 Tako, počekaj sem počekal, da je F2M, kaj je zelo. Tako, imamo Fn, neč negativne, neč negativne, in imamo, da Fn konversuje v F, nekaj nekaj, The last time, we prove that the integral of the point-wise limit is less than or equalum than the limit of the designer'smax infinity is equal to term for the integral of e over听. OK. Now we will see some example, for instance, concerning the fact that here less or equal is really needed. I mean, you cannot put just an equal. So let's see this example. So do you have in mind something, for instance? Can you provide an example of some function, easy one, of some function, of a sequence of function fn, which converts point wise to some function f, but such that this inequality holds with the strict sign. It's always usually the same example, which, OK. One, I mean, there are many, of course, but one of these, yeah. Yeah, exactly. Yeah, OK. For instance, if you take e equal to the all real line, and fn of x equals to the characteristic function of this interval, OK. Then, of course, we have that fn of x, OK, fn converts to f, OK, converts to f, which is equal to 0, almost everywhere in r. But this is equal to 1 for any n, and this is equal to 0. So, of course, you have that, in this case, inequality also with the strict sign, with the strict less. OK, the limit, of course, is 1, so it's less than. OK, then, OK, the limit, of course, in this case, you can write here the limit, no, because the limit exists. But it might be also some example on which the limit does not exist, the limit of the integral of the fn, OK. In this case, it exists, you have the strict sign, but here you can replace the limit with the limit. So, let's see another example. Consider e again is the whole real line, and you define fn of x in this way is equal to 2 plus minus 1 n of key n n plus 1 of x. OK, so basically, you have that fn of x is equal to, 3 of key plus 1, if n is even, and it is equal to just key n out. OK, so in that case, you immediately see that if you take the integral here, OK, of course, the values of the integral depends on which kind of integer are you considering, OK. So you have that, again, fn converts to f to 0, almost everywhere in r, OK, but in this case fn, the integral, you know, does not converge, so does not converge. So you have two different limits, OK, so you have, so the limit in this case of the fn is 1, OK. So this is somehow an example for which you have the strict inequality, and you really need to put here the limit, OK, because the limit does not exist. OK, OK, and then I will give you still another example, but in a bounded domain, OK, here we take e to be equal to 0, 1. So the case, now consider the case where the measure of e is bounded. OK, so just let me define the function, a sequence of function fk by their graph, I mean, it's very, it's clear how you can, if you want to write an analytical definition, but OK. So it's, for instance, it's equal to 1, here, or not, equal to k, actually, and equal to 1 over k, OK, and then they are 0, assume that here is 1, so the graph looks like some, this triangle, OK. So basically the area is always 1, OK. So also in this case you have that fk converges to 0, almost everywhere in 0, 1, OK, because however you fix some point x, then you will find an integer that step out of x, OK. So you are in the domain where fk is, if k is sufficiently large, you are in the domain where fk has value 0. OK, and so you have 0, 1, 0, sorry, the integral of 0, 1 over k is equal to, yeah, yeah, the area here is 1 over 2, I'm sorry, before I said 1 is 1 over 2, 1 over 2, but I mean, it's constant, and OK, so in this case, also in this case you have that the integral of 0 is less, strictly less than the limit, plus infinity of 0, 1. Is that a triangle, OK? So you have, this is the area. This is 1 divided by k. 1 divided by k. You just apply the area, OK. And, OK. OK, now we will somehow provide the other result of convergence under the sign of integral using the Fatouzlem, OK. May I erase this? Sorry, the point was limit. I mean, this is a convergence actually everywhere. I wrote almost everywhere, because it's enough, but it's, no, I didn't get your question. I mean, when k, you're not convinced that this is the limit, because basically the idea is the following now, is that, so you fix, we are talking about point-wise convergence, so you fix one x, even if it's here, you know that if you take k sufficiently large, this would look like something like this, OK. If k is sufficiently large, then the graph of fk would look like this. So this x, the x that you fix, will fall in the part of the domain of the fk, where fk is 0, OK. And then, OK. And this is, OK. OK, now, in state, these are other theorem, the monotone convergence theorem, which is also known under the name of Beppolevi theorem, OK. It's monotone convergence, Beppolevi theorem, which is an Italian mathematician. OK, of course, by the name, you understand that we need to put some requirement on the monotonicity of our function, of our sequence fn, OK. So we start by fn be an increasing sequence in measurable function, OK. Of non-negative, because so far we know how to come to this non-negative measurable function, OK. And let f be the point-wise limit fn, then we have that really, the integral of f can be computed