 Počkaj, pl called the definition of the general-indegral. It is said to be integrable. If the positive part, both the positive part and the negative part integrable. And moreover we define also that in this case as the integral of f ki me nickel imamo pozitiv košnju, očer, pozitiv košnju, bojez, očer, jez obradi pozitiv. Zelo se sezamo pozirte proprate delinjarnost, danes monotonist, meni? Zelo se so drželi, da je obradi pozitiv, Do veče, da se počite vzpe, da je kot se blizno vzpe, da je to či izvajno, da je zelo, da je je vzpe, da je je vzpe, da je vzpe, da je in zelo, da je vzpe, da je in zelo, da je to vzpe, da je reverz. tako našel je niskem tako, bo smo videli, da, nešel njih, je F in G in viči, tako, vsej ne vsožite o vse zač marque, vse pokazujemo zero. V nekaj naprej, da ne pomembimo v in tegrobijške funkcije, ko je vzaj, da se opravimo f plus g. Izgledaj Mark. Tako da je f je in tegrobijšk. F in nekaj, kaj je vzaj, da je vzaj vzaj, f je in tegrobijšk. To je, nekaj, to je od vzgleda, da, if f je integrabil, we know that... F is integrable. We know that both the positive part and the negative part are integrable. So of course you have that f plus, you can express like this. And so you have that this is finite. Ok, the reverse. Ok, if you have that f is integrable, then you have that f plus is less than this. And the same as for f minus. And so of course you have that being the difference between the two. You have that this, this still. Ok, now we see another important theorem of convergence under the sign of integral, which for sure you already know is the dominated convergence theorem or it is also known under the name of the back theorem. Ok, so here you have, for this theorem you have no requirement on the sign of f, ok. So you consider a sequence fn of measurable function, ok. And you have g is a function, which is measurable and integrable, ok. So what we require here is that we assume that the absolute values of fn is bounded by g. Ok, g is clearly non-negative. It is bounded by g for any n, ok. In e, this is also for any n, ok. And moreover we also assume that we have the point-wise convergence of this fn to point-wise convergence almost everywhere to some function f in e. Ok, then what we can say is that indeed under this hypothesis the limit of fn of x, the integral of the limit is equal to the limit, ok. This is f of x is equal to the limit of fn of x in gx. Ok, let me, maybe I should write it more, just write it f of x. Ok, for the proof, may I erase this, the blackboard in the middle? Ok, the proof. So we want to prove two sides of this equality. Ok, we have, of course, that minus gn is less or equal than fn minus g. Ok, so, first we observe that g plus fn is larger or equal to zero. So, and of course we have that, and so fn plus g converges to f plus g almost everywhere in e, ok. Ok, so, what can we use here to deduce at least one first convergence result? Here we use the Fatoulemma, no, because these are positive and so we are under the hypothesis of Fatoulemma. We have that the integral over e of f plus g is less or equal than the limit fn plus g. Ok, so, this is what, this is the limit, the integral of fn over e plus, here it doesn't depend on n, so, plus g e. And, ok, here we have, this is of course, by linearity, is this. Moreover, here we use the fact that g is integrable, ok, so the integral is bounded, so we can cancel out and we can first say that we have this side, ok. And then we consider the other inequality, so g minus fn is larger or equal to zero. Ok, again we use, so we have that the limit of g minus fn is g minus f and again we use the Fatoulemma. Ok, my Fatoulemmas we have that g minus f is less or equal than the limit of g minus fn. Ok, and here you have, ok, the integral of g, of course, does not depend on n, so you can, and here you put minus the limit soup of fn. Ok, for the same argument that I thought before, you have this is g minus the integral of f over e, so these terms can cancel out, and so what we get, we get that the limit soup is smaller or equal, yes, thank you. So we combine these two, ok. So, and we use the fact that of course the limit is always less or equal than limit soup, and so if you ask, we get that indeed the integral of f coincide with the limit of fn. Ok, so we are done. Ok, and now, because you have that the fn is, because of this, so you have that fn is less than g and minus fn, yes, I mean, you have this two inequality, you know, this come from this fact. I didn't get your questions. Ah, you mean to rephrase this hypothesis in the case of g, I mean, here the thing is that you have that fn must be bounded with an integrable function and also the negative from above and from below somehow. And so this is a way to express this. I mean, I didn't get your questions. But they are equivalent, I mean, the fact that you value this or you consider the two inequalities is equivalent. Ok, maybe we can, yes, here. Ok, this is due to the definition of limit soup. You can exchange, ok. Probably we use this, I mean, this trick also when we prove that the limit soup of measurable function is still a measurable function. Ok. Ok, now we see another version of the dominated convergence theorem, which is somehow the hypothesis are a bit relaxed and this goes under the name of generalized dominated convergence theorem, ok. Ok, so we have gn. So we start by somehow, we consider, we replace g by a sequence of function, ok. So gn is a sequence of measurable and integrable function such that we have that gn converts to a function g almost everywhere in e and we require that g itself is the limit is an integrable function. And then we consider let fn, a sequence of function, which are measurable and such that fn converts