AI Assistant
Transcript
00:04:680Paolo Guiotto: Okay, good morning, Gary.
00:10:620Paolo Guiotto: This morning, I will start with some of the exercises I left you.
00:16:920Paolo Guiotto: I will publish the solutions of all the exercises I left.
00:22:250Paolo Guiotto: Today afternoon. So, let's start with this one, exercise 231.
00:29:570Paolo Guiotto: It's, a minimal exercise. So it says N contained in R,
00:39:440Paolo Guiotto: and measure zero set. Null set means measure zero set, so the Lebec measure of N equals 0.
00:48:90Paolo Guiotto: So that, then, this is what we have to prove, be complementary of N is dense.
01:00:430Paolo Guiotto: In R… So this means what? It means that, that is… Whatever is the interval, AB.
01:14:40Paolo Guiotto: In R… This interval contains points of the complementary of N, so the intersection within complementary is non-empty.
01:34:380Paolo Guiotto: Okay? Now, this is, an easy exercise. Assume that the conclusion is false.
01:45:90Paolo Guiotto: If full, sir?
01:47:980Paolo Guiotto: So we put every data. It is not true that for every interval.
01:52:950Paolo Guiotto: AB, the intersection with N complementary, is
01:57:270Paolo Guiotto: non-empty. So it means that there exists at least one interval for which the intersection is empty.
02:04:690Paolo Guiotto: Okay? So this, this is the…
02:08:680Paolo Guiotto: negation of this property. So there exists in interval, AD, such that…
02:15:400Paolo Guiotto: intersection between AD and N on the 3, is empty.
02:22:920Paolo Guiotto: But then, we know that if there are no points of AB in N complementary, they must be in N.
02:32:90Paolo Guiotto: So, AB must be contained.
02:35:460Paolo Guiotto: Ian, N…
02:39:110Paolo Guiotto: And, well, of course, we are assuming that A is less than B, no, otherwise the…
02:47:150Paolo Guiotto: everything would be… so… the property here wouldn't be profitable enough with A equals B.
02:57:560Paolo Guiotto: Okay, but this is impossible, because then we would have that the measure of ABE,
03:04:580Paolo Guiotto: which is, its length, in this case B minus A, which is positive, should be less frequent than the measure of N, which is supposed to be equal to 0. And this is a contradiction.
03:19:360Paolo Guiotto: Impossible.
03:22:860Paolo Guiotto: So this, grosses the RBA.
03:30:330Paolo Guiotto: The next… the size is the…
03:35:360Paolo Guiotto: 232. It is an exercise on the so-called canto.
03:42:110Paolo Guiotto: Sette.
03:43:960Paolo Guiotto: The counter set is obtained in this way.
03:47:500Paolo Guiotto: You take the interval 0, 1.
03:52:600Paolo Guiotto: And you remove, you start at least removing the middle third, so it means you divide it in 3 equal parts, so this will be 1 third, two-thirds.
04:03:950Paolo Guiotto: And you remove this one. So, the step one, if you call this set, let's say, C0, C1 is the set made of two intervals from 0 to 1 3rd.
04:17:269Paolo Guiotto: And from 2 thirds to 1.
04:20:220Paolo Guiotto: Endpoints included.
04:23:490Paolo Guiotto: Then you repeat the same construction. You divide each of these two intervals in three equal parts.
04:29:550Paolo Guiotto: So this is, something like, so one-third, this becomes our nine.
04:36:250Paolo Guiotto: so this is 2 over 9, and so on.
04:42:200Paolo Guiotto: And you remove this part here.
04:46:30Paolo Guiotto: So you produce a new… set C2, which is made of these 4 intervals.
04:59:790Paolo Guiotto: 1 over 3 can be 3 over 9.
05:04:180Paolo Guiotto: And so on.
05:08:60Paolo Guiotto: And you iterate this procedure. Now, from each of these four intervals, you remove the middle third, and you will produce something like 8 intervals.
05:20:440Paolo Guiotto: which, smaller length. You have the set C3.
05:25:260Paolo Guiotto: I'll imagine that we can repeat this operation extremely many times.
05:30:610Paolo Guiotto: What happens?
05:32:80Paolo Guiotto: It seems that it remains nothing, because every time we eliminate a consistent part of each of the intervals, but in fact, it's not true, because, for example, you see that these endpoints, they always remain, so zero remains.
05:48:180Paolo Guiotto: 1 over 9 remains, one… so all the endpoints remain… in fact, at the end, this is the set made of the endpoints of these intervals, okay?
05:57:390Paolo Guiotto: Now, how can we define this operation? Well, we define the set C,
06:03:160Paolo Guiotto: Imagine we have defined a CN for every N with this procedure. There is a formula. We actually don't need to write down exactly what is it to do the exercise.
06:15:30Paolo Guiotto: So CN, however, we can say, is made… is made of…
06:22:610Paolo Guiotto: You see that C0 is one interval, C1 is 2 intervals, C2 is 4 intervals.
06:31:960Paolo Guiotto: C3 will be 8 intervals, so you understand that at step n, you will have 2 to the n intervals, because you see, these are the powers of 2, no? So.
06:43:550Paolo Guiotto: one interval.
06:46:300Paolo Guiotto: to intervals, 4 intervals.
06:52:720Paolo Guiotto: 8 intervals.
06:55:220Paolo Guiotto: And so on. So at step n, you will have 2 to the N.
07:02:360Paolo Guiotto: And what is important here is that they are a closed interval.
07:08:240Paolo Guiotto: Also, it's a remark we can do to the unclosed intervals.
07:21:520Paolo Guiotto: Each of these intervals is going to have a smaller size. You see that at step zero, the size of the interval, we have one interval is 1. Here we have two intervals of length one-third each, right?
07:40:300Paolo Guiotto: Now, we have 4 intervals of length 1 over 9.
07:45:750Paolo Guiotto: Okay, then we will have 1 over 27, and so on. These are following powers of 3. So, they are 2 to the enclosed intervals, of…
07:56:660Paolo Guiotto: length… 1 over 3 to the end each.
08:05:540Paolo Guiotto: Now, this is not yet the Kanto set, but the Canto set is the set made by the intersection of all these sets, CN, when n goes from 0 to infinity.
08:17:670Paolo Guiotto: You may also notice that these sets
08:22:180Paolo Guiotto: Are a decreasing family of sets.
08:24:980Paolo Guiotto: Because C1 is obtained by C0 eliminating something. So C1 is definitely contained into C0. And C2 is obtained from C1 by taking out some subsets.
08:37:710Paolo Guiotto: So you have a… that this is actually a decreasing sequence. So, we may say that that C is the limit set of sets, yeah, in the sense we have C for a decreasing sequence of sets.
08:50:440Paolo Guiotto: Now, the goal of the exercise is that, show that.
08:56:310Paolo Guiotto: C is LeBag measurable, so it belongs to the class N1, and compute Computer each measure.
09:09:700Paolo Guiotto: These are not particularly complicated questions, so let's see the number one.
09:15:590Paolo Guiotto: So here we could do in two ways.
09:19:630Paolo Guiotto: The first way is, of course, we have that each of the CL is a measurable set, because they are made of fine tunes of intervals, so
09:32:800Paolo Guiotto: Each.
09:34:590Paolo Guiotto: CN belongs to the class of measurable sets, and here we are doing a countable set operation on the CN.
09:45:430Paolo Guiotto: So then you know that the family is closed for accountable unions. We have not formally seen that it is closed for intersections, but since intersection unions are obtained by doing the complementary, you can imagine that
09:57:120Paolo Guiotto: If it is closed for intersections, it will be closed for… sorry, if it is closed for unions, it will be closed for intersections as well. So, we can have automatically that this intersection of Siena, it's, of course, a countable intersection, because then it's natural, belongs in M1, and this is just because M1
10:20:320Paolo Guiotto: Is that Sigma algebra.
10:25:420Paolo Guiotto: This is an argument.
10:27:420Paolo Guiotto: Alternatively?
10:33:270Paolo Guiotto: This is, let's say, perhaps a more subtle argument, because we may notice that each of the sienna is made of
10:43:810Paolo Guiotto: Again, a union, a closed… a finite union of closed intervals.
10:51:960Paolo Guiotto: from the topological point of view, this is a closed set, okay? So we have that each…
11:01:210Paolo Guiotto: CN is closed, huh?
11:06:600Paolo Guiotto: If you want. Closed means the complementary is open, no?
11:09:980Paolo Guiotto: Each CN is closed, and there is a property that says whenever we have a family of closed sets, and we do the intersection, we still get a closed set. So the intersection of the CN
11:24:950Paolo Guiotto: is closed.
11:28:890Paolo Guiotto: And, we know that the LeBague class contains all open and closed sets, so this belongs to M1.