as the limit of the integral of fn. Something we gain more with respect to the Fattulema, because here we have the equality, OK, and we have the limit, but of course we also require more, OK, because we have this monotonicity requirement, OK. OK, from one side we use the Fattulema, so by the Fattulema we are under the hypothesis of the Fattulema, we have that this is less or equal than the limit and then we need to prove the other way around, OK. OK, now of course we have to use our hypothesis, our monotonicity hypothesis. OK, so let fn is an increasing sequence, so we have that, OK, since increasing sequence, OK, then we have that fn is less or equal to less than fn plus one and for what we prove fn will be which is convergence to F, so we have this, OK. It's an increasing sequence, so the inequality is preserved to the limit. OK, so just take the little soup, the lim soup on the left hand side and so we have that the lim soup as n tends to plus infinity of n is larger or equal than the F, OK. So if you combine these two you obtain that the t is just by observing that of course the lim soup is larger or equal, but in general it's larger or equal than the lim soup, OK. So you get that. OK, now we see some corollary of this monotonic convergence theorem. So now we consider u n a sequence, so again it's u n b a sequence measurable or negative function and let we define F as series of u n. We have that the integral of F is equal to the sum, OK. So we want to use the monotonic convergence theorem, so it's quite clear how we proceed. So we define Fn as the finite sum and apply convergence theorem to Fn. And so basically you have that F is equal to the limit of Fn. Now you use the theorem. This is equal to the limit, so we take the limit outside the sine v integral and this is equal to the limit of y definition, sorry, of i, which goes from 1 to n of u n. And these are finite, so you can limit, OK. So this is just an application of the theorem. This one? Limit is not... Limit, ah, what is F, you mean? F is this. Ah, OK, yeah, yeah, yeah, OK. OK, you can, yeah, I mean, I just... It's the same, but it's just to emphasize the fact that you can bring the limit outside. Yeah, if you want, you can... Yeah, yeah, yeah, sure. OK, and then... OK, this is another, somehow, consequence. OK, so let F be a non-negative function. We consider some disjoint set. The sequence measure the set and then we take the union of such a sequence, e, and so then we have that. So the proof, OK, this is a countable union, of course, this is... Just if you let define ui has F times the characteristic function as one of the set in the union. OK, then you have that F, q of v is equal to the sum of this ui and from the previous proposition or corollary, previous corollary, pathetesis. OK, now we see another version of the Fatus Lemma in which we, somehow, we require less in the hypothesis but, of course, we also get less. So, which I will not part with as Fatus Lemma in its second version. OK, so we consider, so let fn, a sequence measurable... OK, then we also have somehow of measurable non-negative function. OK, OK, that's it. So we don't require anything else and what we can say, we can say that the integral of the limits of fn, because now we are not even sure that the limit exists, OK, because we do not require any extra hypothesis about point-wise convergence, OK, is less or equal than the limit of the integral of fn. OK, so this can be proved, has... can be viewed as an application. Can I erase this? No, OK. OK, can be viewed as an application again of the monotone convergence theorem. OK, so we have that... OK, I use the definition of limits of fn is equal to the supreme of the theme of fk... k... No, sorry. k is less than n and less than n. OK. And then we want to consider this function here. Gn has this infimum and this is an increasing... this is an increasing sequence of function, OK, for the sequence, because basically I am doing infimum over less and less indexes, OK, in the sequence Gn is an increasing sequence of non-negative and measurable function and so on. OK, so basically we are under the hypothesis of the monotone convergence theorem, but before in this we have that this is an increasing monotone sequence of functions so it admits limit so it admits and we denote with f this limit so limit Gn which coincide with the limit of f of fn. OK, now we apply the monotone convergence theorem to Gn and so we have the integral of f over e is precisely equal to the limit of any tensor plus infinity of the integral of the Gn, OK. And now observe some inequality and then we are done. OK, so being Gk is equal to of fk larger than or equal to n this is lessor equal than fn. So we have that that OK, Gn is lessor equal than, OK, this is lessor equal than fn and from which we have that f is equal to the limit n which is lessor equal than the limit and so what we get, OK. So now we give an example that this is more about the monotone convergence theorem. OK, if you take for instance we require that the fact that we want to take the freezing function, what happens. Is it still valid or not? With an example we will see that it's not valid. OK, so an example to see that the monotone where this theorem need to hold for the freezing function for the freezing sequence of