AI Assistant
Transcript
00:14:680Paolo Guiotto: Okay, so…
00:25:910Paolo Guiotto: Okay.
00:26:890Paolo Guiotto: I, will, start… with the… with the…
00:37:660Paolo Guiotto: some of the exercises I left.
00:41:80Paolo Guiotto: I will publish the solutions after the class, or I will not do all the exercises, but just a couple of them, particularly
01:03:420Paolo Guiotto: is the exercise.
01:06:40Paolo Guiotto: One point, 4.4.4. It's a beat…
01:17:120Paolo Guiotto: Well, it's market with one star. When I wrote the solution, I realized it is a bit more complicated than that, okay? So, it should be maybe a two-star plus size.
01:29:310Paolo Guiotto: At least at this… at this point.
01:33:920Paolo Guiotto: So it says X is an uncountable.
01:42:270Paolo Guiotto: Seth.
01:44:250Paolo Guiotto: So you have to take these exercises just as exercise to practice the definitions. There is nothing that we learn specifically from this example, okay?
01:54:520Paolo Guiotto: In fact, it will be, just exercises on… an exercise on the definition. We take as F,
02:03:790Paolo Guiotto: Family, we already discussed as an example.
02:08:30Paolo Guiotto: of sigma algebra, made of the subsets of X, such that one of A or A complementary is…
02:23:860Paolo Guiotto: Actually, you see that they cannot be both.
02:28:480Paolo Guiotto: uncomfortable, so, comfortable, sorry, because otherwise their union, which is the full space, would be countable. So only one of them is uncomfortable, okay? This is just a remark for the moment.
02:40:830Paolo Guiotto: And we define immune on this family path.
02:45:640Paolo Guiotto: with the positive values. Actually, it takes values 0, 1. In this way, mu of a is
02:54:300Paolo Guiotto: Now, since, when A belongs to F,
02:58:780Paolo Guiotto: we can have only one of these two, A or A complementary, is countable. It says, if it is A to be countable, you give value 0, so if A is countable.
03:17:800Paolo Guiotto: If it is a complementary to be comfortable.
03:21:510Paolo Guiotto: you give value 1. That's it.
03:24:910Paolo Guiotto: Now, the exercise consists in checking that… check… that.
03:32:570Paolo Guiotto: mu is a… let's the line. Well… defined.
03:40:520Paolo Guiotto: measure.
03:46:890Paolo Guiotto: on F.
03:49:320Paolo Guiotto: Okay, so that's the assignment. Let's see the solution.
03:55:80Paolo Guiotto: So, first of all, since we have something which is defined, we should always check with the good position of this thing. That's why it's written well-defined.
04:06:710Paolo Guiotto: Where there can be a problem here.
04:10:640Paolo Guiotto: As I say, mu of A is 0 if A is countable, mu over a is 1 if A complementary is countable.
04:18:230Paolo Guiotto: Now, you see that I give two values, depending on some alternative, but maybe someone could wonder, is it possible that at the same time A is countable and A complementary is countable? Because if this is the case, I don't know what is the value of mu of A. So that's the problem.
04:36:40Paolo Guiotto: with a good position. You see?
04:38:740Paolo Guiotto: So, suppose that for some reason, you can have that for a set A,
04:44:310Paolo Guiotto: both A and AC are countable. You don't know what is the value of mu upon that definition, so we have to be safe that this definition is well posed.
04:54:290Paolo Guiotto: And that's the case. So mu is… well defined.
05:03:460Paolo Guiotto: Because…
05:07:740Paolo Guiotto: for every A in the family F, we have to be sure that only one of these alternatives is, is, is true, okay?
05:18:470Paolo Guiotto: So, we can say that only Wang.
05:24:120Paolo Guiotto: Oh.
05:25:230Paolo Guiotto: A, and A complementary can be countable.
05:34:260Paolo Guiotto: Because if both were comfortable, Indeed.
05:42:00Paolo Guiotto: if both… A and A complementary, Where?
05:48:700Paolo Guiotto: accountable.
05:51:70Paolo Guiotto: Then, you do daily union.
05:53:960Paolo Guiotto: A union A complementary, and this is exactly the space, X. The union of countable sets is a countable set.
06:05:70Paolo Guiotto: So this would be… Countable.
06:11:60Paolo Guiotto: But this is in contradiction with the initial assumption of the exercise. X is uncountable.
06:20:590Paolo Guiotto: But X. Ease.
06:26:330Paolo Guiotto: uncomfortable.
06:29:390Paolo Guiotto: In this… Problem.
06:33:480Paolo Guiotto: So we… we get a contradiction. And this means that, since only one of these two
06:39:930Paolo Guiotto: A by A, for some A, it will be A, for some other A, it will be A complementary, can be countable. There is no problem with that definition. Now we have to check the characteristic properties of any measure, okay? So…
06:55:750Paolo Guiotto: Let's… Now… Check. Good.
07:03:950Paolo Guiotto: Meat.
07:05:470Paolo Guiotto: characteristic.
07:09:150Paolo Guiotto: Properties.
07:12:10Paolo Guiotto: Oh.
07:13:130Paolo Guiotto: And measure mu, which are just 2, So, number one.
07:18:240Paolo Guiotto: measure of empty set must be zero. And number two, measure of a disjoint union of measurable sets is the sum of the measures of the AM.
07:34:540Paolo Guiotto: So these are the two checks.
07:36:930Paolo Guiotto: The first one is trivial, because mu of
07:41:710Paolo Guiotto: Empty set. Empty set is a set.
07:45:250Paolo Guiotto: Where there are zero elements.
07:49:450Paolo Guiotto: So, between him and its complementary, which is the full space, it is him to have a countable number of elements. And so, according to this definition, the measure is zero.
08:01:150Paolo Guiotto: Okay, so this is zero, because… Empty setup.
08:09:210Paolo Guiotto: Oz.
08:10:170Paolo Guiotto: no elements.
08:13:740Paolo Guiotto: So, we can say zero elements, and therefore, it's a countable number of elements. Number two is a little bit tricky, not particularly, but you have to think about one second.
08:26:670Paolo Guiotto: Okay, so we have to check that this identity holds, so let's start trying to compute this thing, no?
08:35:169Paolo Guiotto: So, mu of the disjoint union of the AN. So, in particular, the AN are sets in the family F,
08:45:350Paolo Guiotto: So what we know about this, the unique characteristic about this family is that either each of the set is comfortable, or the complementary is countable.
08:57:620Paolo Guiotto: Okay? But we don't know. They are all comfortable, or are they complementary. Could be someone is comfortable, someone has the complementary, which is comfortable.
09:07:370Paolo Guiotto: Okay.
09:09:330Paolo Guiotto: And they are disjointed.
09:11:950Paolo Guiotto: AN intersection AM is empty.
09:16:520Paolo Guiotto: and different from that. Now, we know that this is a sigma algebra, so this set belongs to F. This is not the point here.
09:26:120Paolo Guiotto: Now, the question is, what is the value of the measure?
09:29:750Paolo Guiotto: But in this case, the measure…
09:31:890Paolo Guiotto: let's say it's somehow simple, because it assigns just two values. So the only thing to have understood is if the set
09:41:930Paolo Guiotto: that union.
09:43:790Paolo Guiotto: is comfortable, or it's complementary, is comfortable, okay? So we can say that.
09:52:410Paolo Guiotto: If a union of the yen He is countable.
09:59:970Paolo Guiotto: then measure of union of AN will be… Dear.
10:06:670Paolo Guiotto: or.
10:08:20Paolo Guiotto: If the complementary of the union, so the complementary of the union, he's comfortable.
10:18:840Paolo Guiotto: Measure of that union will be 1.
10:24:950Paolo Guiotto: Now, let's see, because now… then we have to understand if this is always equal to the sum of the muons of the AN, okay? So let's try to see what is the first case. Let's say case A and case B.
10:41:810Paolo Guiotto: Case A, well, the union of the AN can be countable.
10:49:480Paolo Guiotto: Remind that each of the N is in the family F.
10:54:190Paolo Guiotto: So, either for every yen, either a yen is countable, or the complementary is countable. But if you know that the union of a yen is countable.
11:11:280Paolo Guiotto: What happens here about DAN, necessarily?
11:17:550Paolo Guiotto: Yeah, all of them must be countable. Otherwise, if just one of them is non-countable, the union will contain this set. Okay, so a yen must be countable.
11:31:680Paolo Guiotto: Forever, Yana? Oh, sorry.
11:35:960Paolo Guiotto: But therefore, the measure of a yen is what?
11:41:940Paolo Guiotto: Remind this definition. Measure is 0 when the set is countable, 1 when the set… when the complementary is countable, so the set is uncountable, basically, okay?
