 Ok, zato lahko vedem, da bomo najbolj tudi dobročil o spremistosti z taj spremistosti in zato smo pričel jaznih pristeličkov z taj pričel. Shizel smo pričeliči je pa vzelo za vzelo vzelo v zelo vzelo. Kaj je vzelo vzelo vzelo vzelo vzelo? Tudi je F tudi pristel vzelo vzelo vzelo vzelo vzelo vzelo vzelo, We assume the measures to be finite. OK, that's me, not measurable but bounded. What we prove is that the upper and the lower, kaj je miljene mat, če je opos cevno. A v fimo od kašnji slog, če jaz sem vesele, ki je neč vse, ma Je ľubimo vsuprimo, če jaz sem vsele, če jaz sem vsele, če jaz sem vsele, in kajšča. Zato prva poslednja funkciju psi in psi, kajšča, kajšča, kajšča, kajšča, je tudi prijemna akarakterizacija očko kajšča, kajšča, kajšča kajšča je prijemna. Sposlednje je to, načo ječnje. As we observe, of course now it is trivial how to define the Lebesgue integral. We give the following definition. We will say that if f is a bounded and measurable function defined on a measurable set, defined on e with e measurable, and such that, of course we always require now that the measure is finite, then we define the Lebesgue integral of f as the infimum, for instance, over all the psi, which are larger equal than f. Ok, now we see that somehow this definition is a good one, in the sense that it contains the definition of Riemann integral. So, if you have, if you start from a function which is bounded and which is Riemann integrable, it is also the Lebesgue integral. Ok, so proposition, so let f be a bounded, we have also always to require that f is bounded function defined on an interval, then we have that if is Riemann integrable over the interval on ab, then it is measurable, and we have that the two notions of integrals coincide, so the Riemann integral of ab over f of t, t is equal to the Lebesgue integral, so I denote the Lebesgue integral putting the interval at the bottom of f of x in dx, so this is the Riemann integral, and this is the Lebesgue integral. Ok, so we start by somehow the Riemann integral, and we have that Riemann integral, actually we start, so we know that f is Riemann integrable, so the upper and lower Riemann integral must coincide, so we start by the lower Riemann integral for instance, maybe f of x in dx, this is the lower Riemann integral, so how we can go on now? I want to obtain to a bound from above of this so, I recall you that the Riemann integral is defined as, ok, infimum, so for instance the lower Riemann integral is defined as the supremum over step function, but we know that step functions are particular simple functions, ok, so you can somehow from this observation you can arrange some inequality between the supremum, ok. So this is a supremum, I recall this is a supremum over step function, step function phi less or equal than f, so it must be less or equal than the supremum over a larger class of functions, so simple function phi less than f, but this time phi is a simple function, ok, a, b, c, phi x in dx, ok, this is just because step function contained in the class of the simple function. Ok, then you go, oh, may I raise here? Yeah, yeah, yeah, yeah, yeah, f, yes, psi is a, in this definition psi is a simple function, no, I mean this is the same term that appears in a proposition that we, ok, psi is a simple function, which is, so this infimum is taken over all the simple functions, which are greater than s, ok. Ok, I will raise here, because I need space. Ok, so now we want to go on here. Ok, I will go on here, less or equal. So how we can go on? Take the infimum, now you take the infimum over all the psi, which are larger equal than s, a, b, p of x in dx, this time psi is a class of, still a simple function, ok, is a simple function. But then we, ah, psi, no? Ah, yeah, yeah, yeah, sure, sure, thank you. Yeah, yeah, thank you, yes. Yeah, sure. Ok, and now we want to come back to the remaining integral, ok. So again, observing that this inclusion holds, this is less or equal than the infimum taken over all the psi, this time psi is a step function, so actually I am taking infimum over a smaller class, so I get something bigger in general, ok, something the quantity I get is larger, ok, so f psi is a step function, ok. Ok, so this is a, b, psi x dx, and this is precisely the upper remaining integral, ok, the upper remaining integral is the not. But we know that f is remaining integral, so all these inequality are indeed equality, ok, because the first term and the last term are equal coincide, so even in particular these two coincide, ok. So from this, since, ok, since f is remaining integral, we have that coincide, and so from this we have that also the quantity in between coincide, ok. So the supremum over, ok, phi, phi, phi, simple function of f over e is equal to the infimum over psi, psi larger than f over e psi coincide. Ok, so, what we get is that f from the proposition before we know that f is measurable, then it is the back integral, and then from this we know that the two integral coincide, ok, measurable, then it is f, and we have that the remaining integral of f coincide with the back integral of f, ok. So somehow this should convince you that introducing this new notion of integral will somehow generalize the concept of remaining integral. Ok, now what we want to prove is that this definition of the back integral for bounded function satisfies some, the usual property, so it's linear, monotonicity, and so, ok. Are you ready to copy this part? So we have that let we have f and g to function