 So hi, today we will change, we will start a new argument, which is differentiation and integration. Ok, so I start by giving a definition of a Vitalic covering. So we consider with a note, with this Vitalic high, a collection of intervals. And then we say that high covers a set E. OK, zelo sez, E, dva א, ki sez. V tudi v vitaliji, OK, je v tebelji matematik. Kaj je to, ki je vse epsilon pozitiv, in ki je vse x, in e, zelo sez in interval i, tudi jez, kaj je vse interval. tudi, da je za njih, ki je z njih, ki je z sedem, in vzelo je za njih, ki je z sedem, ki je za sedem, ki je za sedem. OK? OK. Zato letimo, da bilo, da je zelo za v Vitalij, tako, da je za sedem, neč začel, tako, da je za sedem, da bo ne ni, 7, which is a subset of R, such that the outer measure of V is finite. So actually the set needs to not to be measurable. You can take any set. And then you consider I, he's a collection, he's a collection of intervals which covers E, in v vitali sejnji. V svečnih sejnjih. Ok, danes, Lema tači, da, za vse epsilon pozitiv, je vse finajte sekvence in intervali, vse vse finajte vse finajte vse finajte koleksjon I1 koleksjon, I1, In of intervals, of course, in our in our class I, such that you have that the outer measure of E minus the union of this this finite union is small is less than epsilon. Ok, the proof. Ok, so let me so without loss of generality without no loss of generality we can assume that that the intervals in the collection are closed in I are closed because in any case we can switch from closed interval to open interval by means of the endpoint, which has measure zero, ok, because you have that the outer measure up. So we started a new argument which is about differentiation integration and I give a definition of Vitali covering so Vitali covering you start from a set in R which is not necessarily measurable and you say you start by a collection of intervals you say that this collection covers this set he for any epsilon you can find and for any X in E you can find an interval in this collection such that X belongs to the interval and the length of the interval is arbitrarily small, ok. Then now I'm going to prove this lemma which is known under the name of Vitali and it tells you basically that if you have a Vitali covering then you can you can somehow express your set E, which is any set set E can be covered almost by means of a finite sequence of interval, ok, in this sense here. Ok, so we consider O, an open set such that we saw that there exists this open set O such that ok, O contains E and you have that the measure of O is less than the outer measure of V plus epsilon, ok. Ok, the next step is somehow to reduce the Vitali covering of the statement so to modify it to slightly change using this open set O and I will tell you how we modify this Vitali covering so instead of denoting with this capital this Vitali i we will denote with this somehow Vitali j, if you want. Ok, this will be a family the collection of the interval, which is in i such that i is contained in O ok, we claim that this is still a Vitali covering for E so j is a Vitali covering for E ok, so we take, so we have to select a point x in E and some eta ok, eta plays the role of epsilon of the previous definition eta positive ok, then in particular we have that since O contains E we have that sorry x in E x belongs to O and hence being open we have that there exists some delta positive such that you have that you can find delta small so that this open interval is contained in O ok, then by definition of Vitali covering for i for our starting collection ok, what we can infer we have that there exists this i in i such that you have that x belongs to i and the length of i you can choose this so that this length is less than the minimum between eta and delta you can always choose this i so that this is arbitrary small so you can always find this ok, so from this you have that i belongs to this modified covering because i is contained in O and is such that you have x belongs to i and the length of i is less than our eta that we fixed at the beginning so actually we have that also this slight modification leads to the definition of Vitali covering for i ok, within open set we want to define our sequence of interval in the statement of dilemma so we proceed by induction so how we choose the sequence disjoint intervals j by induction so we have i1 be any interval any interval in j you can choose it in an arbitrary way and then suppose that you choose that i1 and in have been already chosen and they are disjoint and then we define this quantity kn in this way is the supremum of the length of i such that i is belongs to the collection j and such that i intersected with one of the previous with all of the previous interval is disintersection is empty i which goes from 1 to n ok, now the question is are we doing the supremum on an empty set or not the answer is not luckily, but why so we have two cases we have two cases so either you have that the set e, our arbitrary set of the statement is already contained in this finite union so there is nothing to prove then the process stop because of course in this case you have that the outer measure of this is zero so there is nothing to prove or not so this is not contained in this union so we have that e minus minus e1 i1 union in is different from the empty set ok, if this is true then we can find