 OK, hej. So, I start by recalling you the last lemma of Thursday. So, we have that, we prove that if we have a measurable integral function and such that we have the integral of f over a, where a is any measurable set contained in e is non-negative, then we show that we prove that f, we can infer something about the sign of f. So, we have that f is larger than equal to zero almost everywhere in e. OK, now, so we want somehow to relax this requirement. So, here we are asking that this, we have to be sure that this holds for any measurable set. So, we take as test set any measurable set. OK, now we want to relax this a bit. And we consider, as lemma 2, f, an integral function, this time we define it on an interval i, interval, and such that the same requirement on the sign on the integral is verified, so f is not equal to zero, but this time u is enough that we have to check for any u open, OK, for any u with u open, and then, OK, then the conclusion is the same, then we have that f is larger than equal to zero almost everywhere in a, in i. OK, so somehow here we relax the requirement, because it's enough to check only for u open, for u open. OK, so, so somehow this is also a consequence of the, an application of the Lebesgue Convergence Theorem. OK, of course we want to use this result, OK. So, we recall that if we have a measurable set, then we can approximate it in a quite nice way, OK. So, we know that then the resist set g, we always call it g, belonging to the class g delta, which are countable intersection of open set, OK. Such that we know that a is contained in g, and the measure of g minus a is zero. OK, so let's, we have that a is equal to this minus g minus a of f, this is zero, OK, because the measure is zero, so we have that this two is i. OK, so by definition we have that of g, we have that we can represent g as the countable intersection of open set, so u and open, and here with no loss of generality we can think that this has a decreasing sequence of, of set, so with loss of generality we can assume that this is a decreasing sequence of open set, so u1 contains u2 and so on, because here, no sorry, u3, un, because here the idea is that for instance if it is not, if it is not fulfilled, then you replace u2 with u1 intersection u2, and so on u3 he would replace u3 with intersection of u1, u2, sorry, u3. OK, now we consider the characteristic function of each open set, un, so we have, we have that characteristic function of un that converts to the characteristic function of g, so in fact we can distinguish two cases if, so if x is in g, then it is in, we have that cg of x is equal to 1, and we have that c un of x is equal to 1, because g is contained in u1. OK, so if x, point wise, in fact, no, no, here is written in fact, here I mean almost everywhere, no, the point wise convergence of n, this is an explanation of why this is true. OK, so if x does not belong to g, then this of course is equal to zero, and if it is not to g, it means that there exists an index, and there exists an index x not, such that for any n larger than n not, we have that x does not belong to un, OK. So basically you have that from key of un of x is equal to zero for any n, OK, larger equal than n not. OK, OK, now, and we have that this sequence is a decreasing sequence, OK, because the set are decreasing. So now we want to multiply for f, this, and we get that this converges, of course, to g, OK, almost everywhere, OK. OK, of course we want to use the back convergence theorem, so we also noticed that, we started by the hypothesis that f is an integrable function, un, so this is bounded by f, and so by the denominator convergence theorem, we have that we can pass to the limit under the sign of integral, so we have that f of key of un converges to, OK, this is, of course, f of un, and, OK, so we have that this is, OK, these are open, OK, so by hypothesis we know that this is non-negative, and so we have that basically, OK, this is, we know that g of f is larger or equal to zero, but the beginning we proved that this was equal to the integral over the test, the measurable set a, to introduce, OK, so from the lemma before, from lemma one, we can deduce that f is larger or equal to zero in i, because, of course, a is an arbitrary measurable set, OK, now we want to relax more this hypothesis, so maybe we can test the positivity of the integral only on a special open set, which is namely on, on open intervals, OK, so now comes the lemma, so again we have f integrable, and we have, and we assume the sign of f, the sign of the integral of f over j is larger or equal to zero for any interval, for any j containing the i, where j is an open