki je zelojte in tako pošljama vzeloj nr. Ok. in zelo vse njih zelo. Ok, potem, po vzeli, da smo videli, nekaj nekaj nekaj bolj funkcij, da smo zelo, je stejnje funkcije, ki je stejnje funkcije, je nekaj nekaj funkcij, z celim njom reprezentacijom. F of x je bilo zelo izpovedan. Samo je v hrošenju kombinacijo inteveljtej. V dne breaks are the important functions for the second step and then we would like to approximate as the second step the second function for the izgledaj, če je ozvrčenja, OK? Zelo sem početnja, da je ozvrčenja teorema. Zelo sem izgleda, da je ozvrčenja, da je ozvrčenja, v nekroze in intervali, vzelo. Zelo smo početnja, da je ozvrčenja včasnih razdajov, in da se učajen, da je zelo, da je vzelo, da je vzelo, da je vzelo, da je vzelo, da je vzelo, da je zelo, da je vzelo. ok, dan we want to show that for any epsilon there exist, we can find a step function h, I mean just in the statement I stress the dependence of h upon epsilon, then I will drop this notation establish the function, of qr from A B to R, step function, which is close to our given measurable function in this way. So, you have that the measure of the set where h epsilon minus F are larger than epsilon, is less than epsilon, des Zo arise, if you want. And we can also find for any epsilon, we can alsolastno drugi igočaj ti vet Jurop added Vojejo in lapsilon. In ta – pa obadneвajte ročne präigane работe pohlaimede b okih možerov� vsak v supportersu,одеassembljične podjakje z napiveč, Outdoor, at least continuous function, and you have the same kind of closeness. So you have J epsilon minus F larger than epsilon is less than equal then epsilon. Okay. Maybe yesterday I put it in that way. I think yes, okay. V admitted 1943. So we will proceed somehow by step. We first will provide ourselves Ruff approximation of F by means of bounded function. Then we will somehow do better and better step by step. Zato vidimo, da se pošličim sekvencijem setem vzajem vzajem vzajem vzajem vzajem vzajem vzajem. Tako, imamo zelo z X in A, B, da je f of X veliko nekaj, da je m. Zelo, da imamo zelo, da je in interval. na pojavo in cestiv. OK, jesteśmy v vse intervali,no tato, ko nam tem, vse izlajte, vse izleso bolje izleč, izleso posledaj iz vse kaj je zelo, je všeša, na plček več, The limit is of course a decreasing sequence of set Hence, by a previous result, we can say something about the limit of the measure of this set because we are under the good hypothesis. So what we have? We have that limit as m goes to plus infinity of the measure of this je, da smaka vse je noga samo naprej, ali je pri vse izpolanske inkredi. Znažemo taj poškot, ne, ne bo, da je, da pa je hoditi za tré se. Zato, da tako vse je, da ne nіє, ne, tako, da so, da, ne, ne. Znažemo, da samo je, ne, to je, da, ne. ki so ide vse, razkončijo ta, na kratku, je to, ki je, ki je več del, ki je tudi morduljale da je pistelo vzelo vzelo. Nokia tudi je tudi, ki vse videl, ki pistelo vzelo vzelo vzelo vzelo vzelo vzelo vzelo. Ok, zato vidimo, da je rečen, da je nekaj zelo, da je to, da je to, in da je nekaj zelo, da je nekaj zelo. V rečenu, da je rečen, da je nekaj zelo. Ok, zato ešte kaj zelo povede ovo št über niko je nekaj zelo? Ešten do rečenu, da je nekaj zelo, da je to, da je nekaj zelo, da je nekaj zelo, da je nekaj zelo, tako, da je noh nedaj, kaj je zelo, da je zelo, da je nekaj, da je nekaj, da je nekaj. Teraz smo zaznačili, da so zelo zelo tudi zelo način. Zelo način je zelo vsevala. Način smo zelo Fm, da je produk vzelo v F in tudi karakteristik, način oto je zelo, tako zelo, da je to nekaj komplement, karateristik način oto je zelo, da je modulj s f of x neč nekaj, da je m. To je zelo, da je zelo, da je to zelo, da je to zelo, Taj je zelo vzelo v počke, tudi je vzelo vzelo vzelo in vzelo vzelo. Tudi je to vzelo vzelo in vzelo vzelo. OK. OK. Now if you, what I want to use now is a result that I state yes. No, no, it's the product. Yeah, I was saying that I want to use a result that I, that I prove yesterday that you can approximate a bounded function by means of a sequence of step function, if you remember, and the convergence is uniform if you start by a bounded and measurable function. If you check in your notes, you can find. OK. So we know that. So this would be the function that I want to approximate. OK. So we can find, OK. We can find. So I write you by a previous remark. I don't know how I call it. Yes. I think it was remark. OK. So we can find a sequence of simple function that call them phi n. OK. Which converge to this fm uniform. OK. Proved. Yes. OK. So just to fix the idea, let n, the integer n be such that you have that phi n of x minus fm of x is less than epsilon over 2. OK. For any, of course, for any x, a, b. OK. Again, by a previous lemma, it was the last thing that we proved yesterday, we have that. We can split. We can somehow estimate the measure of what, of our starting function f minus phi of n. This is, this gap is larger than epsilon in this way. Let's reequal then the measure of f minus phi m, larger or equal than epsilon over 2, plus the measure of phi fm minus, no, this is, sorry, make a mistake, m minus phi n, larger or equal than epsilon over 2. F minus f capital M, f capital M, so you got the answer track minus phi n. OK. OK. OK. Then, OK. We can observe that f minus fm, larger than epsilon over 2, is contained in the set where, for larger m, f is different from fm, which is equal to the set where f is larger than m. And phi