 So hi, so we have to complete the proof of the lebek theorem, won't you remember, just remind you the statement. Okay, so we have a function f defined on an interval a, b, known decreasing, and so then we want to prove that f is differentiable in a, b. Okay, so almost everywhere, of course. And the prime is measurable, and non-negative, and moreover, we have the following inequality. Okay, so we already proved that f is differentiable almost everywhere. Okay, so this is okay. Okay, so basically we have that for almost every x in a, b, we have that limit s, y, 10 to x, f of y minus f of x divided by 1 minus x, we define it as g of x. It exists. So g is defined almost everywhere, and so we have that this belongs to r, and f is differentiable when g is finite. Okay, so we know that the limit exists, so we can consider the limit along as a sequence. So we have that g of x is equal to the limit of x plus m, the limit of x plus 1 over n minus f of x divided by 1 over n. Yes, we will prove that all the other statements that we have done. So we have that we can define this incremental quotient g of n of x. Okay, so it is that we extend the function f on the right of b, of b in this way. So we have f of x would be equal to f of b, so to the values of b on the right of b to the values of f of b here. Then we have that g n of x tends to g of x almost everywhere in a, b, and we have that g is measurable. Okay, now we use again the fact that f is increasing or not decreasing, so we have that since f is not decreasing. Okay, then g n is larger or equal to 0. So by the fact that m so we have the following, we have that a, b, we have x in dx is less or equal than the limit of f of n, sorry, of g n. So now we use the definition of g n, this is the limit as n tends to plus infinity of n a, b, okay, f of x plus 1 over n minus f of x. So this is equal to the limit of n. Okay, now we have, we translate plus b over n, b plus x, okay, and since the limit minus, okay, so we consider 2 separately. Okay, so you have that n, b plus 1 over n of f is equal to n. Okay, here it is constant, okay, this is f of b and so here you have f of b times 1 over n times n. So this is equal to f of b and then you have that a, a plus 1 over n, f is larger or equal than f of a plus 1 over n, this is why f is not decreasing, okay. Decreasing times n, but this is equal to f of a. So at the end what you obtain is that all this stuff here is less or equal than what we want. So we have that, so we conclude the proof by noticing that g is integral to this finite of everywhere and so thus we have that f is differentiable almost everywhere and, okay, g is equal to f prime almost everywhere, okay. Okay, so what remains to prove just to be precise. So you can prove by yourself that a monotone function is measurable, okay. So for instance you can consider, so assume that f is increasing, you can consider, you study this set, you took the infimum of this set, no, take it, for instance it is c and basically you try to prove that it is either c plus infinity or c, like this, okay. In any case it is a measurable set. Try to do this by yourself, okay. Okay, now we introduce a new class of function which are the function of bounded variation, okay. So do you already know this function? No. Okay, so we start by a function f defined in a closed interval with values in r and then we take a partition an arbitrary partition of a, b, a subdivision, okay. So we denote it x naught coincided with a up to the index k, x k with b and we define these two numbers. So we define with b the sum of somehow the positive variation of f along this sub sequence, this partition. So it would be the sum of i, which goes from 1 to k of f x i minus f x i minus 1 and I take, so I consider the sum of this gap and then I take the positive part. Then n would be defined in analogous way. I just take the negative part instead and then t which would be the sum of n and p and this is just the sum of the absolute values of this gap. Okay. So we have that minus f of a is equal to p minus n which is equal to i k. So we are we are considering telescopic series. Okay. Okay, then we want to consider the supremum of this of this set of three numbers over all the possible subdivision of of the interval a, b. Okay. And we call it with the capital letter. So set p equal to the supremum of of small p and n would be the supremum of small n and capital T would be the supremum of small t and all these three supremum are taken take the the soup over all the possible partitions subdivision of the domain a, b. Okay. So somehow this gives you an index of the variation of f. So now we introduced the definition of function of bounded variation. So we have that if this supremum the one that takes into account both the positive and the negative parts is finite. So if this big t is so I get rid of the dependence of t upon f, but it's it's trivial. We say that f is a function of bounded variation. Is of bounded variation over the interval a, b. Okay, in for sure usually we use this this notation, b, v. Okay, f is in b, v. Okay, we start to see some some consequence of this definition. Okay, so if we have that f is in b, v of a, b then we have the following equality. So the first one is that f of b minus f of a is equal to p okay, p, a, b here I meant this quantity where here I stress the dependence over a, b minus n, a, b and second you have the t, a, b the total variation is b, a, b plus n, a, b okay, so we start by proving one that we have that for any subdivision of a, b we have the following we have that f of b minus f of a is equal to this telescopic sum n2 k2k fx i minus fx i minus 1 okay, now we have okay, you understand how the things is going on, you have just to express this in a proper way plus take