bounded, the measure of e must be finite bounded and of course measurable. Ok, then we have that the following properties also. Property i is the linearity, take any a, b in r and you have the integral of the linear combination is the linear combination of the integral. Then you have two i is, ok. We have that if these two functions coincide almost everywhere in e, then also the back integral must coincide. The third one is monotonicity property. Ok, this time you have instead of the equality of that if f is less than g but it's enough that it satisfies almost everywhere in e. Ok, then the inequality is of course is preserved by the by the integral, the back integral. Ok, moreover you have from this you have also that the absolute values of the integral is less or equal than the integral of the absolute values of the function. Ok, ok. Four, there are points of four, you have that if you know that our function the function f is in between two constant capital A and capital B then what you can imagine is that the integral of f is in between a times the measure of e less or equal the integral of f less or equal b the measure of e. And the last one tells you that if this time with AB I do not to set here are two constant, if AB are disjoint measurable set with finite measure then we have that the back integral over the union of these two set is equal of the sum is equal to the sum of the back integral on each on each set, ok. Here are set yeah, yeah, yeah yeah, yeah, yeah with finite measure but I mean if e is yeah, yeah, of course yeah, yes, sure. Here, I don't remember maybe in Royden this part is they say that maybe e is zero outside of a bounded set no, this is why maybe sometimes I I stress that in these cases you can take AB in R and a part of them is this part outside this set of measure where they are zero ok, but just think that they are in E so you don't need as you observe to require that they are it's automatically verified ok, so we start by point I and ok, we trivially observe that if you have a simple function so we need to start by our building blocks that are simple function still a simple function ok, so start for the case where A is positive ok ok, so we have E A times F is equal to the infimum C and here you have the C that you recall in F is equivalent to the fact that HC is larger equal to AF ok, then for simple function you can take this outside the infimum A the infimum of E psi psi larger equal to F ok, this is by definition this A F ok, when you have A negative this is the case positive so for A negative ok, so you have E by F is equal to of an infimum we start by the infimum of A C larger than an F in the quality is robust because A has is negative this is P F ok, you can see this infimum you can see this infimum as minus ok, minus the supremum of F S E minus AF ok, you just had two minus sign and you exchanged the infimum with the supremum and here it is equal to the supremum of F less than F F and this C is for the proposition that we already prove this is equal to A the infimum of psi larger than F E and psi this is for formal proposition ok, and this is precisely this part is the definition of of the back integral ok ok, then ok, we have to take care of the sum not always within point I, but now we are considering the sum of two measurable functions so we have that if you have that psi 1 and psi 2 are simple functions such that psi 1 is larger than F and psi 2 is larger than G ok, then trivially you have that of course the sum of the two simple functions is larger than the sum of the two F and G ok this is trivial ok, so we start by considering this part we prove as often we prove this equality by showing the two inequality ok, so this is less or equal than E psi 1 plus psi 2 ok, here you know that for simple function the linearity also so you can divide this ok, and then you take the infimum the right hand side you take the infimum and I mean of the right hand side so this is remains what it is and so you get I mean you take the infimum over all the possible psi 1 and psi 2 that satisfies this so what you get is that at least one side is confirmed ok, then we consider the other side which is completely somehow analogous so you start by 2 psi 1 and psi 2 are two simple function ok, this time psi 1 is less than f psi 2 is less than g ok, you have that triviality psi 1 plus psi 2 is less or equal than f1 plus I don't know f and g f plus g ok, now you consider instead of infimum you consider the supreme so you have that f plus g e is larger or equal than e psi 1 plus psi 2 we already prove the linearity for this function so take the supreme on the right hand side ok, this is an abbreviation for right hand side and you get what you want ok, so at least now we prove the step i ok, now ok, let's prove step 2 ok, so we start by 2 function f and g which coincide almost everywhere in e and we want to prove that this is true ok, but in view of step of the previous step this is equivalent to prove that f minus g is equal to 0 ok, because of the linearity ok, so we have that since f minus g is equal to 0 almost everywhere in e ok, then of course if psi is larger or equal than f minus g then we have that ok, psi is larger than 0 almost everywhere in e, which is completely trivial so what follows it follows that psi is larger or equal than 0, which implies that f minus g must be larger or equal to 0 ok, and then you can do the analogous for for the function phi for phi, so if phi is less or equal to f minus g which is