function. OK, so here we are within this domain in real line and we define the candidate to provide a control example are this function the characteristic function of the f real line of this portion of the real line. So we have that fn Is it the freezing sequence? OK. Is it the freezing sequence which converts pointwise to zero. So almost everywhere actually we wrote almost everywhere because it's enough, but the convergence of course it's it's everywhere in R. OK, but of course when you compute the integral of the limit function and the integral of fn you have that here. This is infinite this is plus infinity and the integral is zero. So here is the same convergence goes to zero for the same reason before. You fix some x if fn is sufficiently large then x fall into the domain where fn has value of zero. Now we introduce the concept of integrable function so we have that a non-negative measurable function of course f is called integrable over e for instance over the main of the set e is its integral e is finite this result OK, we start by f and g to non-negative measurable function we have that if f is one of the two is integrable over e and if the other one can be bound from above by this integrable one and g of x is less equal then f of x is raining x in e then OK, almost n e then g is measurable as well e is not measurable, it is also integrable and you can compute and you have that e of f minus g is equal to the difference between the two OK, so we have that we just start that we can split f as f minus g plus g and we pass to the integral and then we have that integral of f is equal OK, to the f minus g plus g and then we use in productivity this is equal to the integral of f minus g plus the integral of g but we know that we use this fact we have that f the integral over e over f of x minus g of x is bounded it is non-negative also g, the integral of g is positive and we have two and so since we need to use our hypothesis this one is finite then we have that by this relation we can deduce these are both positive some give rise to something which is finite so it must be separately finite then we have that it's finite, so g is integrable and moreover we are within finite quantities so we can take this and move it on the other on the other side of the equality OK, so we have that minus g and so we are done we will use all this fact to prove a theorem that is called the absolute continuity of the integral so it's somehow a version of continuity for integrals so you have that we start by non-negative measurable function which is integrable so we must know a priori that the integral is finite and then what can we say we have that in the definition of continuity we use epsilon and then delta then given for any epsilon positive that exist in delta positive such that if you pick a subset of e, a subset a of e which small measure less than delta such that for every set a continuity in e with measurable set a with measure of a less than delta then you can say something about the smallness of the integral of this integral function of a this is less than epsilon ok, so we start ok, we start we divide it into cases first we consider an easy case when f is bounded ok, so if f is bounded so we know that f is bounded the integral it is so we know that there exist some m positive such that f ok, we can get rid of since it's non-negative this is true ok, just to stress that it's bounded and so we have that f is of course is less or equal than m times the measure of a which is less or equal than m times delta so if you pick epsilon would be m delta so if you choose delta equal to epsilon divided by m we are done then we consider the case in which the general case which in principle f can be integrable but not necessarily bounded and so the idea is that we want to approximate general function an unbounded function by bounded one ok, so consider this sequence this sequence of bounded function that are defined in a piecewise way so they are they coincide with f of x if of x is in between 0 and n or you it's a kind of cutoff you put them to be equal to n otherwise so if f of x is larger or equal than n then maybe sometimes it's useful to observe that this one you can express this as the minimum between f and n so this is also a way to see that for instance it is measurable because it's a minimum between measurable function ok, fn converts f pointwise so almost everywhere in e ok ok, if you want it you can see that ok, if x is a point such that if so you fix x because we are talking about pointwise convergence if x is such that f of x is finite ok, then you have that that exists some n that depends on x such that f of x is less than this n and so you have that for any n larger than n bar fn of x is coincide with f of x for any x or otherwise if x is a point where f is not bounded is such that f of x is equal to plus infinity ok, then you have that fn of x is equal to x is equal to n and then you have that again fn converts to f ok and then again we want to use the monotone convergence theorem so we observe that it is increasing sequence of functions ok, so moreover we have that fn is an increasing sequence functions which converts to f so by monotone convergence theorem you have that the integral of f over e is equal to the limit of the integral over e over fn ok, then