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00:07:500Paolo Guiotto: Okay, good morning.
00:23:350Paolo Guiotto: Let's start, from the establishment of inequality.
00:29:150Paolo Guiotto: And we will, now, draw some, Interesting consequence from this.
00:37:520Paolo Guiotto: The first one is the following, proposition… That we will use often.
00:49:110Paolo Guiotto: Suppose that we have, A measurable function.
00:53:410Paolo Guiotto: positive.
00:55:560Paolo Guiotto: Almost everywhere.
00:59:900Paolo Guiotto: And, such that, the integral on the domain E of F,
01:06:190Paolo Guiotto: Respect to the measure, mu is zero, so we assume that there is a measure space, blah blah blah.
01:11:490Paolo Guiotto: Then, what can be expected on this app? We have a function which is, positive, in large sense.
01:21:80Paolo Guiotto: Non-negative, technically speaking, and the integral is zero.
01:26:870Paolo Guiotto: We should expect that this function cannot be really positive, otherwise the integral should be positive. So, in fact, what we can prove is that this function must be equal to zero almost everywhere.
01:40:970Paolo Guiotto: Well, let's see the proof, and let's do some remark about this.
01:47:970Paolo Guiotto: The proof is,
01:53:270Paolo Guiotto: Basically, an application of the Chibisian inequality. So if you apply the tradition inequality here.
02:00:630Paolo Guiotto: by, Chebyshev inequality.
02:05:500Paolo Guiotto: We have that the measure where f is greater or equal than alpha, we proved that this is less or equal than 1 over alpha, the integral on the domain of F emu. And since this last is 0, we get that this is 0.
02:20:390Paolo Guiotto: So, in particular, we get that the measure where f is greater or equal than alpha is equal to 0 for every alpha positive.
02:31:900Paolo Guiotto: No? So the measure where F is, even if this number is smaller.
02:38:670Paolo Guiotto: F is, let's say, away from zero, ease the measure of 07.
02:44:60Paolo Guiotto: What we want to prove, this is not yet the conclusion, we want to prove that F is equal to zero almost everywhere. That means F equals zero almost everywhere.
02:57:190Paolo Guiotto: means that the set where this is pulse, where F is different from zero, is a measure 0 set. So, literally speaking, this means measure of set where… of the set where F is different from 0 is 0.
03:12:70Paolo Guiotto: But since F, we know, is greater or equal than zero, it cannot be negative, so saying that F is different from 0 is equivalent of saying that F is positive, so the set
03:26:80Paolo Guiotto: where F is strictly positive is a measure 0 set. So that's what we want to achieve, what is the thesis. Now, is it the same with this, or not?
03:39:220Paolo Guiotto: So, can we… can we draw this conclusion directly from this one, or no?
03:45:670Paolo Guiotto: Well, the answer is, not entirely, because this is the same. The measure of this, of this infographic, but it's still possible.
03:59:510Paolo Guiotto: So, it means, 1 out of 1 million, the F greater than 1 out of the 1 million value, very low value, that is low value, but not zero. This is zero. Why here we want to… F is whatever greater than zero.
04:17:110Paolo Guiotto: And there, we cannot put the alpha equals 0, because we cannot do that, or alpha equals 0, we cannot take the B quality. And also, F greater than 0 wouldn't be the conclusion, because F is greater equal than zero, that set cannot be an N plus 0 set, no? Unless the middle of the set equals 0.
04:38:80Paolo Guiotto: So now the point is, you see that this, this, this cannot be obtained as a special case, yeah.
04:45:560Paolo Guiotto: Now, the point is, how do we get these from that?
04:50:560Paolo Guiotto: So the point is, what is the relation between set F greater than or equal than alpha, alpha positive, and set F greater than zero?
04:58:990Paolo Guiotto: So, we noticed that.
05:05:930Paolo Guiotto: So… What is the relation between this set And this.
05:13:320Paolo Guiotto: for… Alpha positive.
05:43:690Paolo Guiotto: That's warm.
05:45:220Paolo Guiotto: Yes, if you are going to equal, then… Alpha is positive.
05:50:320Paolo Guiotto: you are definitely positive, so this is containing this one. Okay. At left, we have infinitely many set, no?
06:00:330Paolo Guiotto: So it's… for each alpha, we have a different sector.
06:08:920Paolo Guiotto: Okay, okay, so… No, is it? It is contained.
06:16:430Paolo Guiotto: Okay.
06:17:590Paolo Guiotto: But I'm saying, at left, there is not just one set, there is a family of sets, but I ain't got…
06:25:210Paolo Guiotto: You should just take the intersection, because
06:29:460Paolo Guiotto: What happens if you change alpha?
06:32:150Paolo Guiotto: So, for example, suppose that these numbers are positive, zero, say, alpha, you take a beta, which is less than alpha. What is the relation between the set F greater or equal than beta, F greater or equal than alpha?
06:53:10Paolo Guiotto: Is having used to continue?
06:56:440Paolo Guiotto: So, which one is contained there?
06:58:730Paolo Guiotto: Okay, this one. So it means that when we decrease this number, alpha, This set, increases.
07:13:430Paolo Guiotto: It's not just intersection, but rather…
07:20:570Paolo Guiotto: So I want to establish an exact relation between these and these sets.
07:26:250Paolo Guiotto: So we said these sets are contained into this one, and that's one observation. Second, we noticed that when we reduce this alpha, the set gets larger.
07:42:60Paolo Guiotto: So now, if you have to write an equality, F greater than 0, what would you bet to do here?
07:51:980Paolo Guiotto: What should be the… The union. The union… Over the alpha positive.
08:00:210Paolo Guiotto: Okay, let's see if this is true. And in fact… Indeed.
08:06:590Paolo Guiotto: Well, we already have automatically one inclusion, because we know that these sets, are contained, no? The sets where F is greater than or equal, then alpha are contained in the set where F is positive.
08:18:890Paolo Guiotto: So, they are all subsets, these are contained in the set where F is positive, and therefore their union will be still contained. So, indeed.
08:30:930Paolo Guiotto: greater than 0 contains the union of sets where F is greater or equal than alpha for all possible alpha positive.
08:41:220Paolo Guiotto: Because of, because of this, no?
08:46:220Paolo Guiotto: Fact here.
08:48:640Paolo Guiotto: And now, Let's see… The other inclusion is true, no?
08:55:150Paolo Guiotto: So let's see what happens when we pick an X in the set where F is positive.
09:00:970Paolo Guiotto: Well, it happens that f of x is positive.
09:05:00Paolo Guiotto: Right?
09:07:200Paolo Guiotto: Then… Can I say that it belongs to that union?
09:16:390Paolo Guiotto: Yes? Why?
09:19:320Paolo Guiotto: Because we can always find them out of 5. Yeah, that number is positive, no, so it is somewhere here.
09:27:910Paolo Guiotto: IOLO is everything.
09:30:100Paolo Guiotto: some alpha which is between zero, strictly positive, but smaller than that F of X.
09:37:390Paolo Guiotto: So I can say that definitely my X will belong to the set where F is greater or equal than that specific alpha. If you want, let's put an alpha hat, no? So, it means that there exists
09:55:110Paolo Guiotto: An alpha hat with this property.
