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00:10:140Paolo Guiotto: Okay, so the second property is a sort of… is a linearity, but conditioned to have positive things. So if you have two measurable functions.
00:22:810Paolo Guiotto: And the two positive constants, alpha and beta positive, then their linear combination is measurable.
00:30:850Paolo Guiotto: and positive. This we know from the properties of measurable functions, and because of the coefficients, and the two functions are positive.
00:42:610Paolo Guiotto: And what turns out is that the integral of the linear combination
00:50:590Paolo Guiotto: alpha F plus beta G is the linear combination of the integrals, something that
00:59:20Paolo Guiotto: We, of course, expect from any integral operation.
01:05:950Paolo Guiotto: Okay, so now to define the integral, this was the first step in the process of defining the integral.
01:15:660Paolo Guiotto: which is the definition of the integral for positive functions. So, so far, We defined the…
01:27:580Paolo Guiotto: the integral on E of a function F with respect to measure mu for a function F,
01:35:840Paolo Guiotto: Which is, of course, measurable.
01:39:290Paolo Guiotto: and positive.
01:43:820Paolo Guiotto: Now, we want to first remove the restriction about the sign. We want to define integral for a real valued function, so…
01:56:180Paolo Guiotto: wheat.
01:57:960Paolo Guiotto: and want… to extend… these… to a case where F is just a measurable function.
02:12:850Paolo Guiotto: So, F is a function defined on the real value.
02:19:100Paolo Guiotto: Now, the idea is the typical idea of what we do with the concept of integral when we have a function with positive
02:30:750Paolo Guiotto: and negative values. So this is X, this is the codomain R, we have a function, positive, negative, on some domain.
02:39:720Paolo Guiotto: Ye.
02:40:720Paolo Guiotto: Now, as you know, the integral does not count the geometrical area between the function and the axis, because it counts positively areas when f is positive, and negatively areas when F is negative.
02:54:180Paolo Guiotto: So, the idea is that, basically, we should sum the black area and subtract the black area when the function is negative. How can we do that? Well.
03:06:630Paolo Guiotto: We do that introducing a couple of auxiliary functions, so we define the function F plus.
03:17:830Paolo Guiotto: Which is called the positive part.
03:23:110Paolo Guiotto: of F…
03:25:450Paolo Guiotto: So we take this function, which is F when F is positive, and 0 when F is negative.
03:32:240Paolo Guiotto: So if you look at this figure, the positive part is when F is positive, is the green function.
03:39:400Paolo Guiotto: the positive part is F itself. When F is negative, you cut off at zero the value.
03:46:110Paolo Guiotto: And then you define the negative part, F minus, Which is the negative… But… of F,
03:59:00Paolo Guiotto: Also here, actually, despite name, the negative part is positive.
04:04:660Paolo Guiotto: And this is 0 when F is positive.
04:09:260Paolo Guiotto: And minus F when F is negative.
04:13:150Paolo Guiotto: The effect on this example is that here we have 0, and when the function is negative, we have minus F, so this is a positive thing. So the red thing is the negative part, the green thing is the positive part.
04:28:630Paolo Guiotto: So in this way, we basically reflect the negative area into a positive area. And so the idea is that the integral is now made by
04:39:300Paolo Guiotto: the sum of the integral of F, which is this positive green area, minus the integral of F minus. This is what we want.
04:50:30Paolo Guiotto: Now, to do that, we have to first check a few facts. So, when F is measurable, both F positive, F, negative are measurable.
05:00:770Paolo Guiotto: Notice that, huh?
05:08:110Paolo Guiotto: if, a function f is measurable.
05:13:430Paolo Guiotto: on E, both F plus, F minus are measurable on E, and both are positive.
05:23:140Paolo Guiotto: You can see the server.
05:26:780Paolo Guiotto: Actually, it is an if and only if.
05:32:380Paolo Guiotto: You can see this because,
05:35:690Paolo Guiotto: If, let's say, if you look at this implication.
05:40:620Paolo Guiotto: F is measurable, then, for example, F plus is what? Is F when F positive, and 0 when F negative. We can write this thing in this way. It is the maximum
05:55:260Paolo Guiotto: between F and 0, when F is positive, the maximum is F. When F is negative, the maximum is 0.
06:03:230Paolo Guiotto: Okay? So this means that our function F plus at point x can be seen as a composition, you see?
06:15:120Paolo Guiotto: There is a function that I call phi of Y that is a numerical function that gives a maximum between y and 0.