just rephrase this fact by using the epsilon delta definition of limit so you have that for any epsilon positive there exist some n some index n such that for any n larger than n you have that the difference let me be minus e fn this difference is always positive because this is increasing and which is equal for instance is less than epsilon over 2 for any ok, I already wrote so, now we choose our delta in the statement of the theorem to be equal to epsilon divided by 2 times n and we consider a set a and let a, such that the measure of a is less than delta so we want to estimate the integral over a of our function f and so here we use with sum and we subtract the same we use this this trick ok here you use the fact that these fn are bounded and here you use the fact that this difference is small this for instance ok, we can precisely take this index to be this capital N or in any case any index N larger than this capital N so this is less than the integral ok, overall fn plus this is n this is bounded by n times the measure of a no and so at the end what you get is epsilon over 2 plus n for our choice of delta this is epsilon divided to n so ok of course you end up with the fact that this integral is less than epsilon ok, so now so far with all this theorem of for instance of convergence under the sign of integral I mean the strategies was we know something about the point wise some point wise information about the function and we deduce some information about we transfer some of this information to some information about the integral of this function ok, so from point wise we study the integral so now the question is can we do the reverse so if we have some information about the integral of such a function can we deduce some point wise property some good hypothesis the answer is yes and you can you can this is a first example of this correspondence between the two information so you start by f again be a non negative measurable function ok and then we have this equivalence so we have that if f is equal to zero almost everywhere in our domain e so if and only if if and only if the integral of f is zero ok ok, one side is completely trivial no ok if you want you can formally say it as you consider in any h you can take the definition with those h so you will have that the integral this h of course is zero and so you have you have that the supremum of this function h that are involved in the definition of the integral of non negative function is zero so you have that also the integral of f is zero ok, this is the trivial part and so what about the other so we start by knowing the fact that the integral we know this ok, so we define this set a n as the set where f is larger or equal than one over n and so you can somehow reformulate this we have that that f is larger or equal than one over n characteristic function of a n ok so this is zero is equal to f larger or equal to one over n this integral and then this is one over n times the measure of a n and so you have that the measure of a n is zero for n n moreover you have that you take the union you have that the set where f is strictly positive can be viewed as the union we already saw this the composition the union of a set like this f larger than one over n and ok, this is our union of a n is a countable union and so we have that the measure of this set is less or equal than the measure of less or equal than the sum of the measure of this a n which is zero so we have that indeed f is zero almost everywhere in here ok because f is larger or equal to zero ok is not negative if we prove that f the set where f is strictly positive as measure zero means that f is zero almost everywhere and now ok, now we see somehow a related result which goes under the name of Chebyshev inequality probably you already know this ok, so again f is from e from set e to r non negative basically we have the same hypothesis before function not negative ok, measurable function ok, then for any for any t, for any number t positive we have the following inequality that you can estimate from below the measure of this set is kind of not level set but this set by one over t times the integral of f over e ok, the proof is quite is quite fast so how would you prove this we need first to consider something like this so we have that f is larger or equal than t times key f larger or equal than t ok, we start by observing that this inequality all and then as before we just have to pass here we have just to pass to the limit and you have that this is so equal than t times the measure of this set ok, so we are done the measure of ok, now always in the spirit to deduce some point wise property from some fact that are stated as integral property we prove this corollary, which is a corollary of the Chebicev inequality that tells you that if you have that f ok, it's again a function ok, no negative ok, measurable and you know that such that the integral of f over e is finite so f is integrable then you know that in this point wise fact that f is finite almost everywhere in e ok, so to prove this we want to estimate the measure of the set where f is equal to plus infinity and to prove that this is equal to ok, so we consider this set ok, again we already saw that we can express this set in this way so you have more than express you