09:58:630Paolo Guiotto: And so it means that your X is in one of these sets, so it belongs to the union.
10:04:710Paolo Guiotto: So X belongs to the union for all alpha positive of F larger than alpha.
10:14:70Paolo Guiotto: Puico.
10:16:30Paolo Guiotto: Okay, we are, now, we have the, equality, so…
10:22:700Paolo Guiotto: The set where F is greater than zero is union of these sets.
10:29:90Paolo Guiotto: We know that this union is not a disjoint union, because they are one into the other, no?
10:36:280Paolo Guiotto: So… Remind the goal… of F positive equal to 0. So I want the measure of that set.
10:48:340Paolo Guiotto: I know the measure of this set. The measure here is zero.
10:55:450Paolo Guiotto: I know that, the… sorry, this one. The measure equals zero.
11:02:540Paolo Guiotto: For example, we could use, well, to do that, we have to put a discrete sequence, so let's take alpha equal 1 over n, in such a way that it gets
11:17:330Paolo Guiotto: Close to zero, so it goes to zero. It is positive.
11:21:870Paolo Guiotto: we have that the sequence of sets where F is greater or equal than 1 over N
11:29:100Paolo Guiotto: is an increasing sequence of sets.
11:33:540Paolo Guiotto: So we apply the continuity from… continuity from… Below, which is always true.
11:44:140Paolo Guiotto: And we get that the limit of the measures of these sets, F greater or equal than 1 over M,
11:51:810Paolo Guiotto: Which is zero, because these are measured zero sets.
11:56:290Paolo Guiotto: by the Chevyshev inequality, this turns out to be the measure of the union on n of the sets where f is greater or equal than 1 over n. But of course, this union is still the set where F is strictly positive.
12:15:800Paolo Guiotto: And therefore, we get that the measure of this set is, zero.
12:25:250Paolo Guiotto: So we can say that.
12:27:360Paolo Guiotto: Integral of a positive function equals zero. The function is not necessarily flat, constant equals zero, but it must be zero almost everywhere.
12:39:260Paolo Guiotto: This is similar to a property that, perhaps, You have seen, huh?
12:46:500Paolo Guiotto: I say perhaps, but I don't know, for the, say, for the Riemann integral.
12:51:780Paolo Guiotto: So, this is a generic measure, so there is not any particular structure. So, this property.
13:06:900Paolo Guiotto: Is, let's say, an extension.
13:13:870Paolo Guiotto: of, important.
13:24:60Paolo Guiotto: property.
13:26:70Paolo Guiotto: of the… Riemann.
13:32:540Paolo Guiotto: integral. I will mention this in one variable, there is a version for several variables. Of course, it is limited to this kind of integrals, which are very particular in this context. It says that proposition
13:49:440Paolo Guiotto: Suppose that you have a function f. Now, since we are talking about a Riemann integral, we are discussing of an integral defined on intervals for functions of very special type. So here we consider a continuous function on interval AB.
14:07:310Paolo Guiotto: positive.
14:09:400Paolo Guiotto: in a large sensor on AB.
14:15:50Paolo Guiotto: And… such that, huh?
14:18:310Paolo Guiotto: the integral on AB of F, so F of X dx.
14:23:890Paolo Guiotto: The first, calculus integral is equal to 0.
14:28:290Paolo Guiotto: Then, in this case, what happens? You are integrating a positive function, and the area below the graph is zero.
14:36:810Paolo Guiotto: The fact that the function is continuous intuitively forces this to be 0. And in fact, here you get a stronger conclusion, which is F is identical equal to 0. So it means F of X equal to 0 for every X in DB.
14:56:130Paolo Guiotto: So, as you can see, it is a strong conclusion, because it says F… F is constantly equal to 0. The… the… this general property that I repeat
15:08:210Paolo Guiotto: It involves any integral with respect to any measure. It says F is zero almost everywhere, so it's a mild version of that result.
15:19:860Paolo Guiotto: This is, can be proved, easily.
15:27:580Paolo Guiotto: Because suppose that there is a one point, and you can do a figure here, we have the interval AB, our function is positive. Suppose that there is a point where the function is strictly positive, so we want to do a contradiction argument. Take a point of zero where the value of the function is positive.
15:45:740Paolo Guiotto: Okay, so suppose that there exists if there exists a point X0 in the interval AB.
15:53:720Paolo Guiotto: Such that F at point X0 is positive. Now, what do you know? By continuity, the function around point X0 will be positive.
16:06:310Paolo Guiotto: No? That's the key point. So, I can say that by continuity.
16:13:830Paolo Guiotto: there will be an interval centered at point X0, so something like X0 minus R, X0 plus R.
16:25:420Paolo Guiotto: Such that,
16:27:950Paolo Guiotto: Let's say that FA is… if this is the quarter… sorry.
16:34:840Paolo Guiotto: If this is the quote F…
16:39:240Paolo Guiotto: the value F at point X0, I can say that in a certain neighborhood of X0, maybe this function is greater than half of that value. So there will be an interval down here.
16:54:560Paolo Guiotto: So that's the neighborhood in red, IX0.
16:58:760Paolo Guiotto: Where the function is larger than that value.
17:02:910Paolo Guiotto: value positive less than fx0, and you have the same conclusion, such that f of x is greater or equal than FX0 divided by 2 for every x in that interval IX0.
17:18:569Paolo Guiotto: But then you see, it means that there is a positive contribution to the area between the graph of F and 0. I don't know what happens outside of that interval, maybe the function falls down.
17:33:100Paolo Guiotto: But definitely, I can say that, then, in here, there is a continuation. 0 is the integral between A and B of F.
17:42:40Paolo Guiotto: Now, since F is positive, if I reduce the integration domain to a smaller domain, the integral will decrease in general, so I limit my integration.
18:08:40Paolo Guiotto: Bye.
18:15:710Paolo Guiotto: So, if I reduce the function, I will reduce the integral, so this will become greater or equal than this integral, like 0 minus R2.
18:24:520Paolo Guiotto: Because of the constant.
18:27:800Paolo Guiotto: FX0 divided 2.
39:53:490Paolo Guiotto: I'm sorry, I'm sorry, I'm sorry.
39:57:490Paolo Guiotto: But, so you, you are not allowed to talk?
40:01:450Paolo Guiotto: What's going on?
40:03:570Paolo Guiotto: Oh, okay, I will… I cannot unmute the participants.
40:08:940Paolo Guiotto: Okay, you have to check the settings, I'm sorry.
40:13:440Paolo Guiotto: I hope that you now hear me.
40:17:160Paolo Guiotto: Let's see on the chat.
40:20:300Paolo Guiotto: And, yeah.
40:22:20Paolo Guiotto: I hope that you hear now me, okay.
40:24:780Paolo Guiotto: Sorry.
40:27:310Paolo Guiotto: Let's internal here. So, moreover, it says that F is Riemann integral, If, and only if.
40:39:530Paolo Guiotto: Well, the condition is, is, is even nice, because it says if and only if, F…
40:47:760Paolo Guiotto: Is almost everywhere continuous.
40:55:300Paolo Guiotto: That is, F is continuous, except on a set of points which must have measure equal to 0.
41:04:340Paolo Guiotto: Okay, so F must be continuous in interval AD, except the set N.