11:56:340Paolo Guiotto: So quick. Here, the measure is… Zero.
12:01:30Paolo Guiotto: And therefore, what about the sum of the measures or the sets AN?
12:06:760Paolo Guiotto: We're just summing zeros, and therefore we get zero.
12:10:720Paolo Guiotto: So we have that the measure of the union is zero.
12:15:900Paolo Guiotto: The sum of the measures is 0. As you can see, they coincide.
12:23:720Paolo Guiotto: So in this case, we can say that it is true that the measure of the union is the union of the measures.
12:30:810Paolo Guiotto: Okay, so let's go on case B.
12:34:890Paolo Guiotto: The case B is the case where the complementary of the union is a comfortable set.
12:46:930Paolo Guiotto: Okay.
12:48:220Paolo Guiotto: Well, or issue 1 case B is, is,
12:56:610Paolo Guiotto: Well, we can actually say, if you think about a second, that this is really an if and only if, you know? It is here that the union is comfortable.
13:08:400Paolo Guiotto: Vice versa, the unionist government will just say that each of the ends must be counted.
13:16:400Paolo Guiotto: Because we can see that that condition is not refined, so the intercept, the complementary is comfortable. If and if what should happen?
13:30:20Paolo Guiotto: So, complementary of union of a yen, is countable.
13:38:340Paolo Guiotto: If and only if… well, this is the opposite, right, of the case A.
13:45:630Paolo Guiotto: So, what is the condition?
13:49:660Paolo Guiotto: The Union of Agents, North Carolina?
13:54:830Paolo Guiotto: Yes, sir? Yes, sir, of course,
13:58:730Paolo Guiotto: But then I wanted to say something on the yen, the sets AN. Oh, wait.
14:03:390Paolo Guiotto: What would you say on the AEN?
14:06:610Paolo Guiotto: So, in the first case, they must be all countable.
14:10:550Paolo Guiotto: So this case means that… It's the opposite.
14:15:320Paolo Guiotto: If it is not true that all of them are countable, it means that…
14:19:390Paolo Guiotto: At least one of them is uncomfortable, okay? So the rooks is, let's say, an N, such that AN, if you want AN complementar, is comfortable, so this one is uncomfortable.
14:36:720Paolo Guiotto: Or, again, complementary.
14:39:480Paolo Guiotto: Countable, because… In this case, if one is uncountable, they're complementary.
14:45:560Paolo Guiotto: is, is comfortable, because they belong to that family, no? The family says.
14:51:340Paolo Guiotto: Either one of these two is countable. Since a yen is not countable, the other one is countable. Okay.
14:58:490Paolo Guiotto: Now, this says what? If you look at the sum.
15:03:340Paolo Guiotto: of the measures of these AN, What can you save?
15:10:720Paolo Guiotto: With this information about this son.
15:14:300Paolo Guiotto: Remind our goal, our goal is to prove that the sum is the measure of the union. We already know that the measure of the union in this case is 1, right?
15:26:00Paolo Guiotto: Because it's in the case we are discussing. So we want to prove that the sum of these measures is 1 as well. So what can be said at this point?
15:37:140Paolo Guiotto: Yeah, it's larger than 1 because it is larger than the measure of that set, which is equal to 1, because of the definition.
15:50:390Paolo Guiotto: So if we want to hope that this sum is 1, we must prove that
15:56:530Paolo Guiotto: Exactly. So the goal now becomes, let's try to see that only one AN can be uncomfortable.
16:06:550Paolo Guiotto: or the complementary can be comfortable. Okay, you see the pointer? Because to get one, as soon as there is another one here.
16:16:580Paolo Guiotto: Another AI from which AI is uncountable, so AI and complementary is countable, because we are in the family.
16:24:890Paolo Guiotto: We will add another 1 in the sum, so the sum will be at least 2, no, and so on. So, if you want to add 1, you need to deduct all the other terms are 0.
16:36:50Paolo Guiotto: And this means that when the temperature, when OKNR come.
16:42:240Paolo Guiotto: So let's assume for contradiction, that one of them is such that AL is uncountable for the complementary discount.
16:50:860Paolo Guiotto: Got you.
16:52:120Paolo Guiotto: So… Assume, huh?
17:00:310Paolo Guiotto: Bye.
17:01:380Paolo Guiotto: contradiction.
17:07:00Paolo Guiotto: So, there exists another.
17:09:510Paolo Guiotto: say M, different from N, such that the corresponding AM is uncountable, or better, the complementary is countable.
17:25:290Paolo Guiotto: Now… We are now too convinced that this is impossible.
17:37:560Paolo Guiotto: So, what do we know about AN?
17:42:50Paolo Guiotto: It is here, and AM.
17:49:630Paolo Guiotto: We are assuming they are? He's joined. He's joined.
17:53:330Paolo Guiotto: So… We know… That's right, this, that the intersection of the two AM intersection AM is nothing.
18:07:310Paolo Guiotto: You see what you have to do?
18:14:140Paolo Guiotto: We know something on the complementaries, so take the complementary of this relation.
18:19:310Paolo Guiotto: What happens if I do the complementary?
18:24:100Paolo Guiotto: The complementary you have empty is… space.
18:30:130Paolo Guiotto: The complementary of the intersection is… Exactly.
18:36:310Paolo Guiotto: And what this relation says.
18:38:650Paolo Guiotto: Exactly, because both the two are comfortable, their union is comfortable, so it implies that X is comfortable.
18:53:570Paolo Guiotto: And I thought this is a contradiction.
18:57:950Paolo Guiotto: So, there cannot be another M for which a M complementary is countable, and this means that for every M different from N,
19:10:810Paolo Guiotto: The… well, let's say for every little m…
19:15:290Paolo Guiotto: different from N, the set AN is countable.
19:21:970Paolo Guiotto: And therefore, its measure is 0.
19:26:250Paolo Guiotto: And therefore, when we sum.
19:29:220Paolo Guiotto: all the measures of the set AN, this sum is a sum of zeros, except for one single term, which is 1. So that sum is equal to 1, and that's precisely the measure of the union in this case.
19:45:150Paolo Guiotto: And this finishes the proof.
19:51:440Paolo Guiotto: Okay?
19:57:150Paolo Guiotto: So, the tricky part was clearly this part here.
20:01:970Paolo Guiotto: the part A was, let's say, relatively easy. The part B, no, because you have… you need the certain point, there is this, this, this,
20:12:680Paolo Guiotto: This remark that becomes the crucial step.
20:16:960Paolo Guiotto: Okay, let's also do the,
20:24:140Paolo Guiotto: the 149, which is another… this is a 3-star, classified 3-star. Actually, I realized that, well, you know, a part of the exercises, I invented. The other part, I copied somewhere, I borrowed somewhere.
20:40:780Paolo Guiotto: Okay, I'm not the…
20:45:50Paolo Guiotto: I'm not stealing anything, because you can find them in textbooks, these kind of exercises. But when I was preparing the material, I have not done the solution, so I sold, let's say.
20:56:690Paolo Guiotto: By giving a look, and I can say, this is a 3-star, this is a 2-star. So sometimes you will find, like in the previous example, an example which is classified, apparently, one star, so it should be elementary, which is not the case here, as you can see.
21:11:550Paolo Guiotto: This case here, conversely, is a theorem that we will see later, because it's interesting in his probabilistic application. However, since it's a property of measure theory, we will see it here. This would be the Borekan telema. It's called Borel.
21:29:670Paolo Guiotto: Well, not in this version. The Borel can tell you, it's, however, it's basically the same thing. Borel can tell you.
21:40:300Paolo Guiotto: Lemma.
21:42:960Paolo Guiotto: It says we have a measure space, XF… mu… measure.
21:49:830Paolo Guiotto: space.
21:51:630Paolo Guiotto: We have a family of measurable sets.
21:56:490Paolo Guiotto: Such that we know that the sum of the measures of this EN is found.
22:05:290Paolo Guiotto: Of course, they are infinitely many, otherwise,
22:08:450Paolo Guiotto: It's a trivial stuff. Now, it defines this set. Take S, the set of points, X, in the
22:19:60Paolo Guiotto: set capital X, which is a type of set which is interesting by its own. It's the set of points that belongs to infinitely many
22:30:710Paolo Guiotto: sets EN, so X belongs… to Ian?
22:36:230Paolo Guiotto: 4… infinitely.
22:42:720Paolo Guiotto: Manny.