06:27:370Paolo Guiotto: And you evaluate this function on F of X.
06:32:710Paolo Guiotto: If you do phi of f of x, you get your maximum between f of x and 0, which is another way to write DF plus. Now, the function phi of Y as a numerical function, so this is a function from R to R,
06:49:820Paolo Guiotto: So, as you can see, our F plus is a composition. You do, first F, and then C. So we are going to use what we know about the composition with respect to the measurability. What we know is that this guy is measurable.
07:09:70Paolo Guiotto: We know that, at least we told that, in general, composing measurable functions is not,
07:16:580Paolo Guiotto: Safer with measurability.
07:19:630Paolo Guiotto: However, if the second function is a bit more than this, so it's a continuous function, and this is the case because the maximum between Y and 0, if y is the variable of this function, so Y here, what is this? It is 0 and Y is negative, so this is the plot of the function.
07:39:690Paolo Guiotto: And it is Y when Y is positive, so it's…
07:43:500Paolo Guiotto: a continuous function with just a kink in a zero, but it is continuous. So this function here is… the phi is continuous.
07:55:70Paolo Guiotto: So we are composing a measurable function followed by a continuous function. Here we know that there is a general property that says that this F plus is measurable.
08:10:420Paolo Guiotto: And similarly for F-, F- is the maximum between minus F and 0, and you get the same point.
08:20:970Paolo Guiotto: So this shows that if F is measurable, both F plus F minus are measurable, but also the vice versa also.
08:28:990Paolo Guiotto: So, if the two F plus minus are measurable.
08:33:600Paolo Guiotto: then also F is measurable. And this is because we have a relation that allows us to write F in terms of the two functions F plus F minus.
08:47:230Paolo Guiotto: You can understand this relation by looking at the figure here. Remind that the black line is the F, is the function F.
08:59:350Paolo Guiotto: Okay?
09:00:600Paolo Guiotto: So now, if you look carefully, you may notice that if you do the red.
09:06:910Paolo Guiotto: sorry, the green minus the red, you get F, no? Because when F is positive, F plus is F.
09:15:960Paolo Guiotto: And, F minus is 0, so if you do the difference, you get that.
09:20:250Paolo Guiotto: When F is negative.
09:22:660Paolo Guiotto: F plus is 0, minus… F minus is minus F, so it is still F. So that's the relation. F, which is equal to F plus minus F minus.
09:35:930Paolo Guiotto: This is the difference between two measurable functions. We know that the sum difference of measurable functions is measurable, so this is a measurable function.
09:46:130Paolo Guiotto: Yeah. Shouldn't be a minimum between 9 and zero.
09:52:130Paolo Guiotto: At times, people are leaving between the meeting.
09:57:710Paolo Guiotto: No, you see, because when F is negative minus F,
10:05:920Paolo Guiotto: minus F is positive, so you take the maximum is minus F.
10:10:640Paolo Guiotto: When F is positive, Minus F is negative, the maximum is 0.
10:16:30Paolo Guiotto: then probably that's minus the minimum, something, I don't know.
10:24:520Paolo Guiotto: Is it… I don't think it's got yours. Minus the minimum of F0, is that true?
10:32:110Paolo Guiotto: Because if F is positive, the minimum would be 0, and the function is 0. If f is negative, the minimum would be F.
10:40:670Paolo Guiotto: we… so we would have minus F, that is positive, so it's the same. So it is minus the minimum, okay?
10:48:470Paolo Guiotto: Okay, so this shows that F is measurable if and only if both F plus, F minus are measurable. So now this, gives the way now, so the idea is, so…
11:00:950Paolo Guiotto: the idea…
11:06:510Paolo Guiotto: should be… So, do you want to take the break, or you want, we continue. We can continue.
11:16:230Paolo Guiotto: Okay, should be… integral, on e of the function F,
11:23:170Paolo Guiotto: As we said, this should be the definition, no?
11:26:110Paolo Guiotto: We said we sum…
11:29:340Paolo Guiotto: the green area minus the red area, so we subtract the two areas. So these are the integrals, so we should say, the idea is, let's define this operation in this way. Integral on e of f plus
11:44:760Paolo Guiotto: This is the green area.
11:47:150Paolo Guiotto: minus integral on E of F minus, and this is the red area, hmm?
11:54:30Paolo Guiotto: green… area. Of course, it's not an area, and this is the red.
12:03:240Paolo Guiotto: Amazing.
12:06:120Paolo Guiotto: Respect to the figure.