ok, phi is less or equal to 0 almost everywhere in e and then you have that less or equal to 0 and then ok, so at the end you get that f minus g is 0 ok, then we have to prove the monotonicity ok, so we start by two function which f is less or equal to g almost everywhere in e, and to prove that the integral of f over a is less or equal to than this integral as before this is equivalent to prove that g is equal to 0 ok, so if we have a function psi which is less or equal to f minus g ok, being minus g less or equal to 0 almost everywhere in e then phi is less or equal to 0 almost everywhere in e and so we have that for what we prove this integral is less or equal to 0 and then you get from this that also the integral of f minus g is less or equal to 0 about the absolute values also observe that somehow is a consequence of this fact ok, we are always within 3 I think ok, so you have that this is always true ok and so this implies that is or minus f and from those 2 you get that ok, then step 4 you have this a and b ok, we start by this hypothesis that a you have constant a which is less or equal to f of x ok for almost everywhere is enough ok, then you have that a times the characteristic function of x is less or equal to f of x but for what we prove for the monotonicity but step 3 you have that a, q, v, x is less or equal to f of x then you use step 1 no, actually no you use just the fact that we know that for simple function you can do this and then you know that this implies that the integral of the characteristic function of v is the measure of v and so you are done ok, for the other inequality is completely similar so the sign for ok so what remains to prove is 5 ok, so we know that a and b are disjoint ok, measurable of course so you have that maybe ok, you can prove this is very easy but the characteristic function of the union of two disjoint set is just the sum of the characteristic function of a plus b ok so compute the back integral over this union and then you know that this is key a union b ok, you consider first you can think that this has integral over e just to be this is equal use this fact to the e f key a plus e this inequality and then for linearity we know that this is equal to the sum of the two and then we can come back to the integral ok, and then we prove all the 5 all the statement ok ok, now we will see ok, we will see now a first example of theorem about convergence under the sign of integral ok, which is one of the main purpose of this course ok, the name is bounded convergence theorem because you will see that we need to have this maybe quite strong a priori hypothesis that our sequence of functions must be uniformly bounded by some constant ok, bounded convergence ok, bounded supposition but I call it bounded convergence theorem ok, so we start of course by a sequence of measurable function f n be a sequence of measurable function we want them to be defined on a set e of finite measure so define it define it upon e set e measure v finite we are always under the same hypothesis and as I told you we assume we require this this bound we assume that there exists a constant m positive such that f n of x is less or equal than m for any n ok so it's a bounded sequence and for all x then we have that if f of x the point-wise limit exists if of x is the limit of f n of x almost every x in e ok, then we can infer a third convergence under the sign of integral or namely we have that the integral of f over e is equal to the limit as n tends to plus infinity of the integral of f n ok ok, we start by just observing that if we have more than than a point-wise convergence for instance we know for some reason that this convergence is stronger if for instance it is a uniform convergence this is somehow automatic it comes at once so you have that if a particular case if f n in addition converges to f uniformly this f n f which is less or equal than for what we prove f n minus f ok, in this you can bound this by epsilon the measure of e and recalling that if this uniform convergence holds you have that for any epsilon positive there exists an index and such that for any n larger than n bar f n of x of x for any x in e ok but we will see that it is not necessary to acquire such a strong convergence this is enough ok, now to prove this we have to use what we prove before if you remember at some point we prove a theorem that has that the point wise convergence is nearly a uniform convergence outside set of small measure I mean the theorem was I got of Severini maybe here we would use the lemma which was stated before but basically the concept is the same ok, so use the fact ok, we proved we proved that we have that for any for any epsilon positive and for any delta positive there exist measurable set a the measure of a is small is less than delta this arbitrary number and there exist an integer n such that for any for any n larger than this integer f n of x minus f of x is less than epsilon for any x which belongs in e minus this set of small measure ok this is now how we can apply this so we want to estimate the difference between these two ok so we take it the absolute values these two difference first we fix some eta positive which will play the role of of an arbitrary small number ok, so as I told you we want to estimate this difference ok this is trivially is less than f n minus f so now how would you how would you go on how would you split this integral e I mean we know that there exist some this set a so we somehow we have to