the positive part minus the negative part in that way at least we prove one side of the inequality so this is equal to b minus n, so we have that this is less or equal than a, b plus f of b minus f of a take the supremum and we have that this is b, a, b this is equal to so since we are under the hypothesis we are assuming that the function is a bounded variation and then we have that also these two quantities are finite in particular this one is less than t, a, b is less than infinity and then we have that you can take this on this side and obtain this call this with these two so now obviously we have that and so you reverse plus okay, minus f of b minus f of a this is less or equal than b a, b minus b minus f of a so take the soup here for the same reason we have that p, a, b is less than t, a, b which is finite so we obtain the other equality and so okay you gather the two minus b is larger or equal than f of a and so we have done we have that 2 plus implies 1 for the third point 2 so we have that p plus n is equal to t and this is of course less or equal than the supremum, t, a, b okay so we have that t, b is larger or equal than p plus n and without this sum has okay, p plus b minus p plus n is equal to 2 p minus f of b minus f of a here we use the step 1 point 1 and here we can conclude that this is minus p, a, b plus n, a, b so we have that any partition 2 p is less or equal than t, a, b plus p, a, b minus n, a, b and then so here we can take the supremum and then we prove one side okay, so finally what you obtain p, a, b is less or equal than t, a, b we know that they are finite okay and then we know also this but we have also the following so again we take the supremum and call this to star this was star so I think that now we are done we combine the two and we are done we have the t's okay, now okay, now we will see a characterization of b, v function okay which is the following theorem it tells you that the v function is basically can be expressed by the difference between two moroton function so a function f which is in b, v of a, b in b, v if this is a characterization if and only if that exists is just is the difference of two moroton real value function so somehow the fact that we can express a function in b, v has linear combination of difference of monoton function provides us with a bridge with the Lebesgue theorem okay, so this is important okay so we prove this this side so we start by assuming that f is in b, v okay okay, this is a trivial remark probably if f is in b, v of a, b f is also in b, v of each of these subinterval a, x where x is in between sorry, couldn't be okay because I mean you will have that the total variation within this interval can be bounded by the total variation the full interval and this is finite okay, now we want to apply 0.1 of the previous lemma so by 0.1 previous lemma applied here we have that f of x now x plays the role of b minus f of a is equal to p, a, x minus p, a minus n, a, x okay and so now we are done, okay let's create a y because basically for instance you can consider so express f of x as f of a plus p, a, x minus n, a, x these two are increasing function with respect to x that if we decide so you have x because they can only increase because you add no, you add other gap so increasing function okay, we are done right now we want to prove the other the other implication so we start we have that if f is such that f of x okay, actually here you should put non decreasing okay so now we assume that it is the difference between for instance take g of x minus h of x okay, with g and h non decreasing function okay, now we want to see what is the total variation so we fix we imagine to fix a partition so we want to establish what is t so okay, you have xx minus fx i minus 1 this is equal I use this this property so I have g, this is gx i minus gx i minus 1 plus h minus i plus h minus 1 okay, this is less or equal then okay so this is basically is less than the total variation of the two okay, this is tg plus th tf but these two are increasing so this is just gb minus ga plus hb minus ha and then you take the soup and so you have that indeed the ab corresponding to f is finite okay, we can immediately state a corollary which follows from this theorem this is the following you have that if f is in bv over ab then we have that f is differentiable almost everywhere in ab okay, is it clear why? is it clear why? exactly, so you apply the the back theorem, okay fine yeah, you just have to mean so you have to the difference of two increasing functions so they are differentiable almost everywhere you take the union of the of the set where the two are not distinguishable, this is still a set of measures zero and so you remove it, it's okay then this is just a remark maybe you can prove by yourself bv of ab is a vector space so maybe you can prove that tab if you have the total variation of two the sum of two functions is bounded by tbf plus g and the total variation of cf is less or equal than the absolute values of the total variation of f so some example of bv function the trivial example of for instance non decreasing function ab in r decreasing okay another example might be for instance leap sheets function so you have that fab in r leap sheets so it means that there exists positive such that for any x and y in ab you have that of course f of x minus f of y is less equal than x minus y so basically here the idea is that you take so t is the sum of you fix a partition of course fx i minus fx i minus y and this is less or equal than the sum of l x i minus x i minus 1 and this is less than l b minus i so it's finite super okay and then you think there is a relation between function in bv and continuous function so one is containing the other or there