n minus fm, is larger than epsilon over 2, is equal to the empty set, because of this. OK. So again, we can continue this, this estimate, saying that this is less or equal than epsilon. I use this fact here also, OK. OK. Now we know that, now we focus on phi n. So this is a step function. And we know that we can represent it in this way. So this is, because here there are three blackboards. You can just see phi n is a step function, simple function. OK. From now on, let me drop this index, because otherwise it becomes too, the notation is too heavy. So we can represent phi, phi of x, as the finite sum ai kei, with, of course, we still have ai less than m. OK. And ai, there with this joint, this joint measurable set. OK. And now comes the theorem of characterization by approximation that we prove. OK. And so we saw that, by the theorem of characterization by approximation measurable set. So we have that. Since we are under somehow, under the hypothesis that the measure of ai are finite, so if you remember we had at some point an additional hypothesis in that theorem. OK. We have that being the measure of ai finite. What we prove? We have that. We prove something about the symmetric difference of this set and the union of this joint open intervals, if you remember. OK. So what we prove is that we have that for any x of i that exists. A finite union of intervals of, they were also open intervals in this union ui, such that the measure of the symmetric difference between ai and this union is small. OK. It's less than, for instance, epsilon. It's arbitrarily small. So you can think that it's less than epsilon over m. OK. M was the index here at the top. OK. So how we can define? Let us define h. This step function h has the sum, as i goes to from 1 to m of ai characteristic times, the characteristic function of this union ui. And we have that. H is different from phi. The set where h is different from phi is contained in the union from i, which goes from 1 to m of ei symmetric difference of ui. So basically you have that. It leads to the fact that the measure of this set is less or equal than the measure of this union. OK. i from 1 to m e i symmetric difference ui, which is less or equal than the sum of epsilon divided by m, which gives you epsilon. OK. OK. Then we have, we use again the lemma, the last lemma of yesterday. So by previous lemma, OK. We want to estimate the measure of f minus h, larger or equal to epsilon. This is less or equal than the measure of f minus phi, larger or equal than epsilon, plus the measure of h divided by m, different from phi, which is less or equal than 2 epsilon. OK. So we prove the existence of the step function, OK. Now I just want to step function to stress that, so we have this step function h, but we can always define, so represent h instead of, as the linear combination of the characteristic function of this union of open intervals, we can, we know that any, these unions can be, they also can be represent as the union of, the finite union of this joint open intervals. OK. So, in turn, you can see h as h, you can represent h of x as the sum for j, which goes from 1 to l, for instance, of aj of ki aj, where aj are open intervals at this joint. OK. So in general l will be greater than m, of course. OK. Now we have to provide a continuous function, OK. OK. So we want to construct this continuous function. First of all, we start by observing a very easy fact that we can approximate a characteristic function of this type. Maybe withdrawing is easy to understand, so we can approximate a characteristic function of the type, you know, cd, open interval cd is a continuous function, so let me take maybe this open, because then here we have open interval. By continuous function that I will call psi delta, OK, xcd, so it depends on many things, so that they differ, these two functions, just on a set of measure to delta. So can you see how you can construct, I mean it's very easy, so you have c, maybe you can also interpret this. You have c, you have d, you have a characteristic function here, this is the level one, this is the level one, and you want to approximate this with this function. I mean, for instance, you can argue in that way, you can put, OK, here I put, we have aj, which are, OK, let me use this, so you have cd, and then here you have c, so you have c plus epsilon, or delta, c plus delta over 2 probably, and c, so you have c plus, oh, it's too small. So basically what we want to do, I mean, you have this interval aj, which have the form aj v aj, OK, so what we want to do, we have this characteristic function of this interval, and then we want to approximate it, this characteristic function, just here by a function like this, OK, which is continuous, OK. So if you take this set small, they differ only on a small set, OK. So here you can, for our purpose, take aj plus epsilon divided 8 over l, and here you take bj minus epsilon divided by 8 over l, and then this would be the point bj minus epsilon 4 over l, and here in analogy, you have this would be the point aj plus epsilon divided 4 over l, OK. OK, then you can approximate the step function by means of linear combination of this function like this, OK, this function like this. OK, so you have that measure of the set x in AB, gj is different from key, these are the function gj, OK. aj bj is equal to 2 times epsilon divided by 4 divided by l, so at the end you get epsilon divided by 2l, OK. OK, so