use it ok so we will split this in the idea is that we split this integral yeah in two so you will split this in e minus a f n minus f and here you use that here there is a kind of uniform convergence ok and in the remaining here here you know nothing about convergence but you know that at least a is small ok so somehow you divide these two properties ok f n minus f ok ok, now it's just a matter to arrange this delta epsilon but I mean the idea is the following so if you take for instance if you choose ok if you choose delta the delta there equal it is you can choose any so you take for instance eta divided for m m was the bound uniform bound on f n and then you take epsilon which was equal to eta divided the measure of v times 2 so here is crucial that the measure of v is finite because otherwise you cannot divide ok, so you go on you have that ok, what you have ok, here you serve that f n minus f is less or equal than f n plus f which is equal to 2 times m ok, this is trigger so you have that this term can be estimated as 2 times m no no, this can be estimated in this way this is less or equal than epsilon because we are there times the measure of v ok, you can take the measure of v instead of the measure of v and in this term you use the fact that this difference can be bound by 2 times m so here comes it is crucial to require that the sequence is uniformly bound times the measure of v which in turn can be bound by delta and delta is precisely equal to eta divided by 4 m so if you put all this choice together this should be equal to eta over 2 plus eta over 2 the fact that this is arbitrary small what we found if you want to put that for any eta positive there exist an integer and such that for any n positive plus equal to ok, so we are done so now we will define the integral the big integral for non-negative function so somehow we will impose a further restriction on the sign of the function but we relax the fact that the function in general need not to be bounded ok, so this is the next chapter is the integral of non-negative function ok, so we consider we start function f defined on e measure ok with values this time on the extended real line so be function on e ok, e must be measurable of course this is already included in the definition of measurable functions ok, how we define the integral definition ok, so we define the integral of such an f over e as the supremum over all the function h, which are less or equal than f where h what is this h so h must be interpreted as an approximate of f from below defined on a with values in 0 plus infinity is not included because otherwise we don't know how to compute this bounded measurable good as the main definition must be measurable and the measure of a must be finite ok, so again we have to prove somehow all the machinery, so the linearity and so on ok, so we have that if we start again f and g to measurable function ok, then to no negative of course ok, then if you multiply this function f by a positive constant because we want to remain within the domain of positive function this holds then ok, c positive ok, c0 is puto is trivial and ok, f plus g is equal to the sum of the integral and finally the third one is the monotonicity if if you know that f is less equal than g almost everywhere in e ok, then the inequality is preserved by by the integral you have cf the integral of cf is equal to the supremum of ch which h are those kind of functions cf so h less than f of so you have a, ch so for bounded function you can take this is equal to you can take c outside the supremum and ok, so by definition as then is ok, step 2 ok, one side of step 2 is easy so one inequality is easy the other one is a bit more tricky ok, so start by two functions h and k which are bounded h of x is less or equal than f of x k of x less or equal than g of x where h and k satisfies all the property that we need to define the integral ok, trivially we have that the sum of hk plus k k of x is less or equal than f of x plus g of x ok, so you have h of x plus e k of x is less or equal than ok, no, it's equal to the sum and is less or equal than the supremum ah, yeah, yeah, yeah ok, yes, take defined for instance ok, or rather you can either restrict or just for a second simplicity think that they are zero outside those domains so I can write e ok, I mean definition is the same, ok yes, maybe I should have told you ok, let me let me write a little digression here due to your comments what I, the definition that I gave before f, so equal to the supremum of h minus less or equal than f over h with h which satisfies what I wrote ok, this definition is equivalent to to define this integral has f has, ok, again the supremum of h less than f e, this time I wrote e where over the function h such that h is bounded ok, measurable bounded and such that the measure where h of x is different from zero is finite, ok this is the set a basically if from time to time I write e in the domain just think at this equivalence ok ok now you just less or equal than f plus g and if you take the supremum take the supremum on the right on the left hand side take supremum then you get one of the two inequality ok, so to prove the other we have to work a little bit more ok, so this is one that's the other point ok, so we consider l a function defined on a this notation which is with value in zero plus infinity and not included so it's bounded bounded measurable and with measure of a finite such that we want that this function f of x is less or