are no connection between them so for instance this is a kind of comparison between cab interval and bv function so for instance you have that bv implies the continuity the continuity the fact that no because so means that the function c0 ab are not contained in bv just consider function like this with a jump as she suggest okay this function is in bv because you have the total variation will be just the jump, the length of this jump here is constant put 1 level 1 of course is not continuous and for the other inclusion maybe a function which is continuous is in bv or not no there is no connection between them I will show you here is more tricky not that much but what no you can just that bv function has to do with the variation if you think of a function that oscillates very much but it is continuous you can construct it with trigonometric function I will show you the function is so the analytical is x cos 1 over x so consider this function consider f just to be precise define it for instance 0 1 with values in r is defined in a piecewise way so f of x is equal to x times cos 1 over x for x positive but not 0 and 0 in x equal to 0 so you immediately feel that this function must oscillate very much near 0 because here you have the cos which goes to infinity so takes all the values between minus 1 and 1 infinitely many times it is a continuous function we have that the limit as x goes to 0 of x times cos 1 over x is 0 so we have that no problem f is continuous in 0 1 and then so I try to draw a picture of the graph I think it would be terrible but I will try to do so the graph looks like sorry this is so as I told you it oscillates very much around 0 near 0 so the graph is more or less something like this so we are interested in what is happening here ok so ok you have that this is true of course trivial ok and then we are interested in those points where the cos of 1 over x is is maximum no is equal to 1 the absolute values of the cos of x ok so we divide into k's so 1 over x is equal to 1 so you would have that 1 over x must be equal to 2k pi and we call this point we define this point xk has 1 over 2k pi analogously we consider the point where cos of 1 over x is equal to minus 1 these are points of so we have to shift of pi this pi plus 2k pi so we call them with yk and these are 1 over I collect pi and so I have 1 plus 2k ok ok so basically I am interested in evaluating the variation along this partition the partition made up of this point so we x evaluated in the point of the type xx is equal to xk and f in the other point is equal to minus 1, yk so we consider the following subdivision so we fix an index k so we have they are put in that way, order in that way xk minus yk minus 1 less than xk minus 1 and you have y1 less than 1 ok ok ok let's consider the total variation of f in the interval 0, 1 and it would be larger or equal than they increment along a fixed partition ok so you have xi minus xyi with i which goes from 1 to k and this is equal from this this is equal to i1k xi plus y and then we substitute values and we end up with the the harmonic series basically ok this is larger or equal k is fixed 1k then one of them for instance this one ok now here perform a little bit change of index for convenience this is j which goes from 2 to k and this is 1 over pi times j and this is indeed the general term of the harmonic series with divergence ok this as k tends to plus infinity this tends to plus infinity so basically we prove that this f is not in bv so apparently there are no no connection between continuous function and bv function ok ok now step by step we will be interested in somehow provide generalization of the fundamental theorem of calculus ok we will need some lectures to do this and ok ok so we call this lemma 2 so we have that if we start from an integral function that is ab and then the function defined in this way with capital f of x has the integral between a and x over f of t in dt this is a continuous function and it is a bounded variation ok first of all we want to prove that it is continuous ok so we have we want to estimate this difference and ok we use the definition so of course we have of t dt minus ay dt ok then by additivity this is x y dt ok of course ok x and x I took the absolute values ok so this is less or equal then so the absolute values of the integral is less or equal to the integral of the absolute values we already saw this ok and now which theorem do we do we use to prove that that this goes to zero as soon as x y approach to x of the complex so we it's a theorem of major theory so we have that f is integrable absolute continuity of the integral do you remember this absolute continuity of the integral tells you that if you take the integral of an integrable function and the measure of the set where you perform the integral tends to zero is small also the integral must be small provided of course the function is integrable ok by the absolute integral g goes to zero has the measure of ok as y tends to x so the measure of the interval for instance yx perverse goes to zero ok so f is continuous so capital f is continuous ok now remains to prove that it is also bounded variation ah ja, jez, da jez izgledajte no, no, no, no, jez, da jez tako, tako ok tako ok, so we won't prove that capital f is in b v so we fix partition of of a b, so fix sub division so we have a equal to x naught x1 v d and ok, of course we consider the variation along this sequence this ok over f sorry, big f we use the definition and this is equal to a b t and this is finite ok so you take the supremum on the left on the right hand side you have a quantity that does not depends on on the particular subdivision that you take so you have that supremum of of this sum is less the total