 Ok, zelo da je vse tačne teorema, da je zelo tukaj, da sem je početil, lemma, ok, da se me nekaj nekaj lemma vse. Zelo da smo priželi, da, kako je zbondil and measurable on AB, and if you have, if you define capital F of X as the integral, the lebege integral between A and X of F of T in DT, then we saw that the derivative of capital F is equal to F and small f, almost everywhere in AB. Now, we want somehow to generalize this result, so here we had bound and measurable function, which in particular are integrable functions. In fact, we want to prove this result for starting by a function small f, which is integrable. This will be a theorem. So indeed, we start F from AB to R, the integrable function. And again, we assume and let capital F of X equal to A of X F of T in DT. Okay, then we get basically the same conclusion, then F prime of X is equal to F of X almost everywhere in AB. Okay, now we prove. Okay, so somehow to simplify the proof, we assume that F is non-negative. Otherwise, if it is not true, as usually you consider the positive part and the negative part separately, then you put everything together. In your opinion, how could you... Of course, the idea is that we want to use this, because we already know something. How can you somehow use this? The idea is to construct some, starting from this F, some bounded function which approximate F in some way. So the idea is the idea that we already use. We define some function, a sequence of function. Fn has the minimum between what? Fn, okay? So this satisfies these requirements. So you have that fn is less than n, fn is less than f. And we define the corresponding integral. Okay, so let me... Okay, so we consider Gn, capital Gn of X. Actually, this is the difference between the two. Anyway, this is F of t minus F of n t in dt. Okay, we have that view of this. We have that Gn is increasing. Gn is a function, okay, is increasing. Because this is positive, okay? Not as a sequence. And hence for the Lebesgue theorem, we know that it is differentiable. Almost everywhere. Theorem is differentiable almost everywhere. And moreover, we have also that Gn prime is larger than zero. This is always a consequence of the Lebesgue theorem. Okay, and then by the previous theorem, previous lemma by... Okay, we have that. This derivative of this t is equal to fn of X. Almost everywhere in... Moreover, you have also that f of X. So we use the definition and then we split this f in two parts. We subtract the same quantity, f of n, Ax. So fn, okay, first, for instance, f minus fn. So this is Gn plus... Okay, so I continue to write here, okay? No, Ax, sorry, Ax of a sign, okay? So what we have is that f prime of X is equal to Gn prime plus... Almost everywhere in AP. Okay, now we observe that. We know that this is positive or rather non-negative. So we have that f prime of X is larger or equal than fn of X. Almost everywhere. And then since we know that if we pass to the limit with respect to n, we know that the sequence fn converge pointwise to f. So passing to the limit, we have that f prime of X is larger or equal than f of X. Almost everywhere. So we have that minus f of a. And, okay, we have this. And moreover, we have that since f is an increasing function larger or equal than zero, we have by the back theorem. Okay, so we are almost done. So we have that. And then we have that AB. And, okay, since we know that the integrator has a sign, then we have that this difference must be zero almost everywhere. And so we are done. Okay, so now let me just remind you a bit what we did before a little repetition somehow. You have f, when we introduce the bb function in R. For instance, you have X and Y in AB. And for instance, consider X, the case when X is less than Y. And you have a subdivision A equal to X naught, less than X1, less, less, less than Xk, less than X. And then you have X naught, X1, Xk, Xk plus 1. Or rather put it X, and this is equal to Y, for instance. Okay, we define the positive variation and the negative variation. And with this we define the total variation. So just let me, so you have that pAX is equal to the sum of the positive part i to k of X i minus f, X i minus 1. Okay, this is, and the negative part pA, Y, you have plus 1 of the same. In the sense that you start from A. Okay, you can also get rid of this. Plus fY minus fX positive part. So we already saw that this is increasing. And okay, the same, you can repeat the same for the negative variation. And so you have that, if you remember, we saw that we can split f into increasing function. So you had g of X that was equal to pAX f plus fY and h of X, which was just the negative variation. Okay, now we prove this theorem. So it's a function. We assume that f is in bV, so it admits the composition as the difference between two increasing functions. And then what we want to prove is that if you take the integral over aB of the absolute values of the derivative, which, of course, we know that exist almost everywhere, this integral can be bound, can be controlled by the total variation of the function f over aB. So can you see an example for which it holds the strict sign? There is always the same example that... Okay, for example, if you take a jump, yeah, so proof. And okay, we start by considering this fact that you have f prime of X is equal to, as we said, so we have that f is... f, I remind you, is g minus h, okay? We know that it is in bV, so there exists this g and h, this one, which are increasing, so we have that also, the derivative, you have g prime of X minus h prime of X. Okay, I recall you that they are increasing, so the g and h are increasing, so the derivative has a sign, so this is, let's see, g prime of X plus h prime of X, okay? So the idea here is to use, to use the Lebesgue theorem, okay? Okay, so