 Ok, značo vedno malo odpravljamo posledno poslednje, vznikam je, da imam odličenje, se, da je poslednje in poslednju ovrčenje s etalijkem, z srednjem poslednjem poslednjem, If for any set A in the set of parts of R, we have that the following somehow the composition also. So we have that the outer measure of A is equal to the outer measure of A intersected E plus the outer measure of A minus E. Ok, then we also observe that by the sub-aditivity property indeed in general mean it's always enough to prove just one side of this equality which is this one. Ok, the other one come directly, no? Ok, so we can always just refer to this also in the definition. So another outcome of the other of the lesson of last time is the following is that we prove that this collection is a sigma algebra. This somehow was the main result that we prove which means that the empty set belongs to M, that if we have a measurable set then also the complement would be measurable. So this definition is symmetric with respect to the operation taking the complement and then if you have a sequence, a collection, a countable collection of measurable set, we prove that indeed also the countable union of the set in such a collection is measurable. Ok, so the lesson of today will concern to somehow mainly to provide some example of measurable set because so far we just said that the empty set is measurable and the full real line is measurable. And so indeed we will prove which is quite somehow obvious that the interval are measurable and we shall see a theorem that allows you to approximate measurable set by means of nice set. So we will say that the set in the collection that introduced last time you remember this f sigma and this g delta. We will see that measurable set is nearly a set of that kind up to a set of measure zero. Ok, so we have to look, we will use this theorem of approximation many times. Ok, so the first theorem is the following, this call it lemma if you want. Ok, we have that, the interval, we start with this interval, this open interval a plus infinity is measurable. Ok, ok, ok, as usual we take any set, test set A, so let A be any set. Ok, ok, and we define by means of this any set, this test test and our interval these two set, the set A1 given by A intersected. A plus infinity and set A2, which is given by A intersected minus infinity and A closure. So this is basically the set that you use when you have to prove the measurability of this interval, because this is the intersection and this is A minus this set here, or A intersected. Ok, ok, what we want to prove? So we want to prove the following things. We want to prove, we observe that this is enough to prove this inequality, so that m star of A is larger or equal than m star of A1 plus m star of A2. Ok, ok, we immediately observe that one case is very trivial, so if m star of A is equal to plus infinity, then we are done. Ok, there is nothing to prove. This is automatically satisfied. So we consider the other case. Ok, now we use the definition of outer measure by means of the infimum and so on. So we know by definition of outer measure that for any epsilon positive, there exist a collection, there exist countable collection of open intervals. Such that we have, ok, which cover A, so we have A in intervals, and such that this cover A, ok, and for which we have that the sum of the length of this open interval are less or equal than m star of A plus epsilon. Ok, let me call it 2 star. Ok, so by means of this interval in the collection we define other set. So we define in analogy with what we did here, we define this collection, this sub-sequence En prime has, In prime, sorry, In intersected our set, our interval A plus infinity, and similarly In 2 prime equal to In intersected the complement, ok, so minus infinity. Ok, so these are the intervals. So we know that the outer measure coincide with the notion of length on the intervals. So we have the length of In and they are disjoint, of course. This is equal to the length of In prime plus the length of In second. Ok, we know this coincide. This is because, ok, they are disjoint. We have that, this is why, this is because the length coincide, the notion of length coincided with the notion of outer measure over intervals. Ok, so we have this inequality and call it star. Ok, and now we are able to prove this. Ok, we observe, we have that since our, ok, A1 is contained in the union of In prime and analogously we have that A2 is contained in the union of In 2 prime. What we get? Ok, we have just simply by monotonicity that M star of A1 is less or equal than the sum over N of the length of In prime and M star of A2 for the same reason is less or equal to the sum of the length of In second. Ok, this is, this comes from by the monotonicity and the countable sub-aditivity property. You combine these two facts, we proved. Ok, ok, so we try to collect all these facts and if we sum up, ok, we get that M star of A1 plus M star of A2 is less or equal than these two sum, second and now we use this fact, this is equal to the sum of In and here we use this, ok, and this, ok, just, this is for 2 star and this is count by star. Ok, this is equal to M star of A plus epsilon and of course as usual this is true for any epsilon positive arbitrary small so we get what we want, ok. So, we finally get that M star of A1 plus M star of A2 is larger or equal than M star of A. Ok, so this concludes our proof. Ah, ah, ah, you are right. What I… Shofshofshofshof, thank you. Dove, where? Here. Kaj? Here. This is 2 by 2 star. Because I called and this is by star. According to this. With this lem, we provide one example of measurable set. May I erase. Če so tega, da z tvoj lemo možemo zelo vzvečiti vzvečitje in zelo vzvečiti. Zato je bilo početno vzvečitje vzvečitje, vzvečitje, in zelo vzvečitje