22:49:580Paolo Guiotto: Now, the first question is,
22:52:390Paolo Guiotto: check that this set is measurable. This is non-trivial. I told you, most of times we don't check measurability because it is more or less evident the set is obtained immediately by some measurable set, by doing union intersection, and that's all. Also, this one will come by an operation like that, but the interesting part of this question is that you have to really show that you can write
23:16:640Paolo Guiotto: S as a set operation of the ADF. And this is non-trivial.
23:21:990Paolo Guiotto: Because it's non-trivial to write this property. What does it mean for infinitely many N?
23:30:700Paolo Guiotto: And number two, the question is, compute the measure of this set.
23:40:290Paolo Guiotto: Okay.
23:41:720Paolo Guiotto: Let's start with question one.
23:44:400Paolo Guiotto: We have to check that this is a measurable set.
23:47:680Paolo Guiotto: It is not evident immediately, because it is not the end.
23:54:300Paolo Guiotto: It is X, the elemental x, must be in infinitely many EN.
24:01:910Paolo Guiotto: But not all the same, so I cannot say it belongs to EN for N even, something like this, intersection of E to K. So this is a set where, if you are there.
24:14:220Paolo Guiotto: you belongs to infinity and EN. The EN is an even of the intersection of all the E and, you belongs to all the EN, okay? But when you say, EX belongs to EN from infinity many N.
24:28:320Paolo Guiotto: It means that there is a sequence of indexes, I don't know which one, say N1, N2, N3, N4, etc, such that the Xer belongs to these sets, EN1, EN2,
24:40:940Paolo Guiotto: Etc. So, I should say that…
24:45:200Paolo Guiotto: X belongs to S, if and only if there exists a sequence, say, N1, less than N2, less than N3, etc.
24:57:810Paolo Guiotto: Such that our X belongs to the intersection of D, E, and, say, J, when J goes to… from 1 to infinity.
25:12:50Paolo Guiotto: Now, here there is a little problem, because I will say, okay, if I fix this sequence, N1, N2, etc, this set is made off an intersection of GSP sets E, N, so it belongs to F.
25:26:240Paolo Guiotto: So, if this is fixed up, this thing is a separate neon solution yet.
25:33:520Paolo Guiotto: But here it says there is just a sequence of N1. So in this data, if I take these objects, and I do the unions of all possible sequences, I get my set S.
25:48:770Paolo Guiotto: Okay, so S can be written like that.
25:54:330Paolo Guiotto: X belongs to the union.
25:56:940Paolo Guiotto: of all possible sequences NK, such that n1 is less than N2 less than intersection of E and J, J1 from 1 to infinity.
26:13:870Paolo Guiotto: Oh, yeah, you see, it looks like a set operation, huh?
26:18:950Paolo Guiotto: Here, you do intersection of sex again, and then you do union of these things.
26:23:700Paolo Guiotto: You forget what we are, because once you know that this is in the family, union should be the family.
26:29:810Paolo Guiotto: Yes, but there is a competition. You are allowed to do only capital improvements.
26:35:180Paolo Guiotto: And that's not accountable, you know, because you know all possible subsequences of integers, which is not at all accountable set.
26:44:190Paolo Guiotto: It is like… it is more or less, like, part of an insurance, which is real, actually.
26:50:820Paolo Guiotto: Exactly. So you are doing a union over all possible, let's say, you find sequence of NK. That set and the set of the sequences and K is to be…
27:08:330Paolo Guiotto: So that's not accountable union.
27:10:610Paolo Guiotto: So, I cannot accept like that, saying that it is a set operation of a negative set, because this operation is not a countable operation. The sigma algebra contains countable unions. If you're incremental, it contains countable intersections.
27:26:590Paolo Guiotto: So if you do units of intersections, it's fine, provided that you are doing a comfortable operation.
27:32:750Paolo Guiotto: So we haven't fixed this point, but we can do that.
27:37:480Paolo Guiotto: We can, we can say this, this is… The problem is, It is not.
27:50:540Paolo Guiotto: countable.
27:53:310Paolo Guiotto: operation.
27:56:50Paolo Guiotto: accountable union.
27:57:860Paolo Guiotto: Yeah.
28:00:920Paolo Guiotto: So, let's say I'm talking about,
28:07:60Paolo Guiotto: let's say… a detail, okay? It's not the fundamental concept you have to learn, but you have to be aware that sometimes to check that a set can be measured, so you can copy the measure, is non-trivial.
28:22:200Paolo Guiotto: Okay, so sometimes certain sets must be checked that they belong, so you can compute the measure of that set, because if your family is not the family of all the subsets, you are not necessarily allowed to compute the measure of anything. You have to check that you are inside the family of measurable things.
28:41:900Paolo Guiotto: Okay, so now we need a different way to rewrite this set that involves only comfortable operations. And the way is the following.
28:51:430Paolo Guiotto: We could also…
29:02:390Paolo Guiotto: Say… That.
29:06:10Paolo Guiotto: Our X belongs to the set S, which is the set of elements that belongs to infinitely many EN.
29:15:100Paolo Guiotto: If and only if this happens. So you want to say that you belong to infinity of the many of these EN.
29:23:270Paolo Guiotto: So you can say that, for example, for every N natural.
29:31:100Paolo Guiotto: there will be at least a K, natural.
29:35:830Paolo Guiotto: Such that your X belongs to EK.
29:40:100Paolo Guiotto: Let's see why this is, what we are looking for. Because, it says this, sorry.
29:48:560Paolo Guiotto: It's not exactly, it's wrong, this. There is a K. Yes, natural, but larger than N.
29:56:170Paolo Guiotto: Because this ensures that no matter how metic the error.
30:00:910Paolo Guiotto: you will find a set of decay with the K bigger, bigger than your N, such that X belongs to that decay. So this means that it must necessitate that pure X belongs to infinity dec. Because suppose that index n equal 1.
30:19:540Paolo Guiotto: This says you find a K1 such that your X belongs to that AK1.
30:25:340Paolo Guiotto: Okay?
30:26:680Paolo Guiotto: Now, you don't know what is K1, but keep sending numbers in your mind. Now you take N greater than K1, so you take K1 plus 1.
30:35:750Paolo Guiotto: This X, you find a k such that greater than K1 plus 1, so it's greater than K1, such that X belongs to that pK. So it means that this is settling a second set, because it will be an AK2 with K2 greater than K1.
30:52:580Paolo Guiotto: It is exactly what we were saying here with the sequence, no?
30:58:250Paolo Guiotto: And this set is better, because now we can say, This part here.
31:03:490Paolo Guiotto: exists k larger than N such that X belongs to the K, means that X belongs to the union.
31:13:460Paolo Guiotto: offsets EK when K is larger than N, because if you belong to the union, it means that you are in one of them for some K larger than n, which is exactly what is really real.
31:27:670Paolo Guiotto: And vice versa, if for some K greater than n your X belongs to be K, it would belong to the union of the K with k greater than n. So this is the formal way to write. This is the set operation way to write this data.
31:44:310Paolo Guiotto: And now, this is for every N. So, for every N.
31:49:860Paolo Guiotto: In natural, X belongs there. It means that X belongs to the intersection when n arranges from 0 to plus infinity, of the unions when K is larger equal than n of the set EK.
32:07:00Paolo Guiotto: You'll see?
32:08:200Paolo Guiotto: Because if you notice intersection, intersection means you belong to all the elements of the intersection.
32:15:380Paolo Guiotto: It means that you stay in all these sets for MEM, which is exactly what is written in here.
32:23:150Paolo Guiotto: So, we can say that, huh?
32:25:640Paolo Guiotto: We will formally return later. However, our set S, which is the set of X in capital X, such that X belongs to EN4,
32:38:760Paolo Guiotto: infinitely.
32:44:70Paolo Guiotto: Meaning.
32:46:560Paolo Guiotto: N, which is a literal way to describe this set, is actually this, is the intersection in N, here ranging from 0 to infinity, of the union when k is larger than n of the set EK.
33:05:360Paolo Guiotto: This set will deserve a special name, we will return later. We not… we do not need the truth.
33:11:340Paolo Guiotto: Give a special name now. Now, here, the exercise, the first question asks to prove that F is measurable. And we can conclude, because…
33:22:580Paolo Guiotto: The EK are measurable, so any union, of course, countable union, of the K is measurable, so this set is measurable. It depends on N, but because if you change N, you change this union, okay? So you may call it, let's say, FN,
33:42:580Paolo Guiotto: Let's give a name, FN, and we know that this FN is measurable as union, comfortable union of measurable sets. Now, after we will use the section which is countable in the section of sets which are measurable, and that's why it will be measurable as well.
33:59:590Paolo Guiotto: So also this… so this is the intersection in N of the F, and this belongs to F, because the family is closed for unions, but because of the complementary, it's closed also for intersections.