an element in this set ok, so there exist some x which belongs to that set so belongs to e minus this union this finite union ok, so we notice that since this union is a finite union of closed interval is still closed since the union the set that we remove is closed then the complement is open closed set ok, so this is open so this is means that we can we can find delta positive such that the open interval of radius delta is contained in this set x minus delta x plus delta is contained here so its intersection with this finite union is indeed empty ok, now we use again the definition of italic covering for for x and for delta ok x and this delta positive and what we can infer we have that there exist an interval i belonging to this modified collection j ok, such that x belongs to this interval i and the length of i is less than delta as we have that the interval i is contained in this interval and so we have that i is contained in o and so i belongs to the class where we are taking the supremo to the collection where we we perform ok and so we have that k n is indeed strictly positive ok, then we can by definition of supremo we can for instance decide to choose this i n plus 1 the n plus 1 interval in the finite collection so we can decide choose i n plus 1 in j the length of n plus 1 is larger than k n divided by 2 this is possible, ok and of course this is of course we have that i n plus 1 this is just to remind you intersect the finite collection of the previous and interval is empty ok, this is now we notice that we constructed we construct a sequence so a sequence of um perverse disjoint interval ok, such that what about their length such that we have that the sum of i of the length of the interval i is measure, they are disjoint is the measure of this countable union and this is less than the measure of the open set so somehow this sequence is finite is converged so then comes an argument that we already used so we are in presence of a convergent sequence so we can say that the tail of the sequence goes to zero ok, so ok, the general term i n, let me use this um tends to zero and then we have also this supremum because of this also k n tends to zero as n goes to zero as n goes to infinity because of this ok ok, this is one fact and then the other one is about the tail of the sequence of the series so we can also say that there exist some index capital N such that the sum is small so I choose this strange factor epsilon over five and I decide to divide epsilon for five ok, now we claim the following we do this claim which is indeed the thesis that we want we have that this index N is indeed the one that we are looking for so we have that consider this sequence, finite sequence of reintervals constructed in a way just told you are such that the outer measure e minus this finite union is less than epsilon ok, now I prove the claim ok, let me call for convenience this set e r so we denote by r the difference between e and this finite union and we take some point x in r now somehow we deal has before so we have that so x is in r so we have that there is this some delta positive such that the interval x minus delta and x plus delta intersect this finite union is empty ok, and then by definition of Vitali covering for x and delta we have that there exists an interval i in the collection j such that x belongs to this i and the length and the length of i is less than delta so in particular we have that this i is this joint from this previous interval ok, now no, I mean no, it's quite tough it's proof no, this is a claim it's a claim in the whole proof but I mean, but once that you ok, if you want just a raise claim it's just say that here we are within the proof of the dilemma of Vitali ok, so ok, we arrive here ok, you notice that this series is convergent from this we can deduce the tail of the series is goes to zero ok, so you can find some n, such that this is true so you consider this capital N, which is indeed we want to prove, this is why I wrote claim, we want to prove that this capital N is indeed the one in the statement of the Vitali lem so we want to prove this we want to prove so it was somehow a statement of the Vitali lem we want to prove this ok, we are still within the proof of the dilemma of Vitali, we want to prove this and, yeah, now I'm going, ok I know it's tough this proof, but ok, so we found this interval I, which we satisfied this and we also want to prove that we want to prove the following we have that that I intersected IN is different from the empty set for some N for some N, which of course must be larger than N, ok ok, why? this is true, in fact in fact assumed by contradiction assume contradiction, that is not true ok, then what we have then we have then, ok, we would have that I intersected IN would be equal to the empty set for any for any N, ok so here that I would belong to the collection to the set where we perform the supremum ok to the collection where we perform the supremum, ok and this is for any N so we would have that the length of I would be less or equal than KN for any N but this goes to zero, ok so we would have that length of I would be zero, ok and this is not possible, ok ok, so this is true ok, so there is this N which verifies this so we consider just to fix the idea consider N the smallest number the smallest index that verifies this property P ok, then it is the