 Ok, značiš, da bomo smo izgledali. Ok, zato je tudi a n. Enšta sekuenja izradi r. Ok, ki... Ok, tako še, to je inšta sekuenja. in zdaj gaugejte A nasiljenje kaj dobroklju dva unijana s vziv leti. Ok, nu zrešảičim pa o ihljamo, da je radnja u obey vej vziv delelji vesel o zetnih, asetna in ta delovost kaj nekaj stvar leti. Ok. Ok. We already prove this statement under the hypothesis that AN are measurable. Ok. So we just start by removing that hypothesis. Ok. So just observe that by the monotonicity ok. Just let me tell you this by the monotonicity of the outer measure we know that the limit exist. Ok. Ok. Then we have this inequality. One side is trivial. Ok. It's trivial again by because A of course contains AN for any N. Ok. So now we look the other side. Ok. So the idea is to prove what we to use what we prove for AN so to construct a suitable sequence of measurable set to make this AN. Ok. So for instance what we saw is that there exist HN measurable in particular a border set if you want belonging to this class g delta. Ok. Such that you have that AN are contained in HN the outer measure of the two coincide. Ok. Here I might remove the star but let me keep it. Ok. Ok. Then we define HN this vk which would be the candidate to approximate AN in this way. Ok. This way and this vk which is the union of this vk. Ok. Ok. For a theorem that we prove we have that the limit of the measure of vk is equal to the measure of v and now we observe some inclusion so we have that ok. We have that ak is contained in HK contained in ak is contained in ak plus 1 which is contained in HK plus 1 and then you have that ak is contained in HN for any N larger or equal than k and from this you have that ak is contained in vk so at the end you end up with this chain of inclusion you have ak contained in vk and in HK. Ok. So what about the outer measure? So you can say that the outer measure of ak is less or equal than the measure of vk less or equal than the measure of HK which is equal to the outer measure of of ak Ok. So finally what you find ok. So what you find is that the measure of vk is equal to the measure of ak so this limit exist and we have that the limit of m star ak is equal to the measure of v and then since v is equal to the union of vk which is contained in the union of ak which is a we have that the measure of v is larger or equal than this and then you have that also the other the other inequality involving the limit is achieved Ok. Sorry. So basically the fact is that you have to use this because the outer measure coincide but you cannot use them automatically because you need an increasing sequence so you define this vk in order to obtain this increasing sequence Ok. Then if you have to provide a counter example for which then another this this is more concerned with in a counter example Ok. You find a sequence bn of bounded subset Ok. Of course such that bn contains bn plus 1 and m star of b is less than the limit of less where b is intersection of such bn Ok. Ok, to solve this exercise you have first to observe the fact that you have to provide a counter example Ok. So you cannot look for a counter example in the class of measurable set Ok. Because we prove that for measurable set the quality holds so you have to look outside measurable set and moreover we are required that we want bounded subset. So even the counter example that we give during the lecture is not a good one Ok. So the idea is that if we have to look outside the class of measurable set we will use the non-measurable the only example of non-measurable set that we saw. Ok. So here I will be you can for instance you need a decreasing sequence so you can you can for instance consider t, so let t be the non-measurable non-measurable set that we introduce using the termelo action and so on and consider you can define ok. the ti you know you remember the translation model 1 and we can set bn has for instance the union of tk for k which is larger than n. Ok. Ok. If you can check that b which is the intersection of this bn 20 set and what else then you can also prove that m star of bn ok. will be for sure larger than m star of t. Ok. which is which is positive because t is non-measurable because if it is 0 would imply that t is measurable but we know that it is not measurable so this is so we have these two and so and so we are done. Ok. because we have that the outer measure of b is is 0 and so so ok. Ok. and then the last exercise ok. it was easy ok. so we consider en a sequence of this joint measurable set measurable set ok.ovaro of course and let a be any set so take a set belonging to the set of parts of arrow ok. then we want to prove the m star of a intersection union this time was the countable union ok. of EI is equal to what is equal to the sum of the intersect EI yeah. outer measure, thank you. ok. we prove this for finite union and so we want to use this ok. one side come is easy, just come from subeditivity so you just have to ok. let to just see that this fact that the intersection this intersection at the end if you do some observation is equal to is equal to the union of the intersection ok. ok. then you just consider ok. by by subeditivity ok. subeditivity we have this side of the equality ok. on the other hand for the other for the other inequality we use what we prove so you have that of A intersected ok. this union is larger or equal then the same intersection but with finite union ok. and then here we use what we prove so former result ok. so this is equal to the sum of this sum of the intersection of EI