tvoj set. Zelo početno, da početno zelo, da je to vzelo, da je to vzelo. Kaj bi smo počeli vzelo? Zato si se vzelo, da je tudi vzelo, da je to vzelo. To je zelo, da se vzelo, da je to vzelo, da je to vzelo, ...z choreography, çešča se zelo pridem, ...z the set of this kind that we know that are measurable. So we know that this is measurable, because we just prove, ...we take the countable intersection, we are within a sigma algebra, ...so no problem, this is also measurable. Okay? And, okay, then of course you have that minus infinity b, this is very easy v the sense that you can see it as the b plus infinity complement, so we know that this is, we just prove that, this is immejorable, you take the complement, you stay within the sigma algebra and then what you have, you have okay the same, if you get this you can see this as b plus infinity zdaj je bilo očustvo, in ki to vazvo izgledaj, ki si bo, ki je izgledaj,ga je skazati si, ab, ko tega, ab seрачih vzgledaja ab plus infinity zgledaja b minus infinity, ne, ne, čekaj. Minus infinity b. In, okej, to je... Danes je... a b... je a... infinity... b... in, okej, vse je... vse je infinity b. To je a... plus infinity b. ... in so, quzut s infinity... intersected, infinity b. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... Vseh, če je početan po vsej mene, da ima se vsakom zaštavljala, da razvijamo, da ima se vsej, da jaz je bilo vsej vsej. Na vsej, nezavršal, ali obršal, da se nezavršal. A če se vsej, da je izvršal. A če se vsej, da je izvršal. A če se vsej, da je izvršal. asketj of the proof. Ok. The theorem. So this tells you that every open set so this is valid in R. So it can be written as a countable union of this joint open interval. So let me first raise here. Ok, so we start by a g, an open set in R. Ok. And then consider for any element x in g we consider ix. So let ix the largest somehow the maximal open interval the maximal interval in g containing x and such that ix lies in g. Which is in g. Ok, is in g. So you can you can see at ix has the union of all open intervals that contain x and which lies in g. Ok. Ok. We see that Ok. We observe the following that if you have x and x prime in g and you consider the corresponding maximal interval ix and ix prime then we have that they are disjoint ix intersected ix prime is equal to the anti-set. I mean let put in that way either are disjoint or they are identical. Ok. Because why? Because somehow if they intersect you still end up with an open interval which contains both. Ok. So the intersection must be otherwise you get a contradiction. Ok. Because if you take the union you would obtain a largest interval and this means that you are not you are not started by the maximal one. Ok. Ok, because ok, so we have that clearly we have that that we can express g has the union of this ix in x over the x belonging of g. Ok, now what remains to prove is that, I mean, this is too much so it is enough to consider a countable union and then we are finished. Ok, so it is it is sufficient to consider so Eh? This one? Ok, so you have that assume that this intersection is not empty. Ok. It will give you another open interval which contains x, x prime and x and x prime. Ok. So if you take the union you obtain a larger set a larger a larger interval Ok. But you are assuming that ix, a larger interval which contains both x and x prime Ok. A larger interval which contains x but you started by the hypothesis that ix is the largest one Ok. So this leads you to a contradiction Ok. So consider the union, imagine the union the union would be if this is not empty would be a larger open interval, ok? So you get a contradiction because you obtain an open interval which still satisfies this fact which x belongs to to this new interval but you started by the hypothesis that ix was the biggest one which satisfies this. Ok. So it is sufficient to consider a countable union because you know that q rational number is dense in R Ok. So we have that for any for any interval in this union ix ix in the union it is possible to select to associate x alpha rational number x alpha rational Ok. x alpha over belonging to ix and so you can prove that this map which associate ix to x alpha is is injective is injective so basically you can consider the union over this index over this x alpha in q and so you have that g is the union over this x alpha x alpha over ix where the x alpha satisfies this property and so this is a countable union Ok. Let's catch up this proof. Ok. And now we will have the next step the maxima the largest the largest, I mean or you consider it as the maxima in the sense that it is the union of all open interval containing x and such that ix is in g. This is why I tell you that if you consider a union of contradiction because you obtain something which is even larger and Ok. So with this theorem what we get we saw that any open interval any open set any open set is measurable because because we saw of course that open interval are measurable and we just prove that we can express any open set as the countable union of open interval. So we just use the fact that the measurable set is sigma algebra. Ok. So we can introduce a new sigma algebra which is the sigma algebra of the Bore set that will be the sigma algebra generated by the open set. Ok. So I will denote it as this b will be the sigma algebra sigma algebra of Bore set so this is a definition and this is the way you construct it is the sigma algebra generated by the open set. Ok. So I don't know how familiar you are with sigma algebra but I just give you briefly what is sigma algebra generated by a collection of set. You can see it actually in two ways which are of course equivalent. So we have that even this definition we have that a sigma algebra generated by a collection of set in this case the open set