41:13:410Paolo Guiotto: At most, of course, if f is continuous everywhere, it's continuous almost everywhere, no? Because the set of bad points where the function is not continuous is just empty, okay?
41:25:560Paolo Guiotto: But this says it can have discontinuities, but not too many. So the set of points where this thing is not continuous must have a Lebec measure equal zero.
41:39:70Paolo Guiotto: Now, if you look at the case of the Dirichlet function.
41:43:690Paolo Guiotto: No? For example, this function here, if you think about it, this function is never continuous. So it means that the set of points where this is discontinuous is the interval 0, 1. So it's definitely a non-measure 0 set. Okay, so remark.
42:06:290Paolo Guiotto: the… derelict.
42:11:870Paolo Guiotto: function.
42:14:820Paolo Guiotto: So the indicator of erasionals is never continuous.
42:23:830Paolo Guiotto: So the set of points, where it is continuous is empty, or the set of points where it is discontinuous is everything.
42:33:180Paolo Guiotto: Okay? And intuitively, you can understand, because imagine that you pick a point X, no, whatever. Now, this point X can be irrational or irrational. So let's say that it is irrational for a second.
42:48:720Paolo Guiotto: We can do a similar argument for the case of a rational point. The indicator there takes value equal 1, no? So that's the value of the Dichler function at this point.
43:00:80Paolo Guiotto: Now, can you say what is the limit when, let's say X0? When X goes to that X0 of the indicator of the irrational.
43:10:440Paolo Guiotto: Does this limit exist?
43:13:910Paolo Guiotto: No, because it depends on how you get to point X0.
43:18:780Paolo Guiotto: If you go to the point at zero, traveling along rationals.
43:24:530Paolo Guiotto: So imagine that this is a sequence QN of rationales going to the point X0, and there is such a sequence, because we know that the rationales
43:33:670Paolo Guiotto: As well as your rationals are dense, are everywhere, so you can always find a sequence of rationals that go
43:41:200Paolo Guiotto: to your X0, then if you evaluate the indicator of the duration, so the directional function on that QN, you will get constantly the value 0, so this thing will go to 0.
43:56:190Paolo Guiotto: But, if you now go to zero along a sequence of irrationals, so take numbers here.
44:06:210Paolo Guiotto: Excel, still going to the point X0, but now these are irrationals.
44:12:560Paolo Guiotto: You know, as we say, the erasionals are dense as well as erasional, so you always can find the sequence of erasionals going to your point X0, and you evaluate the indicator of erasional on these points, XN,
44:30:270Paolo Guiotto: Now you will get always value 1, so this quantity goes to 1.
44:37:380Paolo Guiotto: So this means that the limit does not exist, because if you go to the point of zero along errationals, you get 0. If you go along errationals, you get 1.
44:47:530Paolo Guiotto: And if the limit exists, it doesn't matter how you go to the point X0, you must go… you must get the same limit value for the function, okay? So the definition of limit says, no matter how you go to X0, the function will go to the same value, the limit. Now, here you are that going to point X0 in two different ways, you get two different limits, so this means that
45:10:720Paolo Guiotto: the limit does not exist. So this implies that this limit does not exist, and therefore the function cannot be continuous. So this function is never… this, of course, is an argument for irrationals, but in fact, zero is rationals. The same, exactly the same argument.
45:28:200Paolo Guiotto: walks.
45:30:500Paolo Guiotto: Okay, so this explains why the Richelie function, if you want, is not Riemann integral, because the set of discontinuities is too big, no? While this theorem says the two integrals coincide.
45:47:750Paolo Guiotto: And you have also a nice test for the Riemann integrability. It says it must be continuous, except for a set of points that must be negligible for the LeBague measure.
46:01:110Paolo Guiotto: Okay, so, this explains also why So, after… this… PRM.
46:15:600Paolo Guiotto: We… will… use… denotation… integral from A to B, F of X, DX, also, for…
46:30:900Paolo Guiotto: the Lebec integral, integral from A to B, F, the lambda 1. So we will never write the Lebes integral in this way, okay?
46:40:840Paolo Guiotto: We know that there is no ambiguity, and this also means that if you want to compute the Lebesgue integral for nice functions, you can use the rules of ordinary calculus that you know. So, whatever you have learned about the Riemann integral still works for the Lebesgue integral.
47:00:250Paolo Guiotto: Actually, some of these formulas can be extended for the… have extended versions for the LeBague integral. I'm talking about, for example, the integration bypass formula, no? Integration bypass formula demands something a little bit heavy for the Riemann integral, so…
47:20:40Paolo Guiotto: In particular, all… Ordinary.
47:31:90Paolo Guiotto: calculus… formulas.
47:38:60Paolo Guiotto: can… be used.
47:45:330Paolo Guiotto: to compute the…
47:52:310Paolo Guiotto: A little bit into glass.
47:55:770Paolo Guiotto: Let me just say an example, even easier. How do you compute the integral from A to B of a given function F, DX? Now, from first year, what you do is, let's assume that this F is the derivative of something, so you compute a primitive. So, if this is integral of G prime of x.
48:20:70Paolo Guiotto: If, so, if F is equal to G prime with… with the…
48:33:60Paolo Guiotto: this G that must be continuous with derivative continuous, so that's why I'm saying it's quite heavy, because the requirements for… to use this formula, technically speaking, are very strong. You must have a function which is good, continuous, with derivative continuous.
48:53:230Paolo Guiotto: So this means that your F is continuous to apply this theorem, and it is the derivative of someone that must be a C1 function, then you say this is G of B minus G of A.
49:05:560Paolo Guiotto: Now, it can be proved that we will not enter into this, I will mention later, that actually this formula holds a much, light sensor with respect to this one, for the 11 integral. We will return this later when we, will,
49:24:210Paolo Guiotto: discuss about, a bit about weak derivatives, but you will see mostly this concept, for example, also, okay? So let's say that this formula can be used,
49:36:600Paolo Guiotto: not necessarily with this kind of heavy machinery, no? With this kind of heavy hypothesis on G. But it can be used with the very weaker assumptions.
49:49:640Paolo Guiotto: that are not applicable for the Liemann Interval, because they deal with discontinous staff and things like that, but they can be indirect. Okay, this is just to say, you, that…
50:04:50Paolo Guiotto: We should not have the impression that LeBague integral is just, sort of, I would say…
50:14:30Paolo Guiotto: It's a mathematical exercise to have some extension of the operation of the integral. That's not completely true.
50:21:110Paolo Guiotto: It's true that when we do exercises, it seems like if we are using the Riemann interval, because we computing the Gaza, and more or less, we use the classified.
50:33:370Paolo Guiotto: But there are lots of differences. Some of them we will see in the next classes. In particular, respect to the problem of computing
50:43:390Paolo Guiotto: limits of integers. This will be the next topic.
50:47:440Paolo Guiotto: We want to take the BRCA, we continue.
50:55:320Paolo Guiotto: Okay.
50:56:940Paolo Guiotto: Okay.
50:59:540Paolo Guiotto: Now… This is about the connection between Le Besgue and Riemann.