to a function f almost everywhere in e and we have that fn is bounded by this function gn. So we don't need to have the same function here for any in e for any n and we have that if we know that the integral over e of gn is equal to g here we can assume that they are positive. Ok, then. Yeah, yeah, yeah, sure. Ok, then we can say that also the limit then goes to plus infinity of fn converts to the integral of f over e, ok. Ok, so somehow the proof is a bit analogous of the previous. So we divide it in so you have that gn plus fn is larger or equal to zero. Ok, is it clear this, no? Ok, maybe later we can. Ok, and gn plus fn we have that it converges of course to g plus f almost everywhere in e. Ok, so we have again by the fatulema you have that g plus f the integral of g plus f is less or equal than the limit. Ok, so in this case for g we know that the limit exists so this can be written as as the limit plus here we have to to maintain the limit of e fn and then here we use our hypothesis and here you keep, ok, again since we know that g is ok, g is finite so we can cancel out the g the integral of g and so we get one side. Ok, ok, the other side it's analogous less or equal ah, no, less now this is this step is equal because you know that the limit of gn is equal to g, no? And here, ah, before this ah yes, this is this is equal, no? because in principle you can split the limit of gn plus I mean, I think I mean, this is you can split between the sum of the the two limit, ok? And then this is equal to the limit you know that the limit exists and here you just have to get the limit ok, so the other ok, so you have that gn minus fn converts to g minus f almost everywhere in e and again by the fatule you have that g minus f is less or equal to the the mean of fn and again ok for me it's equal, but this is equal to the ok, the integral of g minus ok, here you can use the trick that we used before ok, so finally what you get is that f is the integral of f is larger or equal than the limit sup and so again if you combine the two you get you get the thesis may rise here ok, now we will see an exercise so we have fn, a sequence of of integrable function some sequence ok, and we assume such that fn converts to f almost everywhere in e and what we want to prove is that with f measurable then we want to prove the following equivalence so this goes to zero if we have that this goes to zero so one side so this implication the fact that the modulus of this difference goes to zero implies this fact is the easy part so the other part is somehow the difficult one and we have to use also this hypothesis so proof ok, so so this part so this side comes from the fact that in general you have that this is equal to fn minus f and so when you get so in general so when you pass to the integral pass to the integral we obtain the convergence of the integral and then for the other for the other implication so if you start by knowing that this convergence holds this one we have to use some theorem of convergence under the sign of the integral ok, so we pass to the integral ok so if you take the integral ok ok, so what would you use in this to prove the other the other way around maybe we can use the generalized theorem that we just proved ok and so we have to find for instance somehow this difference the modulus of this difference plays the role of the fn ok, and so we have to find a sequence that bound this this modulus here ok how would you one way is that ok, so we want to apply generalized the back convergence theorem ok ok, so first of all we notice that so we have to infer something about the point wise convergence of this first of all ok, but this is easy so we have that fn minus f goes to zero almost everywhere in e ok by hypothesis ok and then how can we bound this in terms of a sequence that satisfies the hypothesis of the generalized the back convergence theorem so one way is for instance if you observe that this is less or equal than this ok so this is somehow plays the role of the gn ok of the gn, they are integrable and we have that gn converges point wise almost everywhere in e to some g where this g is where g ok, gn are this ok g is two times the modulus of f what we have is that what we need also to check is that the limit of gn for the limit of the gn is the limit plus f but we know by hypothesis that this first part of the limit converts to the integral of f so this is indeed two times f ok ok, so we are done because now we are under the hypothesis of the generalized the back convergence theorem and so we are done ok ok, now I want to show you another way to prove this second implication the most interesting without using the generalized the back convergence theorem but only the the back convergence theorem the first one that we state it's a bit more it's a bit more lengthy but it's interesting to see ok, just to simplify the things we let me reformulate so you have fn and f some function so assume that they are non-negative because otherwise we state the result with the absolute value so just assume that they have non-negative integrable ok, and we assume that fn converts to f point twice most everywhere and assume that that we have the convergence of the integral ok, so we want to prove that we have also the convergence of the integral so the integral of the absolute values of this difference goes to zero and we want to prove this without using the generalized convergence theorem but just the classical the back convergence theorem ok ok, we observe this fact that if you consider this sum you can express this sum has the maximum between f and f minus plus the minimum between the same ok, so let me call this denote this with with a dot ok ok, this is because this is