variation of f ok, so we are done ok, now we want to do an example so we consider the following function which is which is somehow is a slight variation of the previous one ok, f maybe is defined between 0 and 1 on the original 0 if x is equal to 0 so basically here we have x to the power 2 instead of the only x of before so, and the question is is f in bd ok, I mean the idea is of course, I mean I am doing this example after the lay so probably you need to use no, theorem no, dilemma 2 ok, so probably you need to use so the answer is yes and to give a rigorous proof you need to use the previous lemma so the previous lemma tell you that a function you are sure that a function is in bv integral of an integrable function so now the question is that we have to provide this integrable function to put into the integral I mean, the intuition says that the function that no, so what would be the function that is in the integral ok, maybe just probably is better if I denote this with capital F would be easier to ok, to compare with the previous ok, so what we want to do is the following yes, this f ok, which is, by the way, is continuous I mean, this is f of x is continuous because you take the limit f of x tends to zero as x tends to zero ok, this is just and so we want to, what we want to do we want to express f of x in terms of where f is integrable, ok if we are able to do this, we are done so what would be this f the natural candidate the derivative the derivative so we just have to check if the derivative is integrable, ok so x is equal to 2x x times plus I think at the end you obtain this and so we have that everything there is a square it's ok and we have that everything is bounded so indeed and then you can also check for example that f prime is if you do the incremental quotient you can also prove that f prime is continuous ok, also in zero so we have that f prime so we define this f as this f prime, which is integrable because everything is bounded and so we are done, ok so we are sure that big f is in bv see, yes, yes if you do the you have to do the incremental quotient ok, I mean not just and ok, now ok, now I state lemma 3 and then ok, this is f if f is integrable then we have that f of t then ok, then if ok, this is integrable and we have and this holds this is equal to zero for any x ok, so take f in a, b as you know we have x with x in between a and b ok, then what you expect intuition this is a consequence actually of a former result f is zero almost everywhere in a, b so y, you remember that we prove actually with that if for any interval this is larger equal than zero then f is larger equal to zero almost everywhere so you use this that result for the two inequalities and you get this ok, so now lemma, I mean I call this lemma 3, lemma 4 we have that if f is bounded and measurable a, b and if we define ok, if f sorry, if f, ok and we define capital f of x as f of a plus ok, you can get rid of this a, x f of t then we have that f prime of x is equal to f of x for almost every x in a, b ok ok, so just let me observe that it makes sense to define this because we have that we start by a bounded function so we are in a bounded interval so it is integrable so there is no problem but this is finite ok, what we can say is that by one of the previous lemma this f, large f is in b, v because it is the indefinite integral of an integrable function ok ok, so we have that from this we have that that exists f prime almost everywhere and ok, now we want to quantify what is the derivative ok, so f is bounded so we mean small f is bounded so we have that there exists some m such that f of x and it is equal to m for any x in a, b ok, so we want to study the incremental quotient so we set f of x as these incremental quotient we consider it a long a long sequence this should be n, we use definition of x x plus 1 over n f of t in v t ok, so since we know we already observed that f is differentiable because it is in b, v we have that ok, since f is is differentiable then we have that this the limit of this sequence converts to the fn of x converts to f prime of x almost everywhere in a, b and then we want to use another hypothesis ok, so this is less equal than m, no, because you have n times length 1 over n times m so this cancel out so this is m so this is so we have that, we have a sequence which is uniformly bounded by m which in particular can be u has an integrable function so which which convergence theorem we can use the boundary convergence theorem ok, also the dominated convergence theorem, no ok, by the but also the boundary convergence ok, we have that we have the convergence of the integral so we have that f prime of x by the theorem is equal to the limit of a, c of fn of x in the x which is equal ok, to the limit of n, a, c and we use we can use the definition fx plus 1 over m minus f of x and ok, if you if you do some computation at the end what you get so at the end you obtain the following you have m c, c plus 1, because you have some time that cancel out, ok so f of x vx minus n of a a plus 1 over n of f of x ok and so we know that this large f is continuous because we prove in the previous in the previous theorem so this converges to f of c minus f of a which is equal to f of x in the x so what we found, we found that the limit as n tends to plus infinity of a, c f of n over x in the x is equal to f of c minus f of a then we have that a, c f prime of x in the x is equal to a b f of x in the x c and so we are done because we just observed that ok, we have that equal to 0 and this is integrable so we have that by the lemma 3 this is true for any c so by lemma 3 we have that we can deduce the point wise equality and so we are done so this concludes the proof so for today I think we can stop