generated of set is what? Ok. Is the smallest sigma algebra contained in such a collection? Or hey there you can think of it such a collection Ok. You can see it in that way or even has the intersection of all sigma algebra containing such a collection or is the intersection sigma algebra that contains that contains such a collection. Ok. This is a general definition you can apply this for the sigma algebra generated by the open set. Ok. So what immediately comes from this? So you have that we just see I will denote with you the collection of open set collection of open set. And so B is the sigma algebra generated by U we use this this Ok. B are the border set over R I mean of course but just to stress. So we just saw that U the open set are contained in the collection of measurable set so automatically from this definition we have that the sigma algebra of the border set which is precisely sigma U because of this is contained in the sigma algebra of measurable set Ok. In the following we will see that this inclusion is indeed a strict inclusion so it exists a measurable set which is not a border set Ok. No, B of R is the sigma algebra of border set this is a definition Ok. This is not just a terminology Usually in the book you find this B to the not the sigma algebra of border set this is italic. Sigma means sigma algebra the sigma algebra generated by U this is yeah I mean sigma algebra generated by by U, ok. This is just a definition terminology Ok. Ok, so just put it as a theorem just to be to fix the idea so you have that every border set is a set in the sigma algebra of border every border set is measurable Ok. Ok, now we can define so we are in position to define the Lebesgue measure because so far we just define a set function we just define the outer measure now we define the the Lebesgue measure Ok. Ok, so of course we start by a measurable set let E be a measurable set so we define the Lebesgue measure of E measure, ok. We shall denote it as with M without any star the Lebesgue measure of E we will call it M of E Ok, to be to be the outer to be the outer measure of E so M of E by definition will be the outer measure of E so basically what is the Lebesgue measure? Is nothing that somehow the restriction of the outer measure over the set of the measurable set, ok so you have that M is defined here so it takes values, I mean the domain is the sigma algebra of the measurable set the co-domain is the extended real line to each measurable set you just define it as the outer measure of E, ok so you can see it you are restricting M star to the set to the measurable set, ok ok, now we will prove some property no, doesn't satisfy the the outer measure doesn't satisfy the countability additivity pro we will prove now yeah but I mean we will prove that the Lebesgue measure satisfy if it is countable additive now we will prove countable additive I mean we will construct some contra example it's not easy probably in on Monday let me see, probably on Monday we will introduce a non-mejorable set which require quite tricky argument a non-mejorable set so a set which is which belongs to the set of parts ok I will anticipate, we will construct a set, we will call it just not to introduce so many which a set T which belongs to the set of is a subset of R but T doesn't belong to M so a non-mejorable set ok, and by this we can provide what you ask yeah, exactly this is the difference between the two Lebesgue measure and Lebesgue measure the domain yeah, yeah because you always have in mind that when you define a function define a function is not always to define the law is also to define the domain and the co-domain this is a part of the definition of a function ok no, as the converse, P of R is larger than the measurable set so usually you will end up with this set of inclusion you will have P of R will strictly contain the measurable set and this will strictly contain the bore this bore set is not accountable additive property this is one reason but I need the time to introduce I didn't get your remark because in next lecture I will provide you an example of a sequence of set that doesn't satisfy this property but it takes time ok, now we will see this proposition ok, let EI be a sequence of measurable set ok, then we have the following fact so the first one is the easy K, is the easy one so you have that measure of the union of this EI now I take any set no matter in the hypothesis any set no matter if they are disjoint of not ok this is less or equal than the sum of I of the union of EI then if in addition you assume that this set in the sequence are also pairways disjoint so if EI in addition pairways disjoint then we get indeed the accountable additive property which is equal to the sum of EI ok ok, last time we prove a kind of this but for finite set for finite union ok, but we will use this in this proof ok, this fact is very easy ok, proof ok, this comes from what why this is true so this is this part one because we know that it is also for outer measure and so one is trivial because since plus sub additive accountable property property ok, 2 so 2 is the interesting part ok, first consider only a finite sequence of set EI so consider only sequence of set EI let me change that ok, we know that we have the following that what we prove we prove that A union EI from 1 to k let's k is equal to m star of A intersected EI so do you remember this property we prove it last time so you take A over A equal to R so we can replace this one with M union of EI so we can get rid of this star because we are doing with some of EI ok, so this is a finite additive property ok, so ok, we want to prove this, this equality we will prove the two side one is of course this one is this is less than this is easy, the other one we require a bit still easy but so we have that