 OK, hi, my name is Eva Cintic, sem lekturer for the course of real analysis, so this is my email, so I am currently research scientist at the University of Nova Gorica, which is in Slovenia, just behind, through the border, but I'm Italian and so basically the main, I would like to outline the topic of this course. Oče vedeljamo, da je častak, častak, častak, častak – nekaj najbičaj pri Sobo-Amplijnev skupite. Se nam izop revi verbila del, da sem daji izop vredila kaj distinguishanje in učinjanje, načo ne kadunje, nič je nekaj površal. In, nekaj, častak, častak, častak, častak, častak, častak, Ok, maybe some of you already studied this subject, do you already know this? No, no, I mean all these things, have you already? Ok, ok, in case it will be a repetition. All this argument will be carried over in one space dimension, so we know. I will recommend you some book, so the book of Reuden, which will be somehow the book that I will follow, so real analysis. Basically we will study the chapter from three to chapter six of this book. So you will find some copy in the library here, ok. So another good book is the book of Reuden and Siegmund. So the title is measure and integral. And finally a very classical book, maybe a little bit tough, but it's a good text, book of Reuden. It's real and complex analysis. Reuden, yeah, it's an H and Siegmund. These are two authors. So as I told you, somehow the main topic of this course will be the Lebecki integral, which somehow the aim is to extend the notion of Riemann integral and especially to extend the class of functions which are integrable. And this also will require to introduce the domain on which this function are defined, ok. So just to give you, before starting with giving the definition and the notion, I will just somehow motivate why we will need to enlarge this space for instance of continuous function. So for instance, consider a function f in a just closed interval with values nr, a continuous function, ok. So in consider the integral, you can think about it as a very elementary notion of integral, as for instance the Riemann integral. And in the most elementary case, you can think about it as the area between the graph and the x-axis, ok. So we shall see that this notion will be not enough when you deal for the application to ordinary or partial differential equation, ok. So we will have to enlarge, because when you deal with application with PDE or ODE, you will need to somehow to define a norm which is defined by means of an integral, ok. And also what you need is that the space which is defined with this norm must be complete, ok. We will see that if you define for instance the space of continuous function and though the space of continuous function with a norm which is defined by an integral, this will not give you a complete space. So a space which is not complete is not very useful for application, ok. And by somehow we will need to enlarge the set of the function, ok. Ok, so, ok. As I told you, consider this norm. So defined in a very simple way. Ok, you have f is continuous, so the modulus, the absolute values of f is still a continuous function. And it's easy to see that this function is a norm, ok. Because it's not negative, it's zero if and only if f is zero. If you multiply it for a constant, the result will be the integral times the absolute value of the constant and the triangle inequality also, ok. This is very easy. Ok, so, no, we, and though the space of continuous function domain defined in AB with values in R with this, with this norm, ok. Ok, of course, starting from this norm you can always define, you can always define a metric, ok. Take the difference between f minus g. So, as I told you, as I already anticipated you, this definition of the metric will not give rise to a complete space, ok. And we need completeness in the application because usually you have to pass to the limit so you want that good property must be precept when you pass to the limit, ok. So, ok. I recall you that a metric space is a space which has this property if you have a Cauchy sequence, it converts and the limit has to belong to the space itself, ok. So, now I will just briefly exhibit a sequence of continuous function whose limit does not belong to the space of continuous function. So, I just will introduce this sequence of function by the graph, which maybe is the best things. So, you have your interval a, b. So, took a point in the middle, for instance point c, and then took a point c plus 1 over n and consider the function f and the finite as follows. So, this is the root, the y axis, for instance. So, is 0 between a and c, here is 0, between c plus 1 over n and b is equal to 1, ok. And here is defined by a fine interpolation, ok. So, probably the analytical definition should be this, just to be 0 between 4x, between a and c. Here should be nx minus cn when x is in between c and c plus 1 over n n is equal to 1 for x between c plus 1 over n and b, ok. Ok, so, this function is continuous because the values at c and c plus 1 over n matches if you take the limit on the right and the left. Ok, so, let us first check that this fn is a Cauchy sequence in the metric that we just introduced, ok. Ok, just I will call you the definition of Cauchy sequence. So, we have that for any epsilon positive, there exists an index n bar such that for any n and then written n bar we have that fn minus fn fn is less than epsilon, ok. Ok, then consider for the computation, for instance, consider that n, ok, for computation just to fix the idea. Assume that n is larger than m and we want to estimate fn minus fn which by definition is nothing but the integral between a and b of fn minus fn. Ok, we can, of course, we can split this integral in three parts. So, the part between a and c of fn minus fn c plus fn plus 2 plus 1 over m and b fn. Ok, so, basically, you have that this is 0, this part is 0 and this