 Ok. Ok. So on Thursday we introduced the notion of a measurable function. So we added f from a to r is measurable if the domain a is measurable and for any alpha in our set of points in x. X in a is such that f of x is larger than alpha is measurable as well. Ok. And probably the last proposition that we prove is the following. I just want to recall you. f from a to the extended real number. Ok. a in m, then we have that f is measurable if and only if two facts holds. We have that for any set b in r, which is open. f minus b belongs to the sigma algebra of measurable set. And then we have that the pre-image of plus infinity and the pre-image of minus infinity is again a measurable set. Ok. So now the purpose now is to say to check if we can say something more about this property. If we can extend this property to more general set then only open. For instance, for bore set if it holds. So the answer is yes. And indeed we prove the following theorem. So what about if b belongs to the sigma algebra of bore set. Ok. So again you start by take a a measurable set. So the domain must be measurable. Then we have that f. We have a measurable function. And then we can show that indeed for any bore set of course of the extended real line we have that the pre-image and the rest of a measurable function of this bore set is again a measurable set. Ok. So I won't just to stress remind you but of course you know that it is the class of the bore set is a sigma algebra. And indeed we will strongly use this fact because the proof of this theorem is not direct. It's not direct. So we will argue in this way. So we define a collection a for which this property holds. Ok. So we define some for instance italic a the set of the set of Nr such that the pre-image of a is measurable. Ok. Now we want to prove that this is a sigma algebra. Ok. So we want to prove. Ok. And then we will see that once we prove this then we are done. Ok. Ok. Ok. So we have that the empty set belongs to a because the pre-image of the empty set is the empty set as well which is measurable. Ok. Then we have that. Assume that you have a set a which belongs to this italic a. And then what about a complement. So we have that a complement still belongs to a because the pre-image of a complement is the pre-image of a which by assumption belongs to is a measurable set complement. So in the end we still end up with a measurable set. And then what remains to prove is to consider a countable collection of set in this family. Ok. So start if we have that a n belongs to a then indeed a union of the countable union of a n still belongs to a since you have that pre-image of that union can be expressed as the countable union of the pre-image of a n which is by definition measurable, by assumption measurable and we have that also the union is measurable. So finally we get from these three property that a is a sigma algebra and then we shall see that the sigma algebra of the Borel set is contained here. Ok. This comes from what we prove on Thursday. So we observe that. Ok. This comes from the fact that maybe you can see this because you have that the family of the open set is contained in a. Ok. By the former proposition. And so also the sigma algebra generated by the open set must be contained in a because it's the smallest one you know. So because you have that the sigma algebra generated by the open set is contained in a and this indeed coincide. So this is by definition the sigma algebra of the Borel set. So finally. So what you have that if a is in Br then you have that a belongs to a and so the preimage under f of a is measurable. Ok. And so we are done. Ok. So this is why I told you that the proof is not direct. Ok. Ok. Now we will introduce another notion which is completely somehow analogous and similar to the notion of measurable function. Ok. Here is this part. So I introduced the notion of Borel function. I don't want to spend too much time on this because as I told you it's completely analogous with what we did with the measurable function. Ok. So basically you start with a Borel set B. Ok. And you consider f, a function which is defined, which has a domain, this Borel set with values in the extended real line. We will say that is a Borel function if you have an analogy for any alpha in R. The set is kind of f of x and sorry larger than alpha is a Borel set. Ok. So basically you can somehow repeat all what we did for measurable function. For instance the fact that the sum of two measurable functions is a measurable function replaces replacing the term measurable with Borel function. So the sum of two Borel function would be a Borel function and so on. Ok. Ok. So repeat, you can repeat what has been done by replacing measurable set with Borel set and measurable function with Borel function. Ok. And for instance in particular you can also prove this theorem. If you have f function from a Borel set to the extended real line, assume that the term is a Borel function, then for any Borel set B tilde you have that the preimage under f of this Borel set is a Borel set. Ok. Ok. Now the question is, so what about if we compose a measurable function with a Borel function? What we obtain? And moreover, is it important the way that we compose the order of composition? Ok. So