34:14:440Paolo Guiotto: Okay, now we have verified that this set is really a measurable set.
34:23:270Paolo Guiotto: What about the measure of S?
34:26:530Paolo Guiotto: Well, of course, since the formula for the measure is not given in the exercise, no, you see, there is not any mu of something equal, you don't know how to compute. So, what would you expect is the answer of this question?
34:44:730Paolo Guiotto: compute.
34:47:690Paolo Guiotto: What numbers could you compute?
34:51:00Paolo Guiotto: Could you say that the measure of this S is 3 alpha, or 5, or,
34:58:450Paolo Guiotto: even plus inf… maybe plus infinity. Why not? Plus infinity could be a set that is big, so big that measure… or the other alternative is…
35:10:290Paolo Guiotto: Measure 0, okay? Now, you may think, well, it's measure zero because it is empty.
35:17:10Paolo Guiotto: Well, it is not the case, because, suppose that,
35:27:180Paolo Guiotto: Yeah, we cannot do the example easily here, because if you take all the sets equal, the sum of the measures is plus infinity, unless all of them are measure zero sets. But there could be measure zero sets, for example, suppose all the yen are measured zero set.
35:45:140Paolo Guiotto: Now, as you will see.
35:47:210Paolo Guiotto: and we will see with the case of the bag measure very well, measure zero set does not mean that you are empty.
35:54:490Paolo Guiotto: The empty set has measure 0, but the vice versa is false, and there are huge sets in the sense of a number of points with measure 0, okay? So you can well have measure equals 0, but the set is non-empty.
36:09:20Paolo Guiotto: So if you have sets with measure 0, that sum is zero without having, maybe, an empty set S. So we have somehow to
36:17:860Paolo Guiotto: Show that, probably, the measure is zero, that's the unique, reasonable candidate, by showing that it must be zero, it cannot be anything else than zero.
36:26:650Paolo Guiotto: Okay, so what do we know about this S? Of course, we come to see that it is this thing.
36:32:760Paolo Guiotto: Right.
36:33:840Paolo Guiotto: So, we just proved that S is the intersection when n goes from 0 to infinity of the union k greater or equal than n of the sets AK that we just wrote as intersection n0 to infinity of Fn.
36:51:640Paolo Guiotto: Now, the question is, mu of this set.
36:57:430Paolo Guiotto: Does it remind anything to you?
37:00:190Paolo Guiotto: mu of S means mu of death.
37:04:870Paolo Guiotto: intersection.
37:08:280Paolo Guiotto: So far, we have not seen so many properties, but in the few
37:13:880Paolo Guiotto: General properties we have seen, there is something here.
37:20:100Paolo Guiotto: Yeah. Yeah, because we are thinking to the… From above. Okay, that's right, because that thing, if you remind.
37:34:690Paolo Guiotto: Involves something like, the measure of the intersection, right here we have.
37:41:980Paolo Guiotto: So that's not true, of course, that you do the intersection of something, this comes from a decreasing sequence, but if this is the case, we could do something. Perhaps we can transform this problem into computing the limit of the measures of the sets, okay? So this could be an idea, and in fact, if you look at the sets.
38:01:550Paolo Guiotto: Sorry, it's not here. In fact, if you look at the sets FN,
38:06:780Paolo Guiotto: Fn is the set which is a union for K larger than N of the sets EK.
38:14:180Paolo Guiotto: Now, let's see what is the relation between Fn and FN plus 1.
38:19:80Paolo Guiotto: which is the union when k is larger than n plus 1 of the FK.
38:25:60Paolo Guiotto: You see anything?
38:28:110Paolo Guiotto: They look the same, because they are built on the same weight, let's say, but…
38:33:110Paolo Guiotto: You see that here I have… well, maybe if you write in lines, this is like EN union, EN plus 1 union, EN plus 2. Of course, in this way, I cannot do the… all the lists, because they are infinlimary.
38:49:40Paolo Guiotto: And this other guy is EN plus 1 union, EN plus 2, solving.
38:58:820Paolo Guiotto: EN plus 2, and so on.
39:02:60Paolo Guiotto: You see anything?
39:05:110Paolo Guiotto: That means less harm is smaller than that.
39:07:750Paolo Guiotto: Exactly, because you see that this part here Is exactly this one.
39:14:100Paolo Guiotto: So, FN is FN plus 1 plus something.
39:19:380Paolo Guiotto: plus in the set sense. So it means that it is bigger. And therefore, we have exactly that case. So this is bigger than Fn plus 1. So our FN
39:32:720Paolo Guiotto: Is a decreasing sequence.
39:35:660Paolo Guiotto: And now, the idea to apply the continuity from above makes sense.
39:42:150Paolo Guiotto: But we know that the continuity from above, unfortunately, does not hold always. It demands something.
39:49:700Paolo Guiotto: And this something is that we should check that… yeah, we wrote this assumption in red. The measure of the first element of the list must be finite.
40:02:890Paolo Guiotto: It is not, it is not, yeah, not necessarily quite the first element of the lead. But let's say, one of the elements must have defined measures, you forget of the first 1 million of elements, you start your sequence from that point, and clearly don't apply on this time, because you have to complete the limit here.
40:24:240Paolo Guiotto: Okay, so let's see if the first element of the list has finite measure. So, in our case.
40:31:350Paolo Guiotto: So, to apply, to apply B.
40:39:650Paolo Guiotto: continuity from.
40:43:60Paolo Guiotto: Above… we, need… to check.
40:54:250Paolo Guiotto: If the measure of, let's say, the first element of the sequence, F0, in this case, is 5, Now…
41:06:580Paolo Guiotto: If you have a mini,
41:09:580Paolo Guiotto: let's say, you see that this is going to have something to do with this condition. So, for the moment, we never use this. You will see that this is…
41:20:10Paolo Guiotto: Well, let's see what is mu F0. Mu of F0 is, in fact, what is F0? F0 is the union of all the EK, right? So, union…
41:32:960Paolo Guiotto: measure of the union of the UK.
41:37:100Paolo Guiotto: Okay, so now, yeah, I'm doing also this exercise because I forgot
41:43:480Paolo Guiotto: To tell you, an important general property that sometimes it is useful to know. So let's open a little bracket, proposition.
41:55:360Paolo Guiotto: Which is called… this property is called countable.
42:02:180Paolo Guiotto: sub… Additivity.
42:06:910Paolo Guiotto: So you know that if we have a disjoint union of sets, the measure of the disjoint union is the sum of the measures, and this is the basic rule of how the measure works. What if we do not have a disjoint union?
42:24:220Paolo Guiotto: We have seen, I left one to check in the exercises, there are formulas that we could, in principle, write a union, which is not disjoint, with some complicated operation. But if we need a quick formula.
42:40:20Paolo Guiotto: The formula is the following, it's a bound, it's not an identity. Let's say, so, XF mu, as usual, the measure of space.
42:51:50Paolo Guiotto: We have a family of sets, measurable sets.
42:55:990Paolo Guiotto: Now, what happens if we do the measure of the union of these EN? Well, this won't be the sum of the measure of the N.
43:07:130Paolo Guiotto: But there will be a natural bound that, geometically, it easily suggested. You have your set, which is made of a union, maybe they are a bit overlapping sets, okay, so E0, E1, E2, and so on.
43:25:580Paolo Guiotto: So the measure of the union won't be bigger than the sum of the measures. Won't be equal, because maybe there are many overlappings that you count many times, so the general
43:37:660Paolo Guiotto: bound will be this one, less or equal. And this property is called the countable sub additive, sub because it reminds of the inequality, okay?
43:49:440Paolo Guiotto: But it's, it's not difficult to prove this.
43:55:540Paolo Guiotto: We can just prove the formula for two sets, and then we extend.
44:00:750Paolo Guiotto: Because, if you have the union, A union B, let's say, generic two sets, we have seen last time
44:11:650Paolo Guiotto: Well, we have seen. We… we can say that this union, let's say that these are A and B, so there is a part of
44:21:170Paolo Guiotto: overlap between the two. Otherwise, we would use the countability. So we can say that this is the measure of A. We can transform this into a disjoint unit by saying, we take A,
44:33:890Paolo Guiotto: And we add only what is in B, not in A. So, which is, let's say, this part in red, and the red set is B minus A. Now it becomes a disjoint union, no? So, A union B,
44:49:570Paolo Guiotto: is the same of A, this joint union with B minus A. We use this trick for the
44:56:930Paolo Guiotto: Yeah, when we prove the additive formula, the addition formula.
45:05:270Paolo Guiotto: And so this is mu of A plus mu of B minus A.