if x is such that fn of x larger or equal than f of x then you have that the maximum between fn of x and f of x is equal to fn of x and the minimum is f of x ok so if you sum up that things are consistent if otherwise x is such that you have that fn of x is less or equal than f of x so you have the reverse of course fn of f is f and the minimum is fn ok, so is it true? and moreover we use another representation of this form ok, moreover we have that fn minus f values of difference is equal to the maximum of fn of f minus f plus ok, plus f minus minimum of fn of f ok, just because if you have two cases if x is such that fn of x is larger or equal than f of x ok, then this is equal to this and you have that the maximum between f of n and f is equal to f of n and the minimum of n and f is equal to f ok, so so if you replace here the maximum is fn so you cancel these two f and the minimum is f and the other case is completely analogous ok, so ok, just to wrote it in a more simple way so you have that this is equal to of fnf minus the minimum of this ok, so we want to study the convergence of the modulus of this difference and we want to to use this representation of course ok, so we have that this is the integral of the maximum of fn minus the integral of the minimum so this is a, the term a and this is the term b so we start by this one which is the easier somehow ok, so we have by hypothesis b ok, we have that fn converts to f almost everywhere in e and so we have that by this you can also prove that the minimum between the two fn and f converts to f almost everywhere in e I think we already prove something of this type ok, ok, so we can apply the the back convergence theorem so by the back convergence theorem so why we can apply the back convergence theorem because ok, let me also sorry, let's state like this, because this is less we are interested in saying that this minimum can be bound by by f and this is integral ok, so now we can apply the back convergence theorem and we have that the integral of this minimum converts to the integral of f ok ok, now we have to treat the other part, the part a ok, so here we still have the point wise convergence of this so we have still that the maximum the maximum between these two converts to f almost everywhere in e but what we don't have in general it's hard to find a function that bounds this sequence, ok so we have to argue in another way so we recall that by this formula that I denoted with this dot that the maximum of fn f can be represented as fn plus f minus the minimum between the two ok so if we pass to the integral what we have we have that maximum minus the minimum ok ok, so what can we say about the convergence of this part by hypothesis by hypothesis we know that this converts to two times f so here we just prove by part b that this converts to the integral over e of f so at the end what you obtain is that this converts to the integral over of f so at the end what we get here is that this converts to to zero, yes, because you have f minus f ok, so this is ok, so this is somehow another way to prove the same result but by using by using the theorem of the classical lebel convergent theorem ok, and now we see somehow some consequence of the lebel convergent theorem so a classical consequence is that you can differentiate under the sign of the integral but ok, we will proceed by step so we prove this theorem we consider f a function defined in this product space so e is a measurable set and y is a metric space ok, we assume the following we have three three hypothesis basically that for any y in this big y the metric space the function which with y fix associate to f x, y is integrable over e then the second hypothesis is this time we fix x and we consider f as a function of y this times we require that it is continuous in y ok and finally we require somehow the existence of dominating function so resist some g defined from e to non negative function integrable such that this is less or equal than gx this is for any x in e and for any y in y ok, then where in the question why associate f as y is continuous in e yeah, it mean in y, I mean in the variable y, yes of course in y below I mean has a function ok, it is clear because yeah, I know that it is ok in y this was just to stress the fact that thinking at it as a function of y ok, y in y ok, then so what we want to prove is the following f, big f defined from this big y to r is continuous and now I tell you also how it is defined f y is defined as the integral over e of this function f x y e x so here you think that y has a parameter somehow ok so we have that this is for any fix at y this is integrable, so somehow this is a good definition it makes sense to consider the integral ok, in just 2 so we want to prove that this f is continuous this big f is defined on a metric space ok, and on a metric space the continuity is equivalent to the sequentially continuity so it is enough to select a sequence which converts to some sequence yk to some y and to prove that f yk converts to f of y ok so f so since y is a metric space continuity if you want f is continuous if so we take point y small y in this big y and in the sequence yk which converts to y ok, what we want to prove of course is that if f y k converts to f y ok so we use the definition of f so we just maybe it is convenient to define this auxiliary function hk x defined as f x y k and in analogy hx defined as fx in y ok, these are measurable functions and so on so we have since f is continuous in y hk x converts to hk ok so infinity and then we can bound this hkx by this function g ok, so we can apply the dominated convergence theorem we have that this is ok by definition so then we use our notation in this