now the idea, of course, is to the natural candidate is, for a continuous function is a g defined as a linear combination of, from j, which goes from 1 to l of aj gj of x. OK, this is continuous, this is continuous and has also compact support, and you have that the measure of the set where the gap between f and g are larger, equal than epsilon, can be estimated in this way, is less or equal, but we use always the same argument, OK. f minus h larger or equal than epsilon over 2 plus the measure where h is different from g. OK, this is less or equal than epsilon over 2 plus the sum for j, which goes from 1 to l of the measure of key aj, is different from gj, which is less or equal than epsilon over 2. So you combine everything, this is less or equal than epsilon, OK. And so this concludes, OK. So this concludes the proof. Maybe I think that in the following we will see that we can do even better, I mean if you smooth out this part, you can obtain also a better approximation, OK. You can obtain approximation of very smooth function, OK. But we will see, we will see later on. Here, but not, just to say that this is epsilon over 2 plus epsilon over 2, OK. And no, no, no, it's just to say that, OK, if you want it to be more precise so was epsilon over 2 plus epsilon over 2, because no, so you end up with epsilon. OK, but epsilon is arbitrary small, so it's not a big difference, OK. OK, now, OK, in two lemma, with one lemma and one theorem, we will see that the convergence almost everywhere, to, I mean among, of course, measurable function, is almost a uniform convergence outside set of arbitrary small measure, OK. This is, I mean, the final result is the theorem of Egor of Severini, which tells you that basically you can, apart from set which has very, which has, which have small measure, you can think at the uniform, at the point-wide convergence, has a uniform convergence, OK. So, lemma, OK, we will prove this by step. OK, the first lemma is the following, let E, OK, be a measurable set of finite measure, OK. And then you consider fn, a sequence of measurable function, OK, of measurable function, of course, defined on A, on E. OK, then let f be a measurable, real-valid function, OK. It's enough almost everywhere in E. So, it can achieve also plus or minus infinity, but just on a set measure zero, OK. OK, such that fn converts to f almost everywhere in E, OK. So, we have a kind of point-wide convergence almost everywhere. OK, then, given any two arbitrary small quantity, so, given epsilon positive and delta positive, we'll prove that there exist set A, measurable, so, A in E, measurable, and, OK, with the measure of A delta and an integer n, which will depends on epsilon, on both epsilon and delta, such that for any x, which does not belong to A, so, we are outside somehow a set of a small measure, OK, and for any n measure equal to n, we have that fn of x minus f of x is less than epsilon. OK, so, we start by, we fix an epsilon positive and we introduce this set gn as the set of points in x, point x in E, such that the difference, the gap between fn of x and f of x is larger or equal to epsilon. And then, by means of this gn, we define other set fn as the union, the countable union for i, which goes from n to infinity of this gn. OK, we can immediately consider that fn is sorry, fn plus 1 is contained in fn by construction, for any n, and, OK, we have that f1, yeah, gi, thank you. Yeah, sure. f1 has finite measure because of our hypothesis, because it is contained in me, so f1 has finite measure. OK, so, now we observe this. OK, now we observe that think let x be a point in E where f of x is finite and such that we have that the limit of f of n of x is equal to f of x. So, we have convergence almost everywhere, so this is true almost everywhere, OK? OK, this means we can rephrase this that in the following way this means that there exists there exists an n integer n in n such that for any n larger or equal to an n fn of x minus f of x is less than epsilon or equivalently if you equivalently we have that x does not belong to fn for any n, OK? OK, now just let me follow this following, so the following fact OK, since fn converges to f almost everywhere by hypothesis and f is finite but this means the intersection of this fn is contained of measure zero where the convergence does not take place and hence that the measure of the intersection of fn is equal to zero. OK. OK, now we use a previous result about the limit of the measure of the of decreasing sequence of set so we will so yeah, in the measure zero sorry? sorry sorry sorry sorry yeah, now it's unique I mean, it's a set, it's unique, yeah I mean, convergence means that the convergence it converges almost it converges everywhere outside of a set of I mean, everywhere in e outside of a set of measure zero that we introduced before yesterday so it means that there exists a set I mean, a set of measure zero such that it converges everywhere in e minus this set, OK? OK, so we use a previous result about the limit of decreasing sequence of sets OK, so we have the limit of the measure of fn so we are under the good hypothesis is equal to the measure of the intersection of this countable intersection OK? and we saw that is equal to zero OK, now we just use the definition of limit so then the definition of limit limit the definition of limit, we have we have that delta positive no that exists in integer n such that the measure of fn minus the