51:07:160Paolo Guiotto: On one variable, we have another definition of integral. I told before, when we want to integrate on unbounded intervals. I know it's a bit boring, this part, because… but we need to have clear what we are doing when we compute these kind of integrals. So, let's say…
51:26:800Paolo Guiotto: on R… there is… There is… an extension.
51:37:620Paolo Guiotto: of Freeman.
51:39:920Paolo Guiotto: Integral, which is called the generalized
51:49:400Paolo Guiotto: Generalized or improper.
51:55:580Paolo Guiotto: integral.
51:57:430Paolo Guiotto: So, let me discuss only one case, because here there are many cases, okay? For example, we want to integrate from A to plus infinity a function.
52:09:190Paolo Guiotto: How do we do this?
52:12:50Paolo Guiotto: Well, the Riemann integral, we say that as this restriction that the integration domain must be a closed and bounded interval, no? So we cannot just put A to plus infinity in the definition. It doesn't work.
52:26:990Paolo Guiotto: So what we do is, however, something natural. We say, we want to integrate from A to plus infinity.
52:32:950Paolo Guiotto: We stop at some point B, so we integrate from A to B, and then we take the limit. So we compute the integral from A to B of the function F,
52:42:980Paolo Guiotto: And then we send B to plus infinity, provided, of course, the limit exists. So we say that this integral is, by definition, the limit
52:52:980Paolo Guiotto: for B going to plus infinity of the integral from A to B of F of X, Yeah, sir?
53:02:80Paolo Guiotto: Provided… the… limit.
53:08:980Paolo Guiotto: exists.
53:10:260Paolo Guiotto: And it is fine.
53:12:490Paolo Guiotto: Okay, so this is basically how the…
53:15:710Paolo Guiotto: generalized integral is defined. Now, which is the… is there any relation between this integral and the…
53:24:230Paolo Guiotto: So… what… is deep.
53:31:770Paolo Guiotto: relation… Between…
53:40:330Paolo Guiotto: this… generalized.
53:44:70Paolo Guiotto: Integral.
53:46:50Paolo Guiotto: And the…
53:47:340Paolo Guiotto: what should be the Lebesgue integral? For the Lebague integral, there is no problem. The definition of a Lebesgue integral can be applied for what we could expect here, is the integral on the set A to plus infinity of the function f in the lambda 1.
54:03:600Paolo Guiotto: So we don't have to do a new definition. We already have, inside the definition of integral, this, this type of integrals defined.
54:13:920Paolo Guiotto: Now, the point is here a little bit more delicate, because the fact is that there can be functions for which the generalized integral is defined, but the LeBague integral is not defined.
54:28:890Paolo Guiotto: So, it's a bit different from this one. This one says, if you are Riemann integral, you are Lebag integral. It's not an if-and-all-if example, the Dirichlet function. Dirichlet function is Lebesgue reintegrable, but not Riemann integral. So, they are not equivalent.
54:46:20Paolo Guiotto: This one says LeBague is an extension of the Riemann integral. We cannot say the same here. We cannot say that you are generalized integral, even though if you are LeBague integral. That's… or, sorry, we cannot say if you are generalized integral, you are Lebesgue integral, as we make perspective.
55:05:680Paolo Guiotto: And here we have this example.
55:12:630Paolo Guiotto: So, it is possible.
55:19:310Paolo Guiotto: to have… That, the improper integral of F Is well defined.
55:32:640Paolo Guiotto: But… There is no LeBague integral of the same function on the same domain.
55:42:340Paolo Guiotto: Now, the problem is not a question of irregularity.
55:45:730Paolo Guiotto: There is a classical example, I write this, but then I will modify a bit to do the calculation to convince you that this is the case. The classical example is the following. For example.
56:00:830Paolo Guiotto: The integral from 0 to plus infinity of sine x over X, huh?
56:08:200Paolo Guiotto: Exists in a generalized sense.
56:13:870Paolo Guiotto: But the integral from… on 0 to plus infinity of sine x
56:21:150Paolo Guiotto: of a X in the back sensor, Is not the file.
56:27:200Paolo Guiotto: Okay?
56:28:540Paolo Guiotto: Now, this is complicated to do, and it's not impossible to show this, but it demands a lot of, pretty perturbations. However, the idea is this one, no? Is sine X over X is a function, it's like a wave.
56:48:190Paolo Guiotto: Is dump the stuff that… goes to shrink to zero. So, the generalized integral means the sum. Mmm…
56:57:630Paolo Guiotto: Means the sum of all the algebraic sums, so positive plus negative, of these areas.
57:04:550Paolo Guiotto: And that sum is finite.
57:07:520Paolo Guiotto: Now, the generalized… the Lebec integral of this does not exist, because remind always that the function to be integral in Lebec's sensor must be integral in absolute value.
57:21:460Paolo Guiotto: And here it is exactly the case. So I will show you this with a modification of this, actually a simplification of this. So I will take a function which is made like this. Instead of taking this shape curved, I take flat.
57:37:290Paolo Guiotto: So, like that. This is the value on the first,
57:41:800Paolo Guiotto: interval. I will simplify, this should be, where sine is, you know, pi, 2 pi, 3 pi, okay. Here, we will take, 1…
57:52:650Paolo Guiotto: 2, 3, okay, will be a little bit easier. Then I take a negative PR,
57:59:150Paolo Guiotto: positive here, negative here, and so on. Now, let's fix, let's put the exact values.
58:07:410Paolo Guiotto: The key point here, you see that you can see this function as 1 over X times sine x.
58:17:120Paolo Guiotto: So the idea is that you are taking, like, 1 of X and multiplying by sine, which is oscillating between minus 1 and 1.
58:24:160Paolo Guiotto: So we will transform sine into plus 1 minus 1.
58:28:480Paolo Guiotto: So that time that usually we write minus 1 to the end.
58:32:800Paolo Guiotto: and 1 of X will become 1 over n.
58:37:890Paolo Guiotto: Okay? So, we have to fix a bit the values of n. So the idea is that we take this function, f of x. The function f of x is equal to 1 when x is between 0 and 1.
58:55:490Paolo Guiotto: Then this value, so this is value 1. This is the value minus 1 half. Minus 1 half when x is between…
59:03:630Paolo Guiotto: 1 and 2.
59:06:250Paolo Guiotto: Then, between 2 and 3, the value is 1 over 3.
59:12:80Paolo Guiotto: when X is between 2 and 3.
59:15:770Paolo Guiotto: And so on, no? You see, we alternate sine plus minus. We can write this in this form, because you see that we are taking minus 1 to… it depends if we want to start from n equals 0 or n equals 1.
59:32:740Paolo Guiotto: I start from n equal.
59:37:430Paolo Guiotto: Let's start from n equals 0, so the first sign is plus, so I put that minus 1 to n, and then, 1 over… let's see, n plus 1, okay, 1 over n plus 1.
59:51:940Paolo Guiotto: Indicator of interval n plus 1.
59:56:00Paolo Guiotto: So you see that if I plug n equals 0, this becomes plus 1, this is 1, and this is indicator of 0, 1, okay? So this is exactly this value.
00:06:220Paolo Guiotto: If I put N equals 1, this is minus 1, this is 1 half, and the indicator is about 1 to 8.
00:15:70Paolo Guiotto: And so on. Now, if you sum these things, huh?