converts by the dominated convergence theorem ok, so we are done so many of you guys ok, now we see another application of the the back convergence theorem ok, so ok, we have that f e times i r e is a measurable set and i is an interval ok and then we assume the following ok, this time that for any y in the interval i the function which to x associate fxy is integrable in e for any x in e this time we require more than the continuity is we want that this function is differentiable so in y belong in 2 y ok, this is clear and 3 as usual we want that there exist an integrable function that control from above this time the partial derivative of f with respect to y ok, such that this is less equal than this g and this must hold for any x in the measurable set and for any y in the interval ok, now we can imagine what is the proof the thesis then we have that f is differentiable ok, of course and ok, f f is defined as before and we have that so f is f of y is defined as exactly as before v x and what about the derivative we can somehow bring the derivative inside the integral ok, this is the interesting part of the theorem so we can exchange somehow the integral and the derivative ok, so one false question that one may ask is that it makes sense to consider this this integral so is this function measurable ok, so the answer is yes, because you can you can think just that the definition of derivative has incremental quotient so you have that df dy xy take a long sequence so this is the limit as k plus infinity of x yk yk minus f of x divided by yk so somehow this is still a measurable function this quotient, this is the difference and the quotient if we take the difference and then the quotient in the framework of measurable function and the limit is still a measurable function this is we already prove so somehow this this integral makes sense this is a first remark ok we have that say what else I want to say that and ok, it is measurable of course it is integrable because of this ok, we want now to investigate the differentiability property of this big f so we just as before take a sequence yk that converts to y and you have f yk minus f y divided by 1 of yk minus y ok, and now we use the definition so this is the limit as yk tends to y of ok, of this so we have this is f x yk minus f x y ok, then just use the linearity of the integral you can exchange ok, so now we are interesting in study this quotient ok, so we want to apply the dominated convergence theorem so we have basically to check two things the point-wise convergence and if it exist a bounding first a function an integral function that can bound this quotient ok, the first thing is that the limit so point-wise limit x yk minus f x y yk minus y converts point-wise, as k tends to infinity to the y of f is y so we have point-wise convergence and b ok, let me define this as before introduce this auxiliary function hk xy this time is all the quotient and this let me call this hx y so b this is a function of course of x I treat the yk as a parameter here so here you have that at any fixed x you can apply the mean value theorem ok, so this is ok, by at any fixed x we can apply the mean value theorem and so we can continue this inequality by saying that this is df dy in x c and this is less or equal than g of x ok, we can get rid of this g of x, this is by hypothesis so this is true for any x and for any y ok, now we really can apply the Lebesgue convergence theorem ok so here we can continue here, ok we can exchange we can bring the limit inside the integral this is k plus infinity fx k minus fx y divided yk minus y and that's it, now and then we are finished this is indeed so we have that f is differentiable and the quality that we state and we can represent its derivative in this way ok, maybe has an exercise you can prove the following this is somehow a slight variation of what we saw so you have that g is an integrable function and ok, e is measurable and bounded f y be defined in analogous way for instance g of x times the sinus of x times y in dx yes, sin sin in dx, of course ok and y, ok, y, this time y I took as matrix space r and then we can prove that f is sin infinity in r, ok so the proof somehow we have to combine the previous theorem so combine so the previous theorem and the and then you have to prove by induction ok and so and induction ok, the induction on the degree of the derivative ok, in the last 10 minutes let me show you this lemma so we have we start by e by a function f which is integrable ok, e, of course is measurable and we suppose that this integral is non-negative for any a measurable contained in e ok ok, then as we already did we want to extract some point-wise information this time on the sine of f from the information sine of the integral so what can we prove is that f is non-negative almost everywhere in e ok so somehow the proof is basically always the same so a way to rephrase this is to prove so the thesis can be viewers saying that f where f is negative as measure zero ha? sorry? f is thesis ok, so it is ok, we want to prove this is to say that f is non-negative almost everywhere so we want to prove that this set as measure zero so as usual how can we express this set here as the union ok, the countable union of f is less than minus one over n and then ok, by assumption we know that the integral of f of x over this set here is non-negative so this is less or equal by definition of the set minus one n time the measures of f ok so we have that is equal to zero because here I have a minus sign for any n and then and then we are done ok because you have that measure of this union is less or equal than sum of the measure of and so this is zero and so we are done ok, we can stop here