and then we just let n go to infinity so this does not depend on n so ok. this is just to to see this proof ok. now we go on with the topic of last time ok. ok. last time I think we conclude the lesson with the losing theorem ok. so last time as I was saying that we conclude the lesson with the losing theorem which is a kind of approximation theorem which tells you that a measurable function is equal to a continuous function everywhere outside and a set of a small measure ok. so we add the theorem with the following ok. so let f measurable ok. yeah. we start by so f defined in an interval measurable function and then ok then for any delta positive there exist a set A ok. now maybe we can just say for any delta there is this continuous function G such that the set of the X in I where f and G are different is less than delta ok. ok. we prove the theorem but I want to warn you so about the fact that this result so doesn't mean that the point where the two coincide are continuity point for f ok. and I give you an example of this or rather the warning I want to I will write it just to stress this so the losing theorem does not imply that f is continuous outside a set a set of measure of small measure ok. so the fact is that the set where they do not coincide might be dense in I ok. so for instance take for instance just to consider f defined in this way to be the characteristic function of this set here ok. is not yes does not imply that f is continuous outside a set of small measure yeah. we assume that f is a measurable function and we say that there is a continuous function g for which they coincide but outside now I will show you an example this is not me so think at this set at this function here it's a measurable function because it's a characteristic function of a measurable set ok. but of course it's not continuous no? so for instance you have that you can take as a continuous function g c0i and the set where they do not coincide is 0 ok. this is 1 this is equal to 1 on the rational number which have measure 0 the rational number in 0,1 ok. and this 0 outside ok. so g if you take g identically equal to 0 which is of course continuous so you have that the set x i where f of x is different so the measure of the measure of this set 0 it's really small ok. but you cannot say nothing about the continuity of f ok. you're not convinced in i, yeah, yeah, yeah, yeah I mean I take i for instance in this k as 0,1 ok. ok. so now I want to summarize what we did so far in just three fact probably in the book of Reuden they are listed under the name of three little wood principle ok. so we saw this fact that for every every measurable set e finite finite measure is almost, is nearly won't finite union of of intervals ok. so basically you can express a measurable set by mean of very elementary set and when I write this I have in mind one of the point of the theorem of characterization by approximation of measurable set ok. was probably the last statement we tell you that the symmetric difference of a measurable set and this finite union is small ok. and then we saw yesterday remember on Tuesday the Egor of Severini theorem so we tell you that point wise convergence is nearly the uniform convergence ok. so this is theorem of Egor of Severini and this is theorem authorization by and the third one is is losing theorem that a measurable function a measurable function is nearly a continuous function in the sense that I just said ok is nearly a continuous function and this is the theorem of ok. so we can somehow outline and summarize what we did so far with these three results ok. ok. now we are in position to start to introduce the Lebesgue integral and I would like first to recall you the construction of the Riemann integral ok. also to see how the shortcoming of the definition of the Riemann integral and so the need to introduce another another definition of integral ok. ok. so we consider a third function defined on an interval and we assume that it is bounded ok. we know how it goes on. you take a subdivision take a subdivision of A B A B so A equal to X naught so the end point are equal to X naught, X1 Xn Xn is B and then we consider the supremum of F and the infimum of F under this this interval ok. so the supremum of F of X for X in between Ci minus 1 and Ci and with small m I denote the analogus so the infimum of F of X over the same interval ok. ok. then we define the upper sum and the lower sum and so define and the not with capital S as indeed the sum for I which goes from 1 to N of Mi times Ci minus 1 and the lower sum S of X equal Mi Ci minus Ci. ok. we define the upper remaining integral and the lower remaining integral so in analogy we have in this bar I denote the upper remaining integral F of X as the infimum of the upper sum over all the possible subdivision of A B ok. this is in analogy we define the lower remaining integral so you understand how it is defined is the supremum this time of the small s over again all the possible subdivision of A B so basically the definition of the remaining integrals is that when these two coincide we say that F is remain integrable and the remain integral with these values ok. so in general we have that the upper remaining integral is larger or equal and the lower remaining integral but when they coincide coincide F is integrable ok. the remain integral of F is defined this is definition as one of the two decrease sides ok. ok. this is probably the definition of remain integrals that you see in the high school or but there is also another way to