is less than 1 because we are both like this. Ok, the drawing is not the best but you understand. So, this is basically less than 1 over m. Ok, so, it's enough to take n bar larger than 1 over epsilon and you got that, indeed, fn is a Cauchy sequence so this is a very easy result. Ok, but now we want to prove that fn does not converge in the space of continuous function, ok. So, we argue by contradiction. Assume that, so, by contradiction, assume that there exists a function, a continuous function belonging to our space and such that we have that fn converges to f in this null. So, we will see that we will reach a contradiction, ok. So, what we would obtain. So, you consider AC of fn minus f which, of course, is less or equal than this which is precisely the norm that we are considering. Ok. Ok, but we know that fn is identically zero in AC so what you get is that AC depends to zero, ok. Of course, this does not depends on n so this means that this is zero and so this means also that the integrand is zero because it's positive. So, in this piece of the full interval, ok. We consider the other part of the integrand. We consider the part between the last part. So, c plus 1 over n B of fn minus x which again can be bound by the fully integral and this we know is zero by our assumption. Ok. Ok, but we know that here fn is equal to zero suit one, sorry. So, you get this. Ok, here you have that the left hand side integral depends on n but we know that the integral is continued with respect to the end points. So, what you get is that this converts this converts to cB. So, at the end what you get is that cB minus f is equal to zero. So, again by the same in analogous argument you have that f must be identically one in the integral cB. So, at the end what we found is that if f is the limit f is like this so it's zero in AC and this is one in cB which of course tells you that f is not continuous, ok. So, we reach at the contradiction. This means that probably the continuous function is not a good space where to work, ok. Ok, so we need to enlarge this function space and so the first way to do it will be to somehow to enlarge the domain in which those functions are defined, ok. So probably just interval or or elementary set are not enough. So tell me if I so the way to enlarge may arise now or the way to enlarge this space function is to introduce indeed the Lebesgue integral and to introduce the Lebesgue integral we have first to introduce the Lebesgue measure, ok. So you have to think of the Lebesgue measure as somehow at a generalization of the notion of length for intervals, ok. Ok, so I will first introduce some terminology which is probably very easy but just to do it right from the beginning so what do we mean as an interval. So these are intervals so this would be a closed interval this would be an open interval and then you have half open and half closed when you have this situation, ok. And then you have these are bounded intervals and then you have also the offline interval and so on the left you get closed and so on so you have minus infinity B and minus infinity B. So we would like, as I told you to give this definition of measure of Lebesgue measure in such a way that it extend the notion of length and moreover I will list now some property that we would like Lebesgue measure fulfill. We will see that will not be possible that to find a measure that fulfill all the four properties that I will list. So we will be content with just three of them. We have to choose some. So the first one that we would like to have I call it the zero property is the following we would like to have a measure so asset function which will extend the notion of length which will be defined on the set of parts of R with values in the set of the extended real numbers. So PR is the set of parts set of parts which is just the collection of all the subset of R is the subset of R and it contains of course also R itself and the empty set. And this it is we will call it the extended real number. Yes, yes it could be plus infinity. If you think as I told you, the measure that we are going to construct will extend the notion of length for integral. For instance, if you have to compute the length of this interval it will be plus infinity. It's infinity the length. So of course we have to include plus infinity also here. So this is somehow the first property the zero property and we fix the domain and the co-domain that we would like to have. We will see that this is too much to require this. We will have to restrict the domain but we will see later on. So the second property or the first property property 1, let's call it property 1 is what I repeat is that we want that the back measure coincide with the length of an interval for any for any interval. As I told you we can see we can consider the empty set as a degenerate interval of measure of measure zero. This is just a brief remark. So the second property that we want is very crucial because we want to to have good property when we pass to the limit. Because in the application you always pass to the limit. And this would be translated in the requirement that we want that the countable additivity property so what does it mean? It means that if you have a collection ok in set of parts of R of this is a sequence of these joints there must be these joints this joint set then of course for which M is defined ok then what you want you want that the union of the countable union of this E n must be just the sum of E n of course they have to be these joints because otherwise and ok so the countable this is just brief maybe trivial remark but of course this countable additivity property immediately implies the finite additivity property that says that of course if you have this also this will also if you have A and B two sets belonging to the set of parts of R and A and B is joint then you have that also the measure of A