this corollary partly answer to this question, so you have that. Ok. So if you have that f, assume that f is the Borel function. So it's defined on B, which is a Borel set with values in R. This is a Borel function and we have G from A to R. This is a measurable function. Ok. Of course we have that B if a Borel set and A is a measurable set. Ok. So if we... Ok. We have also to assume that G of A, the set of image of G is contained in B. Ok. From this you already understand where we compose, so then we have that f composed G is measurable. Ok. So how can we prove this? We take some alpha, some test number if you want. So you have to check if the pre-image of this composition of a set of the type alpha plus infinity, for instance, if it is measurable. Ok. Ok. So by the law composition you have that this is equal to G to pre-image under G of pre-image of alpha plus infinity. Ok. So this is a Borel function f. So this is a Borel set. Ok. This belongs to... This is a Borel set. This is a measurable function. So the pre-image of a Borel set under a measurable function is a measurable set. Ok. And so we are done. Ok. This is a Borel function f. No. By definition of Borel function you have that this set is a Borel set. Ok. So you have that this set here, the pre-image of... This is a Borel set. So you take the pre-image of this Borel set under G, which is measurable. Ok. And so you end up with the measurable set. Ok. Ok. Then of course now the natural question is what if we consider the other way to compose them. Did we end up still with the measurable function or not? What do you think? So the answer is no. Ok. In order to construct a counter example we need just to introduce, to show something else somehow. And indeed now we also see that so far we saw that the set of parts of R strictly contains the sigma algebra of the measurable set. Because we saw that there exists a set that we call T, which is not measurable. But some lecture ago I claimed that also this inclusion is true. So that there exists a measurable set, which is not a Borel set. Ok. We need to construct this before proving a counter example of what I said before. Ok. So somehow this is related with this question. So is the pre-image a measurable set under a measurable function, a measurable set? No. The answer is no. And then we will see that even if you choose a more somehow regular function, even if you substitute here a continuous function, you can find a counter example that the pre-image under a continuous function of a measurable set, sorry, a continuous function here, a measurable set is not a measurable set. So to do this we consider the homomorphism that we introduced. The homomorphism H, you remember, H from 0, 1 to 0, 2. So do you remember how it was defined? In a very easy way. Exactly. So you take X and you consider X plus F of X. F is the counter's function. Ok. I just recall you some fact that we already proved. We had that the measure of H is equal to 1, where K is the counter's set. Ok. And then we also proved that, I think we proved it, that we proved that any set with positive outer measure, in this case, with positive measure, any set with positive outer, I put it in the bracket, because in this case it's not important, contains a nonmeasureable set. Ok. Ok. So in this case we can claim that there exists a set A, which is not measurable, and such that A is contained in H of K. Ok. Now we define, so let B define this set B as preimage under H of A. Ok. This is, by construction, this is contained in the counter set. So it must have measure 0. Ok. And so in particular this means that it is a measurable set. Ok. Now we consider the inverse of H, which is still a continuous function, because H is a nonmeomorphism. Ok. So consider H minus 1. So now with H minus 1 it denotes a function, not only the preimage. Ok. So it goes from 0, 2 and 0, 1 is continuous. Ok. In particular measurable. Ok. So now consider the preimage of H minus 1 of B, of the set B. This is A, which is not measurable. Ok. So somehow we construct the desired counter example. Ok. Because we know that this is a continuous function for which the preimage of a measurable set B, no, we show that B is measurable, is not a measurable set. Ok. Is it clear? Ok. Now from this, are you fine with this? Ok. Now we see that there exists a measurable set, which is not about a set. Ok. So we will use this construction somehow. Ok. Ok. Ok. Let B, the set introduced before. So we have the measure of B. Ok. Now assume that B is also, B is measurable, of course. And assume, we argue by contradiction, assume that B is also a borrowed set. Ok. Assume B is also a borrowed set. So from this, ok, if B would be a borrowed set, then we have that then. So the preimage under a measurable function H minus 1 is A. So what is the contradiction? So A is not measurable, and so we get a contradiction. Ok. Ok. So with this corollary we show that indeed also this inclusion is strict. Ok. And now we come back to our question about the composition between borrow and measurable function. So let G continues measurable. So is F composed G measurable? Ok. No. Of course no. So about how to prove this. One way to prove this is the following. You can exhibit a counter example. So you fix G to be