define this you can also define for instance the upper remaining integral has the infimum of the integral which is of course an elementary definition over all the step function c such that c is larger than F ok. and in analogy you can define the lower remaining integral has the supremum of the elementary integral of step function phi such that phi is less or equal than F now I will write the supremum will be F of X sorry, F of X over all step function phi such that phi is less or equal than F ok. when we will define the Lebeski integral we rather follow this strategy so somehow we would generalize this way to define the integral ok. but we will see later but I just want to anticipate the step function we will consider simple function ok. now just to understand why the remaining integral is not enough for our application we can observe that I mean it's quite intuitive that we want to give a notion of integral that leads you to the fact that the integral of the characteristic function of a measurable set coincide with the measure of the set ok. so we would like to have something like this this is the measure of the set so give some notion of integral so why the remaining integral is not good for this purpose because again we see this example which is the same that before. so consider this function the characteristic function of the interval intersected with again the rational number so these are the characteristic function of all the rational number in 0,1 ok. when we consider the sum the upper sum and the lower sum ok. by saying that ok this is measurable and the measure of this set is 0 ok. so what about the upper and the lower remaining integral of this function ok. this is equal to 0 and this is equal to 1 so they are different the fact is that in 0,1 so when you take the supreme you will always find point where key is 1 and for the same reason here you will always find point where key is 0 ok. so basically we saw that they do not coincide so this is not a good definition ok. so a measurable function is not always we want to extend this notion and to consider the back integral ok. we proceed by step by step ok. so first we will find the back integral for bounded function over a set e the measure of v finite so we will follow the strategy of Reuden then for non-negative function and then we generalize to the general back integral ok. before I tell you that we want to define the back integral we want to use the simple function so now we need to give to introduce what is called the canonical representation of a simple function so as a definition ok. the canonical representation ok. represent simple function phi is the following here that phi of x is equal to the sum ki ai where ok. this set here ok. let's assume that i is from 1 to n is a collection of numbers which are non-zero such that ai are different from zero for any i and the set ai are disjoint ok. ai are the set where phi of x is equal to ai and ai are disjoint ok. i are disjoint by definition ok. this implies that they are disjoint ok. so we want to define the back integral of this function which is quite phi if you want consider phi which is defined on a set on a bounded set ok. then the very integral of phi of course would be the sum of ai measure of ai ok. where this small a and this set are the one of the canonical representation ok. ok. then there is just a lem which tells you that suppose that you want to express phi not in the canonical representation but in another representation for which the set involved in the characteristic function are still disjoint. so for instance let phi assume that phi is equal to ai where intersection ej is disjoint different from j for instance you can think you have a1 you can think that you can split a1 in 2 and call them u1 and d2 for instance this is what I mean ok. suppose that ai of course are measurable the measure of ai is finite and then you have that the very integral of phi must still be expressed as ai the measure of this ai the proof is easy because you consider the set ai where phi of x is equal to r to a sorry this is the union of ai where ie is equal to i so basically it means that if you consider some repetition you can still define the lebeki integral in this way you have that a the measure of ai is equal to a the measure of the union is equal by additivity because we are assuming that they are disjoint has ai the measure of ai and so when you consider you sum up phi of x sum this is I use the definition with the canonical representation is equal to the measure of ai and this is equal to the sum of ok ok now we want to see that the lebeki integral is linear ok may I write here ai, I mean ai because you already said ai which is the one in the canonical representation are defined as the x such that phi of x is equal to a then I express this in terms of this disjoint set ai for such that ai must be equal to a so in principle you can think that these are more than the set of the canonical representation but they are still disjoint so may I write here ok let's find simple function which vanishes outside a set of finite measure outside a set of finite measure or just think of them as defined on a set of finite measures is basically the same ok then you have that linear combination of ai phi plus b see this is the integral of a linear combination is the linear combination of the integral ok ok moreover ok what you have that if phi is less psi then the inequalities preserved by the integral ok by this so idea to prove this and maybe also to prove other theorem concerning the