we integrate this inequality, we want to use the Lebesgue theorem for those two, okay? This is less or equal than aB g prime of X in dx plus aB h prime of X. Okay, now we use this point, okay, the Lebesgue theorem will less or equal, this is by, okay, theorem, so we know that this is less or equal than gB g evaluated in b minus g evaluated in a plus hb minus hha, okay? Okay, now we have to observe some fact, okay, which are easy, but just, okay, so gB, what is gB? How can you express g of B? Is pAB of F, I mean, this is actually by definition, okay, because we define them in this way, so g of A is just f of A, okay? F of A, while h of B was the negative variation of F, and h of A was zero, no, because it's the negative, you replace B with A, so it's zero. Okay, so you substitute this equality here, so what you get at the end, you have, so pAB, okay, F plus F of A, minus F of A, so those two cancel out, and plus nA of B, because this is zero, so indeed what you find at the end is the total variation, okay, because the total variation was sum of the positive one and the negative one, okay? TAB. So to have the equality here, we will have to introduce more regular functions, somehow, okay? And, okay, so before introducing the notion of absolute continuity, maybe we can see a kind of exercise about BV function. Okay, so you have two functions, we start by a function F in AB with values in R, and you have G, which is a function in AB, which is integrable, sorry, integrable. Okay, and what we know is that for any couple X and Y in AB, the following equality holds. So you have that F of Y minus F of X is less or equal to X, Y, G of T in the T. So what we want to prove is that under this hypothesis we have that, somehow this is enough to prove that F is a bounded variation, is in AB. Okay, what we know is, for instance, the indefinite integral of this function, G in G integrable, is in BV, we prove this, okay? So we know, first thing that we can observe, that, for instance, you should define with big G, G of X, the integral between A of X of G, T in the T, this is in BV, okay? This belongs to BV because G is integral. Okay. Okay, this is a first remark. And then we try to find some symmetry here, okay? So you have that F of Y minus, we know this. Okay, but maybe we can express this integral in terms of big G. Okay, so this is equal to AX, no, actually AY in GT in the T minus AX G of T in the T. So somehow we have here a function of Y, here is also a function of Y, here you have a function of F, here you have a function of F. So I will put this here, because G is integral, so what I found is that F of Y minus AY GT is less or equal than F of X plus AX GT in the T. Okay, so which property has this function? If you define this as H of Y, this is H of X. Okay, you know. So which property has this function? H, I mean. I mean, for instance, X is less than Y, and you see that H of Y is less than H of X. So it's monoptone, it's monoptone. Okay, so you have H, H is monotone. So in particular monotone function is in BV. So H is in BV. Okay, so basically we are done, do you see why? So because we, how can you express F? H plus the integral, so we know that this one, we already see that it is in BV. Now we read. So in this way with this argument we can express F as the sum of two BV function, okay, because you have that in general, you have that F of X is equal to HX plus GX. So this is in BV because it's a monotone function. It's there. And this is in BV because we observe right from the beginning that this is in BV because G is integrable, okay. So F is the sum of two function of bounded variation. So it is also bounded variation. Sure, sure. Thank you. So just to be H of X is F of X minus AX GT in DT. Thank you. Okay. Now we introduce a new concept which is the absolute continuity. You already know this. Okay. Okay, the definition is the following. So you have F from defined in AB with values in R is said to be absolute. Solutely continuous on AB, of course. If for any epsilon positive there exist a delta positive such that you have the following. You have that the sum of I1 and F of X prime minus FXI you consider the sum of this increment is less than epsilon for any finite collection here. Okay, for any finite collection XI XI prime you have F evaluated in the point XI prime it's not a derivative, it's a prime and minus F evaluated in because you have you are considering this of non overlapping intervals, okay? These intervals must satisfy the fact that the sum I N XI prime minus XI should be less than delta. Okay, just I recall you just to give you somehow the link with maybe an ocean which is more familiar to you I recall you that what it is a uniform continuous function so you have that if F may be the values in R is such that for any epsilon there exists one delta positive which in principle this delta depends only on epsilon and on the point okay, such that such that F of X minus F of Y is less than epsilon for any for X and Y such that X minus Y is less than delta, okay? So in somehow okay, this is a uniform continuous function okay, absolute continuity we will go with AC so you have that if F belongs is absolute continuous then F is uniform continuous you just take the definition with for just one one interval, okay? Uniform male continuous sorry, continuous okay Okay, so beside this we will see now other example of function in AC okay, for instance Lipschitz function is in AC okay, so do you do you know what it is a Lipschitz function? Okay, just recall this example okay, so you have that if F is in AB with values in R is Lipschitz okay, then F is in AC okay, the proof so you know that there exist that for any by definition of Lipschitz for any X and Y in the interval you have that there exist okay, let's do it more properly, you