equal to H minus 1. And so I keep the notation that I used before. And F is the characteristic function of B. Ok. So this is continuous and this is measurable. And then we check if F composition with G to minus 1. So somehow if the level set of this composition is a measurable set. Ok. This is what? This is H. Ok. Then you substitute our choice now H minus 1, minus 1 of key B. Ok. Sorry, I don't. Is there a result in A? A, I don't know. Where? Is there a result in A? Here? Here. Ah, you are answering. Ah, ok. So you have. Ok. The final, yes, sure. I am going by step, yeah, yeah, but sure. Ok, this is A, which is, as you said, is not measurable. So we prove. So we show that the level set of this composition will lead to a nonmeasureable set. Ok. Now we change a bit argument. Now we don't change argument but we prove some other consequence of the definition of measurable function. Ok. We consider a sequence of measurable function and we want to say something about the supremum, the limit, the lim soup. Ok. So let fn be a sequence of measurable function with the same domain. Ok. Then, then the function. Allor. So we have the supremum of a finite number f1 fn, the infimum f1 fn. And then the soup fn fn. And then the lim soup fn. Ah, ok. No. Ok. You can. Ok. Just a note like this. Or either you can also use this notation for the lim soup, lim with the bar at the top. And here you can use lim with bar on the bottom. Ok. Ok. Bottom is a measurable, a roll measurable. Ok. Ok. Ok. The proof is simple somehow. Ok. I start directly from here. Ok. So the soup of fn less or equal than alpha is equal to the x, such that for any n integer, f of n is less than alpha. And this is equal to the intersection of n of this set of the type fn is less or equal than alpha. Then what about the inf? You can express the inf in terms of the soup. So you have that. Ok. Of course, these are all measurable now. Ok. Then you have that infimum over fn over n can be viewed as the minus, the supremum over n minus fn. Ok. What about lim soup and lim inf? I recall you the definition of these two limits. You have that soup over fn is what? Is a combination somehow of the infimum and the supremum. Soup of k larger or equal to n of fk. And by the previous, and this is measurable. It's measurable, because of the step before. And in analogy, what it means is you exchange the role here of the supremum of n of the infimum of k larger than n of fk. Mejorable. Ok. Ok. Ok. So now we have to introduce the concept of somehow almost everywhere. Probably you already encountered this notion. So do you know what it means that the property is satisfied almost everywhere? What does it mean? Don't know. Stane. Satisfied. Exakt. Ok. So I just introduce it properly, because we will use it a lot. So p of x, property p of x is satisfied almost everywhere. If we have that measure of the set of the x in e, such that p of x is false, is not satisfied, is zero. Ok. So in particular, just to give you some example, you can say that, for instance, you have that to function f and g are equal almost everywhere. Ok. If f and g have the same domain and the measure of the set where they differ as measure zero. Ok. Another example is that f of n converts to g almost everywhere if and only if. I mean point-wise convergence, of course. If the limit of fn is equal to g of x, as that goes to plus infinity, for any x in col e, the domain of definition, minus f where the measure of f. Ok. Ok. Then we want to prove this proposition. Ok. You start by a measurable function. So if f is a measurable function, which coincide almost everywhere with another function g. So what can we say about g in terms of measurability, ok? And assume that f is equal to g almost everywhere. So then, of course, we have that also g is measurable, ok. So how is the idea here to prove? So basically we have to study, for instance, to say something about the measurability of this set here, ok. So the idea here is to split this set into set. Ok. So the note with e, the set of x, such that f of x is different from g of x. And by assumption we have that the measure of e is zero. Ok. And then we can express this set here as the union of two sets. Ok. Has the set, assume that f and g are defining some set a. Ok. This is the common domain. Ok. X in a minus e, where g of x is larger than alpha. Union x in e, where g of x is larger than alpha. Ok. So we have that. Here we can replace g with f. Ok. Larger than alpha. Measureable set. This is measurable because it is measure zero. So finally we have that. All of this is measurable. So just to be precise you can look at it as, you can still split this as the x in a, such that f of x is larger than alpha, which is measurable minus e. But of course it is. Ok. Now we try to compose measurable function because we will need this somehow. Easy notion of composition between measurable function in order to define the Lebesgue integral. Ok. Ok. So we will say that f from e to r is a simple function. Ok. Simple function if it is measurable, f is measurable and if it takes just a finite number of values. Ok. This is a definition. But what we need actually is this equivalence provided by this remark, is that it tells you that f is simple if and only if there exists a finite number