union B will be just the sum no? as you can see it is just defining E1 equal A E2 equal B and then E n equal to the empty set 3 because we require that ok this is just an easy things but just to just to fix the idea and then we want to have another somehow geometric property to be fulfilled quite intuitively so we want that M is translation invariant so or namely we are requiring that the notion definition of measure will not depends on the location of the set this is trivial for the interval but we want to be preserved also for this extended set ok so let me write it more so basically we have a set E in a set of parts of R and you define take one one element Y in R and you define E plus Y which is the translated set with respect to Y which is X plus Y X belongs to E so what we want is the measure of E must be equal to the measure of E plus ok so we will see that it is not possible to have a set function M that satisfies all the four property ok so the third one the last one no we cannot get rid of this because it's this is too important it's very intuitively M shouldn't depends on the location what we could do is to replace to relax the second one in the sense that instead of requiring the finite additive property ok in this case you won't reach a contradiction but it won't be useful for application because as I told you we want to have good property when you pass to the limit so we have to to maintain this this one of course no because we started by motivated by the fact that we want to extend the notion of the length so what remains is to relax this so somehow we will have to reduce the domain of this set function so we will be we have to be content with somehow a subset of the set of parts of R so we will call this subset has the set of the back measurable set and we will see that this is indeed we see a true subset of the set of parts of R ok so let me so basically now the aim of of the first lecture of this course will be to construct a set function which satisfies the properties one two, three and to define a good domain for this for this function ok ok so we will see that it would be convenient to define the domain of M in a way that it is a sigma algebra ok maybe you already know what is a sigma algebra did you already study yes ok but I will recall you again so you have that give you a definition so you have that given X ok I mean still stuck on the real line setting somehow so you give a set X and you consider A as a subset of the set of parts of X so we have that A sigma algebra if you have these three property so you have that the empty set must belong to A second one is that if A is a set that belongs to this italic A if you want then also the complement of A belongs must belong to sigma algebra and finally if you have a sequence of set A n such that A n belongs to A then you must have that also the countable union of A n must belong to to the sigma algebra so we will construct M the domain of M in this way in a way that it would be we show that it is a sigma algebra ok just a brief brief remark you have that probably you already know that the morgan's law ok from that also the intersection the countable intersection of this set belongs to the sigma algebra because of this union ok, so you do operation that still allows you to stay in the set of in A ok, and now I will introduce another definition ok, so we shall say that M is a countable additive measure ok, if it is ok, a non-negative of course value ok, whose domain of definition is a sigma algebra algebra call it this italic M M of sets such that ok, we have that these two properties are satisfied so the measure of the empty set is zero and then we want the countable additivity property because we call it a countable measure, so we want that for any sequence in the sigma algebra of this joint set of course we have that so what we want that the measure of the countable union of this E n must be just the sum of the measure of the E ok, this is the definition of countable additive measure, which somehow embodies all the requirement that we that I mentioned before not all of them but we still want to have that to construct a function like this, so a countable additive measure, which in addition also is translation invariant ok and and preserve the notion of length ok, so now what we want to do from now on is the following task so we want construct a countable additive measure which is translation invariant and such that preserve the notion of length ok for any ok, for any interval interval yeah, yeah will be defined on set of R on subset of R, yeah and we will call this I mean the domain of M as I told you we will define, will be the back measurable set and we will have values on the extended real line ok, this is why high set that is non-negative so it is from zero on and as values in the extended real line so it is ok we will start to construct this such a function and to do this we will need first to introduce the notion of outer measure ok, so finally we will define somehow this is an auxiliary definition and we will pass through this definition in order to define our back measure ok, so you consider a set A in R and then just think about a collection of open intervals which cover A ok so ok, first of all of course A belongs to the set of parts of R this is trivial and consider accountable collection ok accountable call them IN of open intervals which covers intervals which cover A that A is contained in the union of of IN so consider interval because for the same reason because we know how to deal with interval we have the notion of length so we have a starting point ok, so for each such collection you consider the sum of the length of this interval ok and you define as outer measure so we define outer measure in such a way we have an outer measure and it is M star ok in this way this will be a set function which will be defined actually on the set of parts of R