00:18:340Paolo Guiotto: It's a fake sum, because if you pick an X, if you fix an X, only one of these terms will be different from 0, okay? However, I say sum for n going from 0 to infinity of this thing is our function f of x.
00:36:500Paolo Guiotto: So this function here… is, let's say, the discrete prototype of sine X over X.
00:45:650Paolo Guiotto: Okay, now I will convince you that this has a generalized integral, but it has not a Liberg integral.
00:55:930Paolo Guiotto: Okay.
00:57:50Paolo Guiotto: We check… that.
01:03:770Paolo Guiotto: there exists the integral from 0 to plus infinity of this F, as generalized.
01:13:390Paolo Guiotto: Integrado?
01:15:770Paolo Guiotto: But, sir, at work. And… There is no…
01:20:870Paolo Guiotto: integral on domain 0 plus infinity of f of F d lambda 1,
01:27:680Paolo Guiotto: And this is the Lebesgue integral.
01:34:980Paolo Guiotto: So let's do the check number one.
01:38:600Paolo Guiotto: And this is the check number 2.
01:41:510Paolo Guiotto: The check number 1, this can be done by taking the definition. The definition says that the integral from 0 to plus infinity of the function f is the limit for B going to plus infinity.
01:58:740Paolo Guiotto: of the integral from 0 to B of F of X.
02:02:790Paolo Guiotto: Yes.
02:05:180Paolo Guiotto: Now, we can compute this syntax. Let's see why.
02:09:220Paolo Guiotto: Imagine that we have a B. This B is going to plus infinity, so it would be somewhere.
02:15:450Paolo Guiotto: Now, you know that this function, F, is made like this, no? 1 minus 1F, plus 1 third, minus 1 fourth, and so on, no?
02:25:00Paolo Guiotto: So, what can we guess here? That we have one…
02:31:140Paolo Guiotto: So these are the integers, 1, 2, 3, etc. We understand that that B, as every number, will be between two integers, right?
02:41:670Paolo Guiotto: So there will be a capital N here, and a capital N plus 1 here.
02:48:200Paolo Guiotto: So, we can say that our integral from 0 to B, so the integral from this point
02:55:150Paolo Guiotto: to this one. Can we split it into the integral from 0 to n?
03:01:910Paolo Guiotto: plus the last little piece, which is the integral from n to b of the function.
03:08:970Paolo Guiotto: Let's see what is the first integral.
03:12:360Paolo Guiotto: Well, you understand what is… you don't need to write formulas, because we are computing an integral, no? So, from 0 to 1, the function takes value 1, so we are computing the area, that will be 1, no? Base is 1, base will be always equal to 1,
03:29:340Paolo Guiotto: And 8 is here 1, so it is 1. Then, from 1 to 2,
03:34:670Paolo Guiotto: This thing is minus 1 half, right? So we are doing this area with sine, so it is negative. So base is 1, height is minus 1 half, the area is negative, is minus 1 half.
03:48:950Paolo Guiotto: Then what you expect here, here the height is 1 3rd, the base is still 1, the area is 1 times 1 3rd, so plus 1 third. Then we will have minus 1 fourth, plus 1 5th, and so on.
04:03:740Paolo Guiotto: Until we arrive at the last interval, which is the interval from capital N minus 1 to capital N.
04:11:470Paolo Guiotto: So, it doesn't matter what we have here exactly, because it will be a plus-minus 1,
04:17:580Paolo Guiotto: And in fact, it will be a minus one,
04:23:990Paolo Guiotto: minus 1 to capital N minus 1, to be precise, 1 over n minus 1. This is the last area that we have here. I'm doing the case where it is positive, it is negative, it's the same.
04:37:810Paolo Guiotto: So this is… this guy here, this sum, is this integral.
04:44:540Paolo Guiotto: Plus, we have a last piece, which is the part of integral from n to b.
04:50:750Paolo Guiotto: Okay?
04:53:150Paolo Guiotto: Let's, keep this, written.
04:56:590Paolo Guiotto: Because, as you will see, we don't need to evaluate exactly. We can do, of course, because the function is constant.
05:03:550Paolo Guiotto: So still, the area. If you want, you can do it. So the area is the length of the segment, the base, this time is… you go from N to B, so you will have B minus N for the base times the height will be something like 1 over n times plus minus 1. We don't need to bother too much with this, because as you would say, this one B.
05:27:710Paolo Guiotto: is, is disappearing, I think, finally.
05:32:730Paolo Guiotto: Okay.
05:34:20Paolo Guiotto: Now, what happens when we send B, remind that we have to compute the limit for B going to plus infinity.
05:42:570Paolo Guiotto: So, let's try to send B to plus infinity.
05:46:180Paolo Guiotto: If B goes to plus infinity.
05:49:650Paolo Guiotto: You know that this N we introduced here is such that B is between n and n plus 1.
05:58:250Paolo Guiotto: So what happens to capital N? If B goes to plus infinity, also n goes to plus infinity, because that capital N is nothing but… you know how to express this in term of B,
06:13:890Paolo Guiotto: is the integer for which B is between n and n plus 1. Who is that integer?
06:20:430Paolo Guiotto: It's called V?
06:23:790Paolo Guiotto: It has a name.
06:26:150Paolo Guiotto: If B is 10,345, comma 2 is 10,345, is the integer part of B. That's the integer part of B.
06:38:720Paolo Guiotto: It doesn't matter here, but you understand that when I send B to infinity, I send N to infinity.
06:45:960Paolo Guiotto: Now, because B, it is between n and then plus 1, so if you move B, you move automatically n. So what happens when n goes to plus infinity to this thing here? This goes to…
07:01:140Paolo Guiotto: You see that this is an alternating sum, no? 1 minus 1 half plus 1 third minus 1 fourth plus one 5th minus 1 sixth. So this goes to the infinite sum, minus 1 2N over here, n plus 1, because we are talking about starting from 0.
07:19:720Paolo Guiotto: Actually, we have even the numerical number of this, because this is log of 2.
07:25:170Paolo Guiotto: Who cares? The key point is that this arm is fine.
07:29:610Paolo Guiotto: Your mind of light moves faster, in case it is not. So this is a convergent series.
07:36:480Paolo Guiotto: is not a divergent thing, okay? It's converted, and we actually know that the value would be log of 2. And what about this thing?
07:47:250Paolo Guiotto: I claim that this is going to zero.
07:50:320Paolo Guiotto: Watch.
07:51:420Paolo Guiotto: Because what is this gap that we might understand? I don't know exactly, but…
07:57:70Paolo Guiotto: It is definitely between 0, because this positive V is between N and plus 1, and at the most is
08:07:230Paolo Guiotto: Pre-Web
08:10:770Paolo Guiotto: B is between n and n plus 1, so the distance between B and N cannot be bigger than the distance between N and N plus 1, which is 1. So that number, it doesn't matter what is it, but it's not…
08:24:90Paolo Guiotto: greater than 1.
08:26:300Paolo Guiotto: times this, which is going to zero, because n is going to infinity, so that part definitely goes to zero.
08:33:140Paolo Guiotto: So, we can say that… so, the limit…
08:38:140Paolo Guiotto: For B going to plus infinity, of the integral from 0 to B of our function f.