know that there exist an L positive such that for any X and Y in AB you have that F of X minus F of Y is less than LX minus Y okay and then when you fix if you fix epsilon and you now we want to want to prove that F is in AC so let epsilon positive and fix delta to be equal to epsilon divided by L okay, now we study the increment so and assume that we have that okay, assume that we have a collection of interval such that minus EI is less than delta okay then we are done because you have that if you compute the increment along these points B I minus I I from 1 to N so then by definition of by the Lipschitz property this is less than L EI this is equal to L less or equal then delta okay and then if you want take L equal to by our choice you get actually that is less than epsilon okay and then other example okay so so we come back if if you have G our integrable function so G integral okay, then we consider this function f of x has then definitely integral between I of x U of T in the T so we saw that it is in BV but it is more than in BV it is in AC it belongs to AC so you consider a partition family of this joint in for instance call them x I y I I from 1 to N be the family of this joint interval that appears in the definition of AC function be a family of this joint intervals such that we have that the sum of their length is less than is less than some delta okay, now we want to to study the increment of f along this this family of intervals okay, so you have f I from 1 to to N so f y okay, now you use the definition and this is the sum x I T and then you can say that it is less sorry, G less or equal than sorry to y I okay okay, x I y I g T and then okay, and then these are this joint intervals so you can say that it is equal to the union of x I, y I g T in d T okay, this is the union over I which goes from 1 to N because they are this joint, okay so now which theorem would you use here to say that somehow, if the measure of the set where you are doing the integral is small, also the integral is small provided that g is integral of course maybe the name of a c function should help you so we are trying to prove that this function is absolute continuous so blue tree continuous but at some point when we were doing measure theory we proved that what we call the absolutely continuous continuity of the integrals, okay, you remember okay, so by by the absolutely continuity of the integral so, what does it tell you this property, it tells you that if you have that for any epsilon positive there exist a delta positive such that you have a set a which has small measures less than this delta then the integral over a of g g must be integrable of course, no in dt is less than epsilon okay, this was the theorem and so basically you apply that this theorem, so a becomes this union and so you fix the delta you fix the epsilon at the beginning and your delta will be the delta that will be the delta given by this theorem okay, so a is the union of a i b i okay, and then just a brief remark um um so, somehow the the absolutely continuity is a is a weaker concept with respect the leap sheets one, okay so a function which is leap sheets is in ac, but the converse is not true and how would you can you construct an example of a function which is in ac I mean in view of of course of what we already done so take this into account how would you construct an example of a function which is in ac but is not leap sheets so I recall you that a leap sheets function is differentiable almost everywhere and the derivative is bounded but I mean just observe that a function to be integrable g need not to be bounded so basically if you define a function f and you take some g which is integrable but not bounded there exist functions which are integrable but not bounded, okay then you are sure that your f is in is absolutely continuous in view of this result but this function cannot be leap sheets because the derivative would be g and if you take g not bounded cannot be leap sheets, okay this is just a remark okay okay then prove the following okay so you have you start by a function which is in ac and we want to deduce that ac is stronger than being in bv so we want to prove that if ac is in ac then f is of bounded variation okay so we start by okay we consider for instance just to fix the idea we consider so we fix for instance epsilon equal to 1 and but then of course you can do it for any epsilon and delta the delta corresponded to this epsilon and these are the epsilon and delta which appear in the definition of ac function okay would appear of ac okay and then okay we want to prove that f is in bv so how we proceed we take a subdivision of ab and we want to see that it is bounded so let a equal to x not minus x1 minus okay xn equal to b be a partition of ab okay so here the idea is that we want somehow to make thicker the subdivision adding some extra point which are choosing in a suitable way okay so we add to this partition point point of the type of the type a plus j delta prime okay take delta prime less than the delta there and with j which goes from for instance 1 to n okay where n is the greatest integer okay it's the greatest integer such that a plus n delta prime is less than b okay a plus n delta prime is less than b okay I call this new partition obtained by adding this point with the letter y with x is the original one with y I call the partition obtained by adding these extra points so let just to fix the idea so let a equal to y0 y not y equal to be the new subdivision okay so we compute the total variation but I mean referring to this third partition so let t equal to sum from i which goes from 1 to n of f of x i minus x i minus 1 okay then we use the triangle inequality and this new partition is less than equal to k so we can divide