interval, OK, and then, OK, we have the same conclusion, we have that f is larger or equal to zero, almost everywhere in i, OK, this time to prove lemma three we want to use lemma two, and again here the strategy is analogous of the one that we use here in the sense that we want to approximate an open set to use lemma two by an open interval, OK, so we saw, we know that, we know that if you start by, let's call if u is open, then u can be represent, as a countable union of this joint open set, so i and one to infinity of, called i n, i n, and this joint open interval, this is, we prove this one of the first lectures, OK, so now we want to say something on the sign of this integral over u, OK, now we use of course the countable additivity property, so we have this is equal to i n of, yes of f, and this we, OK, we know that this is positive, each of them is positive, and so by lemma two we can say that this is true for any arbitrary open set, so we can, we have that since u is any arbitrary open set, then we can infer that f is indeed larger or equal to zero, OK, I must everywhere in i, OK, OK. And now instead this theorem, you consider again a function, an integrable function over an interval, define an interval with values in a, it should be, must be integrable, assume that i is an open interval, and assume that the integral of this product, now I told you it is phi, phi is always zero, for any, this is true, for any phi, which is c infinity with compact support in i, OK, phi is c infinity with, OK, this must be true for any phi, OK, and then what can we say about phi, f in compact support, compact support, you know what it is? Ah, OK, OK, phi, compact support means that the set where f of x is different from zero, the closure of this set is strictly contained in i, the, the, no, no, no, you just, you consider the set where phi of x is different from zero, take the closure, so it's, and you have that, this must be contained in, strictly contained in i, OK, this means compact support, OK, so what we want to infer is that a point-wise property again, that f is equal to zero almost everywhere in i, OK, so this is a very fundamental theorem, especially it arrives in distribution theory, probably we will see later. OK, so let's prove it. OK, so we consider an opening interval, j, an opening interval, such that j is contained in i, OK, just, for instance, j is equal to a, b, just to fix the idea, and then, no, just believe me, we can construct, we can construct a sequence phi n, in, so, we satisfy this property, such that we have that, yeah, non-negative, and they converge increasingly to key a, b, j is integral from zero to b, from a to b. J, j is n, b, j is integral from a to b? Yeah, it's an interval, it's an opening interval. I don't know, it's called a to b, not from zero. Ah, no, but a to b is, and this is the same, OK, just to fix the idea, OK. OK, so by hypothesis, we have that, we have that, the integral over i of f times phi n is equal to zero. OK, and moreover, again, we want to apply the dominated convergence theorem. OK, so we have that, this converts pointwise, b, n. Yeah, which is increasing. Yes, pointwise, I mean, maybe it is not necessary, just consider this, this is enough to what we have to do, but in addition, it's also good. OK, so f times phi n is less or equal than f, just to be, and she's integral, and then, I give the next convergence theorem. OK, we have that, of course, that n converts to f, this is equal to zero. Ah, OK, let's, so this converts to f, to the integral of f over i. OK, now you can think, so basically you have that, so you have that, this is zero, OK? And then, think of this as the fact that this means two things, but you can see this as these two facts satisfies it tangentially. This means that f is, must be larger or equal to zero, almost everywhere in i. This is j, OK? f is less or equal to zero, almost everywhere in i, so basically what you get is that, indeed, f is equal to zero, almost everywhere in i. OK, now we'll try to, somehow to outline how we can construct this such a function phi n. Is it clear or not? I mean, here is that, I mean, you are here, this is zero, no? And you can think of it, as these two inequalities satisfy, OK? This is the equivalent to say that the integral of f over i is equal to zero is equivalent to say that these two are satisfied, OK? Yeah, yeah, yeah, you played the last lemma, no? This is, no, we said it with the larger or equal, the last year. Yeah, we said it like this, like this part, the integral over j of f is larger equal to zero, then f is larger equal to zero. And then, of course, you can prove that it's the same for the less or equal. OK, so