measure of fn is less than delta OK, and then if we define as a the set in the statement the theorem has fn, for instance then we get the t, this is because we have that a is what? a is a set of the x in e such that there is a k larger than n, such that fk of x minus f of x is larger or equal to epsilon and then we have we have that for any x in e minus a and for any n larger or equal to n fn of x minus f of x is less than epsilon OK, and so we are done so basically the dependence upon delta and epsilon takes it up in n and depends on delta and epsilon and now we use this theorem to prove another fact which goes under the name of, is known as Igor of Severini theorem may I erase? no, OK so again we start with fn as here is Severini it's an italian matematik OK, it's a sequence measurable function which converts to a real valued measurable function almost everywhere on a measurable set set e of finite measure, OK? OK, then for any small arbitrary small number eta OK, there exist subset a of e such that the measure of a is small you can find some a with measure less than eta OK, such that fn this sequence converts to f uniformly on the set e minus OK, so basically you have uniform convergence outside set of very small measure, OK? so the idea is to of course is to apply the previous lem, OK? OK, so we apply with special choices of epsilon and delta, OK? we apply the previous lem with for instance you can take epsilon equal to 1 over m and delta take delta equal to eta times 2 minus to the minus m OK, so we know can infer that there exist a measurable set this is from the previous lem which in principle we depend on eta and m such that the measure of a eta m is less than delta, no? so then eta times to the minus m and moreover again for the previous lem there exist an integer n which again will depends on eta and m such that OK, for any k larger than n fk minus f is less than 1 over m on e minus this kind of set, OK? OK, here we want to get rid of the dependence of m and we do this so we define a set a eta has the union over m, so the countable union of this set a eta m OK OK, the first thing to do is to to estimate the measure of this new set OK, so we have that the measure of a n just use the countable subactivity property so you have that this is less or equal than n so sorry n infinity of the measure of each set which is equal to sum for m which is goes from infinity of eta times to n to the minus m, which is precisely eta, OK? And moreover, what about the convergence? Moreover, we have that OK, we have that for any m there exist n such that for any k larger or equal to n you have that fk of x minus f of x is less or equal than 1 over m on e minus a so you have that basically this means that the limit of this norm the norm that leads you to the uniform convergence as k goes to plus infinity over x belonging to e minus a of f of x fk sorry, fk of x minus f of x is equal to 0 and so this concludes the proof fn conversion uniform to f so now we just want to to ask you one thing do we prove this under the hypothesis that the domain so the set e, so which is the domain of definition of all this function is finite, OK? So what about if you have for instance a domain which is not bounded do you think that the theorem is still true or you can provide a counter example now my question is that I prove this the limit is theorem hypothesis of the boundedness of e so of the domain of definition but do you think that we can remove this hypothesis and still obtain this kind of result or it is really needed or you can provide a counter example for which for instance you can find function defined on the real line which converts point wise to another function so you can never have uniform convergence outside set of small measure of course otherwise I wouldn't ask OK, think at the very I mean when you somehow when you are in the real line you are allowed to move with set of type nn plus one characteristic function of set nn plus one OK, this is sorry this was again so counter example so the fact that measure of e was finite is necessary OK, OK why, because take e for instance the whole real line so of course the measure is plus infinity and define fn of x has the characteristic function of of this interval OK you can see that fn of x converts to f converts to zero almost everywhere actually everywhere on r because no matter how you fix x you will find some hand which will step out but what about so what about the measure of this set fn minus zero is equal to one is what, is the measure n so this measure is fixed is equal to one so you cannot make this arbitrary small OK so here we do not we do not have uniform convergence outside set of arbitrary small arbitrary small measure OK, so it is needed OK, now we will prove another theorem which goes under the name of losing theorem which tells you that a measurable function is almost a continuous function OK goes under the name of losing OK losing so what in capital letter is so letter f a measurable variable value function in interval ab then even delta positive there is a continuous function this function we call it phi on ab OK, such that measure where f is different from phi is less than delta so this is the way I mean almost that a measurable function is almost a continuous function OK so we start by using the first thing that we proved today so we saw today OK, that for any epsilon we can find, we can construct a continuous function g epsilon OK, which