11:40:650Paolo Guiotto: Okay.
11:42:240Paolo Guiotto: Number two, let's discuss about the measure.
11:46:340Paolo Guiotto: Now…
11:49:400Paolo Guiotto: I told you, normally it's difficult to compute measures exactly, no? Sometimes we can prove that the measure is zero because we bound the measure with something that can be shown to be extremely small, a bit extremely small.
12:03:520Paolo Guiotto: Now, here what we notice is that, since we already noticed that this sequence, CN, is a decreasing sequence of sets.
12:14:180Paolo Guiotto: Okay, so we may say, can we use the continuity, of the measure? Here, we should use the continuity from above, so you know that that's not always true, but it becomes true as soon as we have that one of the sets has final measure. And this is the case because of, for example, the C0 is the interval 0, 1,
12:37:680Paolo Guiotto: It does measure one. So we can say that Since,
12:42:920Paolo Guiotto: the measure of C0 is the measure of 0, 1.
12:48:830Paolo Guiotto: It is equal to 1, so it is finite.
12:52:980Paolo Guiotto: And, the sequence CN is going down to C, we apply it, The… continuity… from… Above, huh?
13:15:740Paolo Guiotto: And therefore, we say that the measure of C is the limit
13:20:590Paolo Guiotto: in any of the measures of CN.
13:25:80Paolo Guiotto: Or, in alternative, since at the end, we will get value 0, we don't need to use the continuity, necessarily. We could also observe that lambda C
13:36:700Paolo Guiotto: since it's contained into CN, it is less or equal than lambda CN, and since lambda CN, as you will see, goes to zero, you will get the same conclusion without the need of this problem. However, since we have it, why don't we use
13:53:750Paolo Guiotto: Now, what is the lambda of CN? Now, we said that each CN is made of
14:02:140Paolo Guiotto: 2 to the n closed intervals, each of length 1 over 3 to the n.
14:08:500Paolo Guiotto: So this is this joint union of intervals. We can say that the lambda of CN, so we keep the limit, yes.
14:16:890Paolo Guiotto: limit.
14:18:740Paolo Guiotto: In N of what? So, since they have all the same measure, 1 over 3 to the n, and they are 2 to the n, this will be 2 to the n times 1 over 3 to the n.
14:32:00Paolo Guiotto: And it is clear that this is a limit of, two-thirds.
14:36:890Paolo Guiotto: to Vienna, so it's a geometric sequence with a ratio less than 1, between 0 and 1, and the limit is 0. So this says that the measure of C is 0.
14:50:330Paolo Guiotto: In alternative, as I said, in alternative.
14:54:510Paolo Guiotto: we could have noticed that lambda of C is less or equal than lambda of CN.
14:59:670Paolo Guiotto: Because C is, of course, contented into CN, and since this is going to zero, the other one is a constant, which is less than something that goes to zero, it's greater or equal than zero, it cannot be anything else than zero. Okay?
15:15:690Paolo Guiotto: So this is the kind of argument that does not need the continuity from above, okay?
15:23:530Paolo Guiotto: So, in any case, we have the conclusion.
15:31:130Paolo Guiotto: Okay, now… Another exercise I left you would be 235.
15:42:230Paolo Guiotto: Here we have a set E,
15:46:170Paolo Guiotto: which is defined in this way. It is a set of points, XY in R2, Such that,
15:56:580Paolo Guiotto: There exists a pair of integers, M and N, both natural, Different from 0, 0.
16:08:410Paolo Guiotto: Such that,
16:10:410Paolo Guiotto: MX plus NY is equal to 0.
16:16:680Paolo Guiotto: This, so this is sort of linear combination of X and Y is equal 0, so they are linearly dependent.
16:25:410Paolo Guiotto: Since the coefficients are integers.
16:29:00Paolo Guiotto: We can say that's not exactly linear dependence, but we say something like, we can put the rational coefficients in front, not just any real numbers, because if you divide by N,
16:42:340Paolo Guiotto: you have a rational number times X plus rational number times Y. So, it means that they are linearly dependent if we use the coefficients in rational sense. This is…
16:54:860Paolo Guiotto: Justin.
16:56:50Paolo Guiotto: terminology. However, this is not important for the exercise. It has to prove that the measure
17:06:710Paolo Guiotto: The measure that in R2 of this is equal to 0.
17:13:00Paolo Guiotto: Now, you understand why there is not the pair 0, because if I put M and then equals 0, 0X plus 0y is 0, so that becomes 0 equals 0, which is always verified, and that set would be the planar 2, okay?
17:30:120Paolo Guiotto: Okay, so, what can we do here?
17:37:860Paolo Guiotto: We may notice that if we fix a pair, M and N, so if I pick a pair MN in N2,
17:48:830Paolo Guiotto: Different from 0, 0.
17:52:660Paolo Guiotto: D… Seth.
17:55:900Paolo Guiotto: let's say EMN, made of points where MX plus NY is equal to 0, What is this set?
18:15:270Paolo Guiotto: And then I'll fix. So, 3X plus 2Y equals 0 is…
18:20:660Paolo Guiotto: A straight line, no? So, it's a straight line by the origin, so something like this.
18:28:250Paolo Guiotto: So this is a typical EMN, set.
18:33:550Paolo Guiotto: for certain combination of M, and then we can get to the X and Y axis, no? So if I have 1x plus 0Y means X equals 0, no? It is the y-axis, and so on. So we have all straight lines passing through the origin.
18:51:890Paolo Guiotto: If you think about this as not all the straight lines, but they are straight lines with the rational angular coefficient, basically.
19:01:590Paolo Guiotto: Okay, so in our set, E is the union of these sets BMN.
19:10:550Paolo Guiotto: on all pairs, MN… in N2… Except the pair 0, 0.
19:21:20Paolo Guiotto: Now, of course, this is a countable union, because the set is, you understand, is countable.
19:27:540Paolo Guiotto: And therefore, they are not disjoint because there is a common point, which is the origin, so we… but in any case, we could say that the measure of E will be less or equal by sub-addivity of the sum of the measures of the sets EMN.
19:46:720Paolo Guiotto: And now, what about this measure? Here, we are talking about the two-dimensional measure.
19:52:890Paolo Guiotto: So the Lebec measure of R2, no? What about the LeBague measure of these sets?
19:59:870Paolo Guiotto: is zero, no? The measure of a line in plane is zero. Let's say the area, no? That's the area. And so we have a zero. So the measure…
20:11:330Paolo Guiotto: And of this side is equal to 0.
20:15:970Paolo Guiotto: Okay, now, I don't know if you've tried…
20:20:140Paolo Guiotto: But let's take you at least one or two of the three-star problems, like the 237.
20:31:180Paolo Guiotto: Let's say this is, they are elementary from the point of view, especially the 237, of what you need to do, but they can be tricky, okay? So here we have this. We have 3 sets, A, B, C,
20:48:600Paolo Guiotto: in 01.
20:52:750Paolo Guiotto: With the following property, such that… for… Every action.
21:04:70Paolo Guiotto: Of the interval 0, 1, X belongs to…
21:15:290Paolo Guiotto: 2… At least.
21:22:740Paolo Guiotto: Two sets.
21:25:770Paolo Guiotto: The mom was like… A, B, C.
21:32:170Paolo Guiotto: So, you pick an X, and this says it belongs to at least two of these three sets. Maybe A and B, maybe A and C, maybe B and C, okay?
21:44:200Paolo Guiotto: Now, the problem asks to prove that, check that, necessarily, the measure of each of these three sets, so the measure of A, as well as the measure of B, as well as the measure of C, must be greater or equal than 2 thirds.
22:04:20Paolo Guiotto: And,
22:05:870Paolo Guiotto: it gives you a hinder, which is basically nothing but a rewriting of what is this thing. So it's not such a big hinder. But maybe…
22:18:360Paolo Guiotto: It's just interesting, how can we write this box properly?
22:24:720Paolo Guiotto: Okay, so this is written in words, no?
22:27:730Paolo Guiotto: It seems easy, but we have to put in a formal way, because we have to say something about measures, so this is an information about measures, some sense, that we have to translate.
22:42:20Paolo Guiotto: So, okay, we know that every X in 01 belongs at least to two of these sets.
22:49:250Paolo Guiotto: So, this means that the set, the solution We know.
22:59:30Paolo Guiotto: That, 01,
23:04:300Paolo Guiotto: So every point of 01 must be in at least 2, so it must be either in a intersection D, so ina and in B,
23:14:400Paolo Guiotto: So let's write this intersection, or it can be in A and C, or it can be in B and C.
23:26:30Paolo Guiotto: Now, since you must be in at least the…
23:28:930Paolo Guiotto: two of these three sets, it means that you must be in at least one of these intersections, no?