08:44:710Paolo Guiotto: At the end is the infinite sum.
08:48:300Paolo Guiotto: for n going from 0 to infinity of minus 1 to n divided to n plus 1.
08:55:620Paolo Guiotto: Which is, by the way, is log of 2.
08:59:740Paolo Guiotto: It's the logarithmic series, this one.
09:02:590Paolo Guiotto: But it doesn't matter. The point is that this is a finite number, and therefore, this means that there exists the generalized integral from 0 to plus infinity of this F.
09:16:830Paolo Guiotto: as generalized integral.
09:21:189Paolo Guiotto: Okay?
09:22:670Paolo Guiotto: So this function is integral in generalized sense.
09:27:350Paolo Guiotto: Now, let's see why it is not integral in layback's sense.
09:31:790Paolo Guiotto: Number 2.
09:34:29Paolo Guiotto: Now, remind that the LeBague integral We introduced formally last time.
09:43:210Paolo Guiotto: To be defined, we have always a key condition.
09:47:770Paolo Guiotto: It must be the integral of the absolute value, finite, okay?
09:53:40Paolo Guiotto: And that's what we are going to check here.
09:56:130Paolo Guiotto: So, in order… D… Le beg Integra.
10:05:430Paolo Guiotto: From 0 to plus inf… let's say, let's keep still this notation that sets So this,
10:12:590Paolo Guiotto: For the moment, keep separate the bag from other stuff, to do not make confusion.
10:18:650Paolo Guiotto: In order that the backend to have to be defined, We need the…
10:25:660Paolo Guiotto: that the integral, which is still a Liberg integral, from 0 plus infinity of the absolute value of F,
10:33:980Paolo Guiotto: B… defiled.
10:38:470Paolo Guiotto: But what is this? Why do you take the absolute value of F,
10:44:460Paolo Guiotto: Now, you remind what is F. Now, F is… we divided the line in intervals.
10:51:150Paolo Guiotto: With the endpoints, the integers.
10:54:230Paolo Guiotto: And we have this function that it is plus minus something, no? When you take the absolute value, you get, in the first interval, you add 1, it remains 1. In the second interval, you add minus 1 half, it becomes plus 1 half.
11:09:90Paolo Guiotto: So let's say that this is the value. So what I'm now plotting is the absolute value of that.
11:14:890Paolo Guiotto: In the third interval, we have 1 third, plus 1 3rd, it remains 1 third.
11:20:230Paolo Guiotto: Then we have minus 1 fourth, then we have plus 1 5th, minus 1 sixth, and so on.
11:26:610Paolo Guiotto: So actually, when you take the absolute value of F,
11:30:750Paolo Guiotto: I know that, formally, I cannot say that the absolute value of the sum is not the sum of the absolute values, right? I cannot say that, okay, take absolute key, absolute put inside it to eliminate this sign, yeah? But here, I can't do, because this is actually a fake sum.
11:48:650Paolo Guiotto: Only one of these themes on.
11:52:170Paolo Guiotto: all the other are off, because for every X, X production in one interval, NM plus 1, and not in two of them, no, because these are consecutive intervals. So, in fact, when you fix an X, this is not a sum. This is a sum of zeros plus 1 term, which is different from zero.
12:12:60Paolo Guiotto: So, in fact, we are authorized to say that the modulus of the sum is the sum of the models, because it's… all terms are 0 except 1. So I can say that my F becomes the sum for n going from 0
12:28:500Paolo Guiotto: But if this is disturbing for you, you look at this now, put the absolute value, what you see here is that models of n is 1 here, 1 half here, 1 third here, 1 fourth here, and so on, you get the same formula. So you have that this is 1 over n.
12:47:400Paolo Guiotto: Indicator of interval n plus 1.
12:53:730Paolo Guiotto: And now, what is the Lebec integral of this thing?
12:57:860Paolo Guiotto: So…
13:00:730Paolo Guiotto: If you want, we can help you in this way. It is, the bag measure of this set. I'm dashing here.
13:11:70Paolo Guiotto: So it will be the sum of the areas of these rectangles. So the integral on 0 plus infinity of the absolute value of F in the back measure is the sum for n going from 0 to infinity of these areas.
13:29:160Paolo Guiotto: So if you want, I write in extended form. The first area is 1, the second is 1 times 1 half, so 1 half.
13:36:150Paolo Guiotto: Then we have 1 third, 1 fourth, and so on. So this is the sum of terms 1 over n plus 1, or if you prefer, sum for n going from 1 to infinity, 1 over n, which is in a harmonic sum, divergent harmonic sum, this one.
13:54:450Paolo Guiotto: So this is equal to plus infinity.
13:57:500Paolo Guiotto: So, this means that this integral is not finite, and therefore the function cannot be integral.
14:05:520Paolo Guiotto: So… F is not the… Lebecca integral on 0 plus infinity.
14:15:350Paolo Guiotto: With respect to the badge match.
14:17:910Paolo Guiotto: Okay.
14:19:790Paolo Guiotto: So… This, as you can see, is something which is not related, in fact.
14:27:330Paolo Guiotto: from, to the fact that the function is, not nice enough. This has to do with, actually, with the definition of,
14:38:710Paolo Guiotto: of, generalized integral. Because if you look at this example.
14:44:890Paolo Guiotto: At the end, the generalized integral turns out to be the sum of these two years.
14:50:420Paolo Guiotto: And the vet integral for the absolute value is the sum of the modules of this series.
14:58:140Paolo Guiotto: So you can have that acidity is convergent, but thus not convergent with the absolute values that's done.
15:06:610Paolo Guiotto: Maybe you remind that when the series convergence with the absolute value, you have something which is called the absolute convergence, which is convergence which is stronger. So if you have convergence with the absolute values, then you have convergence without the absolute value, but not vice versa.
15:24:660Paolo Guiotto: Now, if we take the generalized integral, and we impose that the function f
15:33:610Paolo Guiotto: B, generalized integral, but in absolute value, then we get the connection between the two integrals. So what can we prove there?
15:45:870Paolo Guiotto: But let's say that this is…
15:47:780Paolo Guiotto: So this example shows that you can have the generalized integral existing and the function not being Lebesgue integral.
15:56:830Paolo Guiotto: If we add something to F, to be generalized integral, which is the following, so… Proposition.
16:08:560Paolo Guiotto: Let's… F… B.
16:16:480Paolo Guiotto: Absolutely.
16:21:360Paolo Guiotto: Integrable.
16:23:530Paolo Guiotto: in generalized… sensor.
16:30:230Paolo Guiotto: So this means the integral from A to plus infinity of modulus f of x dx, this thing, with the definition of generalized integrals, so this is the limit from… of the integrals from A to B when B goes to infinity.
16:46:900Paolo Guiotto: suppose that this is finite, then this function is the Beck integral, also, on A2 plus infinity.
16:56:230Paolo Guiotto: And the two integrals coincide.
17:00:910Paolo Guiotto: And… the integrals from A to plus infinity of F byx dx, that's the generalized Integral.
17:11:820Paolo Guiotto: coincide with Lebeg integral.
17:15:520Paolo Guiotto: A to plus infinity of FD lambda 1.