smallest we have that in particular just to fix the idea we have that I intersected I J is equal to the empty for the J which goes from one to N minus one, ok because of the choice that we make so basically what we have is that just to stress that I belongs to the set where we perform the supremum corresponding to the index N minus Y supremum corresponding ok, then by definition by consequence as a consequence of the definition of the supremum ok, we can say something about the length of of I we have that the length of I is less than N minus one which is less by the choice that we make two times the length of I N and so I call this two inequality with this dot ok now ok, now comes the five so now would be clear why I decided to choose this five ok, now fix the knot with Y N the midpoint of this interval I N so let N midpoint of I N ok and I take take an X point X in I ok, and let I then we want to estimate the distance between this X and this Y N so you have that Y N minus X so I remember that I N overlap ok, so just visualizing in this way this is less equal than the length of I plus the length of I N divided by two and then we use this this is less equal than two times on the length of I N plus the length of I N over two so we clear so we end up with this ok, so what does it mean geometrically so it means that it means that X belongs to an interval having midpoint Y N but five times larger than I N so means that so as we have that the point X belongs the interval call it J N having the same midpoint Y N ok, the same midpoint Y N of I N and five times the length in symbol you have that I N is five I N ok, so we have I remember that X was a point in R so we have that R is contained in of N N plus one infinity of J N and so at the end we finally found that the outer measure of R which is the difference between this finite sequence of this joint intervals is less or equal than this is five times the length the sum of which is less than epsilon and so we are done so may erase as a consequence, yes as a consequence ok, now we state we state and prove the Lebesgue theorem which is a consequence of this lemma, of course so may erase this blackboard ok, so this is a fundamental theorem this part of the course was under the name of Lebesgue theorem ok, and it tells you that you consider it a monotone function so you have F function which is defined in the interval with values in R so non-decreasing function ok, then we have that F is differentiable almost everywhere in the interval AB it means that you understand what it means but to be precise that there exists derivative in almost at almost every almost every every point of AB ok, then ok, it is differentiable prime is measurable and non-negative and moreover the following equality also you have that the integral the beggy integral of F prime of T in T is less or equal than F of B minus F of A F prime so you know that it is differentiable almost everywhere ok, so just before proceeding with the proof which is still quite hard today so can you provide an example so the answer is is it really needed a strict inequality or we can just put equality can you provide an example of a function for which this is verified with a strict inequality a function with a jump for instance for instance, if you have equality with H which is usually known as the every side function as defined in minus 1 1 ok, so you have that the function is defined like this, you have this is equal to 1 here and equal to minus 1 here ok, the important thing is that he has a jump that must have a jump so of course you have that it is constant almost everywhere ok, and so F prime is 0 almost everywhere but you have that H put it in 1 and minus 1 so you have that H1 is equal to 1 H minus 1 is equal to minus 1 so the difference is 2 so you have that minus 1 of H prime this is 0 and this is strictly less than H1 minus H minus 1 which is ok but this is you can object that this non-continuous function there is a jump so if we look is there another example of the fact that inequality my holds with a strict sign within the continuous function within the class of more regular function for instance continuous function so we saw the example of a function which has many interesting properties which is continuous which is constant almost everywhere but is increasing the counter function the counter function so this is another example this is an example another example is the counter function so the counter function F was expanded in 0,1 in 0,1 if you remember it was takes the value 0 in 0 takes the values 1 in 1 was increasing and we have that it was constant almost everywhere so F prime of the counter function was 0 almost everywhere in 0,1 but it was it is also continuous so F the counter function is continuous because if you remember we constructed it as a uniform limit of continuous of continuous function and so you have that also in that case you have that 0,1 this is 0 and so this is less than F1 minus F0 which was 1 so you have the strict sign ok ok, now we need to introduce a set of of four derivatives which are called Dini derivatives derivatives ok, so ok, I will with the plus on the top F of X is the Lim soup H H tends to 0 plus of Fx plus H of the incremental quotient minus Fx divided by H then I take the minus so here you have the limit no, sorry the Lim soup but H you reverse the order so one is performed from