23:36:30Paolo Guiotto: Because staying in the first one means you are in A and C. In the second, you are in A and… sorry, in A and B. In the second, A and C. In the third, B and C. So it means that 01 is… that assumption says that 01 must be contented into the union of these three sets.
23:56:270Paolo Guiotto: And actually, since they are contained in 01,
24:01:700Paolo Guiotto: So A is contained in 01, B as well, so the intersection will be a subset of 01. All these are subsets of 01, so this is contained in 01,
24:13:160Paolo Guiotto: This is contained in 01.
24:16:280Paolo Guiotto: And this is also contained in 01. So everything is in 01.
24:22:320Paolo Guiotto: So, I deduced that 01 He is,
24:30:00Paolo Guiotto: the union of A intersection B, union A intersection C, union B intersection C.
24:39:730Paolo Guiotto: That's, the way to rewrite that assumption, no? Every X belongs in at least two sets of these three. Okay, so this is the assumption written in…
24:56:590Paolo Guiotto: set for.
24:58:360Paolo Guiotto: And so, in particular, this means that if you computed the measure, and this is the hint, you apply lambda to both sides, you get one equal measure of a intersection.
25:12:220Paolo Guiotto: Sorry, A, intersection B.
25:15:410Paolo Guiotto: Union, maybe we put some parentheses, A intersection C,
25:21:790Paolo Guiotto: union B intersection C. You know that we don't need, normally, to use parentheses with these set operations, because it is like sum and product.
25:33:100Paolo Guiotto: So whenever you have product and sum, if there are no parentheses, it means that you do first the product, and then you do the sum, okay? But just to be clear, let's write this in this way.
25:45:490Paolo Guiotto: Okay, so this is basically the assumption.
25:51:470Paolo Guiotto: Well, in fact, since everything depends on this, the problem could have been stated in a slightly weaker form. Instead of saying every X belongs into at least two, we could have said almost every X
26:06:980Paolo Guiotto: Okay, so the set of X for which you are not into is a measure 0 set, okay? However, let's take this as initial fact.
26:17:790Paolo Guiotto: So this is, basically the hypothesis.
26:22:850Paolo Guiotto: And, let's write what we have to discover, that the measure of A, measure of B,
26:30:880Paolo Guiotto: a measure of C, they are all greater or equal than 2 thirds. The problem suggests you to argue by contradiction, so assume that this is false, and let's see what happens.
26:44:710Paolo Guiotto: Okay?
26:46:350Paolo Guiotto: You look at me, I see that. Okay, I haven't saying anything strange?
26:52:630Paolo Guiotto: Okay, so, by contradiction, Let's start writing, this time we see what we can do. Contradiction.
27:06:910Paolo Guiotto: we assume, huh?
27:11:610Paolo Guiotto: This is false.
27:14:760Paolo Guiotto: And this means that if it's not true that all of them
27:21:510Paolo Guiotto: I'm sorry, sorry, the thesis is actually at least one of them.
27:27:820Paolo Guiotto: Right, I'm, I, I… Let's say that this is one off.
27:37:100Paolo Guiotto: So the thesis is one of…
27:42:500Paolo Guiotto: So, this is false, means all of them less than 200.
27:48:800Paolo Guiotto: So, lambda A… Lambda B.
27:54:00Paolo Guiotto: lambda C is 33 less than 2 thirds. Let's see what happens. Now, so far we have just,
28:02:640Paolo Guiotto: Written the hypothesis, or at least the consequence
28:07:170Paolo Guiotto: Of the assumption, and basically written what does it mean that the thesis is false. Now we have to try the way to
28:15:990Paolo Guiotto: Find a contradiction.
28:21:870Paolo Guiotto: So Do you have any idea?
28:26:60Paolo Guiotto: the intersection of, the center, the Union is equal to Guana, so the intersection is contained in,
28:37:840Paolo Guiotto: intersection of A and B, it's moving in A.
28:42:160Paolo Guiotto: Yeah. So, the sun must be one.
28:48:630Paolo Guiotto: So, you're doing union, some… Okay, let's try, but, you know, when we start, we never have a clear idea, okay? We try, and then we correct, and we arrive a solution.
29:01:170Paolo Guiotto: So what are you saying? Because you are saying union, some… Italian Italian, okay.
29:14:400Paolo Guiotto: will be… una, boister.
29:37:80Paolo Guiotto: Okay, just, it is not completely clear what is the idea.
29:45:480Paolo Guiotto: At least I do not understand what you…
29:52:850Paolo Guiotto: what you want to say. Perhaps there is something
29:57:430Paolo Guiotto: if I do not understand the…
30:01:190Paolo Guiotto: if I understand correctly, is that from this, let's say, star.
30:07:170Paolo Guiotto: We know also another information, because what about the measure of the union A, union B, union C?
30:18:100Paolo Guiotto: What do we know?
30:20:270Paolo Guiotto: Spray the book.
30:23:970Paolo Guiotto: It's greater than 1.
30:27:930Paolo Guiotto: It's 1. That's, why? Because clearly, all of them are contained in 01. So this is a subset of 0, 1, and therefore, it is, definitely less or equal to 1.
30:42:570Paolo Guiotto: But of course, it contains… this thing contains this one.
30:48:440Paolo Guiotto: No? Because A intersection B, for example, is contained in A, A intersection C is contained in C, and B intersection C is contained in B. So that intersection is contained into… sorry, that union, the union you see here, is definitely contained into this union.
31:06:720Paolo Guiotto: And therefore, the measure of this will be larger than the measure of that, which is
31:11:580Paolo Guiotto: Actually, equal to 1, larger of the measure.
31:14:910Paolo Guiotto: we have here, which is equal 1. So we deduced that the measure of the union
31:22:440Paolo Guiotto: A union B, union C is also equal to 1.
31:28:580Paolo Guiotto: Okay. Now, what would I try? Since we have a measure of the union, let's see, and we need to get some information on AB…
31:40:520Paolo Guiotto: Let's try to write the measure of the union in terms of the sum of the measures. We know this is not a disjoint union, so we cannot say it is the sum of the lambda A plus lambda B plus lambda C, but let's see what comes out. So we have that 1 is the measure of A union B, union C.
32:00:770Paolo Guiotto: Okay, so how the addition formula works when the union is not disjoint?
32:07:10Paolo Guiotto: maybe… we have done this, I don't… I don't mind if we have done.
32:12:760Paolo Guiotto: Have we never done this?
32:14:920Paolo Guiotto: We have done for 2, no? You reminded that for 2 it happens that lambda of A union B, well, that's actually for any measure. Mu of A union B is mu of A plus mu of B
32:28:330Paolo Guiotto: And this is the formula when they are disjoint.
32:31:150Paolo Guiotto: But if there is an intersection, you have to subtract the measure of the intersection.
32:37:190Paolo Guiotto: Now, what is… what happens to this formula when we have a union of three?
32:41:900Paolo Guiotto: Now, it happens that this is the measure, the sum of the measures, so lambda A plus lambda B plus lambda C.
32:51:400Paolo Guiotto: Imagine that we have two, so there would be a subtraction of the measure of the intersection, two by two. So we have minus the measure of A intersection B, minus the measure of A intersection C, minus the measure of B intersection C.
33:12:190Paolo Guiotto: But then we are subtracting too much.
33:15:460Paolo Guiotto: And we have to re-add the measure of the intersection of the three, A, intersection, B intersection C. This is the formula. That was an exercise. I don't… I do not remember if I knew to do, and I wrote the solution at this moment.
33:32:960Paolo Guiotto: But, it's, but probably, we have done in Placer.
33:38:590Paolo Guiotto: Do you remember anything of this? Oh, however.
33:41:880Paolo Guiotto: It can be obtained by this one,
33:44:790Paolo Guiotto: you apply this, you do A, union B, union C. You start doing this, that's a union of two sides, no? A union B, and C. You apply this formula with A, union B,
33:57:710Paolo Guiotto: This is an element, union C. Then you iterate, because you will have a union here, you will iterate the formula of union of 2, and you will get this one. This is for any measure, not for the LeBerg measure in particular.
34:11:610Paolo Guiotto: So, as you can see here, we have lambda A+.
34:17:219Paolo Guiotto: Lambda B plus.
34:19:739Paolo Guiotto: Lambda C,
34:21:290Paolo Guiotto: Then, that term with the minus, so let's put in a parenthesis, is lambda A intersection B
34:29:460Paolo Guiotto: plus lambda A intersection C plus lambda B intersection C,
34:38:739Paolo Guiotto: And then we have a plus lambda A intersection B intersection C.
34:45:889Paolo Guiotto: Now, if you look at this guy here.
34:49:409Paolo Guiotto: It is the sum of the measures
34:52:429Paolo Guiotto: Of the sets we have inside exactly this relation.