A i measurable. Ok. And corresponding values c i and r. So take i, which goes from 1 to n. Ok. Such that you can express f of x, so is what you expect, as the sum from i, which goes from 1 to n, c i key a i of x. Ok. So it is a linear combination of characteristic function of measurable set. Ok. So let's prove this. Ok. Ok. So gesture from this. Ok. This fact is easy because, ok, f is measurable. Ok. It's measurable because it's a linear combination. It is a linear combination of measurable function. Ok. Linear combination of measurable function. Ok. And then the fact that f takes only finite number of values, you can see this because it can take the values that it can take, is, ok, c1, cn. Then if, for instance, in principle these are not disjoints, so they can overlap, so you can also have this situation. You can have c1 plus c2, c1 plus c3, and you can have actually all the possible sum, but still they are finite. Ok. Takes, and takes only those, can take only those values. Ok. Ok. So the reverse implication, we have a simple function, and we want to prove that we can express it in that way. Ok. Ok. We know that f is simple, so it takes only finite number of values, so we can order them. So just, so, since f is simple, we have that, the codomine of f is, for instance, b1, b2, bn. Ok. Fine. And then we consider the preimage under f of these values, and we define it as bi. Bi would be the preimage under f of this single point. Preimage, sorry. Ok. In particular, these bi are also disjoint. Ok. These bi are disjoint. The reunion gives you the domain. Ok. B i gives you the domain is e. Ok. And so this tells you basically that you can express f of x as the union of bi key bi of x. Ok. Ok. Then we want to somehow consider some particular class of simple function that we shall call step function. Ok. So step function are the simple function for which, in this representation bi are in particular intervals. Ok. Ok. So given h, we say that h is a step function. Ok. If it is a simple function, and it admits the following representation. Ok. With the following representation, we have that h of x is equal to a i key i i x into n, where i i are in intervals. Ok. So then consider the following lemma. So now the aim will be to somehow to use this step function or the simple function to approximate measurable function. Ok. So the lemma is the following. It tells you that f from ab to r, r bar actually, is a non-negative measurable function. Ok. Then we can construct or exist sequence, an increasing sequence, simple function, or call them fn, such that for any x in ab, we have that limit of fn of x is equal to f of x. Ok. The point wise limit of course. Ok. So we have to construct this increasing sequence of simple function. Ok. One way to do this is to define fn in this way, fn of x as, for instance, a function which takes the value i minus 1 divided by 2n for the point such that f of x is in between i minus 1 to n, so the same values, and i divided to n for, ok, for some i among 1 and n. Ok. n be equal to n if f of x is larger or equal to n. Ok. So here if you substitute the last values of i here, you get n. Ok. So this is the limit somehow of that. Ok. These are simple function. Ok. These are simple function because they take only finite values, finite number of values and moreover they are measurable because this set here, the set of x for which this satisfies are measurable because f is measurable, ok. So fn simple function because the set takes constant values are measurable because f is measurable. Ok. So then what about the convergence? Ok. The easy case is that if f of x in the point for which f of x is equal to plus infinity, then it's true, ok. Then you have that f of n of x are equal to n and as we have convergence. Then if f of x is a point, so if x is a point for which f of x is finite, ok, then by construction you have that fn of x minus f of x is less or equal to zero and moreover you can check it. Moreover you can say that you can see that there exist. So this is somehow a bound from below and moreover there exist, now we check the bound from above and there is some n such that for any n larger than n bar you have that f of x minus fn of x is less than 2 minus n. So basically put them together so you have that f of x minus fn of x is less or equal to minus n. So you have convergence and then by construction they are an increasing sequence, ok. n times 2 to the power n, ok. And then you have that fn is less or equal to n plus 1. This is by construction, ok. And then we have done. Ok, now this is a very easy remark. If you remember here we start by with f which is a measurable function and non-negative, ok. If we remove this hypothesis on the sign we can still say something about a possible approximation of this f. Ok, so how can we argue? So this is, if you want, put the remark. Ok, so if you have f, let f be a measurable function, ok. Then you can construct the sequence of simple function, ok. Such that these times you have that the absolute values of fn is increasing, ok. And you have that the limit of fn of x is equal to f of x, ok. And, ok. This is quite easy to see. So how would you argue? I mean, we prove in the lemma before we prove that if you start by a function which is measurable and