the left the other from the right then you have D with the plus on the bottom F of X is the limit as H tends to 0 plus of the same Fx plus H minus divided by H and at the end you have the D minus of F of X is equal to the limit which comes to 0 plus of reverse divided by H ok, ok, of course we say that the derivative exist when this four number exist at finite and they coincide prime of X exist if these four quantities these four Dini derivatives are finite and they must coincide ok, ok, so now we are interested in investigated when when they coincide ok, what you can for instance you can an inequality which comes straight forward is for instance the fact that this Lim soup is always larger or equal than this Lim inf so you have that D plus of F of X is larger or equal than D plus plus of F of X and in analogous way you have that this D minus of F of X is larger or equal than D minus of F of X ok, now we prove we start to prove the Lebesgue theorem proof of the Lebesgue theorem ok, so to somehow so we want to prove that our function F is differentiable so this means to prove that the set where any two of these four quantities do not coincide has measure zero so for instance ok, we just select two of them and we prove for two of them and then for the other is completely analogous ok, so ok, so for instance just to fix the idea consider D plus F and D minus F ok, now we want to prove we want to prove the following that this inequality is satisfied almost everywhere that ok, D plus F is lesser equal than D minus F almost everywhere in AB so we prove this to prove the reverse is analogous ok is analogous you just said to exchange the rule of the truth but it's analog and then the same is true for all the other couple ok ok, so we want to introduce this set A as the set of X in AB such that the converse is true so that D plus is strictly larger than D minus F so in the claim is to prove that this claim is that the measure of A is zero ok I mean actually ok, actually this is not correct so in the sense that we we still do not know if the set A is is measurable because because when we define so when we define the the derivative we use uncountable operation ok, so we are not sure that doing uncountable limit now which because H tends to zero is not along a sequence so we are not sure actually now that this is a measurable set ok, but this is not important so we want to to express this set by means of a countable union of more nice set if you want so we want to express this set here as in this way as the union of when you have two quantities V and U with values among the rational number and such that U is less than V ok, and we have that this set is like this and we have that this is D plus of F is larger than V larger than U larger than D minus of F ok, just call a set of this type E ok, and now what we want to do now is that if we define with S the outer measure of E because we still don't know that even E might not be measurable we want to prove we want to see that S is zero ok, this will be our purpose now ok, so sorry, excuse me think at a fix V and U just fix it now we focus on a set of this type just think ok, so as usual now we have this set E and we know that there exists an open set U open such that you have that U contains E and and we have such that for any epsilon positive of course such that the measure of U of this open set is bounded by the outer measure of E so plus epsilon so S plus epsilon ok, then fix a point take D a point in E and consider this collection of intervals which I call I which has which have this form so they are C D D, I mean I stress that belongs to E and you have that this closed interval are contained in the open set and such that we have the following inequality, so you have that FD minus FC divided by D minus C is less than U than this U here ok, so now you understand what we want to to use we want to prove that this collection I is a Vitalik covering for E ok so we have that is a Vitalik covering for E, so why so this is how comes from the definition of from this fact and the definition of liminf actually so ok, in fact fix a point X in E and in arbitrary and some eta positive ok, then by definition of liminf here definition of we have that there is some C such that if C is in between D minus eta less than C less than D this C less than D the quotient such that we have that FD this quotient the quotient in the definition is less than U C is in between D minus eta and D ok the point C is in is in U ok so basically we have that the interval I defined as C D is what we are looking for is the one that comes in the definition of Vitalik covering ok, because this is trivial because D our D is in I and the length of I is less than eta ok and then if U, this U because because I use this is capital U is the notation for open this is the this module ok ok, so what I want and then if you take C sufficiently close to D you have that if C is sufficiently close to D then you also have that C D is contained in U so indeed we are within this collection here and this capital G provide a Vitalik covering for you ok, now now we can apply the lemma the Vitali lemma because we did all this effort to prove it so we can apply the Vitali lemma so we have that by the Vitali lemma that exist a finite sequence that interval resist a sequence that they have in this specific case they have this form C1 D1 C capital