34:57:50Paolo Guiotto: So now I would try to find a connection between this formula here, this one, and the sum we have done here.
35:07:500Paolo Guiotto: In fact, you see that that formula says that one is the measure of the union of three sets, which are exactly these three intersections.
35:18:630Paolo Guiotto: Now, I apply, once again, the formula for the measure of the union of 3, but not for ABC.
35:25:920Paolo Guiotto: but rather for A intersection B, A intersection C, and B intersection C. In such a way, I get the connection between this term and the yellow underlined term. So, I have that lambda of
35:39:340Paolo Guiotto: A intersection B.
35:42:420Paolo Guiotto: So, this is the first set. Union.
35:45:310Paolo Guiotto: A intersection C. This is the second set, union. B intersection C, this is the third set.
35:52:640Paolo Guiotto: This is, reminder philosophy, this is not necessarily a disjoint unit, even here we have common parts, probably. So we will have the sum of the measures, so measure of A intersection B plus measure of A intersection C,
36:10:260Paolo Guiotto: plus measure of B intersection C. And that's exactly this thing.
36:17:60Paolo Guiotto: Okay?
36:18:250Paolo Guiotto: Then I have minus… you reminded the… the… how this formula works, no? I have the sum of the measures minus the sum of the measures of the intersection, 2 by 2, but here what happens when you do the intersection, this with this.
36:35:530Paolo Guiotto: So you do lambda of A intersection B, intersection with A intersection C.
36:43:380Paolo Guiotto: You see, what is this?
36:45:250Paolo Guiotto: A, intersection. B,
36:47:910Paolo Guiotto: intersection with A intersection C. So you must be in A, in B, in A again, and in C. You must be in A, B, and C. So that's the measure of A intersection B, intersection C.
37:05:730Paolo Guiotto: Then I have minus… Measure of, we have done a first intersection second. I do first intersection third.
37:14:390Paolo Guiotto: That's A intersection B, in the section B intersection C.
37:18:880Paolo Guiotto: So it's again A intersection B intersection C. So I have another A intersection B intersection C. And then, finally, minus, I have to do second intersection third.
37:32:520Paolo Guiotto: But this is A intersection C intersection B, intersection C. Again, A intersection B, intersection C.
37:41:370Paolo Guiotto: So, as you can see, it's very tricky, but we are not doing anything particularly technical, okay?
37:49:860Paolo Guiotto: It's not yet over, because now the last term is the measure of the intersection of the three sets.
37:57:690Paolo Guiotto: So, plus lambda off.
38:01:50Paolo Guiotto: First, intersection, second, intersection third. But you see what happens when we do the intersection of the three sets here? So, the… the one…
38:12:680Paolo Guiotto: The 2 and the 3?
38:16:320Paolo Guiotto: When we do the intersection of these three, we have A, intersection B, intersection C. So we have plus.
38:23:30Paolo Guiotto: A, intersection B, intersection C.
38:26:640Paolo Guiotto: So this means that we get lambda of A intersection B plus lambda of A intersection C plus lambda of B intersection C, then we are many times this lambda of ABC, which is minus, minus, minus
38:46:660Paolo Guiotto: Blast, sir.
38:48:600Paolo Guiotto: So at the end, we have, minus 2.
38:53:550Paolo Guiotto: lambda of A intersection B intersection C.
39:00:140Paolo Guiotto: What is this? This is the measure of this unit.
39:04:740Paolo Guiotto: That, the cause of the assumption is 1.
39:10:420Paolo Guiotto: Okay, so this is the hypothesis.
39:14:970Paolo Guiotto: This thing is 1. And here we can extract this term, which is what we have in this expression. So we get the data, so…
39:25:670Paolo Guiotto: Let's write this relation. Lambda of A intersection B plus lambda of A intersection C plus lambda of B intersection C is equal to 1 plus
39:42:340Paolo Guiotto: to lambda of A intersection B intersection C.
39:49:650Paolo Guiotto: Now, what we do is we plug this…
39:52:650Paolo Guiotto: into this relation. Remind that this is 1 equal all this thing.
40:01:70Paolo Guiotto: So, replacing it, the term is this one we are going to replace in this formula.
40:10:160Paolo Guiotto: So we get… So…
40:19:580Paolo Guiotto: So 1 is equal lambda A plus Lambda B.
40:26:470Paolo Guiotto: plus lambda C, now we have… minus… These things.
40:34:430Paolo Guiotto: Which is 1 plus 2, So, 1 plus 2 lambda of A intersection B intersection C,
40:46:10Paolo Guiotto: And the last term, you do not forget of this one, plus another lambda of A, the section B, the section C.
41:02:660Paolo Guiotto: So we get, lambda A, plus lambda B, plus lambda C, Then we have minus 1.
41:14:110Paolo Guiotto: Then we have minus 2, plus 1, minus 1,
41:18:240Paolo Guiotto: minus lambda of A intersection, B intersection C. And now we have, all we needed for the conclusion, because from this,
41:29:270Paolo Guiotto: we extract this term, lambda A plus lambda B plus Lambda C.
41:38:350Paolo Guiotto: This is equal to, carry this thing on the other side. It is equal to 2 plus lambda of A intersection B intersection C.
41:51:500Paolo Guiotto: Now, you see that we have not yet used the contradiction assumption, but it is here where it comes.
41:57:770Paolo Guiotto: Because… The contradiction assumption is, if all three are less than 2 thirds, here you see that,
42:08:330Paolo Guiotto: If these… So, one… 2 and 3.
42:16:610Paolo Guiotto: Are less than two-thirds.
42:19:670Paolo Guiotto: You see that the left-hand side is the sum of 3 numbers less than 2 thirds, so I would have that the left-hand side would be less than 3 times 2 thirds, and this is 2.
42:35:620Paolo Guiotto: But the right-hand side is larger or equal than 2.
42:45:850Paolo Guiotto: Because it is 2 plus a measure, which is a positive quantity.
42:49:810Paolo Guiotto: So we would have that left-hand side is less than 2, right-hand side is greater than equal 2, it cannot be equal. And here, they are equal.
42:58:840Paolo Guiotto: You see? So this is a contradiction.
43:07:430Paolo Guiotto: And therefore, it means that the, the contradiction assumption cannot be verified.
43:17:570Paolo Guiotto: Me.
43:19:900Paolo Guiotto: Is that tricky? Have you got any solution of this problem?
43:26:300Paolo Guiotto: It's another way, but…
43:31:150Paolo Guiotto: I thought that being said, the X belonging to A only is empty and disabled for B and C, and then… And why, why assume that in this end?
43:46:100Paolo Guiotto: They are all in the intersection.
43:50:280Paolo Guiotto: The pitch as close to at least
43:53:550Paolo Guiotto: Okay, you show me your solution. I want to see if… No, no, okay, show me next. These, if you want,
44:05:560Paolo Guiotto: Because what you say, it's not…
44:09:710Paolo Guiotto: So much for convincing me. But, like, the last lines of my conversation, I would say, it's just that I use some different types.
44:20:230Paolo Guiotto: Okay, let's see here, okay? An important thing is that, of course, for example, you could…
44:27:360Paolo Guiotto: try to do this a little bit, much easier. If you assume that ABC are intervals, no, because in that case, intersections are intervals, and you can… you can do that. But of course, that wouldn't be,
44:42:650Paolo Guiotto: the requirements, because here it is not specified if APC are in the labs, okay?
44:50:830Paolo Guiotto: Okay, in any case, if you have any solution, that you want to check, please show me, or you can send me by email, I will give a look. Okay.
45:02:600Paolo Guiotto: Okay, so, so we spent the first hour on this. I would say that at the moment it's okay. If you have not yet done, do exercise 2.3.8, which is still
45:18:470Paolo Guiotto: A bit complicated.
45:22:130Paolo Guiotto: at least… at least the first part, that's the important part, you know? You have a subset of 0, 1 with the measure equal to 0, and you take the set made of squares of these numbers, no?
45:41:280Paolo Guiotto: So, it says you have an N contained in 01,
45:46:720Paolo Guiotto: with the measure of n equals 0.
45:50:870Paolo Guiotto: I… I… a typical error working with Measure Zero Set is to think that they are necessarily countable.
45:58:750Paolo Guiotto: So, typically, people say, measure zero means that the set must be comfortable. It is true that the set comfortable, at least for the back measure, will have measure zero, but that's not an if and only if. The counter set that we have,
46:15:620Paolo Guiotto: discussed here, it can be proved that it is in correspondence with reals.
46:22:510Paolo Guiotto: So it's not a comfortable set. Nonetheless, it has measure equals zero. So that's a remarkable example
46:30:180Paolo Guiotto: of a one-dimensional uncountable set with measure equals zero. That's non-trigger, I told you. In dimension 2, it's trigger. As you can see here, we take a straight line, it's a measure zero set, there is no head. It's much easier.