how we can construct such a function phi n. So consider just to fix the idea i to be an interval p of n points of a, b. And then we have that there exists a sequence phi n of function in c infinity, c has infinity with complex support in phi n on i, in sej that they are in between zero and one, OK? And they converge. We have that, what we used before, they converge increasingly to the characteristic function of a, b most revenue. OK, so the proof is a constructed one, so we have to proceed by step. So, first of all, let me introduce the following function, f, defined in a piecewise way. So f is defined from r to zero, one as follows. OK, it is zero for x, if x is less or equal to zero. Otherwise, is equal to e to the minus one over x, if x is positive, very strictly positive. OK, I mean, with elementary argument, you know that f is continuous, take the limit, OK? It's continuous on r, because if you let x tend to zero from the right at the limit is zero. OK, so what about the derivative of this function? So what about the derivative? So we have that f prime of x is equal to zero, of course, if x is less or equal to zero, otherwise, is equal to one over x square, so do the computation times one over, OK? Again, you can prove that f prime is also continuous. Square, yes, x square, x square, f prime is continuous, also in zero. OK, you can prove this either by considering the incremental quotient, or just by observing that the two derivative exist in the two pieces, and they coincide in zero. OK, now, if we go on, so the second derivative and so on, what do we obtain? So you would have that second of x is equal to zero, x is, and you get, or here you get minus, no, plus x to the power minus one over x, x positive. And basically here, if you look, the idea is that if you go on with the derivative, here you obtain a polynomial of one over x, OK? But of course, we have to prove that this is true, and we will do this by induction. So again, here, second is continuous, and so on. So we claim that the k derivative k of x is equal to zero, if x is less or equal to zero, otherwise, it is a polynomial pk of one over x times e minus one, one over x, so x is positive, pk is in the variable one over x, OK? The step k equal to zero is true, for the step k equal to zero and the claim is true. We have that, so we prove, we want induction, OK? So we have that if is equal to zero, then OK. And so we assume that the claim is true for k, it's true for the index k, OK? So we have that if you do the computation, what you obtain? It's zero, it's x is less or equal to zero, otherwise, you get something of the type minus one over x square, pk prime one over x, e minus one over x plus pk one over x, x positive. So basically, if you collect e to the minus one over x, you get the t, so you have that pk plus one would be minus x square pk prime one over x plus pk one over x, OK? By analogous consideration, you can prove that this is continuous in zero. So indeed, we finally prove that we have that such an f is c infinity r, OK? And another mean related property is that such a function, the function f, provide an example of a function which is c infinity, but not analytic, OK? Because you have that the derivative of any order are zero, so the series converts to zero, near zero, but the function is not zero, OK? OK. And now we want to construct using this f, this function phi n, which converts point wise to the characteristic function of an interval, OK? OK, just to visualize the things, this function is zero here, and here it's very flat, because the derivatives of any order are zero, so it's something like this, as as an asymptote in correspondence of one, OK? So at level one, you have the dot, OK? So how we can modify, we can handle this function f. We define a function g to r in such a way. G is defined by means of f, like this, f of x times f of 1 minus x. OK, just to observe that if x is non-positive, you have that f of x is zero, and then g of x is zero, and if x is larger than 1, f of 1 minus x is also zero, then, again, g of x is zero, OK? So just to, so the graph of g looks like something like, something like this, OK? OK, now we want to use g to proceed, and we want to construct this function, so we want to construct function g, capital G, which is c infinity r, and which satisfies the following properties, that g of x is equal to zero for any x less or equal to zero, and it must be in between zero and one, if x is in between zero and one, and must be equal to one for x larger than one. So a way to, so to provide such a function capital G by means of this small g is the following, and we define g of x as the quotient between the integral, between