this sum into two parts somehow so we can divide we can divide this sum into n capital n parts which are the points y i which lies in for instance in a in the interval of this type so then you have a plus delta prime and a plus 2 delta prime and so on until a plus n minus 1 delta prime and a plus n delta prime so we consider now the sum over this point okay so consider the sum over the point y minus 1 y i which are contained in a a plus delta prime so now I focus on the first interval okay y i minus 1 y i which are those point of the type which are contained in this interval okay and so you have that f y y minus f minus 1 plus then 1 because because we use the absolute continuity because by the the absolute continuity we have that the measure of this is less than delta prime in particular less than delta so if we repeat this is been done for the first interval of course you can repeat this for any of these intervals so repeating the argument for any j probably from 1 when and summing up we have that this is less so let me write here then the sum y i minus 1 contained in a plus j delta prime plus j prime and let j goes from 1 to n now actually j goes from 0 to n because you have to include also a ok ok we prove I mean the sum this is less than 1 ok then 1 times n plus 1 so I mean it's not correct to write it's not really correct I mean what I want to say is that if you fix j if you take this sum for j fixed we prove it is less than 1 we prove before ok and then when you let j vary from 0 to n you get all the sum is less than in any case this is finite and n this big n doesn't depends on the partition that we the one we dex doesn't depend of the original partition that we fix ok it was as more to do with the definition of absolute continuity so when you pass to the supremum if we take take the soup we have the total variation of a b of f is finite ok and this concludes the proof so as a corollary so you have that if f is in bb ok then f has a derivative almost everywhere no sorry f if f is in bb has a derivative ok ok so if you have function which is in is absolutely continuous and if in addition you know and if such that we know that it is derivative so you can we know that it is 0 almost everywhere in a b what we can deduce we know that f is constant ok so if f is in c1 is trivial but now we have to extend ok this is equivalent to show that ok so this means reasons to show that for any c in a b we have that f of c is equal to f of a ok so consider the set e so it is a set of a c such that f prime exist and it is equal and it is equal to 0 so it is through almost everywhere we consider this set here then we fix a point x in this set e so let x a point in e and then definition of derivatives so what we have that for any eta positive arbitrary small we have that that exist infinity many h positive such that you have ok arbitrary small ok such that you have that x x plus h is in a c and you have that fx plus h minus f of x is less than eta times h ok so the reason why is that if you have this if you divide this by h you have that incremental quotient is less than eta and this incremental quotient can be arbitrary small because f prime exist and it is equal to 0 ok and then here maybe this recall you something the vitally lemma so we have that the family the family x x plus h covers e, the set e in the vitally sense so what we can infer so by the vitally lemma there exist some for instance call it that we have that for any delta positive there exist a finite collection in intervals collection we call it so of the type x i x i plus h i for instance with index i which goes from 1 to n ok, such that they almost cover e ok, such that you have that the measure of e if you remove from e finite union the measure of this set is still small, is less than delta and moreover we also have that for this point you have that f x plus h i minus f x i is less than eta times h i which goes from 1 to n ok, assume that just to fix the idea that we can order them they are disjoint so you can assume that they are ordered in this way this point collection you have for instance x i less than x i plus h i plus i plus ok, then you take this partition so x2 you have x2 then you have xn xn plus hn and b and c, sorry this is a partition of ok, now we use the definition of ac ok, so fix some epsilon positive and let delta the corresponding and let delta the corresponding parameters in the definition of ac function ok, of the definition ok, then we have ok, consider this sum x1 minus a plus h i so somehow here we have that of small measure so we are in the complement of the finite covering so this is less or rather this is ok, less than the measure of ac less than this finite union which is less than delta ok, then we have that since f is in ac you have that the corresponding increment fx1 minus fy plus the sum plus 1 minus fx is less than epsilon ok, this comes from the fact that it is in ac and call it star, so here we use the fact that it is in ac for the other point of the partition we use the fact that f is 0 and you have that for the other point the one of the Vitalilema ok, this is less than eta times the sum of h i and ok, they are disjoined this is less or equal than eta times c minus a and call this 2 dot then you sum up the 2 sums so this one and this one no, sorry, this one and this one so sum up ok, we have that ok, plus plus you have fxn plus hn minus f sin plus fc minus fx xn plus hn ok, and this is less than here here was from 1 to 1, no, it's the same probably, ah ok, it's n minus 1 ok, if you sum up then what you obtain is less than epsilon plus eta times c minus a but this can be chosen arbitrarily small so at the end what you have that indeed f of a is equal to f of c and so f is constant ok ok, think that for today we can we can stop here