46:45:880Paolo Guiotto: While it's much more complicated to show that there is no length, and to have an uncountable set. So, I'm saying that if you do the exercise in this way, okay, measure set equals 0 means the set is countable, then when you define the set N squared as the set made of X squared.
47:04:140Paolo Guiotto: with X in N.
47:06:220Paolo Guiotto: It is clear that…
47:07:890Paolo Guiotto: That set will be countable as well, no, because you just are in correspondence with the numbers X and X squared, no?
47:17:200Paolo Guiotto: For positive number, it is even a big section, so if n is countable, n squared is countable, and therefore the measure is 0. This would be correct, but the point is that it is not necessarily true that a countable measure set is countable set.
47:33:590Paolo Guiotto: So, in fact, this is a bit more complicated. The goal is to prove that the measure of N squared
47:39:850Paolo Guiotto: is equal 0. This is, let's say, the essential part of the exercise. Then there are some… a couple of extensions. It says, what if the set N is a bounded set? What if the set N is unbounded? But these two are extensions that can be obtained by the first case. That's good.
47:59:820Paolo Guiotto: So this one is the… really the… let's say, the core of the problem, okay, to solve this one. And here you have to think about, what does it mean to have measure zero?
48:15:610Paolo Guiotto: So I just give you the stark of this. What does it mean to be a measure zero set?
48:22:430Paolo Guiotto: Well, we have to go back to the definition.
48:25:90Paolo Guiotto: We have to remind what is, at the end, lambda of something. Remind the infamum of blah blah blah, no? So it means that the infamum of sum of intervals, I, K, A, I, N, that cover
48:43:170Paolo Guiotto: I was at N.
48:47:830Paolo Guiotto: This thing is zero.
48:49:820Paolo Guiotto: And since the interim is 0, it means that this is the best lower bound. Whenever you take a positive epsilon, you can find a number of these which is strictly less than epsilon. So, for every epsilon positive, there exists a covering, IN,
49:09:960Paolo Guiotto: such that, union of IN Kova. N, huh?
49:16:20Paolo Guiotto: With a sum of sizes of, in this case, of lengths of the intervals iN less or equal than epsilon.
49:26:80Paolo Guiotto: A final premise is that since we are covering a set N, which is a subset of interval 0, 1, we don't know how this N is made. We know that it is something, a measure 0 set inside 01, which is covered by intervals.
49:46:860Paolo Guiotto: Okay, so if these intervals are outside of 0, 1,
49:51:970Paolo Guiotto: are com… let's say that part of these intervals outside of 01 are useless to cover N, are really excess, so you have… you have… suppose that you have an IN like that.
50:04:790Paolo Guiotto: You don't need this part here to cover the yellow set, no? So you can always assume that all these IM are contained in the interval 0, 1 themselves.
50:16:940Paolo Guiotto: Otherwise, you replace them by
50:20:710Paolo Guiotto: them intersected with 01. Intersection between two intervals is an interval, so you are sure they are still interval, they cover, and the length sum of sizes will be less than epsilon. It's here, important, now, to assume this, assume that this is an interval AN
50:38:880Paolo Guiotto: Bienne, and now you have to think about what to do.
50:42:990Paolo Guiotto: So the goal will be to prove that N square will have measure 0. So we have to try to do the same with N square and show that this value must be necessarily zero. To try to do,
50:57:770Paolo Guiotto: these.
51:00:110Paolo Guiotto: Okay.
51:01:490Paolo Guiotto: And, okay, so about measurable functions…
51:11:570Paolo Guiotto: Maybe we do problems, next week. I leave to do these exercises. Do… These are exercise…
51:25:560Paolo Guiotto: Okay, 3, 4, 1…
51:33:470Paolo Guiotto: If you want 3, 4, 2…
51:41:500Paolo Guiotto: 3, 4, 4.
51:43:300Paolo Guiotto: This is the proof of the product of two measurable functions is measurable.
51:49:370Paolo Guiotto: We have not seen a group, but I told you already, try to extend what we have seen for the sun.
51:56:820Paolo Guiotto: The 3-4-6.
52:02:500Paolo Guiotto: Here in the 347, which is,
52:06:200Paolo Guiotto: A nice exercise is something that must be known, even if it is an exercise. I have not written a proposition, but it's an interesting result, because it says that you have a continuous function, G, which is 0 almost everywhere. What would you expect from this function?
52:26:50Paolo Guiotto: It is continuous. It is zero almost everywhere. It means it is zero except for a measure of zero set of points. The question is, can be different from zero? No. You have to prove that G is constantly equal to zero.
52:41:540Paolo Guiotto: Or, if you want, the function is continuous. It is constant, except for a measure zero set of points.
52:50:790Paolo Guiotto: then, since it is continuous, it must be constant everywhere. So, the continuity forces that, you, you must be, flat.
53:03:20Paolo Guiotto: So I would say that these are the exercises for the next… Bye.
53:09:300Paolo Guiotto: Okay, I'd say also that we basically finished the part on measurable functions. Well, actually, as you will see, there will be a remarkable property that we will introduce with the definition of integral that we are going to start now, so…
53:28:720Paolo Guiotto: Let's start talking about, abstract.
53:35:390Paolo Guiotto: Team to go out.
53:40:630Paolo Guiotto: So, the goal is, clear.
53:46:390Paolo Guiotto: So, let, X… F.
53:51:360Paolo Guiotto: mu… B.
53:53:780Paolo Guiotto: a measure… phase.
53:58:680Paolo Guiotto: F, a function defined on some domain, E, of X,
54:05:560Paolo Guiotto: Real value, though.
54:08:60Paolo Guiotto: With some property, of course.
54:11:970Paolo Guiotto: we… And… to define
54:19:690Paolo Guiotto: Oh, so are you okay that we don't do any brack and we go till the end? Okay.
54:25:210Paolo Guiotto: We have to define the integral on X of function f with respect to measure mu. We use these notations because they are familiar with the classical notations
54:37:340Paolo Guiotto: of integrals, you have studied for integrals of functions of one real variable or several variables.
54:45:390Paolo Guiotto: So that's the integral… I'm sorry, integral on the main, of course.
54:52:10Paolo Guiotto: Integral.
54:55:880Paolo Guiotto: Oh.
54:57:00Paolo Guiotto: F.
54:58:930Paolo Guiotto: on. E… With respect to the measure mu.
55:07:420Paolo Guiotto: Of course, you'll expect that there will be some requirement on the function, and the right condition is the function must be measurable.
55:17:830Paolo Guiotto: So this, is going to extend a lot the definition of integral we have seen, we studied so far, because, first of all, the space XF mu can be whatever.
55:31:460Paolo Guiotto: So it's not necessarily R or RM with a class of sets that we know, and the measure is the Lebesgue measure. Even if X is RM, F can be even the Lebag sigma algebra, but mu is not necessarily the Lebesgue measure. We can define for other measures.
55:51:220Paolo Guiotto: But most importantly, we can define for other type of environment.
55:56:750Paolo Guiotto: Okay, so it's important because in probability, we do not necessarily work on any Euclidean space as the probability space.
56:06:930Paolo Guiotto: Okay, now, there are… Let's say,
56:16:90Paolo Guiotto: At least 3 or 4 different paths that lead to the definition of integral.
56:28:890Paolo Guiotto: I, I will, I will do a, let's say, an unconventional path, for two reasons. Number one,
56:38:770Paolo Guiotto: As you may imagine, the construction of the integral follows a very long sequence of steps, and we're not going to prove anything because
56:49:810Paolo Guiotto: proofs are of no interest, and the construction in itself is not particularly relevant. We will be happy knowing that the interval is defined, we know that we can integrate these kind of functions.
57:04:610Paolo Guiotto: And then will be another problem, how to compute the integral, as always, okay?
57:10:560Paolo Guiotto: And secondly, because I want a definition that, in, let's say, in, in…
57:17:920Paolo Guiotto: More or less immediately yields to the definition of integral, without passing through these long steps.
57:24:850Paolo Guiotto: And the difficulty is that this definition is basically what is actually the original LeBague definition.
57:32:430Paolo Guiotto: that was given, you know, mathematics works in this way. It's like a town, and you start building,
57:42:170Paolo Guiotto: You start,
57:44:960Paolo Guiotto: constructing buildings, and with maybe a sort of erratic path. Then, at the second moment, mathematician starts to clean up, to do better, to fix things, to make it easier, and so on.
58:01:990Paolo Guiotto: So at the end, we have theories that are well established, clear, so the, for example, differential calculus of Newton was not the differential calculus we have today, nowadays.