ok, we ok, we observe that of course this countable union this countable union contains the finite union ok and so by the monotronicity property we have that the back measure of the union of this this countable union is larger or equal then the measure of the finite union we use this use the property star ok, we observe that the left hand side of this inequality does not depend on k, we use the same trick we can get to the limit so since the left hand side of the above inequality does not depend on k then we have that if we pass to the limit we obtain the following ok, ok, the reverse inequality comes from the countable sub additive property this is one this is A and B you have that the fact that the reverse you have that measure of E so this is the countable sub additive property ok, so you combine A plus B and we get the thesis ok and now we prove the proposition which tells you that somehow how can you use some monotonicity property of a sequence of set if you have a set and if it is a increasing sequence of set how can you compute the measure of the intersection of such a set is it ok? so, proposition is the following ok, so let E n decreasing sequence of measurable set so or namely this means that you have that the E n plus 1 set is contained in E n for an E n ok, in addition we assume that for instance the first one set in that collection E 1 has finite measure ok I mean actually it's enough to get what we want to prove it would be enough to require that a set in the sequence is finite, but I mean for the sake of simplicity I choose the first one because this is ok, we will want to state a property that the notion of limit so it's enough that from one point on this property is satisfied ok, anyway take this ok, then what we have then we have that the measure of this the countable intersection of this set is equal to the limit as n tends to plus infinity of the measure of E n ok this is countable intersection ok, start to prove ok, so consider just introduce this terminology this notion call E this intersection intersection of E n over n ok, and define some auxiliary set call them f ok, fn equal to E n minus 1 ok, or maybe E n minus E n plus 1 for any n and we claim the following fact that will help us to prove that we have that E1 so the first set is equal to E union the union of this fi ok ok, we start by one inclusion we prove this claim we prove this claim ok, we have that ok, we have that E is contained in E1 and ok, this is this is the easy this is the easy part E1 plus 1 is contained in E1 so we have that E this gives you that E union the union, this countable union is contained in in E1 ok, now we prove the other inclusion ok so we pick one X in E1 ok, at that point we have two possibilities two possibilities so so either we have that X belong to EI for any index I in that case we have that if this is true X would belong to E which is the intersection of this EI countable intersection or we have another possibility there must exist some index I bar of course in N such that we have that X does not belong to EI bar but X belongs to EI for any I less than this I bar so you can always choose you can always find this smallest integer and in that case ok, in that case we have that being EI decreasing sequence sequence of sets ok, we have also that X does not belong for EI for any so we can say that X does not belong to E but X belongs to which is I recall you this is equal to EI minus 1 minus EI and even more so X of course belongs to the the union of FI so at the end what we show is that indeed to inclusion we have that E1 is equal to E union this union of FI1 to infinity ok, so now we will use this decomposition raise here ok, now we used the fact that they are measurable so we want to compute the measure of that set and they are disjoint ok, so we have that since E, so this countable intersection and this collection of set EI minus EI plus 1 are measurable and disjoint we have that by what we prove before we have that the measure of E1 is equal to the measure of E plus the sum of I of the measure of this set here EI plus 1 ok ok, this just come from the fact that from the countable additivity property now we use the fact that they are measurable and we use EI as test set ok, we have this, we have these two fact we have this and we have that the measure of EI can be expressed as the measure of EI intersected EI plus 1 plus the measure of EI minus EI plus 1 ok so you use here you use the measureability of EI plus 1 and you consider EI just as a test set ok so this will give you that since this is equal to the measure of EI plus 1 plus the measure of EI minus EI plus 1 ok ok, now we observe that all of because of this hypothesis that the first set in the sequence as finite measure and we know they are decreasing sequence of set we know that all this quantity are finite ok, here on the left and on the right side this inequality are finite so we can bring this on the other side so we have that the measure ok, since all if you want these measures are finite because of this hypothesis first one is finite we can exchange we have that the measure of EI minus EI plus 1 is equal to the measure of EI minus the measure of EI plus 1 ok and we use this we insert this inequality here ok we go on here so we use this put it here and if we continue this is equal to the measure of E plus the sum of the measure of EI minus the measure ok ok, you can see it as the measure of E plus the limit as for instance N goes to plus infinity of 1 to N and the measure of EI minus the measure of E plus I ok, then of what you have so basically I will skip some ok, but because it's very easy so you will end up at the end you will have that N plus infinity minus 2 to N plus I so it's telescopic so you can this measure of E plus the limit as N goes to plus infinity of measure