12:08:600Paolo Guiotto: Of course, yeah, we have to be careful, because,
12:13:910Paolo Guiotto: In the definition of the integral of a positive function.
12:19:630Paolo Guiotto: We have not said anything about the nature of the value of the integral. We know that the integral, by definition, is a limit of IN, which are definitely both in numbers, but they could be infinite, and the limit could be even infinite. Why not?
12:38:10Paolo Guiotto: So this number, in general, is positive, and it could be even equal to plus infinity.
12:45:990Paolo Guiotto: Okay? So this creates some problem here, because…
12:50:660Paolo Guiotto: What if, well, if the two quantities are both fine, there is no question, no? It's clear that that operation makes sense. If one of them is infinite, we put two, no? The difference, this is infinite, this is fine, we can make the difference, and vice versa.
13:09:920Paolo Guiotto: But if both of them are infinite, that would be a problem, because, of course, they would be both plus infinity, now that these are positive countries.
13:19:370Paolo Guiotto: And so, we would have plus infinity minus plus infinity, and this would be a problem. So, to have a proper definition here, we need… at least one of them must be finite. Actually, we prefer to give the definition with both of them finite, okay?
13:38:210Paolo Guiotto: So, to… 2.
13:42:490Paolo Guiotto: Me.
13:45:250Paolo Guiotto: Peace.
13:47:160Paolo Guiotto: Proper.
13:50:360Paolo Guiotto: definition.
13:52:300Paolo Guiotto: we… require… That both of them, so the integral of the positive part.
14:01:810Paolo Guiotto: And the integral of the negative part
14:05:600Paolo Guiotto: that I remind you are positive.
14:08:850Paolo Guiotto: This is not a requirement, so let's put in parentheses, this is always true.
14:13:550Paolo Guiotto: But the requirement is this one, let's mark in red.
14:17:980Paolo Guiotto: They are both fine.
14:22:430Paolo Guiotto: Now…
14:23:510Paolo Guiotto: If we think about it, this means that the sum of this, referring to this figure, the green and the red area, the sum of the two must be fined, because since they are positive, if one of them is infinite, the sum will be plus infinite.
14:42:840Paolo Guiotto: If one or both, okay? So, saying that the two are fined is equivalent of saying the sum of the two is fine.
14:52:500Paolo Guiotto: And now here, if you look in this figure, the sum of the two is still the area between F12, which is the green.
15:02:180Paolo Guiotto: and red. And that function, let's color with another thing, with, A blue, maybe light blue.
15:11:950Paolo Guiotto: This function here…
15:16:40Paolo Guiotto: is F when F positive minus F, when F negative. That function is the absolute value of F.
15:22:520Paolo Guiotto: So, the fact is that to have these two positives is equivalent of having their sum.
15:29:690Paolo Guiotto: positive.
15:36:790Paolo Guiotto: Which is, again, equivalent of having, if you want, you can also see this by using the linearity of integral with positive functions. So, integral of F plus plus F minus. If you do the sum, if you do the difference, as we noticed, we obtain F.
15:56:20Paolo Guiotto: But what if we do the sum? If we do the sum, we obtain the absolute value of F.
16:00:810Paolo Guiotto: So, in fact, this is the equivalent of saying integral of F plus plus F minus the mu finite.
16:09:900Paolo Guiotto: And this thing turns out to be the absolute value of F. So we arrive to this condition, integral on e of modules of F, d mu.
16:22:330Paolo Guiotto: Fine, Nita.
16:24:430Paolo Guiotto: Notice that, huh?
16:26:430Paolo Guiotto: Remarker.
16:29:190Paolo Guiotto: If the function f is measurable.
16:34:150Paolo Guiotto: then the absolute value of F is measurable.
16:40:120Paolo Guiotto: If you want, this implication can be seen as a consequence of the fact that, in fact, it's measurable, both the plus and minus are measurable, the sum is measurable, so the multiplication is measurable. Or if you want to consistently start the composition, it is F…
16:56:640Paolo Guiotto: Followed by absolute value. The models of F is a continuous function, so we are composing measurable with continuous. But this is not an if and only if.
17:11:740Paolo Guiotto: So this thing is not correct.
17:15:170Paolo Guiotto: And then, you remember that we said that, we can, we can build an example of a function, which is 1 on a set, minus 1 on each component, and if that set is not measurable, that function cannot be measurable.