zero and x of g of t, dv t divided by the integral between zero and one of g t. Because you have that, if x is less or equal to zero, then g is zero, OK, and also g is zero. If you have that x is larger than one, then you can look at this, the numerator has zero, plus one x, and this is zero, and so g is one. In this case, you have that also g is in between zero and one, so, OK, if you have x between zero and one, then g is positive, and so this quotient is between zero and one. OK, now we define a contraction of this g, so how it looks like, this g looks like this, you have, OK, this is zero, so here is zero, and then you have a transition from zero to one in this interval zero, one, OK, this is the level one. So now we make a contraction of this bg in order to make this transition from zero to one somehow faster, OK, so we define g epsilon of x, g epsilon of x, has g of x divided by epsilon, so it's a contraction, and so how it looks like, the graph, you have zero, this times transition happens between zero and epsilon, so you have something like this, and of course g epsilon is still c infinity, OK. The first graph is for capital G. Sorry? The first graph. Yeah, the first graph is for capital G, and this is for g epsilon, OK. See, the only thing that changes this part, this transition. OK, and then basically what we want, so we have to define this function phi n, which approximate the characteristic function of a, b in a c, of course they have to be c infinity. OK, so we are interested on somehow on this transition, on this part of the graph, OK. So we will use this g epsilon, how it is defined in this part from zero and epsilon to approximate the characteristic function, so OK, you will have basically the level one, and here I can construct this function phi n, I just show you, with drawing, in by using this transition between a plus epsilon, for instance, and a minus b, b minus epsilon, OK. And I define this function phi n in between a and a plus epsilon has g epsilon is defined here, OK. So it would be like this, and here it's like, you can argue, you can reflect the graph, and this, the dashed line represent the graph of the phi n, OK. So basically what I found is that the phi n are c infinity function, they have compact support. OK, actually you can, no, to have compact support, but actually I had to, so let me consider, OK, let me move a here, and here is a plus epsilon over 2, OK, so a is here, and here is b minus epsilon over 2, and b is here, OK. So there must be some room to put then phi n equal to 0, OK. So they have compact support. From here, epsilon is 1 over 2, I took a plus epsilon over 2, I mean, OK, a plus epsilon over 2 plays the role of 0 here, OK, while a plus epsilon plays the role of epsilon, OK. I define it by the drawing, I mean, just you understand that, I mean, the delicate point is how to define a transition from 0 to 1 in a c infinity way, OK, this was somehow the difficult part. So then you can, OK, you can do the details, but the meaning was this, OK, OK. Yeah, and also let me observe that the construction of such phi n, I mean, also, I mean, with the construction of this phi n, we can also improve a former result that we prove, just when we introduce the first result in major theory. And so more precisely, we can mistake this theorem. So basically, we prove at some point that we can approximate a measurable function by means of a continuous function, if you remember, in some, OK, in some proper way. And then we can say more, we can say that we can approximate a measurable function by means of c infinity function with compact support, OK, with the, if you do the, sorry, so if you have let f be a measurable function function, so you must have something, you have to require that it must be, it must take simple values, but we have to assume that the measure of the set where f is plus minus infinity is 0. OK, then we can approximate such a measurable function in this improved way. So then, given any epsilon positive, we have this time that there exists a sequence, OK, there is some h, h, which is c infinity with compact support, OK, in a b, such that the measure of the set of the, OK, the measure of the set of the x in a b, where we have that f, OK, f of x minus this h of x is larger, larger than epsilon, is less than epsilon. OK, if you remember, we proved this theorem and we proved that there exists an h, which is only continuous, do you remember? And at some point, if you remember, we had this problem arise that we had to approximate with a continuous function, the characteristic function of an interval, OK? And we do this, so we did it in a quite, if you want, rough way. We just