 Zgledaj. Tako očasno, naredaj smo počasno odvihčili vseh, da je zavrčil posponil vseh odvihčili. Zato priberimo, da je to priberila. Zatimamo, da je zatečnjamo zbičnjak vzelo in v 0-1 vzelo v 0-1 vzelo, da je zato vzelo. Ok, skupnja. Ok, f is continuous is increasing. Oh, non decreasing actually. Ok, it's continuous because we saw that it is a uniform limit of continuous function. Ok, we saw that f is 0 is 0, f of 1 is 1, and we saw that it is, by definition, it is constant on the interval, on the open interval that we remove in the construction of the counter function, ok? On the counter section. Ok, now by means of the counter function we can construct anemomorphism between 0, 1 and 0, 2. Ok, so let us consider h, we define it by h, the following function between 0, 1 and 0, 2, such that 2x associate x plus f of x, where f is the counter function, ok? Ok, so we immediately saw that h is continuous because to continuous function is strictly increasing, ok? So h is continuous, strictly increasing, and so in other words h is anemomorphism between 0, 1 and 0, 2, ok? Ok, so somehow the final purpose of this first part of the lecture would be to see that the notion of measure is not preserved under anemomorphism, ok? Ok, so we will see that h can map a set of measure, a measure 0 into a set which has a positive measure, so the notion of measure is not preserved under anemomorphism, we will need to require more in order to preserve a measure. Ok, so consider, ok. Consider k is the counter set, the counter set. Ok, then we have that the complement of the counter set is the union of this opening interval that we removed in the construction, ok? So is the union just to recall this joint interval. Ok, ok, we know that. As I recall you before, we know that f, the counter function is constant on each such an interval, ok? Ok, so by construction the counter function is constant on each i n, and so, what about? We have that, this means that for any x in this open this joint interval i n, we have that h of x, and for omeomorphism what looks like as x plus some constant c n, ok? Of course the constant depends on the interval. Ok, so what is the image under h of i n? Ok, so we have that h i n would be i n plus c n. We have that the measure of h i n is the measure of i n, ok? Because we saw that the measure is invariant under translation. Ok, ok, just let me call this equality star. Ok, we know that since h is omeomorphism and so it's 1 to 1, we have that, we have that the image of the union of this joint set will be the sum, the union of the image of the set. Ok, we have that. The union, the image of the joint interval is the union of the images of the interval, ok? Of the images, and they remain this joint, ok? And they remain. Ok, so you have that. Ok, call this complement a. So you have that h of a is equal to the union, countable union of h of i n, ok? Ok, now we want to compute the measure of h of a. So we have that the measure of h of a is, we use this, is the measure of the union of h i n, which is equal, because we saw that they are disjoint, it's the sum of the measure of h i n. Ok, this is because of this remark, and this is because of star. Ok, so this is equal to the sum of the measure of i n, which is equal to the measure of a, that we saw on Tuesday that this is equal to 1, ok? Ok, so again we use the fact that h is 1 to 1, it's a normal morphism, so we have that. Ok, again, so since h is a morphism, ok, the complement, ok, the complement of the image is the image of the complement. Ok, so in particular we have that, we consider h of k, where k is the contour set, this will be equal to 0,2 minus h of a. Ok, so we compute h of k, the measure of h of k is equal to 2 minus 1, which is equal to 1, and so we saw that finally what we prove is that h map a set of measure 0, which is the contour set, ok, into a set of measure 1. So basically this is to say that the notion of measure, of the bag of measures in this way is not topological invariant, ok? So it's not preserved under homomorphism. Ok, now we start with another argument, with a new argument. Ok, so the next argument will be the measurable function, ok? Can you see? Ok, so before introducing the definition, I would like just to anticipate you that when we will introduce the notion of the bag integral, we will have to deal with the set of, sorry, the bag integral for function f, for instance, we will have to deal with some function f, put like this. So we will have to deal with this kind of set, ok? And so somehow we would like to know, to classify, to characterize the function f for which this kind of set is measurable, ok? Otherwise it won't make sense to compute the measure of this kind of set. Ok, so from this somehow remark, we define what is a measurable function, ok? So consider a set A, which is a subset of R, of course, and the function f defined on A, which values in the extended R, so which is the set of the extended real number, so you have R if you want union minus infinity and plus infinity. So we will say that this function f is measurable if two facts are fulfilled, so is measurable, ok, right here, if the domain R is a measurable set, is measurable to I, if you have that for any alpha in R, the set f larger than alpha is measurable, ok? So two, somehow there are two requirements. Ok, so from this you can see that if this set is measurable, also there are many other set that we can infer are measurable, so we see the following proposition. So let f be defined from A to R be a measurable function, and, ok, so of course it's already included in the definition, but just to stress