is continuous so belongs to this continuous function over an interval such that you have that the measure where f of x minus g epsilon of x is less than epsilon is less than epsilon and then OK, now we want to fix this epsilon somehow to discretize so let fix epsilon to be equal to 2 to the minus n and so let g n the corresponding g and let g n equal to the continuous function OK, then we define a sequence of set a n this way the x in i such that f of x minus g n of x are larger than 2 to the minus n OK, we know that the measure of this set a n is less 2 to the minus n and moreover moreover this also means that f n of x minus g n of x no f n, sorry f of x minus g n of x is less to the 2 to the minus n for any x outside a n in i minus a n OK, now we consider the union of this a n so we consider we define as b k the union for n, which goes from k to plus infinity of a n and we call and we define as b infinity what? intersection of this b k over k no b b k is a union for n, which goes from k usually I use the other notation OK now I consider the intersection this countable intersection I define this set b infinity OK, so by construction we have that b k plus 1 is containing b k and the measure of b 1 is finite and so again we use a former result then you have that the measure of this intersection which I call b infinity is equal to the limit of the measure of b k as k goes to plus infinity OK, so what about this measure the measure of this b k is less or equal than the sum less or equal than n, which goes from k to plus infinity to k minus 1 which goes to 0 as k tends to plus infinity so basically from this we deduce that the measure of b infinity is 0 and you can prove that for any x it does not belongs to b infinity you have that g n of x converges to f of x OK so basically what we have is that g n converges to f almost everywhere OK and we found a sequence of measurable function which converges pointwise almost everywhere to a measurable function f so we are under the hypothesis of the theorem of Egorov-Severini OK so we are somehow under hypothesis of Egorov-Severini theorem OK so what can we say we can say that for any epsilon positive there exist measurable set g epsilon for instance g epsilon measurable such that this measure is small is less than epsilon and such that g n converges to f uniformly on i minus g epsilon so at this stage how would you proceed so you have these are continuous function you have uniform convergence this is a uniform limit so what would be the natural candidate to be would be for instance the uniform limit of g n on i minus g epsilon but but what is the trick here is the point is that you can deduce that the uniform limit of continuous function is continuous if you are over a compact set on a closed set so you are not sure that this is a closed set so we have to argue a little bit more OK so this is not but we did a lot about how do you can approximate measurable set by mean of open, closed bore set and so on OK OK so we know that we can find we can find an open set o epsilon so it's open OK such that what o epsilon contains g epsilon and such that the measure of o epsilon is less than the measure of g epsilon plus epsilon which is less than 2 so you put less than 2 epsilon OK so basically you have that e i, the interval i minus o epsilon is contained in i minus g epsilon of course and this is indeed a closed set OK closed set in the relative topology of i OK now we need to use so we want a continuous function but which is defined on all i is not enough on i minus o epsilon so to extend this continuous function on all i so to extend the uniform limit of g n over this closed set which we know that this is a continuous function over all i we use the t-z extension theorem so we recall extension theorem what does it say it's in a quite general framework OK let's take it so let x be a matrix space and let y be a closed subset of x and then if we have g a function defined on y with values in r continuous unbounded then there exist a g defined from x to r which is still continuous and it coincide with the small g over y and we have that g so the restriction of g on y is equal to this small g OK we want just to apply this theorem OK so we can infer that that exist OK there exist some capital F which belongs to which is continuous on the whole interval in the matrix space c0i which of course plays the role of x such that F restricted to the set i minus o epsilon is equal to f restricted to i minus o epsilon and we also saw that and we proved by construction actually that the measure of o epsilon is less than 2 epsilon so but in any case it's arbitrarily small OK OK so you can conclude the proof I mean c0i plays the role of x c0i minus o epsilon plays the role of y OK it's bounded because it's continuous on a compact and OK and then we are done OK OK so for today I think I will stop here maybe we will discuss more in detail tomorrow but just a little bit of warning maybe you can think about it this losing theorem the losing theorem OK does not mean that that any x in i minus o epsilon is a point of continuity for f so you cannot infer any property of this kind in this the fact that they coincide that a measurable function coincide with a continuous function outside a set of small measure this doesn't mean nothing about the fact that f is continuous or not maybe you can think about about this for instance take think about it in any case I will discuss on Thursday OK