45:10:210Paolo Guiotto: And since now B minus A is, of course, contained into B, this quantity is less or equal than the measure of B, and therefore, you have that, at the end, the measure of A
45:25:00Paolo Guiotto: Union B, is less or equal than measure of A plus measure of B.
45:34:770Paolo Guiotto: So this is the formula for the…
45:37:640Paolo Guiotto: not comfortable, but finite, let's say, the measure of the union is less than the sum of the measures. Now, as you can understand, this formula extends in a straightforward way when we have a finite union. So, if I do measure of union.
45:55:880Paolo Guiotto: for K going from 1 to N of sets.
46:01:90Paolo Guiotto: Let's return to the letter E.
46:03:910Paolo Guiotto: decay, this will be less or equal
46:07:340Paolo Guiotto: do an induction argument. Sum for k going from 1 to n of the measure of the… well, let's start from 0.
46:16:150Paolo Guiotto: measure of EK.
46:19:440Paolo Guiotto: Now, I want to pass N to infinity to arrive to my desired formula, because that's for a countable union. Can I do that? Yes, because on the right-hand side, I can notice that if I put the entire sum from 0 to plus infinity.
46:37:630Paolo Guiotto: I clearly have a big number.
46:42:10Paolo Guiotto: Because, I'm just summoning possible quantities, so the sum increases.
46:48:80Paolo Guiotto: Right.
46:49:80Paolo Guiotto: So this quantity here, which is now independent of Na, it's something like a constant, is a bound for these measures. So I get this measure of unions, k going from 0 to N, EK,
47:06:180Paolo Guiotto: Less or equal than a quantity.
47:10:780Paolo Guiotto: sum for k to infinity, mu of EK.
47:14:820Paolo Guiotto: You see that this quantity is independent.
47:20:120Paolo Guiotto: of N. It's a constant in N.
47:24:250Paolo Guiotto: Okay? Why? At the left-hand side, we're asking this N.
47:29:750Paolo Guiotto: Now, I want that some of you gives me the idea, how can I get… the countable union.
47:37:380Paolo Guiotto: So, how can I go from this to get that?
47:43:230Paolo Guiotto: The measure of the union for k going from 0 to plus infinity of the K is still less or equal than the sum of a K of the measures EK.
47:56:340Paolo Guiotto: What is the argument I would justify this?
48:09:580Paolo Guiotto: Don't be precise, don't be worried to tell the right argument. We need ideas, then we can write the arguments. Yeah, we want to delete it, because we want to send this little anti, so…
48:28:640Paolo Guiotto: ill-limited. The issue is limited.
48:31:290Paolo Guiotto: Live of what? Live of measures.
48:35:70Paolo Guiotto: I have two properties for the dimensions, which is the continuity from below and from above.
48:41:100Paolo Guiotto: Which one is important for PDF?
48:44:600Paolo Guiotto: From below, because if you look at this and all the unions, look at this, call them FN.
48:53:430Paolo Guiotto: These sets are unions.
48:55:850Paolo Guiotto: And, every time you enlarge N, you add a new set, so this is an increasing sequence of set. So that, this is the increasing sequence of a fan. So occasionally continuity from below, because it works in any case.
49:12:900Paolo Guiotto: So it means that since these are less than this, I passed the limit, so if you want the formal argument, you could say, then.
49:21:690Paolo Guiotto: the limit… in N of the measures of the unions, k go from 0 to N of the AKE,
49:32:180Paolo Guiotto: Now, this limit exists because we have a sequence of numbers. The measures are e to n, which are increasing, so this is an increasing symbol, so I am allowed to write limit, otherwise it's something which is a nonsense. Of course, will remain less or equal than this.
49:51:20Paolo Guiotto: Because all the elements of the sequence are bounded by this.
49:55:380Paolo Guiotto: The force of the limit will be bound by that. And now, by continuity, I can say, get inside the limit, and this is… this is where the continuity theorem comes.
50:07:360Paolo Guiotto: So, this equal is the continuity from.
50:13:380Paolo Guiotto: Below.
50:15:760Paolo Guiotto: And that finishes the proof.
50:19:800Paolo Guiotto: Okay, so remind of this property, because maybe we use it sometimes. When we are desperate, we don't know how to assess a measure, and that set can be somehow decomposed. We could use this. Of course, it's a bound.
50:35:510Paolo Guiotto: So, it could be useful if you want to prove that a measure is finite, and you are able to prove that the sum of the measures at the right-hand side is finite. If that sum is infinite, it does not tell you anything.
50:48:960Paolo Guiotto: It could be good if, from this part, you are right to say that maybe these measures are small, like what we are doing, the exercise. So let's go back, I want to finish the exercise, then we do…
51:01:460Paolo Guiotto: Maybe a little back if you need, or we continue.
51:06:380Paolo Guiotto: Okay, so we were returned on the problem. The groundwater completed the measure of S. S is that sector. We realized that that sector is an intersection.
51:17:660Paolo Guiotto: of sets which are decreasing, following a decreasing sequence. We want to apply the continuity from above. This time, we need to verify if the measure of F0 is fine.
51:28:330Paolo Guiotto: F0 is the total union of the UK, so we need to bound that margin. So you see here, I just need a bound, I don't want the exact value of that 1D.
51:39:140Paolo Guiotto: This is not the case, not the exact time, but this is a typical example of a general situation that may happen. I need to know that something is fine, so I need a bound to bound this one, and I can use that subtitivity, because
51:55:960Paolo Guiotto: Now, returning… to… D.
52:02:50Paolo Guiotto: Question 2.
52:04:530Paolo Guiotto: So we say, the measure of FCO
52:07:490Paolo Guiotto: is the measure of the union of all the K, for k ranging from 0 to infinity, by countable
52:17:650Paolo Guiotto: sub-additivity.
52:20:840Paolo Guiotto: This is less or equal than the sum of the measures of the K, and now, finally, you see that this was the assumption
52:29:800Paolo Guiotto: assumption problem.
52:32:860Paolo Guiotto: Assumption, okay?
52:35:410Paolo Guiotto: So finally, we know that that set us finite measure. We can apply continuity from above, and let's see what happens.
52:43:770Paolo Guiotto: So… continuity.
52:47:390Paolo Guiotto: from… I moved.
52:51:980Paolo Guiotto: applies…
52:55:950Paolo Guiotto: Therefore, what were we doing? We were saying that, okay, mu of S is the mu of the intersection, now becomes the limit of the mu, of the FK. So I have…
53:10:250Paolo Guiotto: let's write here. So, mu of S,
53:13:790Paolo Guiotto: equal mu of the intersection of these sets of K,
53:19:260Paolo Guiotto: Now, this is the point where we invoke the continuity. This becomes the limit
53:24:550Paolo Guiotto: in case of the measures of DFK.
53:29:930Paolo Guiotto: And now the problem becomes, can we tell anything about these measures
53:36:510Paolo Guiotto: Well, this is basically similar to what we have done a minute ago on the measure of F0.
53:41:940Paolo Guiotto: So we can do only a bound, but look at what bound. Now, mu of FK… FK is similar to F0. It is just a union for K… or sorry. Let's use a exchange letter.
53:58:520Paolo Guiotto: So, let's go to the meeting now of the FN.
54:02:590Paolo Guiotto: Maybe we could use also here.
54:04:990Paolo Guiotto: the letter N.
54:07:660Paolo Guiotto: So, this is the measure of Fn.
54:12:900Paolo Guiotto: is the measure of the union for K larger or equal n of the EK.
54:22:610Paolo Guiotto: We still do not know anything about the K, so we can apply the additivity, but we can estimate this will become lesser equivalent than the sum, of course, for the same K. So this is the K that goes from n to plus infinity of measure of EK.
54:44:470Paolo Guiotto: Okay, so this quantity, for which we have to take the limit when n goes to plus infinity, is controlled by these sums, infinite sums.
54:58:900Paolo Guiotto: Now…
55:00:850Paolo Guiotto: This is what, if you are familiar, very familiar with the medical fields, it would be, let's say, automatic, but if you are not even familiar, estimating. What happens when you send energy?
55:16:10Paolo Guiotto: However, this is not the entire sum from 0 to the infinity, but it is what we call the tail of the Sun, no?
55:23:160Paolo Guiotto: The sum of wood elements starting from that one at place n. Then there is a third at place N plus 1, and that's 2, and so on. But when you say that n, this starting point to infinity, the tail goes to
55:41:70Paolo Guiotto: costum?
55:43:710Paolo Guiotto: Zero. Zero, because…
55:47:70Paolo Guiotto: It's not always true that the tail of an invited son goes to zero, but it is true because…
55:54:580Paolo Guiotto: I'm negative.