58:15:930Paolo Guiotto: So you work with the Taylor formulas, directly, we compute the derivative as if everything is a polynomial, so it's, if you go on,
58:27:920Paolo Guiotto: on Newton books, you see that the calculus, or if you try to read, because it's quite hard, read letters between Newton and Leibniz, it's like if they are talking about something that you don't realize is calculus, so it was very different.
58:45:950Paolo Guiotto: So, as you may know, Gene LeBerger, who invented this definition.
58:49:630Paolo Guiotto: used paradigms and, let's say, a road that was, completely different from what we use today. But the point is that we have immediately a definition of integral.
59:03:550Paolo Guiotto: And it makes also clear what is the difference between the LeBague approach and the Riemann approach.
59:11:620Paolo Guiotto: So, we will take this one.
59:15:70Paolo Guiotto: Okay, so, let's start, as usual, for any kind of approach. We consider, initially, the case of positive functions. Okay, so let…
59:27:190Paolo Guiotto: F… defined on some E containing X, B positive.
59:35:790Paolo Guiotto: So, the integral is what you might expect. Now, I will do this figure as if X is the real line, but of course, you have to imagine that this X can be whatever.
59:48:880Paolo Guiotto: So we say we have a domain here, where the function is defined, and we have a function. Now, I will draw a function as if this function is very regular, but that's not the case, okay?
00:02:210Paolo Guiotto: In any case, this is F, positive.
00:06:700Paolo Guiotto: Any idea of integral is you want to have a method to compute, let's say, this area. Let's call it the area, okay?
00:17:670Paolo Guiotto: Now, what is the Riemann approach to this problem?
00:23:330Paolo Guiotto: Now, Riemann…
00:29:220Paolo Guiotto: approach says, of course, we want to use elementary figures, like rectangles, to do that. So, since that will be, say, a curved shape, we… the idea is we try to fill with rectangles in the best possible way. How do we do this?
00:48:100Paolo Guiotto: We divide the domain. If you remember, we divide the domain, that case for the Riemann integral, is an interval. You divide the interval into subintervals, and you use each sub-intervals as the base
01:04:150Paolo Guiotto: Of a rectangle.
01:07:50Paolo Guiotto: Inscribed into this, set, huh?
01:12:590Paolo Guiotto: Which is actually called trapezoid.
01:17:140Paolo Guiotto: trapezoid of the function F. We will give a…
01:21:890Paolo Guiotto: formal definition in a moment, okay?
01:26:390Paolo Guiotto: And, so once you have done this.
01:33:40Paolo Guiotto: So you fill the area with these rectangles. These rectangles, they have an elementary area, base times 8. The base is the length of these intervals.
01:46:50Paolo Guiotto: on which you divide the set E, and E height is what? Well, E is the best possible height, such that the rectangle is, for example, inscribed in that figure. So, you have to take as here the minimum value of the function.
02:06:790Paolo Guiotto: Now, talking about minimum-maximum values is, of course, natural if functions are good, like continuous function.
02:15:360Paolo Guiotto: But if the function is discontinuous, it could not have any minimum or maximum. That's why we replace with infimuma, infimum.
02:26:890Paolo Guiotto: of, the function F on that interval, let's say, IK.
02:36:740Paolo Guiotto: Now, the problem is that…
02:43:80Paolo Guiotto: This imposes some restrictions. First of all, it imposes the restriction that the function f must be bounded
02:51:500Paolo Guiotto: And second, these definitions suffer a lot from discontinuities. There is the famous example of the Dirichlet function, which is the function equal 1.
03:02:840Paolo Guiotto: on irrationals and zero on ratios. For that function, that function is not integral according to this, to this sense.
03:11:900Paolo Guiotto: Now, the LeBague idea… Is, instead of dividing the domain, huh?
03:21:160Paolo Guiotto: We divide the codomain.
03:23:570Paolo Guiotto: So, it says, we want to compute this area still, so we have…
03:29:40Paolo Guiotto: the, say, the base domain E. So we have… still, the object is to compute the area inside here. But instead of dividing the domain, we divide the y-axis, the codomain, into a number of parts. So let me exaggerate a bit the graph of the
03:46:810Paolo Guiotto: Of air for the while, we do not get a nice picture.
03:52:590Paolo Guiotto: Now, DVI…
03:57:810Paolo Guiotto: In this way…
04:11:290Paolo Guiotto: And, so what we want to do is, we want to approximate that area by computing these areas.
04:21:560Paolo Guiotto: Now, how these red sets are defined? Let's imagine to do a zoom and consider just one single slice of this figure. So, what I do here is…
04:35:540Paolo Guiotto: sort of enlargement. Suppose that you divided with the two, we call this the y-axis, so let's say, YK, YK plus 1.
04:49:330Paolo Guiotto: So you will have the function that,
04:57:470Paolo Guiotto: This is E.
05:00:890Paolo Guiotto: So, you have a… let's assume that the function does something like this. What we have to sum is a…
05:07:190Paolo Guiotto: About these, parts of, these areas.
05:12:590Paolo Guiotto: How can we measure these areas?
05:15:220Paolo Guiotto: Now, you need to take the, you say… you see that this is a measure of something like base time height, where the base is down here.
05:28:900Paolo Guiotto: And the height is just this interval here.
05:33:340Paolo Guiotto: Now, the base here is the set of points in the domain.
05:38:850Paolo Guiotto: This is the set of X that belongs to E, such that the value of F is between the two values Y.
05:47:280Paolo Guiotto: So, you see, so F of X is larger or equal than yk, and it is less. I put strictly less to have the next one that takes the next point. YK plus 1.
06:01:180Paolo Guiotto: So this set here is where? This set is in the domain E, because it is a set of points of E, where F is between these two. So this is a subset of the domain E.
06:15:380Paolo Guiotto: And never mind that the domain E is in the space axle where we have a measure.
06:20:330Paolo Guiotto: So we will have a measure of this guy, A measure of this sector.
06:26:740Paolo Guiotto: And the measure of this set will represent law.
06:30:240Paolo Guiotto: Let's say that it represents the area of area, in a large sensor, let's say, the area of this wave, yeah.
06:40:720Paolo Guiotto: Then we multiply by the size of this length term of this interval, which is the difference between YK plus 1 minus YK, so times…
06:53:620Paolo Guiotto: YK plus 1 minus YK, and the idea is that we have an approximation of the black region of the area of this flat region.
07:05:130Paolo Guiotto: So then we sum up, okay, This,
07:11:10Paolo Guiotto: And we get an approximation of the total area under the graph of F. Of course, this will be an approximation that will become more and more precise by showing these intervals
07:28:360Paolo Guiotto: as exactly we do with the Riemann construction. You know that this is an approximation. To get the exact value, we have to shrink these intervals to zero.
07:40:170Paolo Guiotto: to send the number of points to infinity, and the length of each other to zero. That's it. Roughly gave you.
07:48:900Paolo Guiotto: Now…
07:54:460Paolo Guiotto: So, let's now construct precisely this quantity, okay? This will be done with a special choice of the values YK. So, we start from the y-axis.
08:13:710Paolo Guiotto: So this is value 0.
08:16:120Paolo Guiotto: We have to divide the y-axis in, say, small parts.
08:21:140Paolo Guiotto: And we will increase the number of parts to infinity. So these are, let's say, equal intervals of the same length.
08:32:380Paolo Guiotto: And here we do a specific choice. We use these parts not of something like 1 over n, but 1 over a power of 2.
08:41:790Paolo Guiotto: I will explain you, you will see why later. So we assume that these are numbers of type. This is 1 over 2, let's say, to the,
08:53:10Paolo Guiotto: to the N. This is… the next one is 2 over 2 to the n.
08:58:920Paolo Guiotto: The next one will be 3 over 2 to the N. So, in general.
09:04:520Paolo Guiotto: We will have a point that it is K over 2 to the n.
09:11:319Paolo Guiotto: So the next point will be K plus 1 over 2 to the n.
09:16:859Paolo Guiotto: You see that when this k belong… becomes 2 to the n, we arrive to value 1.
09:24:420Paolo Guiotto: Okay? So, we will vary this K,
09:29:890Paolo Guiotto: From 0, so we take values 0,
09:33:80Paolo Guiotto: The point in that case is 0. 1, 2, etc. We arrive to 2 to the n, and this point, k over 2 to the n, is just 1.
09:43:810Paolo Guiotto: We want to divide all the y-axis, so this K will continue to grow. It will continue to grow up to the value, 2 to the n.
09:56:220Paolo Guiotto: What is the K that makes K over 2 to the n equal to 2 to the n? You can see this is carried 2 to the n. This side is 2 to 2N.
10:10:410Paolo Guiotto: Okay.
10:12:540Paolo Guiotto: So, for that K, we are… yeah.