17:28:930Paolo Guiotto: And so, whenever you have that not all the subsets of the space are measurable, that will… this means that there is someone who is not measurable, and therefore you have non-measurable functions. So, this cannot… cannot be said. It was an if and a if.
17:46:310Paolo Guiotto: Okay, so now we have, finally, the ingredients to put the definition.
17:52:900Paolo Guiotto: So, let, X… F… mu B.
17:59:930Paolo Guiotto: A measure of space.
18:04:890Paolo Guiotto: F be a function which is measurable on E.
18:10:350Paolo Guiotto: We say that.
18:15:660Paolo Guiotto: F is… Integral.
18:19:850Paolo Guiotto: If you want a mu integral.
18:29:620Paolo Guiotto: on E, Ethan…
18:36:770Paolo Guiotto: So the condition that we need is that this one. The integral of the absolute value is finite. If integral on E of absolute value of F, E mu, is finite.
18:51:90Paolo Guiotto: And in this case, if this is true, in this case.
19:01:80Paolo Guiotto: we called.
19:04:120Paolo Guiotto: Integral.
19:05:790Paolo Guiotto: or mu integral, then we never use mu integral of F on E,
19:14:30Paolo Guiotto: simply what we defined. So the integral of F with respect to measure mu is the integral of the positive part of F minus the integral of the negative part of F.
19:28:670Paolo Guiotto: So we have now a definition of integral for function of variable sine.
19:37:760Paolo Guiotto: We introduce also an important notation. We call the class of integral function with this symbol, D.
19:47:480Paolo Guiotto: Class.
19:50:630Paolo Guiotto: of integral functions.
19:55:950Paolo Guiotto: is denoted.
20:00:140Paolo Guiotto: Bye.
20:01:660Paolo Guiotto: But here, there are…
20:03:860Paolo Guiotto: many equivalent notations, so it depends on how much you want to be… to put details. So, let's say that the more,
20:14:870Paolo Guiotto: The most complete notation is this one. L1XF mu. Sometimes you see this.
20:23:970Paolo Guiotto: Well, if, for example, sometimes it may happen that the sigma algebra is clear, but there are several measures involved, so you may be… you write something like L1X mu.
20:38:770Paolo Guiotto: you forget, let's say, the F, because the F is the same, and in your problem, there are more measures involved. Or sometimes, there is one, just one measure, and so the setup, the sigma algebra and the measure are clear, and you just write L1X.
20:58:770Paolo Guiotto: Okay? So, it depends on how many details you want to make clear, no? If you are…
21:09:730Paolo Guiotto: Okay, the building is not collapsing.
21:14:470Paolo Guiotto: So we will mostly use this one, okay, because it will be always clear which is the reference space, and so on.
21:25:190Paolo Guiotto: Okay. Final step… Last step.
21:34:10Paolo Guiotto: Step… It's pure bureaucracy, is the extinction Extension… to, C-valued… Complex value there.
21:54:10Paolo Guiotto: functions.
21:55:380Paolo Guiotto: This will be important for us, because, for example, if we get transformer in analysis, which is the characteristic function in probability, is
22:07:150Paolo Guiotto: the integral of a complex valued thing, okay? So it's important to be able to integrate functions F defined on some E that are complex valued.
22:23:10Paolo Guiotto: Now, what does it mean to be measurable here? Of course, you cannot say F larger than a number is a measurable set, okay? So, we should…
22:33:390Paolo Guiotto: I need to be fixed to three details. So in this case, since the function is C-valued, it means that the number, F of X is a complex number, no? It's a…
22:47:180Paolo Guiotto: A complex number, and you know that algebraically, any complex number can be written, for example, in the algebraic form with the real part and in an imaginary part.
22:56:850Paolo Guiotto: And since the number depends on X, the real and the imaginary part will be functions of X. So we will have something like U of X plus IV of X
23:09:480Paolo Guiotto: where U and V… now, these are functions defined on E, and they are real valued if they are the real and imaginary part, no? So this is the real…
23:21:650Paolo Guiotto: part.
23:24:420Paolo Guiotto: of F, and this is the imaginary part.
23:33:460Paolo Guiotto: But… of… Yes.
23:37:920Paolo Guiotto: So, what will be a measurable fun… a C-valued measurable function? So, we will say that F is measurable on E by definition.
23:50:190Paolo Guiotto: If and if the two real and imaginary parts are both measurable functions on E.