of E1 minus ok, I can put the limit just put the limit just here because the limit as N goes to plus infinity of the measure of EN ok, these are the part that we are interested in ok so just let me erase this one ok, we know that everything is finite so we know that the measure of E1 is equal to this things ok, so everything is finite by hypothesis I can do this so finally so I use so I want to stress this so this measure of E1 is finite and everything is finite so I have that the measure of V of this countable intersection is under our hypothesis the limit is equal to plus infinity of the measure of E of EN ok, and that's concludes the proof ok, so now the question is concerning this this proposition so do you think that this hypothesis that we made on the first set ah sorry so I was saying that do you think that the hypothesis that the measure of the first set of one set in the in the sequence is finite is really necessary we can remove it and obtain the same result what do you think I mean the extra hypothesis that we made that the first set is as finite measure do you think is necessary to is necessary, why? yeah N, or N plus infinity this is decreasing sequence yeah, exactly exactly just to write what she said ok so the hypothesis that measure of E1 or ok, any set is necessary in fact we can construct a counter example so you consider for instance this sequence of unbounded set this is still decreasing sequence so you can define it that way E2 equal to 2 plus infinity so you have that EN is equal to N plus infinity ok, this is the decreasing sequence of measurable set and ok, what else? ok, but of course they do not satisfy the fact that the first one has finite measure because we have that measure of E1 is plus infinity ok, so we ok, so as she said we have that the intersection of this one is the empty set because if you assume that there is a point X in the intersection you will always find an index N which step over the and they all have infinite measure so what you get you would have that zero is equal to the measure of the empty set which is equal to the measure of that intersection but this is strictly less than the limit as N tends goes to plus infinity of the measure of EN which is of course plus infinity so indeed that hypothesis was necessary and now I will start to give you a really fundamental proposition probably we will give I mean the proof is very long so for sure we will have to finish next on Monday ok, this is the name of this proposition is characterization by approximation of measurable set so which provides your way to approximate measurable set by nice set so in the book of Rao then this proposition is left as the next but I think it's worthy to see it during the lesson approximation ok, so we will prove some equivalent proposition so we start by be any set ok so any set then the following five statement are equivalent ok, so the first I is E is measurable so I is given any epsilon positive there is an open set O which contains E such that the outer measure of O minus E is less than epsilon ok, here I use outer measure because you have to read it, let E be so we don't know if O minus E is measured down 3i is similar, but concern closed set so given again epsilon positive there is this time a closed set which is contained in E and such that the outer measure of E minus F is less than epsilon then we can do more we have a sharper approximation if we use set in this G delta and F sigma collection so we have that there is a set G in G delta so let me observe that is a borel set because G delta, do you remember what is G delta ok, so this is a borel set ok, such that ok, E is contained in G and the outer measure of G minus E is zero analogously you have that there is a set F in F sigma such that this is still a borel set such that F is contained in E and the outer measure of E minus F is zero ok and then ok, if you assume, if you do if you put an extra hypothesis on E so the extra hypothesis is that E has finite outer measure so this is one, two ok, if this is five, if ok, if in addition M star of E is finite ok, all above statement statement are equivalent to the following ok, given epsilon positive there is a finite collection of open interval of finite ok, union U open intervals ok, such that you deal here with the symmetric difference the symmetric difference the outer measure of the symmetric difference between E and this union E small U E ok, I recall you that what is the symmetric difference you have AB in set so you have that symmetric difference which is denoted with this little triangle of B is the union of what the difference of the two set this is ok, the name is symmetric difference because of course it is symmetric in A and B ok, so if in addition you know this you can say more you can say that this symmetric difference of E with finite union of open intervals is small ok proof is long for today I just outline the strategy to prove this which requires somehow 5 3 step ok, so in step A in step A you use this extra condition improve things under this hypothesis that M star of E is finite and you will prove that I implies 2I which complies in 6 ok, then in the step B ok, step B we use step A sorry, step A ok let me so we get rid of this hypothesis so we remove we just consider the general case the hypothesis M star of E this, so we deal under any without any hypothesis and we use step A to prove the following implication so we prove that I implies 2I implies 4 implies I ok, and then finally we conclude with step C step C ok, we use B, so we use B to prove to prove which I implies 3I which implies 5 which implies I ok so this would be this is the outline of the proof the strategy that we will prove but I will show you the proof tomorrow so I think we can stop here today so it was clear this factor of the maximum