non-negative, you can construct in the way that we saw a sequence of simple functions which converge to f. Now somehow we just remove the fact the hypothesis on the sign so f might change sign. How is usually the trick to, yeah, exactly. So you can split f as the positive part minus the negative part and apply the proposition before separately for the two, ok. Ok, so you know that f can be decomposed in terms of the positive and the negative part. So both non-negative functions, ok. f plus and f minus are both non-negative. Ok, just briefly recall you what is the definition. So you have that f plus is equal to f of x when f of x is larger or equal to zero and zero otherwise. While f minus of x the negative part is equal to minus f of x when f of x is negative so you still obtain a positive function non-negative function and zero otherwise, ok. And then you apply apply the theorem the lemma before twice before two f plus and f minus ok, so then this sequence would be then would be f n plus of course minus f n minus, ok. Ok, now just remark really quick remark you have that if ok, if f f goes from a to r is measurable and finite finite ok, then you can say a little bit more about this kind of convergence you have that then f n converts to f uniformly uniformly because if you remember at some point we show that f n of x was less than f of x was less than 2 minus n so basically you have uniform convergence if they are finite ok and then I just state probably this theorem that we will see the proof tomorrow ok, you have let f be a measurable function ok, define it on an interval ab and assume that it is finite almost everywhere and with this I mean that it takes that f takes the values the values plus minus infinity only on a set at most on a set of measure 0 only on a set ok, then how can we approximate this measurable function in terms of easier one so we can say that given any epsilon positive arbitrary small we can find a step function step function ok, such that we have this, somehow this new notion of approximation such that step function call it h from ab to r ok, such that the measure of the point x in ab for which these two functions the difference between these two functions is larger than epsilon this measure is less than epsilon ok, and then we can say we can say even more we can perform this kind of approximation in terms of not only step function but continuous function we can say the same and moreover ok, moreover there exist continuous function g from ab to r such that the same is true here actually such that the measure of x ab of f of x minus g of x is larger or equal than epsilon is less than epsilon ok also this statement must be read that given epsilon you can find this continuous function because in principle the continuous function of course ok, the proof is very long and just before concluding I will just prove some lemma that will be useful to prove this proposition, this theorem ok, ok, you consider three function f g and h defined on the interval with values in the extended real line all the three function are all measurable measurable ok, then we would like to to achieve a kind of triangle inequality or something that that looks like and ok, and let epsilon be positive number then we define this set a, which is the set where f of x minus g of x is larger than epsilon larger or equal then you define b as the set where f of x minus h of x is larger than epsilon over 2 and c this set where h and g h of x minus g of x are larger or equal than epsilon over 2k ok, so what we want to provide is that so what can you say can we say about the measure of a intuitively then all right, it is measurable for sure because they are all measurable but I mean, how can you estimate the measure of a in terms of the measure of b and c they are less equal greater or equal what? the measure of a is less or exactly ok, then the measure of a is less or equal then the measure of b plus ok, the measure of c ok somehow it is enough to prove that a, if we prove this we are done ok, so we have that if x so we want to prove this ok if x is in a so we know that by definition that e e no, this is ok see, this is a proof here start the proof ok, this is the claim this is the thesis and now we are starting to prove we observe that if we prove this it is enough, ok ok, so we know that if x is in a then by definition f of x minus g of x is larger or equal than epsilon ok and now we want to prove that this x must belong at least one of these two because if doesn't belong to later of them we get a contradiction, ok so we have that f of x minus g of x is less or equal than this is just the triangle inequality h of x plus h of x minus g of x and we have that ok, this is by definition so we have that this sum is larger or equal than epsilon, ok ok, so we cannot have that both of them are less than epsilon over 2 ok, so at least one of these must be must be larger than epsilon over 2 so we have that at least one of these two terms f of x minus h of x or the other h of x minus g of x must be larger than epsilon over 2 in this, in other terms means that x must belong to the union to be union c because it must belong either to b or to c and so this is done so we conclude the program ok, now as we stop here ok, we can stop