that we already know that A is in R, ok? I mean, ok, this is the set of the x in A, the domain, such that f of x is larger than alpha, ok? Ok, in the following for sure I will denote this set f larger than alpha, ok? Just to be, but, ok. Ok, so we have that, the following, if you start from this, we have that the following statement are equivalent, ok? Ok, then the following statement are equivalent, ok? So we have I for any alpha in R, ok, we start by this, we have that, ok, x in A such that f of x is larger than alpha is measurable, ok? Then we have 2I, ok? So again for any alpha in R, the set of the x in A such that f of x is larger or equal to alpha is measurable, and then we consider the other inequalities, 3I for any alpha in R, the set of x in A such that f of x is less, strictly less than alpha, it's measurable, and finally, for any alpha in R, the set of the x in A such that f of x is less equal than alpha is measurable, ok? Ok, so we start by proving that I implies 4, which is the easy case, it's very easy sometimes, because just you consider the complement and you are done. I don't get your, in the definition you mean, the proposition. So we start by f be a measure of a function, which means that it satisfies these two, it's a function from A to the extended real number, and by definition we know that A is measurable, I write it again, but it was not, no, no, but we are under the hypothesis that f is measurable. Ah, I mean, this is already, ok, we can remove it, ok, or, this is already included, ok, I know what you mean. Ok, let be a function, ok, thank you, yeah, sure. So let be a function, ok, we require this, because if we don't require it, it's measurable, we have to require it, yeah, sure, I agree with you. Then the following statements are equivalent, ok, so if you read this I mean that it is measurable. Ok, yeah, sure. Ok, so, ok, so we start by proving that I implies the last one, ok, so we can just see the set, ok, x in A, such that f of x is less or equal to alpha as a minus the set x in A as f of x is larger than alpha. Ok, so we know that this is measurable and the complement is measurable as well. Ok, then ok, for i is implies i is similar, you can argue similarly, now it's equivalent by exercise if you want. And then we prove ok, 2i implies 3i again, so you just have the f of x less than alpha a minus f of x larger than alpha. Ok, and then 3i implies 2i it's a similar, it's measurable so the complement is must also be measurable and so it remains to prove that i for instance i implies 2i ok, we can write this set here as the intersection for instance so you have x in A such that f of x is larger or equal than alpha has the countable intersection of x in A such that f of x is strictly larger than alpha minus 1 over n, ok? Ok, so we know that these are measurable and the countable intersection of measurable set is still measurable ok, so this concludes the proof ok, so ok, now we want to deal with the level set of measurable function ok and we prove the following ok, so let f from i to r a measurable function ok, then we have that for any alpha belonging to the extended real line so it might be also plus or minus infinity we have that ok, the level set of f so the point in A such that f of x is equal to alpha is also measurable ok, we divide two cases the cases in which alpha belongs to to r so this is finite so you can see it as an intersection of two former set ok, so you have that this case x in A such that f of x is equal to alpha you can see it as the x in A such that f of x is larger or equal than alpha intersected the set x in A such that f of x is less less or equal than alpha ok ok, then the case when alpha is for instance plus infinity you need to do countable intersection so if alpha for instance is equal to plus infinity so you have that avoid the x in A such that f of x is equal to plus infinity intersection over n of the set x in A such that f of x is larger or equal than n ok in infinity you intersect the same so you have f of x less or equal to minus n ok, so this concludes the proof ok, now we give definition of measurable function but we would like also to somehow to see some example of how a measurable function look like ok so think about a continuous function so some example of measurable functions ok, so consider f from A to in n the domain is a measurable set and assume that f is continuous ok and see that f is measurable so do you see you know why the preimage of the set like alpha plus infinity is an open set so the preimage is open so you take the intersection with A it will be open the relative topology so you have actually an intersection of two measurable set which is measurable ok, so just you have that so we want to establish that it is a measurable set you can see it as the preimage under f of alpha plus infinity ok and this is open this is open so the preimage is open is an open set in the relative topology and it is open set of A if you want which means that it exists in U op and I said U open of R such that this set can be view as U intersected A U is open so it is a borrow set we saw that borrow set are measurable set and this is this is measurable so we get that also this set is measurable ok, so this is one example of measurable function then you can construct also another example you start by a set A in R and ok, I don't know if you know what is the characteristic function of A is a function from R to 01 ok, you can take for instance 2 to the full R so this is called characteristic function of the set A and it is defined in this way so you have that set A of X is equal to 1 if X