10:17:720Paolo Guiotto: And then we say, okay, let's define, the set E.
10:24:380Paolo Guiotto: Kn, there are two indexes, as the set of points that belongs to E, where our function f is above or equal k over 2 to the n.
10:40:120Paolo Guiotto: And less, strictly less than K plus 1 over 2 to the n.
10:47:310Paolo Guiotto: So, for k ranging from 0 to value 2 to 2N, or better, technically, I have to stop at 2 to 2N minus 1, because in this case, k plus 1 over 2 to the n would be 2 to the n.
11:05:850Paolo Guiotto: Okay? In such a way that this last value is exactly 2 to the N.
11:12:590Paolo Guiotto: So these are the sets where, if you imagine to have a function here.
11:17:750Paolo Guiotto: So, okay, let's keep these two.
11:22:90Paolo Guiotto: These have… These are the values.
11:27:860Paolo Guiotto: where F… let's say, when F is between these two, exactly here.
11:37:30Paolo Guiotto: You see that on this set here, down here on the X space, and also here.
11:48:680Paolo Guiotto: The black thing is the set EKN.
11:53:920Paolo Guiotto: Because this is the set of points where the function is exactly between
11:59:130Paolo Guiotto: these two values, k over 2 to the n and k plus 1 over 2 to the n.
12:04:460Paolo Guiotto: Okay? Now, we have a set down here in the SpaceX. Notice that this set is a measurable set, because this is written shortly. It is F above K over 2 to the N.
12:21:480Paolo Guiotto: and below k plus 1 over 2 to the n, this is measurable, so it belongs to the family F, if F is a measurable function.
12:33:20Paolo Guiotto: So if F is just a measurable function, these sets are all measurable.
12:39:580Paolo Guiotto: Now, what happens for values which are larger than 2 to the n?
12:44:410Paolo Guiotto: So let me, do this.
12:48:140Paolo Guiotto: So imagine the function goes up there.
12:51:440Paolo Guiotto: The nice thing of this definition is that it allows the function… it removes completely the bound… the requirement that the function must be bounded. So, above the threshold 2 to VN, you have a set of points down here that, let's call it, E…
13:09:780Paolo Guiotto: just EN, the set of points X that belongs to E, where F is larger or equal than 2 to the N.
13:22:170Paolo Guiotto: So, if you are in EKN, you will not be that, because, when the K range from 0 to 2 to 2N minus 1, at most, A plus 1, you see, is 2 to 2N, divided by 2 to N is 2 to the N, this number.
13:42:680Paolo Guiotto: So, the satin in this figure would be this one.
13:48:870Paolo Guiotto: If you don't like, maybe it's, it's confusing, let's change that, I call it F.
13:54:250Paolo Guiotto: F.
13:56:80Paolo Guiotto: This is FN, the set where F is above 2 to the M.
14:02:720Paolo Guiotto: Also, this set, you see, that's F larger than something 2 to the n is a measurable set.
14:11:300Paolo Guiotto: when F is measurable. So all these sets we constructed are measurable sets.
14:21:630Paolo Guiotto: Now, what we do is to compose the area in this way. We take…
14:31:310Paolo Guiotto: the area on the… you set EKN,
14:35:770Paolo Guiotto: We consider this red, say, area, That should be…
14:42:270Paolo Guiotto: The minimum value of the function, which is this one.
14:46:570Paolo Guiotto: K over 2 to the n times the measure mu of the set EKN.
14:56:580Paolo Guiotto: Now, this number makes sense. If mu is infinite, of course, this number will have value plus infinity.
15:04:00Paolo Guiotto: Otherwise, this is a finite value.
15:07:140Paolo Guiotto: And, so we introduced this quantity.
15:10:890Paolo Guiotto: Let's call it, IN for a moment.
15:15:980Paolo Guiotto: So, let's put some formalities. Let F… Be a measurable function.
15:25:290Paolo Guiotto: and define IN… as the, summer…
15:31:260Paolo Guiotto: for K going from 0 to 2,
15:34:780Paolo Guiotto: to 2N minus 1 of k over 2 to the n times the measure of the set EKN.
15:44:490Paolo Guiotto: Now, this would be the sum of the areas.
15:48:270Paolo Guiotto: Until… for the function, until the value, 2 to the n. Then we have a last area, which is this one.
15:55:710Paolo Guiotto: which is plus 2 to DN, the measure of Fn.
16:01:860Paolo Guiotto: Now, we define this quantity. What this quantity represents, well, if you change, imagine that you take the K minus 1
16:12:210Paolo Guiotto: The k minus 1 will be here. This is k minus 1 over 2 to the N.
16:18:580Paolo Guiotto: So the set of points for which F is between k minus 1 and k over 2 to the n is just this one. So, let's say in green here, you see the set EK-1N.
16:34:910Paolo Guiotto: There is no, below there is nothing. So, the area I'm coloring in green is a measure of this set here times this state.
16:48:910Paolo Guiotto: You see that the sum of these areas is going to be what looks like an approximation by defect of the area below the graph of the function.
17:00:400Paolo Guiotto: Now… What we show is this theorem.
17:07:640Paolo Guiotto: the sequence I am.
17:09:950Paolo Guiotto: Which is a sequence of numbers.
17:12:900Paolo Guiotto: what can we expect? What happens when we change N?
17:21:160Paolo Guiotto: The idea is that increasing N, I am reducing the size of the intervals. I have the choice of this 1 over 2 to the n is because when I pass from n to n plus 1,
17:37:110Paolo Guiotto: It's like if every of this interval is divided in two parts. And as you will see, it happens that the sequence IN is increasing.
17:50:260Paolo Guiotto: So, I am… is less or equal than IN plus 1 for every n.
17:59:380Paolo Guiotto: And, so, since it is an increasing sequence of number, So… There exists the limiter.
18:07:660Paolo Guiotto: when n goes to plus infinity of D is IN,
18:12:990Paolo Guiotto: And we call this limit integral of function F on domain E with respect to measure mu.
18:23:330Paolo Guiotto: So the integral is, in this way, defined for every measurable, positive measurable function.
18:32:10Paolo Guiotto: So, you see, the unit condition I'm asking is measurable.
18:36:530Paolo Guiotto: If you think a back measure, this will mean that this thing integrates whatever, basically.
18:44:730Paolo Guiotto: Well, the proof is technical, but it's not extremely complicated, because let's see what happens when we switch from IN to IN plus 1.
18:59:530Paolo Guiotto: So I want to convince you that IN plus 1 is larger than IN.
19:04:590Paolo Guiotto: So IN plus 1…
19:10:50Paolo Guiotto: Is, let's take the formula. And what is nice, in principle, with this, that is that this is even a formula.
19:19:10Paolo Guiotto: So, you see, I'm not passing through complete supremum of this or that, that it's practically unfeasible.
19:29:320Paolo Guiotto: This is a numerical formula, in fact, you know? You have a definition of IN, which is a finite sum. Of course, it depends, maybe these sets are… have infinite measure. In that case, the sum will be infinite.
19:42:830Paolo Guiotto: But, apart from this, this is a numerical formula, and you are constructing the integral directly with a formula, without passing through complex definitions.
19:56:270Paolo Guiotto: Now, if you plug in into this formula n plus 1, this literally becomes sum for k going from 0 to 2 to 2N plus…
20:07:710Paolo Guiotto: one here.
20:10:200Paolo Guiotto: minus 1, K over 2 to the n plus 1.
20:13:700Paolo Guiotto: mu EKN plus 1.
20:18:530Paolo Guiotto: plus 2 to N plus 1 mu F.
20:23:10Paolo Guiotto: N plus 1.
20:27:90Paolo Guiotto: Let's see on the y-axis what is going on. We don't need to do figures, graphs, and things like this. We just look at the y-axis. So this is the value 0,
20:38:700Paolo Guiotto: So imagine that I have fixed NN. So these are intervals 1 over 2 to the n, 2 over 2 to the n, etc. In general, I have k over 2 to the n, and the next one is k plus 1 over 2 to the n.
20:56:760Paolo Guiotto: When I switch from n to n plus 1,
20:59:950Paolo Guiotto: I basically… what I do is I divide by 2 each of these intervals, so this is 1 over 2 ton plus 1, because it is half of the previous one. This is…
21:14:670Paolo Guiotto: This one is, 2 over 2 to n plus 1.
21:22:930Paolo Guiotto: This is, 3 over 2 to n… sorry, 3 over 2 to n plus 1 is down here. This is 4 over 2 ton plus 1. This is 3 over 2 to n plus 1.
21:35:590Paolo Guiotto: You see what happens. I'm just having the same points, but plus another… the same number of points in the middle points of each of these interval.