17:19:790Paolo Guiotto: So let's say that the Leban integral is an extension of the generalized integral when we consider not the class of functions for which the generalized integral is the fact that a restricted class, which is a class
17:36:500Paolo Guiotto: For which, the generic part of the absolute value is found.
17:42:660Paolo Guiotto: This may give the idea that… but then in general, the integral is not an exception, because it seems to be a particular base. Now, that's not true, no? Let's say that we… it is clear that we may lose some…
17:59:160Paolo Guiotto: functions, like, for example, this, by the way, turns out to be an important function, the function, not the discrete one. The sine x over x is a function we encounter. For example, it turns out that it is a Fourier transform of a function, so this is an important
18:14:740Paolo Guiotto: function for us, we will see later, where this comes from. It's an interesting function. And this function is not an N1 function, so it's a bit disturbing with fact. However, we accept there are… there will be exceptions. Not everything can work.
18:33:740Paolo Guiotto: okay, so this is the Lebegg integral.
18:40:410Paolo Guiotto: Of course, there are versions of this for all generalized integral and finite sets. I don't want to do all the list, because it takes too much time, so we…
18:53:120Paolo Guiotto: we just consider this as sufficient. So this is about the relation between Riemann, or generalized integral, in dimension 1, no?
19:06:600Paolo Guiotto: Then, what else? Well, I just leave most of this chapter as a reading, the chapter 5. There are things that you already know, so let's see… so…
19:26:540Paolo Guiotto: what happens to the Lebeg integral?
19:30:370Paolo Guiotto: Les begues Integral.
19:33:410Paolo Guiotto: in RM. So we now give a look to the Lebec integral in RM, so this means that, what if we have an integral on a domain E of a function F, B lambda M,
19:51:640Paolo Guiotto: Where F is in L1,
19:56:660Paolo Guiotto: E. E is a subset of RN.
20:01:160Paolo Guiotto: How do we compute this?
20:03:220Paolo Guiotto: Well, you remind that when you study the integrals for functions of several variables, second year calculus, you have seen that there is a technique to reduce this calculation to one variable integrals that you already have a machinery, no?
20:19:800Paolo Guiotto: And how do you do this? You do this by splitting the integration in the array of variables. For example, suppose that you have to integrate in X and Y, and you can say, I can integrate first in X and then in Y on treatable domains. This machinery is called a reduction formula.
20:39:280Paolo Guiotto: And it, holds also ER, no?
20:42:700Paolo Guiotto: The idea is, suppose that, now I will do this figure. Suppose that we are in RM,
20:53:510Paolo Guiotto: If you… if you want to make this easy, think to R2, okay? In such a way that these are R, the two axes are R and R.
21:05:330Paolo Guiotto: In general, if I am in RM, this means that my Bible will have M conference.
21:12:410Paolo Guiotto: Let's imagine that we can split the design components into two blocks.
21:17:70Paolo Guiotto: No? I have an alright example. In Act 3, I could split a block of one component and a block of two.
21:24:570Paolo Guiotto: In total, 3.
21:26:270Paolo Guiotto: If I had an air 10, I could do 5 and 5, or 6 and 4, or 3 and 7, okay?
21:33:350Paolo Guiotto: So imagine that we can,
21:35:860Paolo Guiotto: If, say, if I have F of X, X here is an array of M variables.
21:45:720Paolo Guiotto: So let's assume that we take just, for example, the first K variables here, and the next M minus K variable.
21:56:580Paolo Guiotto: Okay, so we introduce two subgroups of variables.
22:02:970Paolo Guiotto: So let's say that we call X1 XK.
22:08:630Paolo Guiotto: Which letter? I don't know. U?
22:12:220Paolo Guiotto: And, XK plus 1.
22:16:00Paolo Guiotto: to XM, we call this V, no?
22:21:260Paolo Guiotto: If you reduce to two variables, it means that you have two letters, X and Y, one is X and the other is Y. So it means that on the horizontal axis, you have the axis of U, and on the vertical axis, you have the axis of V.
22:39:400Paolo Guiotto: So this means that U belongs to R,
22:42:930Paolo Guiotto: K, and V belongs to R… M minus K.
22:49:680Paolo Guiotto: Our domain, integration domain, so this figure is not… we are not plotting F, we are just working only on the domain of, integration domain, which is the SATE, which is a subset of RM.
23:04:800Paolo Guiotto: Now, the idea is that if you want, let's say the intuitive idea is that intuitive.
23:15:870Paolo Guiotto: idea.
23:18:820Paolo Guiotto: It's a very rough idea, it's not a proof, but it explains where this formula comes from. Now, think to the integral of F,
23:29:440Paolo Guiotto: Instead of thinking as a quantity that we don't know, actually, how to compute, we imagine that we discretize and it comes… we replace the integral with a sum.
23:38:740Paolo Guiotto: So we are doing a sum of points F of X when X is in E. Of course, I'm not saying that this is in identity, I'm imagining that we can discretize, and this becomes a sum, no?
23:53:960Paolo Guiotto: So it is like if we're doing sum of values F of X when X is in E, no? Well, point X is in E. We may represent X in E as point X in E as a pair, UV, where U is down here.
24:10:730Paolo Guiotto: Envy is on the, wire and the, ordinance.
24:18:30Paolo Guiotto: Now, the point is that we may mention that we could split, so this as the sum on all UVs that are in domain E of F UV,
24:30:150Paolo Guiotto: And the idea is now to split this sum into a double sum. We could say, okay, let's sum up the FUV
24:40:450Paolo Guiotto: summing first on the V, for example, But on which V?
24:48:200Paolo Guiotto: on V, because you are summing on points UV that belongs to E. So, in this sum, U is freezed, you are moving only V, so you sum on the V such that UV belongs to E.
25:05:550Paolo Guiotto: Okay?
25:07:750Paolo Guiotto: It's like, if you are,
25:12:00Paolo Guiotto: Imagine that UVR integers, okay? Natural numbers. So you have F11, F12, F13, no, you can see as in this way, F11, F12, F13, etc.
25:28:940Paolo Guiotto: Of course, you have infinitely many values, so F1N, then you have F21, F22, F… 2, 3…
25:42:510Paolo Guiotto: F to N, and so on.
25:46:740Paolo Guiotto: And you have M lines, FM1, F.
25:52:990Paolo Guiotto: 2, and so on. F, M,
25:57:160Paolo Guiotto: Now, the integral is the sum of all these numbers, okay?
26:02:740Paolo Guiotto: Mehdi, you have saw a table of numbers.
26:07:230Paolo Guiotto: You can decide to sum. For example, isomarker around the first line, no?
26:13:890Paolo Guiotto: I've decided to sum all these numbers. Plus.
26:17:340Paolo Guiotto: the sum of all this number, plus the sum of all this number, and so on. So you sum by lines.
26:23:950Paolo Guiotto: Have you… The total would be the sum of the sums.
26:29:460Paolo Guiotto: Or you can sum by column, for example, you can decide that I keep fix the second component, I sum all the elements like that, and I sum all these ones, then I sum all these ones, and then I do the total sum of all these sums, huh?
26:44:110Paolo Guiotto: So this is the idea behind this. I have to sum on all UV that belongs to E. I decide to freeze the first U and sum over the V. But on which V?