took the function g like this, we call it g, I think. Now we can improve, because now if you repeat the same argument and you replace the function g, the continuous function g, with this function phi n, OK, so the graph will be slightly smoother, OK? Then you can prove this theorem, OK? OK, now I will introduce the notion, just to conclude the part concerning measure theory, the notion of convergence in measure. So we give the following definition. So we have fn and f, where fn is now measurable. OK, then we say that fn converts to f in measure. The sequence fn converts to f in measure. If you have that for any epsilon positive, the limit as n tends to plus infinity of, sorry, n, OK. The limit of the measure of this set, so we have that where fn of x minus f of x is larger than epsilon, OK, this limit must go to 0. And so in analogy, you have the definition of a Cauchy sequence in measure. Give another definition. The sequence measurable function is a Cauchy sequence in measure. OK, if you have that for any epsilon positive, there exists an index and bar, such that for any n, then larger than this n bar, the measure of the set of the x in e where fn of x minus fn of x is larger than epsilon, is less than epsilon. OK, then we prove a theorem. We state a consequence of the convergence in measure, OK. OK, so be a sequence of measurable real valued function, which is a Cauchy sequence in measure, such that, OK. Then what we can say about the point was convergence of this set fn, we can infer that then there exists a subsequence of fn, there is such a function, a measurable function, f subsequence, call it fnk of the original sequence fn, such that we have that for any epsilon positive, there exists a measurable set a with small measure, less than epsilon, such that we have that outside this set, this set of small measure, the convergence is uniform. OK, on e minus a, OK. Of course, e is the domain of the fn and of f. By hypothesis, we know that fn is a Cauchy sequence in measure, so hypothesis, we have that, if we fix, we have that, OK. For any, instead of taking an epsilon, we take a k, positive integer, and we have that, there exists a corresponding nk, such that for any n, for any m and n, larger than n is mk, we can have that measure of the set of the points in e, where fm of x minus fn of x is less, make a particular choice of this epsilon k is less than 2 minus k. OK, again, now we do an extra hypothesis, but this with no loss of generality. We can assume that the dependence of nk is increasing with respect to k, because otherwise you can replace, for instance, nk plus 1 by the maximum of n1 and the form of 1, OK. OK, now we define the set ek as the set of the point of x in e, such that we have that fn of kx minus fn of k plus 1x is larger or equal than 2 minus k, OK. OK, and then we consider the union of this set, starting from an index m, for instance, so you take fm to be the union for k, which goes from m to infinity of this ek. OK, now we consider a point, which is outside fm x, which is in e minus fn. OK, how it behaves, so, and we have that and let 2 index jk, such that, for instance, they are larger than n. OK, then by the triangle inequality, we have that fn k of x minus fn j. Let x, let j, there are 2 index, no, j and k, which belongs to n, and they are larger than our index n, OK. So they are there here. OK, now we use the triangle inequality and we have, we can bound this as taken index l from k to j, for j minus 1, fn l of x minus fn l plus 1 of x. And since we are outside fm, we have that this is less or equal in the sum of 2 minus l, which is, OK, which is 2 minus k plus 1, 1 minus 2 k minus j minus 1. OK, from this we have that, we get that, that if you take the limit as k and j, this 2 index tends to infinity, yeah. This is the summation, I took the index l to k to j minus 1, because I supposed j is larger than k, 2 minus l, 2, yes. OK, so the supremum over x, which belongs to e minus fm of fn k of x minus fn j of x, this goes to 0, OK, we call it star. OK, so we have that r is a complete space, so fn k is a kushi sequence, so there exists a limit in r, so this f, such that we have that fn k, so we set this fx as the limit, has nk tends to plus infinity of fn of x. OK, so f is defined on each e minus fn, so f of x is defined union of v minus fn over n, OK, and this is e minus the intersection of fn. OK, so we have that, OK, by star what we, by this, we know that fn k converts to f uniformly on e minus fm, and this for any m, OK. And then of course we have, we will have to say something about the measure of this fm, if it is small or not. And of course we have also that if we took f, has taken the