55:57:30Paolo Guiotto: The normal series is 5, because the full sum is 5, no? The full sum of the numbers is fine. If I'm taking a day, and I send the index, the start of the index to infinity, it appears that that sum must disappear. This is what is called the reminder of the series.
56:17:440Paolo Guiotto: It is actually equivalent to say that the series is convergent, the reminder goes to zero. They are equivalent.
56:24:460Paolo Guiotto: Okay, so this quantity goes to zero.
56:29:130Paolo Guiotto: when n goes to infinity, I mean, if you are not solving the finest answer.
56:37:360Paolo Guiotto: let's use the intuition. Now, if the total sum is fine, then I have an infinite list of numbers.
56:44:150Paolo Guiotto: Let's sum is 5. Now, I start summing from position N on, and they sent me something to infinity, so I'm summing less and less and less and less. So the total is fine, and summing nothing at the end.
56:57:530Paolo Guiotto: Otherwise, the function cannot be fined. That's more or less the idea. So this means that this limit
57:06:260Paolo Guiotto: is 0, and therefore we have the conclusion. So the measure of S Is equal to zero.
57:15:710Paolo Guiotto: And that's, in fact, the unique thing that could have been… or maybe sometimes could be… you can say that the measure is 5.
57:24:350Paolo Guiotto: Okay, which is not such a big information, but it's better than nothing.
57:29:140Paolo Guiotto: Sometimes. Okay, now it's now 9.45. Do we need a bracket?
57:36:630Paolo Guiotto: Okay, so let's take 5 minutes back.
57:41:460Paolo Guiotto: And then we… restart.
57:49:240Paolo Guiotto: Okay.
58:04:190Paolo Guiotto: Okay, so let's restart now, returning on the LeBague measure.
58:10:980Paolo Guiotto: So, we introduced this at the end of the class of yesterday. I just quickly reminder thinks we define the interval, a Cartesian product of intervals, which is in the planar rectangle, an interval in the real line in dimension 1, a parallel paper in dimension 3, and so on.
58:30:920Paolo Guiotto: We want to build a measure on subsets, for the moment specified, of RD, such that rectangles are measurable.
58:41:500Paolo Guiotto: And moreover, we want that the measure of the rectangle be the natural geometrical measure, as you have in that formula.
58:49:700Paolo Guiotto: And the second goal is that we want also that this measure is invariant by translations, and possibly invariant by rotations, and some other kind of operations. If you do a symmetry with respect to the axis, then you have that the measure should be preserved.
59:08:120Paolo Guiotto: Well, actually, you will see that these two conditions alone are sufficient to determine what is the level measure.
59:14:730Paolo Guiotto: And to determine all the story. So we started saying this, okay, we introduced this first concept, which is the concept of outer measure.
59:23:650Paolo Guiotto: So this is going to be an approximation by excess of what should be the measure of a set. We take any subset, E, in the space, RM,
59:35:100Paolo Guiotto: And we consider, all possible coverings of this E made of intervals, countable sets, families of intervals.
59:48:440Paolo Guiotto: For each of these coverings, we compute the sum of the measures of these intervals.
59:55:310Paolo Guiotto: And this quantity will provide an approximation by XS of the measure of E. Of course, we expect that the measure of E should be somehow the best of the approximation by XS, and in other words, we would expect that to be the smallest among these
00:13:580Paolo Guiotto: Approximations, so the minimum.
00:15:970Paolo Guiotto: The minimum cannot be defined in general for sets of numbers, but we can replace it by the infamum. Now, infamum is
00:24:870Paolo Guiotto: probably the first thing you have seen in Analysis 1, so I don't know what you remind of Infimum. We will need just to remind a bit some of the features of the definition of what is it. I will do in a moment when we will need…
00:39:720Paolo Guiotto: But the point is that, in any case, this way…
00:42:890Paolo Guiotto: We are a theoretical, not practical, of course, definition of a quantity that can be associated to every set P. There is no condition on it for the moment.
00:54:590Paolo Guiotto: So this lambda star turns out to be defined.
01:01:950Paolo Guiotto: So… Let's close this up.
01:07:590Paolo Guiotto: So, we… defined…
01:14:880Paolo Guiotto: outer.
01:17:360Paolo Guiotto: Measurable.
01:19:970Paolo Guiotto: on number.
01:21:260Paolo Guiotto: RM, then actually, this procedure can be really generalized to more abstract setup, so there is an abstract way to do the same thing, but not necessarily on RM. So we do not enter into this, even if it is an important
01:38:60Paolo Guiotto: point, because it allows to construct more general measures, but we just trust the folks in
01:44:110Paolo Guiotto: on the LeBag measure.
01:45:930Paolo Guiotto: That is given by this lambda star E, so let me just quickly rewrite the formula. It's the infamum of the sums of the measures. For the moment, you see, I use a different symbol, because I do not want… I don't want to confuse the two.
02:02:350Paolo Guiotto: lambda star with this absolute value, okay? That's the size of the interval defined by the elementary geometry, such that E is contained into unions of these EK, and AK are
02:18:700Paolo Guiotto: introverts.
02:24:110Paolo Guiotto: Now, this lambda star E is defined.
02:32:240Paolo Guiotto: for every E subset of RM,
02:36:550Paolo Guiotto: And therefore, what we have built here is a function, lambda star, which is defined on parts of RM,
02:45:670Paolo Guiotto: With the positive values, clearly, because these sums are somehow positive quantities, so these numbers are always greater or equal than zero, and this quantity could be even equal to plus infinity.
03:02:10Paolo Guiotto: So, since this is a sigma algebra.
03:08:200Paolo Guiotto: We may think that now we have basically defined a measure.
03:13:100Paolo Guiotto: Yes, but the point is that you have to check if this is countably additive to say that this is a true measure.
03:19:950Paolo Guiotto: We may accept that,
03:22:410Paolo Guiotto: But let's start working on some simple properties of this. What I want to show you here is to show that, of course, as you may expect from this definition.
03:35:640Paolo Guiotto: Even if the definition seems to be natural and relatively elementary, it's not at all easy to do calculations with this thing. And even to prove elementary things, it can be hard. I will give you an example of this. However, the proposition says.
03:54:770Paolo Guiotto: Well, we can say that, number one, the lambda star of empty set is zero.
04:03:750Paolo Guiotto: For example, the empty set, since it is nothing, this is quite easy to be understood, because we can use as covering an interval. Let me do, like this, A1, A1… sorry.
04:22:360Paolo Guiotto: I have to take a closed interpass.
04:26:410Paolo Guiotto: A2… A2.
04:29:480Paolo Guiotto: taken into my like that. It's what we see intimate.
04:33:500Paolo Guiotto: AM… AM.
04:36:610Paolo Guiotto: Actually, this is a set made of one point, no? Because the first coordinate is between A1 and A1, it's A1. The second is between A2 and A2, it's A2. So this is a single tone. It's a single set made of a point A1AM.
04:54:330Paolo Guiotto: So it's not an empty set, but of course it contains the empty set, right? So I can say it is a covering.
05:02:500Paolo Guiotto: here alone, okay, is a covering, and therefore, if I take the sum of the sizes of this, which is just the size of this, is 0,
05:15:500Paolo Guiotto: So this number, 0,
05:18:770Paolo Guiotto: belongs to the set that you see here. Because among all these sums, there is also the value equals 0.
05:27:650Paolo Guiotto: And since they are positive, the infamum cannot be negative. It must be greater or equal than zero. Since there is zero, it means that the infimum is achieved, and it is equal to zero. Okay?
05:39:370Paolo Guiotto: So this is just to convince you that even to prove that mu of empty is 0, it's not a trivial story. So this means that 0 belongs to the set of sums of measures IK is like that, empty is contained into union of IK, blah blah blah. I don't want to write that thing.
05:58:320Paolo Guiotto: So, since 0 belongs there, the infamum of this set.
06:03:200Paolo Guiotto: is… must be equal to zero. And that's the measure of our… Empty cell.
06:10:920Paolo Guiotto: Second property, is, let's say, a coherence property. Well, what happens if I compute
06:20:530Paolo Guiotto: with this lambda star, the measure of an interval, so a rectangle with all the definition. Does it come equal to the measure of the interval that we defined by the geometry? So, is it true that lambda star of i
06:38:30Paolo Guiotto: is equal to measure of i for every i interval.
06:46:230Paolo Guiotto: Yes, of course, but as you can see, the proof is not completely… it's not obvious. You have to do a little army.
06:57:640Paolo Guiotto: Let's see why.
07:00:50Paolo Guiotto: Because my set, is just the interval I.
07:05:610Paolo Guiotto: So, let's do the figure in the Cartesian plane, but of course, the argument is… I will not write any particular formula, but it made you for the intuition that our rectangle is a plain rectangle.