21:48:950Paolo Guiotto: So…
21:54:470Paolo Guiotto: Let's take, so you understand that, in general, this will be…
22:01:310Paolo Guiotto: If I have 2 to n plus 1 here, it's like if I multiplied by 2, the denominator, so I'll multiply by 2 also the numerator, and they are the same.
22:10:480Paolo Guiotto: So this is K over 2 to the n is 2K over 2N plus 1. Then there is the midpoint, which is 2K plus 1
22:21:820Paolo Guiotto: divided 2 ton plus 1, and this one will be 2K plus 2 divided 22N plus 1.
22:32:490Paolo Guiotto: So let's focus on the term of this sum corresponding to this index 2K. So let's take this one.
22:43:130Paolo Guiotto: 4K equal, to, let's say, for the term 2K. So at certain moment, we will have 2K divided 2N plus 1,
22:56:630Paolo Guiotto: mu.
22:58:70Paolo Guiotto: E to K.
23:00:350Paolo Guiotto: N plus 1. The next one will be plus 2k plus 1 over 2 to n plus 1,
23:09:580Paolo Guiotto: mu toe of 2K plus 1.
23:14:560Paolo Guiotto: N plus 1.
23:16:260Paolo Guiotto: So, we split this sum in pairs, the first two terms, the second two tenth, and so on.
23:24:690Paolo Guiotto: Just to, if you want, we can, we can take, if, if this is,
23:33:220Paolo Guiotto: is complicating things. Let's just take the first… the first two terms of this sum. So we start with k equals 0, so we have 0 times the measure, this is 0. Then, for k equals 1, I have 1 over 2…
23:48:980Paolo Guiotto: 2N plus 1… mu E.1.
23:53:680Paolo Guiotto: And last one… So, the first one was zero, no?
24:01:760Paolo Guiotto: No, no, maybe it's better if we take the general case, but this would confuse the ideas. Okay, let's give a look to these two quantities. This is,
24:11:930Paolo Guiotto: The first one, as you can see, is equal to, when you simplify, k over 2 to n.
24:19:190Paolo Guiotto: And this is the measure of the set where F is above what? Well, it is above 2K of M plus 2 to the N plus 1, less than 2K plus 1.
24:33:870Paolo Guiotto: divided 2 to n plus 1, huh?
24:38:280Paolo Guiotto: You see?
24:40:570Paolo Guiotto: So…
24:41:710Paolo Guiotto: the set EKN is F between k over 2 to the n and less than K plus 1 over 2 to the n. My set here is E2KN plus 1, so above 2K over n plus 1, below 2K plus 1 over 2 to the n.
24:58:690Paolo Guiotto: Now, this is the same of k over 2 to the n.
25:03:590Paolo Guiotto: So this is the set where F is above K over 2 to the n.
25:09:80Paolo Guiotto: and below 2K plus 1.
25:12:10Paolo Guiotto: of a 2 to n plus 1.
25:16:330Paolo Guiotto: And here we have k over 2 to the n, the measure of this.
25:22:520Paolo Guiotto: And the second one, this one.
25:27:690Paolo Guiotto: is equal to… we have 2K plus 1, huh?
25:33:820Paolo Guiotto: of a 2 to n plus 1 measure of F above.
25:39:550Paolo Guiotto: this number, so 2K plus 1 divided 2N plus 1,
25:45:820Paolo Guiotto: Above or equal, less than 2K plus 2,
25:50:00Paolo Guiotto: 2K plus 2, divided 2 to n plus 1.
25:55:330Paolo Guiotto: Now, if you divide by 2, this is K plus 1 divided into 2DN.
26:01:720Paolo Guiotto: So, now you see that,
26:07:990Paolo Guiotto: You see that since this number Kia.
26:12:550Paolo Guiotto: Is, less than…
26:19:800Paolo Guiotto: is larger than… so I have to take the sum of these two terms, so this plus this.
26:28:10Paolo Guiotto: So this one is larger than 2k divided 2N plus 1, which is K over 2 to the n. So the sum of these two, if I reduce this coefficient, is larger than k over 2 to the n.
26:45:880Paolo Guiotto: measure F between K over 2 to the N, And, less than…
26:53:590Paolo Guiotto: 2K plus 1 over 2 to n plus 1.
26:58:80Paolo Guiotto: plus same coefficient, k over 2 to the n, measure F above, or equal 2K plus 1,
27:07:90Paolo Guiotto: over 2 to n plus 1, and less than, what is that? K plus 1 over 2 to the n.
27:16:60Paolo Guiotto: But now you look at these two, the coefficient is the same.
27:21:740Paolo Guiotto: So, you have a unique coefficient, k over 2 to the n.
27:26:370Paolo Guiotto: And you have the sum of these two measures.
27:31:860Paolo Guiotto: One is the set of points where F is between value and value, the other is the set of F between these two values. These are disjoint sets, because in the first set, F is below 2K plus 1 divided 2N plus 1. In the second set, F is
27:48:890Paolo Guiotto: Great or equal. So they are disjointed.
27:52:530Paolo Guiotto: So it's like measure of A, measure plus measure of B with A and B disjoint. This can be made measure of the union of these two.
28:02:90Paolo Guiotto: set. But what is the union?
28:05:30Paolo Guiotto: The union is either F is between K and 2 to the n and less than k plus 1 over 2 ton plus 1, or greater or equal to k plus 1 over 2 to the n plus 1, and less than k plus 1 over 2 to the n.
28:19:640Paolo Guiotto: So in any case, when you do the unit, you get F between k over 2 to the n, and less than K plus 1
28:26:360Paolo Guiotto: over 2 to the nth.
28:28:690Paolo Guiotto: And that's the measure of the sector EKN.
28:34:410Paolo Guiotto: So, what we have here…
28:37:930Paolo Guiotto: Is that if you take the sum of these two terms that you have in this
28:43:760Paolo Guiotto: sum, which yields IN plus 1,
28:48:20Paolo Guiotto: The sum of these two guys, which are the two red tens, Is greater or equal than
28:56:70Paolo Guiotto: K over 2 to the n, the measure of EKN.
29:02:910Paolo Guiotto: And when you sum up over K, you get that the entire sum of this, so this sum, is larger than the sum where you have k over 2 to the n, the measure of EKN.
29:15:960Paolo Guiotto: So basically, what we have is that sum for k going from 0 to 2 to 2N plus 1 minus 1 K over 2 to n plus 1 mu EKN plus 1
29:34:960Paolo Guiotto: This is greater or equal than sum for k going from 0 to 2 to 2N minus 1 of K over 2 to the N measure of PKN, which is…
29:49:130Paolo Guiotto: The sum that you use for IN. This is the sum for IN plus 1.
29:56:430Paolo Guiotto: This is the sum for a n. Yes, there is one other term, which is the term 2 to the n, but that's the same. It can be shown in the same way, but here you have 2 to n plus 1 measure of the set FN plus 1,
30:14:40Paolo Guiotto: This is the set where F is larger than 2 to n plus 1.
30:19:500Paolo Guiotto: But if you look at this, this is greater than… 2 to n plus 1 is larger than 2 to n.
30:27:850Paolo Guiotto: And the set where F is…
30:33:90Paolo Guiotto: Above 2 to the n plus 1,
30:35:970Paolo Guiotto: is larger than the set where F is above 2 to the n.
30:43:450Paolo Guiotto: So you have that this term is bigger than that one.
30:48:610Paolo Guiotto: And so this means that IN plus 1 is greater than ion.
30:55:190Paolo Guiotto: And therefore, there exists the limit.
30:58:900Paolo Guiotto: of the sequence of the values IN, and this is what defines the integral.
31:07:930Paolo Guiotto: Okay, I've done a bit of mess, however, the point is that it is,
31:15:990Paolo Guiotto: with a little bit of technical work, it is possible to prove that this IN plus 1, each of the terms, if you
31:26:750Paolo Guiotto: Take 2x2, The sum of two consecutive terms here is larger than
31:35:710Paolo Guiotto: the term you have in the previous sum, in the sum of order, N.
31:41:100Paolo Guiotto: So the pairs of terms in the sum IN plus 1 are larger than a single term in the sum n. So when you sum up, you get that IN plus 1 is greater than IN. So this sequence is a sequence of increasing numbers.
31:56:780Paolo Guiotto: And, they are,
31:59:570Paolo Guiotto: we call it, the limit is what we call the integral of the function. In this way, we have a definition of integral. Of course, it's not an operational definition, but no definition is operational.
32:10:750Paolo Guiotto: The advantage of this is that we have a formula to compute the integral, in principle at least, and secondly, we have done this in sort of short time, with respect to the usual construction.
32:25:610Paolo Guiotto: Okay, so do the exercises, and we will see you on Monday. Hopefully, we should have, solved this, this problem with homes.
32:37:670Paolo Guiotto: Have a nice weekend.