can say that this set is contained in the set which is involved in the Chebicev inequality and then by monotonicity ok, we have that the measure of this set is less or equal than the measure of this set here now we use the Chebicev inequality ok, that tells you that this is less or equal than 1 over t the integral of f this is true for any t and so we know that this is finite by hypothesis of since this is true for any t if you let t goes to plus infinity you get that indeed the measure of of this set is zero ok and so we are done and now we are in position to define the general back integral we start somehow with a new argument ok, so I recall you that so we already saw that you remember what is the positive and the negative part of a function do you remember? ok ok, so we will skip this part so from this you understand that how it can define the general back integral so we can split a function f any function f into the difference between the positive and the negative part which are positive non-negative functions so we know how to compute the integral and the natural way will be to define the integral of f as the integral ok we say that a measurable function f is said to be integral of f ok, if f plus and the positive and the negative part are both integrable over e in the sense that we introduce before ok and we define the back integral of function f over e as this quantity f plus f minus e f minus ok, now we have to prove the usual properties of the integral because when you give a definition then we need to see that the linearity holds and so on, so maybe we will start today and tomorrow we have f and g to function which are integrable over e ok, then we have these 4 fact ok the function c times f is integrable and you can you can take c outside sign of integral and then you have when you consider the sum f plus g is integrable and you have that the sum the integral of the sum is equal to the sum of the integral ok, then there is the monotonicity so if you have that f is less or equal than g almost everywhere in e then the inequality is preserved by the integral and finally the last one or consider to this joint set a and b this joint set then you have that a union b of f is equal to the sum of the 2 ok, we consider the first one if c is 0 of course there is nothing to prove or start with the case when c is positive the case c negative is analogous so you have that cf is now here we use just the definition we just if this is e c f plus minus e cf ok, then we have that for this integral you can write that e this is f plus you can take c outside minus c e f minus and this is equal to you collect c so you get what you want ok, the case c negative is analogous and probably maybe you can do it by yourself you just have to use this fact this is the positive part of minus f is f minus and in analogy the negative part of minus f is f plus ok so you have to use something you have to use this fact at some point ok, to prove the point two we start first by a general fact we will prove a general fact and then we apply it to our case so if we have in general if you have f which is the f1 is given by f1 minus f2 with f1 f2 non negative and integral function then you have that f is equal to f plus minus f minus is equal to f1 minus f2 ok, so by this we have that f plus plus f2 is equal to f minus minus f plus f1 so they are both side of this equality are non negative ok, and then we use previous result about the additivity of non negative function ok, so by previous result additivity non negative function ok, what can we say? we can say that f plus plus f2 ok, is equal to f plus plus f2 which is equal plus plus f1 is equal to f minus plus f1 ok, now we observe that everything is finite so all these quantities are finite because we are talking about integral function so we can write everything also all the integrals are finite all the integrals involved in this equality so you have that f plus minus minus is equal to f1 minus f2 so finally you get that the integral of f is equal to f1 minus f2 so you can use also this decomposition so in call this star ok, this is a general fact so how we use this for our purpose ok, so we observe that we have f plus plus g plus and f minus plus g minus are integrable and positive and satisfies all the good properties and so we have that f plus g is equal to f plus plus g plus minus f minus plus g minus so this is would play the role of f1 and this would play the role of f2 ok plays the role of f1 and this plays the role of f2 ok, so basically you have that f plus g is equal by star ok, again by f plus plus g plus minus f minus plus g minus and so you can collect in the correct way so you add that this is equal to this again because everything is finite ok, just see 3 monotonicity ok, we start by observing that f minus g we start by no, actually it was the reverse g minus f is larger equal to 0 ok, and then what we have ok, this is easy, so you have that just g minus f is coincide with part plus which is so by step we can use the linearity so we have that g is larger than f ok, and the last one is ok, it's very easy 4 ok, we have this this joint set a and b, so you have f you just use the fact that f times key of a union b is equal to the sum ok, f key a plus f key b ok, so we are done ok, for today we can stop here