07:20:10Paolo Guiotto: Now, among all coverings of I, of set I, with rectangles, there is the special covering made by EI itself. So the set, the covering made by just I,
07:34:830Paolo Guiotto: Covering…
07:39:380Paolo Guiotto: made… Bye.
07:44:790Paolo Guiotto: I alone.
07:48:540Paolo Guiotto: is a… covering of I. It's trivial, no? I is contained into itself. And therefore, this means that
07:59:50Paolo Guiotto: Among all these sands that I used to compute the lambda star of i, there is also
08:07:680Paolo Guiotto: So, the, measure of IE is… In…
08:15:630Paolo Guiotto: the set of sums of measures of IK, such that I is contained into union of the IK.
08:27:340Paolo Guiotto: So, if this number is inside this pattern.
08:32:590Paolo Guiotto: What is the conclusion about lambda star i?
08:37:590Paolo Guiotto: Lambda star i is, by definition, let's use this improper term. It's the minimum of this star.
08:44:290Paolo Guiotto: So, if we decide there is this number, the minimum might be
08:52:710Paolo Guiotto: My number is the minimum of a set of numbers, among which I have this. So, the minimum is… Lower or equal. Exactly. Okay?
09:05:210Paolo Guiotto: Oh.
09:06:470Paolo Guiotto: Sorry, wait a second, because I am… They have low battery.
09:17:170Paolo Guiotto: It seems stupid things, but, I mean, they are non-trivial, in fact.
09:24:630Paolo Guiotto: No, that's… Okay.
09:29:859Paolo Guiotto: So, this means that this is less or equal than
09:35:370Paolo Guiotto: this. So, I'm now sure that the outer measure of an interval is less or equal than the geometric measure of the interval, but I want, of course, to get they are equal.
09:48:529Paolo Guiotto: So, it needs that I have to prove also the… the other sign, so that this is…
09:54:790Paolo Guiotto: Larger than this.
09:56:850Paolo Guiotto: How can I do?
09:58:340Paolo Guiotto: Now, from this, In particular, since I'm taking a finite interval, this number is final.
10:08:590Paolo Guiotto: And this tells that from this, the outer measure of i is 5.
10:16:240Paolo Guiotto: This is important because I have here to remind, just,
10:22:160Paolo Guiotto: Second, what is the infamum of a set? What is the characterization of the infamum of a set of numbers when that infamom is finite?
10:32:600Paolo Guiotto: And it is also a positive.
10:35:260Paolo Guiotto: So, what is the infamo… so let's open a parenthesis here. If,
10:41:520Paolo Guiotto: If S is a subset of real numbers, and let's say alpha in the infamom, of the set essence.
10:51:200Paolo Guiotto: What is the characteristic property? The idea is that it is the so-called moral minimum, okay? It's going to be the minimum if the minimum exists, but if the minimum does not exist, it's the best lower bound. So the idea is that…
11:07:20Paolo Guiotto: The idea is that you have your set S, there is some part left, because we are taking lower bound.
11:14:990Paolo Guiotto: This number, alpha, And what is the feature? Well, there are two features, then…
11:24:790Paolo Guiotto: Number one, alpha is at the left of set S, so I have to say alpha less should equal then X for every X that belongs to the set S. Number two.
11:37:610Paolo Guiotto: But there are many alpha that verify this condition. Alpha numbers, there are numbers that are at the left of center.
11:44:900Paolo Guiotto: But the infimum is the best one. Best means that you cannot do it right, say, at the left of the set. It means that whenever you take a number of vector greater than alpha.
12:00:590Paolo Guiotto: So let's say that this is beta, For every beta.
12:06:540Paolo Guiotto: greater than alpha, that beta is no more a lower bound, is no more at left of the set. So it means that there is something of the set, some elements X, that is at left of beta.
12:22:790Paolo Guiotto: So there exists
12:27:620Paolo Guiotto: such that X is less or equal
12:31:970Paolo Guiotto: than beta. If you want, you can put strictly less than beta, it's indifference is the same, no? So these two characteristics are those properties that characterize this alpha, the best lower bound of itself.
12:48:210Paolo Guiotto: Good.
12:49:290Paolo Guiotto: So… Since we know that, alpha…
13:01:570Paolo Guiotto: Well, let's see what is the argument.
13:04:990Paolo Guiotto: Well, let's say this. Okay. Suppose that, by contradiction, that you don't have equality. So that sign that you have there is a strict sign. So…
13:17:610Paolo Guiotto: If the outer measure were Strictly less than the geometric measure.
13:27:570Paolo Guiotto: To suppose that it is not true that the two numbers coincide. Now, you see that this guy is an infamo.
13:35:930Paolo Guiotto: And I have here a number which is larger than the infamone, so they are alpha and beta in the definition I introduced here. So, it means that
13:50:170Paolo Guiotto: What is the case? The case 2 says that whenever you have a number greater than the infamum, you can always find an element of the set which is less than that beta.
14:04:60Paolo Guiotto: So, it means that there would be an element of the set. The set is what?
14:11:550Paolo Guiotto: You remind that the set is made of sum of measures of intervals, such that the union cover I. So, there would be a covering, a family IKE,
14:26:940Paolo Guiotto: Such that I is contained into the union of the IK.
14:32:760Paolo Guiotto: And at the same time, the sum of the measures of D and K
14:38:430Paolo Guiotto: would be strictly less than the measure of i.
14:45:120Paolo Guiotto: But now, here, the situation is the following. You have I, We are saying…
14:53:580Paolo Guiotto: I is covered by a number of rectangles, they can be infinitely many, of course, I cannot do such a picture, it would be a mess, but let's assume that it is covered by two, okay, just to fix ideas. Imagine that there are two rectangles, say this is I1,
15:12:250Paolo Guiotto: And this is I2. I'm doing bigger, it's not necessarily the case, but the point is that the unions cover I.
15:20:950Paolo Guiotto: Such that the sum of the measures of the areas of the red rectangle is strictly less than the area of I. You see the figure? It seems to be impossible, right?
15:34:350Paolo Guiotto: Now, how can you be sure of that? You see that when I do the intersection between an I, the black rectangle, and each of the red rectangles, I get a rectangle, and that's true. You can prove that when you do the intersection between two rectangles, it is always a rectangle, just in general.
15:52:810Paolo Guiotto: So I color this intersection, only the intersection. So this one.
16:00:390Paolo Guiotto: This is, I intersection I2.
16:05:270Paolo Guiotto: And, let's do in green, the other one.
16:09:70Paolo Guiotto: This is, in green, I intersection I1. In general, I will do I intersection.
16:19:670Paolo Guiotto: IK, and this guy is a rectangle.
16:24:40Paolo Guiotto: Okay, let's call it JK.
16:28:150Paolo Guiotto: is a rectangle.
16:32:310Paolo Guiotto: And now we know, since we have done the intersection, that the union of DJK is exact. So when you do the union of the JK, you obtain exactly I.
16:45:600Paolo Guiotto: Okay.
16:47:940Paolo Guiotto: Now, you understand that the area of I will be definitely less or equal than the sum of the areas of DJK.
17:01:410Paolo Guiotto: Because there are the overlappings also. Now, you see here, there are zone of overlappings that I'm counting several times.
17:10:590Paolo Guiotto: They are not a disjoint union. If you want, you can even reduce a bit and transform into a disjoint union, okay? Okay, so this will be less, and since the JK are smaller than the AK, this is less or equal than the sum
17:25:330Paolo Guiotto: of the IK, right?
17:28:420Paolo Guiotto: But for this particular family, IK, This holds
17:34:580Paolo Guiotto: The sum of the AK is less than the measure… the area, let's call it area, not the outer measure, because it's a different object for the moment, of I. And look at the initial point and the final point, you read.
17:48:280Paolo Guiotto: the size of I is strictly less than the size of I, which is, of course, impossible.
17:55:800Paolo Guiotto: So this is a contradiction to what?
17:58:620Paolo Guiotto: To the assumption that, This is true.
18:02:940Paolo Guiotto: The outer measure is strictly less. Since it is less or equal, it cannot be less, it must be equal.
18:09:130Paolo Guiotto: You see what the really tricky argument we have to do to prove just this stupid thing?
18:16:370Paolo Guiotto: Okay, we are just at the basics, no? We are saying the measure of an interval, an object rectangle that we see there, is exactly base time A, but that's why we are given a complicated definition of area.
18:29:450Paolo Guiotto: You can do the same experience if you go back in the Riemann integral, and you try to compute, by definition, the area of a rectangle.
18:38:710Paolo Guiotto: Without knowing that there is the fundamental theorem of calculus. But using the definitions, you will see that it's complicated. You will get the formula base time H as well.