23:56:490Paolo Guiotto: Okay? Because at the end, that F will be a linear combination of measurable functions, doesn't matter if the coefficients are not real, but
24:05:530Paolo Guiotto: The point is that some linear combinations of measurable things is still measurable. So, we say that, by definition, F is measurable if this happens. And now, what do you expect be the operation of integral? So, we expect…
24:25:890Paolo Guiotto: that, the integral on X of a function f
24:31:360Paolo Guiotto: will be something like, since F is U plus IV and UAV are real-valued things, for which we already have the integral, so we expect that… and then the integral is also linear, so if something is a linear combination, the integral of the linear combination is the linear combination of the integrals. So without no surprise, we may expect that this will be the definition of integral.
24:56:960Paolo Guiotto: Sorry, let's put the domain in here as well.
25:06:140Paolo Guiotto: Of course, to do this, you must have that U and V are integral, So, provided…
25:18:400Paolo Guiotto: U and V are in L1E, so that these two integrals make sense, okay?
25:26:850Paolo Guiotto: Now, clearly, you would like to have a direct condition on F, rather than having… it's not so elegant to define the integrability of F in terms of the integrability of the real and the imaginary part. The end, for practical purposes, we have to pass through real and the imaginary path, but…
25:45:150Paolo Guiotto: What is the condition for F? Is that equivalent that, this is the question, modulus of F is integral on E?
25:56:780Paolo Guiotto: Because modules of F is a real valued function, huh? That's the absolute… the complex modulus of this. Let's see if this is the case, and in fact, we will see that this is the case.
26:08:980Paolo Guiotto: Because, models of F is… the algebra is the square root of u squared plus V squared.
26:19:680Paolo Guiotto: No?
26:20:960Paolo Guiotto: Now, I want to convince you that this is an if and all if. So, if u and v are integral, then modulus of F is integral, and vice versa, if modulus of F is integral, then both U and V must be integral.
26:36:240Paolo Guiotto: So… If U and V are in L1, E?
26:44:840Paolo Guiotto: Then, can I say that the integral on e of modulus of F d mu is finite? This means the function modulus of F is in L1, because the modulus of the modulus is the modulus itself, no?
27:02:980Paolo Guiotto: So I need to use the fact that integral on e of the modulus of U, D mu
27:11:860Paolo Guiotto: and integral on E of the modulus of V, D mu, these two guys are finite, because this means integral for U and V.
27:21:250Paolo Guiotto: So now, the point is, can I control this quantity through these two quantities?
27:29:320Paolo Guiotto: Let's see, models of F, we said it is the root of u squared plus V square.
27:38:480Paolo Guiotto: Now… Can I say something like,
27:43:730Paolo Guiotto: This quantity is controlled by a modulus of U plus modulus of V. Well, if the root of the sum were the sum of the roots, this would be the case, no? Because you see, root of u squared is this, root of V square is that.
27:59:810Paolo Guiotto: But of course, the root of the sum is not the sum of the roots, so I cannot say that.
28:05:60Paolo Guiotto: But perhaps, it seems to be reasonable now, because this is the root of a square, so it must be a first-degree star, no? So, the question is, I don't necessarily need to have this with
28:19:760Paolo Guiotto: An exact constant 1, but maybe there is a constant for which this becomes true.
28:25:340Paolo Guiotto: So, let's see what happens. Let's determine that constant in such a way that this is true.
28:31:540Paolo Guiotto: Now, this is a simple inequality, because it's positive less or equal than positive. We square everything, it is equivalent. This is equivalent saying u squared plus V square is less or equal than constant modulus U plus modulus V squared.
28:49:280Paolo Guiotto: Now, developing the square, we get U square plus V square plus the double product, and in fact, we see that
28:58:240Paolo Guiotto: As an inequality, it is correct with constant even equal to 1, no? Because if I put constant equal 1, I'm asking if u squared plus V square is less than U square plus V squared plus 2 times modulus U times modulus B.
29:14:900Paolo Guiotto: Which is, of course, true, no? So, this is true with constant even equal 1, okay? So, since modus of F, which is root of u squared plus V squared, is actually less or equal than modulus of U plus modulus of V, I see that
29:35:10Paolo Guiotto: integral on E of modulus of F, d mu, will be less or equal of integral on e of modulus U plus modulus V,
29:48:700Paolo Guiotto: Which is the sum of the two integrals.
29:55:230Paolo Guiotto: You're reminded that,
29:58:170Paolo Guiotto: Here, you should remind that we are exactly using those properties. I told you, if you want, you can prove, as a maybe not easy exercise, that are the fact that if you increase a function, you increase the integral when they are positive.