is in A 0 otherwise so actually it takes 2 values ok, so we can show that this kind of function is measurable if and not if the set A is measurable ok so we have that then it's a kind of characterization A is measurable if and only if key of A is measurable as a function is measurable ok, let's prove this ok, we start by assuming that key of A is measurable if key of A is measurable ok, we have that we can express the set A has the set for which key of A is equal to 1 so it's a level set ok, this is always holds it doesn't if key is measurable, we have that go inside with the set for which which X in R if you want key of A of X is equal to 1, so this is a level set of a measurable function so this means that that it is measurable, ok this is a level set of a measurable function ok, the covers assume that A is measurable and we want to key of A is measurable so we have to study the set of the type key of A is larger than some alpha ok, so ok, we start by just assuming that the domain adjust a raise, but the domain of key of A was R because this is the check is that the domain is measurable so the domain of key of A is R and R is measurable, ok ok then we fix we pick some alpha in R take some alpha in R and so let us study this kind of set the X in R such that key of A of X is larger than alpha ok, then we have that this set would be actually three, somehow three cases three possibility ok, this would be the empty set if alpha is larger or equal than one ok, because it's never it's never greater than one, ok then it would be then it would be it would coincide with A if alpha is less than one and greater or equal than zero ok, and would be R if in the remaining case, if alpha is larger no, sorry, is less than zero ok, it's negative ok, any case we end up with the measurable set in any case in any case excuse me I mean, if you put here a number which is less than zero this is always it's always true, ok ok, yeah, because I have a the strict inequality ok, in any case we end up in any case means either empty set A or R we end up with with a measurable set ok, now we we study the operation between measurable functions the sum, the product ok, so ok, so let we take two function f and g from A to R ok, for the moment we just assume that the codomain is just R ok, not the extended real line line, so then we will we will study also the general case ok, R measurable function and let's see a constant point in R ok, then we have that also the following the function f plus c c times f f plus g f minus g and f times g R are also measurable ok, so if we perform this operation we still remains in the class of measurable function ok, so we start by f plus c ok, this is f plus c this is f plus c larger than alpha pick some alpha now ok, this is of course is equal to f larger than alpha minus c so, we are done c times f ok, we ok, c times f larger than alpha we can distinguish three cases according to the sign of c ok, for example if c is positive then we have that c times f larger than alpha is equal to f larger than alpha divided by c if c is equal to zero then you have that zero is larger than alpha so, also in this case you have to distinguish some cases, of course the result will depend on how alpha looks like ok, so it would be all a actually, no, you have three and two cases, of course this is all a if alpha is less than zero and will be the empty set if alpha is larger or equal to zero in any case, you yeah yes yes, but f is continuous ah, a zero, ok yeah you can see us in that way yeah ok, just let me finish in this way c is less than zero ok, so you observe that if c is zero the function is constantly equal to zero so it is continuous and we prove before that a continuous function is measurable ok, this is zero less than no, sorry, c times f is ok, larger than alpha, so in this case inequality ok, this case, inequality is reverse so you have that f is less than alpha divided by c ok, and then we have to study the case ok, f plus g ok, the case f plus g ok, we have to work a bit ok, so consider f plus ok, we again we pick some alpha in r we have that f plus g is larger than alpha and call this set b, for instance, ok ok so now since to deal with these cases we will need to to use countable operation between set and somehow to use countable operation we have to use the fact that the rational number are dense in alpha ok, so you have that if, just observe that if f of x plus g of x is larger than alpha ok, then this is trivial f of x actually ok, let me deal with the other set I mean, it's the same it's just for for a sake of convenience, but of course it's the same, so f of x is less than alpha minus g of x ok then, as I told you, we observe that since q is dense in r, we have that this is equivalent then there exists there exists a rational number such that f of x is less than r and less than alpha minus g of x so you have the strict inequality so here you have room to put a rational number ok ok, I call call this inequality r this property r and so what we want to use is the fact that if x belong if x belongs to b then x satisfies satisfies this and also the converse is true ok ok, this is to say that we can characterize this set b ok, this means that that we can characterize the set the set b where f plus g is less than alpha has the countable union over the rational number of these two the countable union of an intersection of x ok, x such that f of x is less than r intersected with the set of the x where f of x no, where g of x g of x is larger no, g of x is minus, so g of x is smaller than alpha minus r ok, so this is these are nice set because