so later we will restrict this with values in the extended real line and so we associate to a set A these values so will be the infimum of the sum of the length of IN where IN is a set in the collection of open intervals which cover A ok, this is a definition ok, just some some remark ok, so we have that the length of the interval of course are non-negative number so this sum is well defined so it doesn't depends on the order in which you sum we are not talking about the empty set because of course at least we have R which will cover A for sure ok, so this somehow it's a good definition ok, now we will we will list some property of this outer measure ok, some really easy easy property ok, so it follows that so the outer measure of the empty set will be zero of course ok because ok, you can think as a particular collection as the collection of of the empty set which can be understood as degenerative interval ok ok, then you have ok, monotonicity property or namely if A is contained in B then you have that the outer measure of A is less than the outer measure of B so why this? so you can so actually when you consider this you are considering a bigger bigger set, so the infimum in principle would be smaller ok follow from definition from the property of the infimum if you are doing the infimum in a bigger set of course what you will get would be smaller or equal ok, so just to ok, I just write this because to be complete, but of course it's easy, this is because when you consider this collection of IK IK which covers the collection which covers B this is a subset of the collection of the IK which covers R so then you get property of infimum you get the tazis, ok ok, then ok, another easy things is the following try to raise ok, each set consisting of a single point X has intermeasure 0 and it can be easily seen by the fact that of course you have the single point set X so for any epsilon this this set is contained in this symmetric interval and the length of this interval yeah this one IK and this one the same are contained, yeah I mean, you can use them to cover A, since A is smaller than B you can use them to cover ok yeah, it's the converse so A is contained in B but all the collection which covers B will cover also A so when you you do the infimum of this this will be bigger than the infimum of this, ok yeah, it's the reverse ok and so this is of course to epsilon so since this is true for any ok, already for any epsilon you can make this length arbitrary small so this is we get of this single point set is 0 ok, these are easy things ok, now at least we have defined this notion of outer measure we just list some quick consequence of the definition ok, now we will prove a proposition that tells you that the outer measure did extend the notion of length which somehow is what we want ok, so the outer measure of an interval is its length ok ok, to prove this claim we will argue by step ok ok, we will start by considering a closed bounded interval and then step by step we will extend the notion, ok so we will start with the case the case I equal closed bounded interval ok, one way is quite easy in the sense that ok, you have that for any for any epsilon positive you have that the corresponding open interval A minus epsilon B plus epsilon we have that cover AB so you can think about as a set in the definition of outer measure so we have that m star AB is less or equal than B minus A plus epsilon this is for any epsilon and this is follows from from the definition and so as usual since this is for any epsilon we can this is preserved this must be true even when epsilon tends to zero ok, so at least we proved one one side of the of the equality, we want to prove the other one, which would be somehow more tricky so now what we want to prove is this ok, now again we recall the definition of outer measure and we observe that this would be equivalent to prove another fact, so by the definition so by the definition of outer measure which in turn is defined by the notion of infimum ok, we have that it is equivalent to prove to prove that if you have accountable collection if is accountable collection of open intervals which covers AB we have that the sum of the length of IN which is precisely the right hand side of the definition of of outer measure is larger or equal than B minus A and I call it star for any such a sequence this will be preserved by the infimum so at this point we will make use of the unborel theorem ok so somehow it would be better to deal with the finite sequence now it would be easier so this is why we will use the unborel theorem the unborel theorem we have that ok, that any collection of open intervals which cover, covering AB contains finite which is which is what we gave so we pass from accountable collection from finite collection finite subcollection which also cover AB and of course it would be enough to prove star for this finite subcollection ok because the sum of the length of the finite subcollection would be no greater of the sum of the countable collection ok just just to be enough to prove finite subcollection so basically we have that q is contained in the union of a finite ok, so we have that since the closed interval is contained in this finite subcollection so of course we have also that element A which belongs to this closed set must belong to this union of course ok, from this there must be must be one ends which contains A ok, and let us call let us denote such an interval as as A1, B1 ok ok, then there are two cases the good one is that if if B1 is larger than B then we are done, that is nothing to prove ok, otherwise we have to continue with our argument, so if on the contrary B1 is less than B that is what we call then B then we have to continue then we have that B1 belongs to AB so, and since B1 does not belong to A1 B1 open there must be another interval an interval A2 B2 belonging to this subcollection