30:16:600Paolo Guiotto: and the linearity of integral with positive functions, okay? So that's what we are using here.
30:24:750Paolo Guiotto: So, if we know that the two, U and V, are in L1, these two quantities are finite.
30:33:670Paolo Guiotto: Because,
30:38:30Paolo Guiotto: UV are in L1. So now I know that if U and V are in L1, then modulus of F is in L1.
30:46:300Paolo Guiotto: Vice versa.
30:47:990Paolo Guiotto: Let's assume that it is modulus of F to be in L1.
30:52:810Paolo Guiotto: So this means that the integral on E of modules of F will be fined.
31:00:970Paolo Guiotto: And now we want to deduce that also U and V must be in L1.
31:07:70Paolo Guiotto: In this case, it is a bit easier, because if you look at the modulus of F, which is the norm, the root of u squared plus P squared.
31:17:480Paolo Guiotto: Now, for example, if you, in this argument, forget V,
31:22:600Paolo Guiotto: Since V is positive, the argument of the root decreases, so we have the root of u squared, which is exactly modulus of U.
31:32:140Paolo Guiotto: And the same way, we can discard U and say that this is greater or equal than root of V squared, which is modulus of V. So I see that modulus of U and modulus of V are both bounded by modules of F.
31:48:880Paolo Guiotto: And therefore, when we integrate, still we use the monotonicity property of integral for positive functions, we have that integral on E.
31:59:880Paolo Guiotto: of models of U in the mu.
32:02:780Paolo Guiotto: And also, the integral on e of modulus V in the mu.
32:07:210Paolo Guiotto: are both less or equal than the integral on e of modulus F, d mu.
32:14:450Paolo Guiotto: Okay? So, if the last one is fine, these two are fine, and so U and B are in L1.
32:23:150Paolo Guiotto: So if this is found, it says that U and V are in L1.
32:29:430Paolo Guiotto: Okay, so we are now ready for the definition.
32:34:450Paolo Guiotto: So, let XF mu the usual measure space.
32:43:180Paolo Guiotto: F, a measurable function on E, C-valued.
32:48:360Paolo Guiotto: So, from B to… C.
32:53:270Paolo Guiotto: Let's say F equals U plus IV.
32:58:540Paolo Guiotto: So, we say, that… F belongs to the class L1. Sometimes we write something like L1C,
33:12:40Paolo Guiotto: on E to specify that these are C-valued functions, but most of the time, we would not specify this C, okay?
33:21:790Paolo Guiotto: There will be times for which there is some difference between the fact that functions are real-valued and functions are C-valued, so it will be needed to be clear on that. But in general, you will never see this specification. It is clear from the context.
33:39:490Paolo Guiotto: We say that F is in L1, if…
33:43:50Paolo Guiotto: there exists, or if the integral on E of models of F is finite. So, as you can see at the end, it's the same definition of the L1 for real value.
33:55:190Paolo Guiotto: And in this case, we set…
33:59:500Paolo Guiotto: the integral on e of f…
34:02:470Paolo Guiotto: Since, as we know, that condition implies that both U and V are integral, so this definition makes sense. We define the integral as the integral of the real part plus I, the integral of the imaginary part.
34:18:489Paolo Guiotto: And that's it with all the definitions of integral for the All possible cases, okay?
34:30:50Paolo Guiotto: We know that we can also integrate functions with complex values, taking complex values.
34:36:810Paolo Guiotto: So let's say that the general properties are the usual properties of any integral. General.
34:44:800Paolo Guiotto: Properties.
34:48:639Paolo Guiotto: There is nothing particularly surprising, so we can say, number one is, called the linearity.
35:00:900Paolo Guiotto: It says that if you have two functions which are integral on E, then any linear combination alpha F plus beta G is integral on E.
35:15:50Paolo Guiotto: Here, for example, this holds, depending on the context, for every alpha and beta. So, if you are with the real value, the function.
35:26:80Paolo Guiotto: You cannot take some examples here, otherwise, in general, this will become a C-level in the function.
35:32:270Paolo Guiotto: But if you are on C-valid functions, the alpha and beta can be on complex numbers, okay? So let's say just for every alpha and beta that can be real, or…
35:46:890Paolo Guiotto: complex, in the case you are considering complex variant functions.
35:52:930Paolo Guiotto: And, of course, the integral.
35:57:30Paolo Guiotto: On E of the linear combination.
36:01:800Paolo Guiotto: Will be the linear combination of the integrals.