composition of two of two simple function is that you want to to express both phi and psi as linear combination of characteristic function of the same plus of set ok consider phi is equal to the sum of ai key of ai like this c is equal to the sum of i key of b i ok i goes from 1 to n and i goes 1 to n now i consider all the possible intersection of the set in those two class of course ai are disjoint and bi are pairs with disjoint so consider all the possible intersection of the type ai intersected bj ok if you want i can change the index here if you prefer ok, i will still find a finite collection of disjoint set, maybe it will be a finite collection so i end up consider obtain a finite collection e e k of disjoint set and now i want to represent phi mc with respect to this collection ok, because this is something common between the two ok, so we have that c is equal to the sum of a k c e k and ok k will go from 1 to sum n and c k which goes again from 1 to the same n bk e k ok, this is of course not in general the canonical representation but we saw that it doesn't matter we just proved that we are not obliged to use the canonical representation because the notion of integral for simple function does not depend the important thing is that you take for the moment that e k are disjoint ok, so consider the linear combination so this is the sum which goes from 1 to n of a ak plus b vk key of e k ok, so just compute the integral ok, by definition ok, this is the sum a ak plus b bk of the measure of this set e k ok, here you just use ok, the linearity you just use the fact this is a k ak the measure of k plus b k bk the measure of e k and this is by definition is just a this is the begi integral of the function phi plus b the begi integral of c so, the first part is proved ok, now we have to prove the one concern inequality ok, this is easy in the sense that ok, you have so we start by the positive psi is so trivially here that psi minus phi is positive and then when you compute psi minus phi by definition of integral this is non negative and for what we by the step before, by the linearity so by the step before so concerning the linearity you have that psi is larger or equal than phi ok ok, now we see that we can we can use also other representation of of simple function to define the begi integral so this is a very easy lem and we will use to prove this, we will use the linearity so corollary consider phi equal to the sum of ei ki ei i from 1 to n so here we do not assume any hypothesis on the fact if ei are disjoint or not ok, just you can represent it the begi integral is the same form the measure of ei ok, proof is simple just ok, just use this phi is equal to the sum of ei ki ei ok, use the linearity that we prove so consider this as the function linearity, so you have that this is equal to ei the integral of ki ei and this integral is the measure of and here here use the definition here use the linearity and here use the definition measure of ok, now we want to to define the begi integral function and as I told you we want to use the same strategy something analogous of what has been done for the remaining integral ok, so consider start, consider simple function of course and and we know how define how to calculate the begi integral for this function and simple function of phi and psi and then consider the number this quantity so the infimum of the integral of the begi integral over e of psi of a simple function of phi such that psi is larger or equal than some function f some measurable function f and ok, in analogy infimum of phi where phi this time is less or equal than phi ok, then you have a characterization the following characterization is the following in all this definition I always, I already told you but just to stress that I consider the measure of v finite ok, we have that let f defined on e values on r bounded measure of v finite ok, then you have the following characterization that if these two quantities coincide then f rather, the two quantities coincide if and only if f is measurable ok, so the infimum of psi over all the possible psi step function, which are larger than f is equal to the supremum of phi over all the possible simple function phi, which is less or equal than f if and only if f is measurable so assume so we start by this implication so we assume that f is measurable ok, since by hypothesis it is bounded we have that there exists some m positive such that f is less or equal, is less than m ok, or less or equal ok, then we define some set e k in this way the x such that f is in between two values k minus 1 we already consider actually this kind of set half of x larger than k times m divided by n where k goes from minus n so e k are measurable ok, this joint by definition measurable because f is measurable ok moreover we have that the union of this k, e k gives you for k, which goes from minus n, and then gives you the whole set in u ok, done then by additivity we have that the measure of e is equal to the sum of the measure of this of this set e k ok, now we want to thanks to this set we want to define somehow special simple function of the type psi and phi ok, psi n of x is equal to m divided by n k phi of x is equal to by n, and now instead of k i k minus 1 so basically the end point of this n, n, n minus n no, no, the index probably because I use the index k is in between minus n and then it runs from minus n and then the k here because maybe I use it's the same k because k in capital ok, ok so they are of course like this ok, and then you consider