N the N J ok, this joint ok, interval such that they almost cover E in the sense that E minus the reunion is less than epsilon ok now we want to sum up this gap over the index i so we consider this sum, this is a finite sum of f the i minus f C i so this is less than u u so d i minus C i ok, recoljno that they are disjoint no, so this is equal to u the measure of the union ok, C i d i and this is less or equal than u the measure of big u, this is that and this is less than u times s plus epsilon and call this first bound 1 ok, now we want to use to evaluate the outer measure of e by using subaditivity ok, so we have that s ok, by definition is the outer measure v ok, now we want to split e into 2 set the one is the one appearing here, so this is less or equal than m star e minus this finite union plus what remains so the intersection plus star of e intersected ok, this is less or equal than epsilon plus this, I repeat this intersected so basically we have that at the moment we found that s minus epsilon is less than the outer measure of e intersection this finite union ok, now somehow we want to to repeat the same argument of before within this integral c i and d i ok and ok, so the idea is that we so we define e i as e ok, you can just fix one of these c i d i ok, in this way you can ah sorry, here ok, from this we have that that you have the i from 1 to n m star of e i is larger than s minus epsilon this come from there ok, because we prove that this is less or equal than the sum of m star of of e i ok, so we basically we argue as before but now e i plays the role of the argument before so again, so we can claim that we have that that exist an open interval u i so it depends on the index of course such that you have that c i d i is contained in u i which is and such that the measure of u i is less than m star of e i plus, now I prefer to take epsilon divided by n, in any case it's an arbitrary small number, ok ok, if this is not true I just consider c i d i intersection of u i we just replace ok, so consider take an e e i in this collection in this way so you have e h such that e h is contained in this u i so I'm dealing in analogous way e is in e i and such that I consider the reverse inequality f h minus f e divided by h minus e is larger than than v this v was the number which is involved in the countable union that I started right from the beginning of the proof, ok ok, Arvina, as before you can prove that i is a vitali covering for is a vitali covering for e i ok, so by the lemma of vitali by the vitali lemma we can select a finite a finite number of this joint intervals such that we have that exist of this type e, I use the index gel, ok e1 h1 until em hm ok, disjoint ok as such that the outer measure of e i minus this finite union is less than epsilon divided by by n ok, this finite union j goes from 1 to m of this e j h j is less than epsilon divided by n and call it denoted with these two dots and so we know that this intervals belongs to this collection so they verify this inequality so we will use it ok, just let me ok, so consider fhj minus f e j, j goes from 1 to m sum of h hj minus j ok, dis is larger or equal then larger or equal then v times m star minus e i minus epsilon divided by n ok, why because this is because m star of e i so basically you argue as before, no? is less or equal than m star of the sum of two sets, the one is the one appearing here and then you have the sum so equal then epsilon divided sum divided by n minus plus this, plus this ok, you understand, e j intersect this e j j intersection e j hj ok, now we want to write explicitly this sum to understand how it behaves so this sum is hm minus f e m plus h n minus 1 minus f e m minus 1 ok, no you have fh1 minus fe1 this is less or equal so we have all these terms, because so far we did not use the fact that f is non decreasing, no? now we use the fact that f is non decreasing and we can say that this is negative this terms is negative and so you have that at the end here that this is hm minus fe1 which is less or equal than f di minus fci because you have that hm is smaller than di is contained and e1 is larger than c this is a c, c i and c i ok, now we consider this difference so you have that f di minus fci is f is non decreasing, this is larger than fhj minus fej r equal than v hj minus ej these are all sum from j goes from 1 to m j goes from 1 to m larger or equal than v times star of ei minus epsilon divided by n if we combine the two inequalities we have and we sum up we have f minus f this times is i goes from 1 to n is larger or equal than v times the sum of the outer measure of ei minus epsilon this is by dot then v s minus 2 epsilon ok, finally by 1 we have that u is s plus epsilon is larger than v s minus 2 epsilon epsilon is arbitrary small so this must hold for any epsilon positive so here tells you that u times s must be larger or equal than v times s but if s is different from 0 you get that u if you assume that if s is different from 0 you would have that u is larger or equal than d which is a contradiction because we start by another hypothesis so this is a contradiction so we add that indeed s must be 0 so in this concludes at least the first part so what we prove is that d plus f is less or equal then d minus f almost everywhere in ab so the next time we will conclude the proof ok for today I think we can stop here