36:12:810Paolo Guiotto: Number two, this info… this is, called the monotonicity, or ordering.
36:22:970Paolo Guiotto: This, of course, is a property that we can, we can write only for real-valued stuff, because there is the ordering, and you know that you cannot order complex numbers. So, the point is that if you have F and G,
36:37:120Paolo Guiotto: R in L1E, let's specify that they are real-valued, okay, with F less or equal than G, on E.
36:52:590Paolo Guiotto: then the integral of F on E will be less or equal than the integral of G.
37:04:940Paolo Guiotto: Number 3, which is a derivation of this, but this property holds for C-valued functions, okay? And this is the triangular inequality.
37:22:300Paolo Guiotto: If you have a function f which is integral on E,
37:27:640Paolo Guiotto: Then, it says that the absolute value of the integral
37:32:940Paolo Guiotto: Is less or equal than the integral of the absolute value.
37:37:810Paolo Guiotto: The triangular inequality is that inequality that says the modulus of the sum is less than the sum of the moduluses, okay?
37:45:520Paolo Guiotto: If you think about that intuitively, the integral always has behind the idea of doing some of things, no?
37:53:780Paolo Guiotto: But that's why the modulus of the integral, it's like modulus of a sum.
37:59:560Paolo Guiotto: And this… this holds also if the function is c-valued, because it makes sense, you know?
38:05:120Paolo Guiotto: Dan, verada.
38:10:330Paolo Guiotto: Other evident formula, like the composition. So, if you have that,
38:18:240Paolo Guiotto: You have a function F, which is in L1 on some domain E, it is in L1 on some domain F, and the two are disjoint, E intersection F is empty, then the function is L1 on the union.
38:33:520Paolo Guiotto: And… the integral… On the Union.
38:42:800Paolo Guiotto: of F… Will be the sum of the integrals.
38:47:660Paolo Guiotto: Now, you can decompose, if you need, the integration on two domains. In other words, you usually,
38:57:950Paolo Guiotto: To this, you have to compute an integral and a domain. For certain reasons, it is better if you divide this in two parts, and you can split the integration in two sub-integrations.
39:07:470Paolo Guiotto: An important factor is the role of null sets. So what happens if we integrate on a null set? Well, it happens that if the measure of the domain is zero, then whatever is F, the integral.
39:26:590Paolo Guiotto: On the set will be equal to 0.
39:30:750Paolo Guiotto: So this is not particularly important, because you may imagine integrating or measure zero set won't be such a big deal. Of course, the value is zero, but you are not interested to something which are no relevance, apparently. But this means that
39:47:580Paolo Guiotto: In particular.
39:53:490Paolo Guiotto: This is more interesting. If you have two functions, F and G, which are equal, not necessarily everywhere, otherwise it's clear that the two integrals are the same, but this is true almost everywhere.
40:07:150Paolo Guiotto: So I remind you that this means that the measure of the set where F is different from G is 0.
40:16:200Paolo Guiotto: So it's not empty. See, if it is empty, it means that the two functions are the same thing, no? There is nothing where they are different.
40:23:630Paolo Guiotto: This says there might be something, but this something must have measure 0, must be negligible for the measure. Then, the two integrals coincide, the integral of F
40:36:770Paolo Guiotto: on domain E is the same of integral of G on domain E.
40:43:110Paolo Guiotto: So modifying, if you want, in another… to see this in another way, modifying a function on a measure zero set, first of all, it doesn't change the measurability. We already know this, no? You can modify on a measure zero set any function which is measurable, and you still get a measurable function.
41:01:750Paolo Guiotto: And also now, we know that if the function is also integral.
41:06:440Paolo Guiotto: We modify the function and measure zero set, we not only keep the measurability, but we have the same value for the integral.
41:14:150Paolo Guiotto: This is important because sometimes we have irregularity, bad points here and there, subsets where the function is not defined.
41:23:340Paolo Guiotto: If these things, if these bad points, these bad subsets are measure zero, we can just forget about them, because they have no weight in the integral.
41:37:520Paolo Guiotto: Okay, so these all we will not prove, because we should work out all details. They are long, boring, and bureaucratic, basically, so…
41:47:290Paolo Guiotto: Instead, we show a very important, I'd say, inequality, which is called the Shebyshev inequality.
41:57:330Paolo Guiotto: It's a very simple inequality, in fact, but it turns out to have an interesting role in many circumstances, so this is…
42:08:590Paolo Guiotto: proposition.