this is measurable this is also measurable because we know that f and g are measurable so the intersection are also measurable and now we are dealing just with countable union, so all this stuff will be will be measurable ok, and then ok, f minus g is measurable just comes from combining this step here and the fact that if you multiply for a constant you still obtain a measurable function so just to briefly of that f minus g, you can see it has ok, f plus minus 1 times g no, we are done ok, so what about f times g ok, we we want to express this in we want to express this in this way 1 over 2 of f plus g square minus f square minus g square ok, but we still don't know if this kind of if you take the power 2 of a measurable function is measurable so if we prove this, we are done because then we already prove that f plus g is measurable ok, so consider for instance f square, so what about f square ok, and consider the set f square larger than alpha ok, if you have that if alpha is larger or equal than zero f square larger than alpha can be split as the union of two of this set here larger than the square root of alpha union f of x less this ok, these are two are measurable ok, so the other case if alpha is less than zero this is always true ok, we have that f square larger than alpha is equal to the domain in any case, we know that the domain is measurable ok, so once that we check that if you take the power 2 of a measurable function you still obtain a measurable function then we are sure that while doing this operation we remain within the class of measurable functions so we show that also the product of two measurable functions is a measurable function so I think we span all the cases ok and then we would like to see what can we say about this operation if the co-domain of our measurable function is the full extended real number ok what about if if we have f and g from a to r bar ok so means r union minus infinity and plus infinity ok, we start by this case so when you have s time f ok, somehow this case, the nasty case is when with c equal to 1 and so if you want the interesting case and f is equal to plus and minus infinity ok, in this case there is a convention measure theory in this case in this case to assign this value 0 times plus infinity is equal to 0 which is equal to 0 times minus infinity ok, of course it doesn't always mean in the limit so it doesn't mean that ok, it doesn't mean not mean that if you have if you have a sequence if you have two sequences if you have sequences so there is a cn which tends to 0 and fn which tends to plus infinity in general you cannot say that cn tends to 0 so this is not true so this is just a convention we have precise values c times f ok, what about when you when you have to to consider the sum between two measurable functions in the extended real line ok, so for instance ok, you have plus g so again we use a convention ok, we use we use this convention ok so we have that f plus g is equal to sum constant c fixed when you have the four which are indeterminated, for instance when you end up with this minus infinity plus infinity or the reverse ok, so you put it and c of course when you have plus infinity you put it as plus infinity which is otherwise as I said you put just f plus g ok, otherwise ok, so what about so we can decompose a the domain of this function as the union of two set so somehow there are finite number of possibility for so you have a you see a the domain as a I call it a finite union a infinity ok, so a finite is the finite in this way is the finite union of this set no, actually it's an intersection but is this set is the x in a such that both function are finite ok g of x in r and you can see this has the x in a such that f of x is in r intersection x in a such that g of x belongs to r ok, so why this for instance this is a measurable this is a measurable set because you in turn you can express this has the complement of the set where f of x is equal infinity here I mean plus and minus infinity the complement of the set and we saw that is a measurable set intersection of x in a of g equal to infinity complement, ok, this is measurable as well so the set a finite is measurable what about the set a infinity we will define it in as follows so you have all somehow the other cases f is equal to plus infinity g is equal to plus infinity call this a prime infinity if you want a1 union f is equal to plus infinity g is equal to minus infinity union f is equal to minus infinity g is equal to plus infinity, there are 8 sets so then union f equal to minus infinity and g equal to minus infinity union ok, you have other f equal to plus infinity in r no and maybe you understand how the things are going on actually there are 8 sets then you have f is equal to minus infinity g in r union f in r and g equal to plus infinity union f in r and g equal to minus infinity so there are 8 sets and we call them a2 infinity a3 infinity a4 infinity a5 infinity a6 infinity and a7 infinity ok so when we have to say if our set of the type f plus g larger than alpha whether it's measurable or not we have just to consider a finite a finite number of sets so when you fix alpha in r and you and you study this set f plus g is larger than alpha you can split this in in such a way these are of course x in a x in a finite f plus g larger than alpha union ok union x in a infinity f plus g larger than alpha for this case we can deal as before when we are in the finite case in this case we can deal we can deal as before so when actually we are in this case with r ok we can argue as before more than argue we can directly