18:47:590Paolo Guiotto: So this says it works. It is a good definition because it provides exactly what I expect, no? It's a coherence check, no? If this fails, well, we can throw away everything here, no? Imagine that this concept gives that the outer measure of a rectangle is less than this area, each area.
19:04:370Paolo Guiotto: We are not interested to this, because we want exactly that it measures as we expect.
19:10:90Paolo Guiotto: Okay, so this proves that,
19:12:960Paolo Guiotto: lambda star i equal size of i. This means also that from this point, we… there is no ambiguity to call this the measure of i, because they're two coincide, they are no different, so I'm not one use of two different quantities.
19:32:520Paolo Guiotto: Okay, so other properties… For example, one which is easy to be verified, number 3, lambda star, is monotone.
19:47:840Paolo Guiotto: This means that if you take set E, which is contained in the set F, there is no other condition, because it reminds that lambda star is defined no matter who are the sets.
20:00:420Paolo Guiotto: Then lambda star E will be less or equal than lambda star.
20:05:440Paolo Guiotto: F.
20:07:70Paolo Guiotto: Also, this property, which is, let's say.
20:10:960Paolo Guiotto: A property that any measure will verify, yeah?
20:16:240Paolo Guiotto: Here, it requires a little proof. I will not do. If you are curious, you can find the proof on the notes, but the idea is easy. The fact is that whenever you cover E… whenever you cover F, you automatically cover E, no? Because E is like that, is inside F,
20:37:480Paolo Guiotto: So, any covering made of rectangles of the set F, I will do with 3.
20:44:260Paolo Guiotto: you see that automatically covers the step E. So it means that, among the sums that I use,
20:54:400Paolo Guiotto: for E, there are, basically, all the sounds I use for F.
21:00:290Paolo Guiotto: So it means that the set of Samsa for E, is bigger, and when the set is bigger, the minimum is smaller. If you have a bigger set, the minimum is smaller than J. Not necessarily, but… Exactly.
21:16:480Paolo Guiotto: Okay, so, another interesting property that can be easily proved is that this lambda star is invariant by translations.
21:27:300Paolo Guiotto: So, it's something that you see that
21:30:630Paolo Guiotto: It is like, oh no, they are written in the other class.
21:34:700Paolo Guiotto: This lambda star is going to verify these two conditions.
21:39:190Paolo Guiotto: Because the measure of intervals is their area, volume, etc.
21:43:970Paolo Guiotto: And the measure is invariant by translations, okay? So this lambda star verifies
21:50:830Paolo Guiotto: verifies these conditions. So, an estar is invariant.
21:58:680Paolo Guiotto: Bye.
21:59:960Paolo Guiotto: translations.
22:06:30Paolo Guiotto: And, so this means that if you take a set E, and to each point of the set, you have the same vector V, well, this is the…
22:16:110Paolo Guiotto: Same of the measure of E, so for every E subset of RM,
22:21:970Paolo Guiotto: And for every factor V of RM.
22:26:690Paolo Guiotto: Also, this one is, is, easily verified, because, the point is that if you have your set E,
22:36:370Paolo Guiotto: and you take a translated of him, of course, they should be the same, so imagine that this is E, and this is E plus V. What you see is that whenever you cover E,
22:48:180Paolo Guiotto: you translate the covering, and you cover E plus V, and vice versa.
22:53:470Paolo Guiotto: So they have the same coverings, and since the area, the area, the area, the modulus of I, to be clear, that area is invariant by translations, because if you look at the definition.
23:11:200Paolo Guiotto: Imagine that you do a translation of an interval.
23:15:330Paolo Guiotto: You see that the area? What is the foreground?
23:20:470Paolo Guiotto: The formula for the area depends on the length of the interval, and if you translate the interval, the length does not change.
23:28:590Paolo Guiotto: This one, which is common. So, they do not depend, they are invariant by translation, and therefore, also, the lambda star is invariant by translation. Okay?
23:44:520Paolo Guiotto: Another property that can be verified, so…
23:47:790Paolo Guiotto: If we look at this point, we have that lambda star of empty is zero, no?
23:54:550Paolo Guiotto: If it is countably additive.
23:56:760Paolo Guiotto: Then we know that it coincides with the geometric measure of intervals, it is invariant by translations.
24:04:670Paolo Guiotto: Seeing here, this would be the solution of the problem. We have a geometrical measure on RM that is coherent with the natural measure of intervals, okay? So the missing point is comfortable additivity.
24:22:730Paolo Guiotto: What can be proved relatively easily, but it is non-trivial, there is a proof on the notes I don't want to do here, is that this lambda star
24:34:830Paolo Guiotto: is countably.
24:38:330Paolo Guiotto: Sub.
24:39:670Paolo Guiotto: Additive.
24:41:180Paolo Guiotto: So the property we have seen before. So, if we have lambda star of a union of sets y n, this is less or equal than sum of the measures of union… of sets CN.
24:58:900Paolo Guiotto: But the additivity for the disjoint sets, it's difficult. What can be proved, for example, is that if the sets are well separated.
25:10:630Paolo Guiotto: I don't want to be precise, what does it mean, well separated, but let's say… so the question is, countable.
25:19:370Paolo Guiotto: Additivity…
25:21:640Paolo Guiotto: Well, it can be proved that if you have two sets, like this, E and F, and they are well separated.
25:30:220Paolo Guiotto: So, you see, they are disjointed, but they shouldn't be in touch.
25:36:490Paolo Guiotto: Okay? So, for example, imagine that you have, a disturb.
25:42:550Paolo Guiotto: You eliminate the boundary, and you take the boundary, the circle.
25:46:750Paolo Guiotto: Those two sheds, the disc without the boundary, and the boundary, they are in contact, even if they are disjointed, no? The skin of the orange and the orange inside, no?
25:56:900Paolo Guiotto: If you take two separate sets, they are disjoint, but they are in contact. Those are not good sets. You have to take two sets, they are well separated. Now, I don't want to exactly tell you what does it mean. It means that the distances between points is bounded below in something.
26:12:700Paolo Guiotto: It's something like if they are two well-separated items, okay? Now, in this case, it happens that the measure of the disjoint union is the sum of the measures. So let's say that counterpart activity works when sets are well separated.
26:28:320Paolo Guiotto: But, in general, and now we have… we have, your time… our time is over, but whoever, let's say, in general.
26:39:50Paolo Guiotto: Lambda is not.
26:43:600Paolo Guiotto: Countable.
26:46:50Paolo Guiotto: We do.
26:47:690Paolo Guiotto: Well, maybe we will return on Monday on this. I don't want to do the example, this is a so-called Vitaly
27:01:80Paolo Guiotto: TOM.
27:04:200Paolo Guiotto: That actually shows something more in general. It is impossible to build a measure which is invariant by translations, and the measure of rectangle is the area.
27:14:940Paolo Guiotto: that can measure every subset of RT, RN.
27:19:330Paolo Guiotto: Okay? The problem is that, to show that it is not countable additive, you have to build a family of disjoint sets, such that the measure of the union is not the sum of the measures.
27:32:80Paolo Guiotto: And precisely, it is less, because less or equal is always true because of this, no? So there are cases where you have peace, so it may happen that lambda star of this joint union of yen can be
27:48:450Paolo Guiotto: Strictly less than the sum of the outer measures of the yen.
27:54:470Paolo Guiotto: Here, there is a long story of paradoxes. For example, Banach and Tuske proved that in dimension 3, it is even possible to build a disjoint union of a finite set, like a bowl.
28:12:130Paolo Guiotto: And making such a way that the sum of the sizes, so the sum at right is the double of the sum at left. So it is like, a very strange thing, it is like if you take a ball, you can slice the ball in a finite number of slices.
28:27:440Paolo Guiotto: And putting together, summing the volume of the slices, you get the double of the volume of the bowl, which seems impossible, because if you take an apple and you slice, it seems that if you sum the volumes of the slices, you should get the volume of the apple.
28:43:500Paolo Guiotto: But they say it is possible to slice the apple in a complex way, that you get the double of the volume, which is really strange.
28:51:270Paolo Guiotto: Okay, so this is a very long, complex, technical article, but Vitaly is an example that can be done in few lines. The unique problem is that there is no intuition about that example.
29:04:750Paolo Guiotto: So you don't see what is… how is made this family of sets such that the measure of the union is 3 less than the sum of the measures, even if they are disjoint.
29:15:310Paolo Guiotto: Okay? It's really counterintuitive, and therefore, I don't think we are going to give it a look carefully, but if you are curious on the notes, you can find it, so give a look. Okay, so have a nice weekend, and see you on Monday.