pass to the integral ok, it does the infimum psi larger than f over all the possible in simple function so this is the quantity in the statement of the theorem is less or equal sorry, no, yes less or equal than cn this cn here of e which is equal to m over n sum n to n k ok, and in analogy you have the supremum over the all possible simple function phi larger than f f is larger or equal then the integral over e of phi n which is equal to ok, then we consider the difference of these two here so consider the difference of this and this we want to see that the difference is zero not because we want to see that they coincide so that zero is less or equal than infimum of e psi minus the supremum phi e then we use the equal smaller yeah, yeah, yeah sure, thank you the phi are always smaller and the psi are always bigger ok, phi then we use those inequality ok is equal to m divided by n the sum of the measure of e k ok so we use the fact that e has finite measure so this is indeed the measure of e which is finite and so has n tends to infinity this goes to zero this goes to zero and so we saw that the inequality is equal to zero ok, now we have to prove the other implication may I raise here yes so we assume that these two are equal and we want to prove that f is measurable ok, so again ok, now we use the definition of infimum the property of infimum and supremum ok ok, so by by definition of of sup and inf ok, there exist simple function again phi n and psi n ok, such that we always the same chain of inequality and and we have that zero is less or equal than psi n minus phi n less or equal than 1 over n ok, and call this fact ok, define phi bar on the top has the supremum of those phi n and psi on the bottom has the infimum of those psi n ok, we saw that these two functions are measurable because we prove this and again the chain of inequality is still valid, ok now we introduce this set n as the set of x where psi bar is strictly less than phi sorry, phi bar is strictly less than psi ok so why we need to see so our aim now will be to prove that the measure of n is zero if we see that the measure of n is zero we are done because we see that we see that for instance phi f does not coincide with one of the two for instance phi bar only on a set of measure zero ok, so we saw that if a function coincide almost everywhere with a measurable function then it is still measurable ok so we want to see that the measure of n is zero ok ok, we automatically know that n is measurable because of the definition because these two are measurable ok now we need a countable union to express n as the countable union of the x such that phi x is less than psi x minus one over k and call this set here nk and we define another set which is a slight variation of this nk and I will call them nk n where instead of this I use instead of phi bar and psi bar I will use phi n and psi n so this would be the set of x where phi n of x is less than psi n of x minus one over k and we know we see that nk is contained in nk n now let me yeah this is the upper ah, no maybe yeah, no, no I think I call them both of them with the upper sign I mean this phi bar was the supremum of the phi n that I define this phi bar are you convinced? ah, ok ok, now I know what you mean ok ok, use two notation ok, so put this is the same function yeah, I'm sorry yeah yes yes, I'm sorry now take x yeah, what I wanted to do is to ok, this is cause this is trivial but I need to do this this is the set of the psi n of x is more convenient for me to express this set in this way ok, now let's consider this function ok, this is equal to one over k if if x is in the set if psi n of x minus phi n of x is larger than one over k and this zero otherwise so if on the contrary psi n of x minus phi n of x is larger less or equal than one over k in any case this is a positive this is always positive so in any case we have that that this one over k times key of nk n in any case we have that one over k of key of x is less or equal than n of x minus phi n of x ok because if we are in this case it is zero, zero is always is always less than something which is positive ok and if we are here this values is one over k and here in the in the situation when one over k is less than this ok, so we always have this inequality ok, now we want to pass to the integral so we have that one over k of the integral key n is less or equal than the integral of this ok, we can use the linearity by n which is by what I call dot is less or equal than one over n was one of the first inequality in the proof ok, so as usual we let that this tends to this n tends to infinity so we have that the measure of nk n is less or equal than k over n for any n and then this measure is the measure of nk is less or equal than the measure of this m sorry of this nk n which is less or equal to k nk over n which is equal to zero so this measure is zero ok, so the measure of nk is zero and think that now we are done because n the set where where our function f was strictly in between this phi bar mc bar so the set n of this nk so we have that the measure of n is zero so basically we have that f is equal to phi, for instance phi bar almost everywhere because it's always they coincide outside to a set n of for any x in b minus n n with measure zero so we have that being this measurable we have that also our f is measurable ok, so we conclude the proof so now it's clear how we will define that the baggy integral for measurable function as one of these two quantities and ok, I think that for today we can stop here