26:58:20Paolo Guiotto: Not every V, because the point UV must be in E, so on the V, for which UV will be in E. We can see this in the figure, because here you are fixing U. So, you are fixing the abscessa of these points, like here.
27:13:780Paolo Guiotto: And you have to sum along all the V for which this point belongs to E. So you see that this means that there is a set of points
27:24:330Paolo Guiotto: This one… that you see on the V, on domain of V,
27:30:540Paolo Guiotto: This is the set of V,
27:34:190Paolo Guiotto: for which UV belongs to E. In other words, these are, let's say, roughly, they are the units of points with the chisel U that are in E. It's like if your E is a potato, you slice the potato along the chisel U,
27:53:640Paolo Guiotto: And you take all the originates of these points, okay?
27:59:450Paolo Guiotto: When you change U, you see that this set of points change, because if I take an RW, let's use a different color, for example, imagine that U is this one.
28:10:920Paolo Guiotto: The set of points UV are on this green line.
28:16:830Paolo Guiotto: The set of deordinates is this one.
28:21:210Paolo Guiotto: Is the projection on the… on the vertical axis.
28:27:160Paolo Guiotto: And this is, again, a set of V.
28:31:270Paolo Guiotto: For Witcher?
28:32:990Paolo Guiotto: point UV belongs to E. As you can see, when you change U, this set will have a different shape, will be different. So it's actually a set that depends on U, that we call the U section of E.
28:50:170Paolo Guiotto: So it says that this is exactly this, this set here.
28:57:150Paolo Guiotto: is the set where V belongs to the U section of E.
29:03:890Paolo Guiotto: Now, in this sum, we freeze the U, and therefore there will be a second sum where we have to do the sum of all the possible U's.
29:13:260Paolo Guiotto: What are the U we consider?
29:17:720Paolo Guiotto: So, you see, not all you look at this, you're not all you have little cigarette.
29:23:540Paolo Guiotto: Because if I take this U, for example, take an U here.
29:29:870Paolo Guiotto: Is there any point with UV? So, any point with D sub Shissa that belongs to E?
29:38:110Paolo Guiotto: No. So it means that this type of sector for this U will be empty, because there is nothing for which
29:47:930Paolo Guiotto: So in this section can't end.
29:54:780Paolo Guiotto: So I am not interested in those you for which this set is known.
30:00:770Paolo Guiotto: And this idea has been called EU sections, so I take the sum of the EU, such that EU is different from MD.
30:12:240Paolo Guiotto: Okay, so I'm just saying, you have a… I'm changing a sort of more refined version of this idea. You have a table of numbers to sample.
30:23:90Paolo Guiotto: You have to sum all the entries of that table.
30:29:610Paolo Guiotto: Okay, you can decide to be way, for example, this plus this plus this plus this plus this plus this, you can go at random, but the rational way would be
30:39:650Paolo Guiotto: Either you sum by lines, and then you, do the equation of all the totals, and you get the sum of all the items. Or the other way is to sum by columns.
30:50:20Paolo Guiotto: Of course, these are not the only possible ways, but they are the most natural. So, if we go back here, so we have that integral on e of f in the lambda m can be seen as this double sum
31:04:640Paolo Guiotto: of FUV, This one is for V that belongs into EU,
31:12:870Paolo Guiotto: This is for you, that… such that
31:17:180Paolo Guiotto: The EU section is different from NT.
31:21:300Paolo Guiotto: Now, you restored the integral notation. You look at this as an integral.
31:26:370Paolo Guiotto: This is an integral only in the V variable, because U is frizzy here. So you are doing integration here of what? The function is FUV,
31:37:400Paolo Guiotto: Now, we have to introduce something to specify that we are summing in V, so we will write D lambda V to make… to remind that this is a sum with respect to the V variable.
31:49:920Paolo Guiotto: Now, the V variable is in… in this notation, V is in RM minus K, so that's Y is… this is an M minus K measure.
32:03:800Paolo Guiotto: On which domain? On the domain? EU?
32:07:380Paolo Guiotto: And then you have a second sum, which is, again, an integration.
32:12:640Paolo Guiotto: Once you have done this operation, you don't see any more V, because V has been summed. You see only something that depends on you. So this second
32:23:420Paolo Guiotto: Operation?
32:25:80Paolo Guiotto: must be summed in U, so that's why we write D lambda U.
32:29:890Paolo Guiotto: U belongs to the dual variable, so RK, in total, we have M, M components, we divide it into K plus M minus K. And this is, you have to integrate on the domain of U such that E of U is different from MT.
32:48:640Paolo Guiotto: And so we get this remarkable formula, which is actually a theorem.
32:54:270Paolo Guiotto: This is called the… Foobiny.
32:59:940Paolo Guiotto: ERM?
33:02:620Paolo Guiotto: It says that if you have a function F, which is integral in a domain E, E,
33:10:220Paolo Guiotto: domain of RM.
33:13:970Paolo Guiotto: Then… the integral, on E of the function f
33:21:570Paolo Guiotto: with respect to the Lebesgue measure of RM can be computed as this nested integral operation, which is made by two integrations, two consecutive integrations. I do a first integration in V variables, and these are
33:40:890Paolo Guiotto: So, it is assumed that I split my array of M variables into two blocks. One is made by the first K component, I call it U.
33:51:360Paolo Guiotto: And the other one is made by the last M minus K component, so I call it V. So this will be an integration in M-K variable, in the V variable.
34:02:220Paolo Guiotto: On which domain? On the domain EU. EU is the domain of V. EU is the set of Vs such that UV belongs to E.
34:15:780Paolo Guiotto: It is called the U section of E.
34:19:900Paolo Guiotto: Now, once I've done this integration.
34:23:120Paolo Guiotto: the output will be a function that depends only on U, and I integrate on the remaining variable, so in the measure, in the K variables.
34:34:30Paolo Guiotto: So, and this is an integration on EU.
34:38:320Paolo Guiotto: on which you… on the U for which the EU section is non-empty.
34:45:950Paolo Guiotto: If you add the U for which the U section is empty, in this evaluation on the empty set, you have to go back to zero. So, you can add, or not, these components have been counting.
35:00:330Paolo Guiotto: So this is insane, and this is the way we practically do the integration, because it says, if you have two Bibles, we can split into two integrations in one Bible, because this M is 2, K is 1, M minus K is 1.
35:15:30Paolo Guiotto: So these are two integrals in two variables.
35:18:890Paolo Guiotto: in one variable nature. We know how to handle that with the tools of calculus, and we do the calculation of the double integral in this way. Okay?
35:30:670Paolo Guiotto: Okay, so it's 12 or 6, so let's stop here.
35:36:70Paolo Guiotto: Maybe, last… the next time, I will tell you…
35:45:290Paolo Guiotto: Well, actually, let's, let's, let's leave things like this. I just suggest you, we don't, we don't use too much,
35:55:920Paolo Guiotto: Integration in several variables in this tool, so I will not insist too much, because we mostly do exercises with integration in one variable. Okay, so I will never use the Fabini theorem, just to be clear. But you are aware that there exists this tool. There is also change of variable formula.
36:13:50Paolo Guiotto: I suggest you just to read the, I won't insist on that.