intersection of this fm, this is just by definition, this is the intersection of a union k, which goes from infinity of this e k. OK, so just let's estimate the measure of fn, so we want to infer that this is small in some sense, OK. This is less or equal, OK, it's equal to the measure of e k, k, which goes from m to infinity, OK, then of course this is less or equal to the sum of the measure of e k. OK, but by hypothesis the measure of this e k is less or equal than 2 minus k. So equal 2 minus k, OK, this can be that. So for any m, so basically this of course tends to 0, as m tends to 0. And so what we prove is that the measure of f is equal 0. So actually we prove more, so we prove that. So we add that fn, OK, converts to f almost everywhere e in e, because it converts pointwise in e minus f, OK. It converts pointwise and f as measure 0. So we prove that if we have a Cauchy sequence in measure, then it converts uniformly outside set of small measure, and it converts almost everywhere in e, OK. So we prove two facts, actually. OK, and now in this last 15 minutes maybe we can do some left exercise. So these are exercise on the monofon convergence theorem. OK, so just start from a general fact. So you have, for instance, a function f from 0 to 1, take 1, include it in 0 to plus infinity, which is continuous, for instance. So basically by the monofon convergence theorem we can compute. This integral has the limit as epsilon actually over a sequence, epsilon goes to 0, of epsilon of n f of x, OK, so fine. OK, so, for instance, we can specify the choice of f. So if you have, for instance, f of x is equal to 1 over x alpha with alpha positive, OK, so what can we say? We can divide three cases. So we have the cases when, for instance, epsilon over 1 over x alpha in the x, you have this is, the primitives is x minus alpha minus 1 minus alpha plus, OK, plus 1, so this is over x epsilon x equal to 1, and then this is 1 over minus alpha plus 1 minus alpha plus 1, so this goes to plus infinity when minus alpha plus 1 is negative. So as we have that the integral between 0 and 1 of 1 x alpha is equal to plus infinity for alpha larger than 1. And then when alpha is 1, I mean, these are course things that you already know, but just to see them as an application of the monofon convergence theorem, this is the logarithm of x epsilon over 1, and this minus the logarithm of epsilon plus 2 plus infinity, this is for alpha 1, OK, c, this is a, this is b, OK, so a and b, the integral is equal to plus infinity if alpha is larger than 1, in case c, which is the finite one, you have that 0, 1, 1 over, for the same reason that you have this, this is equal to 1 over 1 minus 1, so it's finite. OK, then we can slightly change, for instance, the denominator alpha x, so what you have is that, for instance, if we, in the exercise, might be to consider this integral of sin x alpha in the x, OK, here, alpha is always, OK, so we have that in this interval sin x is less or equal than x, then you have that sin x to the alpha is larger or equal than 1 over x to the alpha, so we try to use this result, and so we have that the integral of 1 sin x alpha in dx is equal to plus infinity alpha larger than 1, but on the other hand we have also that, we also know this is of k, a standard fact that when x tends to 0, the sin of x divided by x is the limit of this question goes to 1, OK, so we have that, this means that there exist some delta positive as such that, for instance, 1 over 2x is less or equal than sin x, x in between 0 and delta, and then you have that the sin x alpha is less or equal than 2 alpha x to the alpha, we can split this interval 0 delta has sin x to the alpha plus delta 1 sin x to the alpha, OK, this is finite, it's finite in a problem because we are outside and call this a, we can integrate between 0 and delta of sin of x, the alpha is less or equal than the integral between 0 and delta of 1, OK, if you want 2 alpha x alpha, but we already completed this, less or equal than 2 alpha 0 delta, OK, no, OK, less or equal, OK, then this, which is finite alpha less than 1, OK, so now maybe you can do, you can study by yourself a strain, a slight variation of this, and you can use somehow the same argument, OK, what you can, you can, for instance, you can try to solve is the following, so over the interval 0, 1 over 2, you can try to solve this x alpha and log x beta in dx with alpha positive and delta also positive, so you can try to somehow to discuss the convergence of the integral with respect to these 2 exponents alpha and beta, OK, so probably when alpha is equal to 1 beta plays a role, otherwise it's more easier, OK, so for today we can stop here.