equal than f of x plus g of x so l of x should be interpreted as a test function that you use to compute the integral of f plus f plus g, ok ok, then we define the following to function and we define ok, we define f of x as the minimum between f of x and l of x and k of x has l of x minus h of x ok, what we want to see now is that h and k are good test function for f and g respectively ok so what immediately we can say that h of x by construction is less than f of x because it's been defined as a minimum ok, and what about k of x ok, for k of x is l of x minus h of x which is equal so you have two option this is l of x minus l of x if l of x is less than f of x and it is equal to l of x minus f of x and here you use this one is less f of x plus g of x minus f of x ok, this is zero of course if l of x on the contrary is larger equal than f of x in any case you end up with the fact that k of x is less or equal than g of x g is non negative also ok, then they are good test function test function good test function both h and k are bounded because they are bounded by the same bound for l ok so h and k bounded by the same bound and non negative and so we can use them as test function ok, so we have that e, the integral of l over e is equal l is the sum of k and h, ok equal to h plus e k which is less or equal than integral of g, because we saw that they are good test function and if you take the supremum on the left hand side we end up with the converse inequality, ok ok, so we still have to prove the monotonicity ok, we start by considering the back integral of f over e this is by definition we use here the definition, the supremum of e of h h less than f and ok call this class of function af this is less or equal than the supremum of e of function h over this time h is less than g and we call this class as hg because f is less or equal than g almost everywhere, ok so we have that this class is contained in this class ok, and so here we just wrote that ok, now we prove we will conclude with the fatulema ok, so you start by a sequence non-negative function f is a sequence non-negative measurable function and we assume that there is pointwise convergence almost everywhere so f converges to some f of x almost everywhere in the domain then we have the following inequality integral of integral of f over e is less or equal than the limit of the integral of you have to read this the thesis as the fact that the limit of the integral does not need to exist ok, so we want to use the bounded convergence theorem so somehow we have to construct some suitable sequence of bounded function which approximates somehow our function, our sequence fn ok, so we consider bounded and measurable function h function h such that you have that of course is non-negative because we are dealing with non-negative function and h is less or equal than f and assume that h vanishes outside a set e prime of finite measure this time I prefer to use in the other definition set e prime of finite measure ok ok, then we want to construct this bounded function called mhn ok, define hn of x as the minimum between h and fn of x ok, first of all you notice that hn are bounded because h is bounded, this is defined as the minimum between a bounded function and something else but h is bounded ok, so we have that ok, let's be right here ok, this sequence hn is bounded by the bound for h and then, ok, we know that hn of x converges to h of x for any x in prime let me ok for a sake of brevity, I will prove the theorem for assuming that the convergence holds everywhere, but of course in the set of measure 0 where it doesn't also doesn't change much set of measure 0, so the integral over set of measure 0 are 0 ok ok, so ok, this is maybe easy to see because we have that for any epsilon positive ok, first fix some x in prime here so for any epsilon positive there exists some n bar and bar depending on epsilon and on x such that f of x minus f of x is less than epsilon, so we want to prove this and no, this is no, sorry, this is true by hypothesis so if we consider this gap between hn and h of x ok this is 0 if n is such that f of x is larger than h of x and this is you have that this is equal hn of x minus h of x in this case you have that hn of x minus h of x is equal to h of x minus f of x which is less than h of x minus f less epsilon minus epsilon for this is for the other case otherwise ok, this is just to say that indeed you have the 0.2x convergence of hn of hn 2h of x ok, so we can apply the boundless convergence theorem to this sequence of function so the bounded what you have, we have that the integral of h over e is equal we are assuming that it vanishes outside the prime so this is h this is equal by the theorem and the limit as for e prime limit of the integral of e prime of hn and this is less or equal than the limits of e of fn ok, this is um most infinity ok, this is because you have that hn are always less or equal than fn and ok, if you take the supremum take the supremum on the left hand side ok, then we we can conclude that the thesis is filled, so you have that here you will have the integral of f over e is less or equal than the here just rewrite and then it tends to plus infinity of fn ok, and so we are done ok, thank you for today we can maybe stop the next time we will we show you some some example for which the strict inequality holds just to convince you that I mean it's really needed to put less or equal it's not an equality in general ok, so for today I think we can stop