apply what we proved before ok on the contrary we are here we have just to to compare so we should have something like this our function f plus g takes at most a finite set of values actually we have a in a1 plus infinity is infinity then we use the convention in a2 infinity is equal to c a3 infinity is equal to c again a4 infinity should be equal to minus infinity and then a5 infinity is equal to ok plus infinity then you have a6 infinity which is equal to minus infinity a7 infinity is equal to plus infinity and infinity is equal to minus infinity so basically what I want to tell you is that you have to compare c and alpha just for a finite number of cases ok so for the cases a1 infinity you have that infinity is larger than alpha so this is true and then you continue has a2 infinity you have to say if c is larger than alpha ok and so a that is true so you have all the set or is not true so you have the empty set in any case you have to compare with 8 cases and you have so we have a finite number of cases and you always end up when you have to say where this is the finite you end up with with a measurable set ok yeah sure please go and ok so you can do the same let me repeat this argument because it's very long it's nothing too deep but for f times g ok no no you can go I will continue for the minutes but you can go ok when you consider f g r and you have to deal with the product I mean you can argue similarly so you have 0 if use the convention if f times g is of the form for instance 0 times plus minus infinity and then will be plus minus infinity if f and g are equal to plus or minus infinity and then ok this is the easy case f plus f times g if f is less than infinity and g is less than infinity ok ok and also in this case you can define a finite as the case when this case and you can define ok you can see a the union a finite union a0 infinity and this a0 infinity somehow collect this first case just start to write some of this but then you understand how it is f is equal to plus infinity and g is equal to plus infinity union f is equal to plus infinity and g is minus infinity then you have the reverse and then union you have the cases in which f is equal to 0 and g is equal to plus infinity union ok f is equal to 0 g is equal to minus infinity and you continue as you have again 8 cases so when you have to compare alpha the test the test values you have to compare alpha just with a finite with a finite numbers of values and you always end up with a measurable set ok so you argue the argument is completely analogous with what we done for for the sum ok it's very long I don't want to to repeat it again ok so now we have some minutes left who of you has to go to do some of you has to ah no ok may I raise here ok so remark so do you think that there exist no measurable function how can you construct a function which is not measurable you use the no measurable set so you have to use I mean the way is to use the no measurable set that we construct ok but then how could you define has we saw that a characterization about characteristic function of set you have to combine the fact that there is a set T which is not measurable T21 or rather you can use it as the characteristic function of the no measurable set ok ok this is ok this is not measurable ok this is an example of the function ok what can we in the last minutes we can prove this proposition ok so let S A to R measurable then S measurable only if you have that for any B to M sorry pre-image of infinity pre-image of minus infinity measurable excuse me ah sorry ah function ok ok so ok I think we here then to to require A fact that A is in M ok actually somehow this is related with with the for instance she your name Torejla before she suggest this example has a no measurable function the constant function define on T which is completely correct here the fact that for instance F is not measurable is also because the domain is not measurable but this is correct so these are two functions this is why we when we want just to check the second part so the part concerning the the level set if you want we have still to to require this ok so let's see this implication so we assume that F is measurable ok we saw that if B is we start by B open set we saw that we can express this as the countable union of this joint open set of the type type A i B i ok these are this joint open interval sorry open interval ok so if you look at the preimage of B this is the union S minus 1 of A i B i and then we have that because the preimage preserve preimage of this joint set preserve the countable union ok the preimage preserve what is that F minus I know yes yes sorry it's the preimage I'm sorry preserve countable ok so we can understand how it's going on you have that this preimage can be seen as the intersection of of these two sets and so you have that the preimage of B is ok it's the countable union of ok of this set S minus B i and everything is measurable ok and then ok what about the preimage of this plus infinity this is the intersection F is larger than N and again everything is measurable ok ok now we deal with the vice versa ok we want to prove that F is measurable if these two are satisfied so we so we check, we pick some alpha enough and we check if this set is measurable so but how we can express this set we can express it as the preimage of alpha plus infinity union the preimage of plus infinity ok ok but we we assume that these are both measurable and so we are sure that also this set here is measurable and so we this concludes this concludes the proof so i think that for today we can stop here