of course because we are speaking about open interval ok, so there must be an interval A2 B2 always belonging to the subcollection in the finance collection ok, such that B1 belongs to A2 B2 and so we have something like this A2 B1 B2 and also here, A2 B2 is larger than B, then it's ok it's a good case otherwise if B2 is less or equal than B we have to continue with this argument and so if we iterate somehow this argument what we get we will find a finite somehow we get a finite sequence of open interval belonging to the finite subcollection ok from the collection ok, such that they are placed in that way you have A AI BI minus 1 and BI something like this ok, this is enough ok, so the thing is that since our subcollection is finite this process must end at some point this is why we end up with some finite sequence of interval ok and somehow this process is over so the process is over when the endpoint B belongs to the open interval AK and ok ok, so finally what we get we have that the sum of the length of IN is larger or equal than the sum of the length of AI BI this I mean in the finite subcollection these are this this interval that we selected somehow so this is equal to what this is a kind of telescopic sum plus we have BK minus 1 I mean it's not a telescopic sum but you will see some ok, so if you sum up in another way no B B1 B1 A1 so if you observe this you have BK you collect this term AK minus BK minus 1 this is negative and then you have minus AK minus 1 minus BK minus 2 which again is negative minus A2 minus B1 minus A1 so at the end you get that all this stuff here is larger than BK A1 ok, because this is negative but you have a minus in front ok, which is of course is larger than B minus A ok, so at the end what we found is indeed the inequality star that we were looking for larger than B minus A so somehow the statement the thesis follows for for the interval for compact interval the thesis follows for I equal to AB so now we have to extend this equality to the other kind of interval for any finite interval but you have to deal with the opening interval and this measure is B minus A and then the close interval from A to B is there is Julian's union of 3 sets the single term A, the single term B and the open interval from A to B from the counter of this Julian's property you can see that M star of the problem from A to B equals 0 plus B minus A equals 0 equal B minus A I mean I'm not maybe we can discuss later I don't know if you use the countable what do you use, the countable additivity property for what, for length or for what I did not get your maybe we can quickly discuss later because now I can if I don't see the things written I cannot follow okay now we have to extend the proof for so we use what we prove so far to stand this equality for any bounded set any bounded interval, sorry if I is any finite interval then you get what I want to do is that given any epsilon positive there is a closed interval called it J okay, such that J is contained in I and the length of J can be controlled from below by the length of I minus epsilon okay, hence what we get okay, you have that of I minus epsilon is less than the length of J and this is equal for the step before is equal this is step probably I call it A this is step B J J is contained in I okay and this is less or equal now you use the monotonicity property of m star of I less or equal to m star if you want again for the monotonicity property of I, the closure of I which is equal of the length of I closure which is equal to the length of I so we know that this holds for length since this is true for any epsilon positive you get that indeed the outer measure of I is equal to the length of I okay and because it's in between these two so finally what remains to prove if infinite interval then you have that real number find a closed interval J contained in I such that you have that the length of this closed interval J is equal to this arbitrary large m okay as you have m star of I is not equal to m star J okay since this is true for any m then any n positive of course then you have that m star of I equal to plus infinity which is equal to the length of I so with this we conclude we span all the cases we consider all the cases we conclude this proof so the last minute okay just anticipating what we will prove tomorrow okay so tomorrow we will prove the following we will prove the countable sub additive property for the outer measure so this is composition so you have an countable collection of set you don't need to require them to be disjoint because we just get what we get is so weak that we can also take them as any countable collection so what we will prove is that the outer measure of this countable union will be less or equal of the sum of the outer measure of an so for the outer measure we just be we just be we just obtain this less or equal not even if we require a countable disjoint set we cannot we have somehow to restrict the set that we so we will have to introduce another notion of measure okay so this is the sub additive okay sub additive property okay and then okay in the last minute we can just introduce some terminology but tomorrow I will recall you again so we will note with this italic g the collection of open sets of r with the with this italic f on the core where is the closed set with g delta okay the family let's call it the family of sets obtain it as countable intersection of open set with f sigma the family of set obtain it as countable union closed okay so somehow the idea the reason why we introduced this family set is that this somehow nice set easy set set that you can imagine would be to approximate the measurable by means of this easy set treatable set so would be very useful for us to see that you can approximate in some suitable way the measurable set by means of a set pick it from this class of set from this family okay thank you