42:10:500Paolo Guiotto: Both in analysis and in probability. Chebyshev… In a world.
42:21:420Paolo Guiotto: Well, there are so many ways to write Shebyshev. I don't know if anyone here is from Russia.
42:28:310Paolo Guiotto: Okay, so there are at least five times… five ways to write this with Y, with I, with the C, like that, so… or with the T.
42:40:770Paolo Guiotto: This says that, suppose that we have the usual measure space, XF muon.
42:48:70Paolo Guiotto: measure.
42:49:790Paolo Guiotto: Space, huh?
42:51:360Paolo Guiotto: And the F, the measurable function, positive.
42:55:830Paolo Guiotto: I will… I give this version for positive, then there are version for variable sign, stuff. It says that you want to measure the set where F is larger than a certain value, alpha, that will be positive for this.
43:16:130Paolo Guiotto: negative doesn't make sense because alpha is positive, so it's like to take alpha equals 0. Alpha equals 0 cannot be taken because of what you see here…
43:26:150Paolo Guiotto: It says that this is less or equal than 1 over alpha, so let's say that the most,
43:33:800Paolo Guiotto: detailed version of the inequality says that it is less frequent than 1 over alpha, the integral on the part of a domain where f is larger than alpha of the function f, d mu.
43:47:200Paolo Guiotto: Normally, slightly more simple bound is used. That's because you see that here you are integrating on a subdomain of E, no? The set of points where F is larger than alpha.
44:06:330Paolo Guiotto: So if you replace this with all the domain, the integral, since you are integrating a positive thing, the integral increases. So we get the second version, which is this one. Let's equal 1 over alpha integral on e of f denu.
44:23:630Paolo Guiotto: It's a one-line proof, nothing special.
44:30:560Paolo Guiotto: But the interesting fact is that this is a bound that relates measures, no, of sets where f is bigger with integrals.
44:45:460Paolo Guiotto: So, sometimes we have a bound on the integral, and we want to assess and measure. That's the tool we need to use.
44:52:530Paolo Guiotto: There are variations of this, for example, I will show later. Let's see first the proof. The proof is just, I told you, one line. In fact, I take the measure of the set where F is larger than alpha.
45:07:330Paolo Guiotto: where we say that our measure works in this, we use this formula, that if you have any measure, any set S, measurable set, of course, this is equal to the integral
45:21:50Paolo Guiotto: if S, say, is a subset of E, it is the integral on E of the indicator of the set S. Now, this is because you are
45:32:960Paolo Guiotto: There is the idea that this function is 1 on the set S0 L squared, no? S is containing DBA, so this function is… the picture is like this, no? This is domain E, where you are integrating, and there is a subset S, where your function is 1.
45:55:350Paolo Guiotto: Now, the integral of this particular simple function we said before, into our constant indicator, that indicates norms, the area of this plane
46:07:370Paolo Guiotto: say, region, which is 1v8 times the measure of the base, mu or less. So, the formula is this one. This is the formula that we use,
46:19:710Paolo Guiotto: In practical applications, if we are able to integrate this computer-based compute measures, you transform measures into an integral, okay?
46:30:740Paolo Guiotto: Now, so this says that you can say this is the integral on E of the indicator.
46:36:520Paolo Guiotto: of the set, which is the set where F is larger or equal than alpha d mu.
46:43:280Paolo Guiotto: Now, the trick is the following, that on this set.
46:48:00Paolo Guiotto: So when this thing is won.
46:51:380Paolo Guiotto: F is larger or equal than alpha, and you may notice that F greater or equal than alpha, since alpha is positive, is equivalent. Divide both sides by alpha, and you get F divided alpha greater or equal than 1.
47:09:570Paolo Guiotto: So the trick is that you see this one, an imaginary one in front of the indicator, and you say, that one is less than ratio F over alpha.
47:23:80Paolo Guiotto: Since you are increasing the function, if I put here F divided alpha, I increase the function. Because of the monotonicity of the integral, I increase the function, I increase the integral. So this will give less or equal integral.
47:40:610Paolo Guiotto: on E of F divided alpha, indicator of the set where F is large or equal alpha in the mu. Now, it's finished because you carry out 1 over alpha. That's a constant you can put in front of the integral.
47:58:960Paolo Guiotto: Integral, you do.
48:01:540Paolo Guiotto: you are integrating on the subset F, this is a subset of the domain, because these are the points of the domain where f is larger than alpha, so you can integrate only on that subset, because outside of that set, the function is 0.
48:18:740Paolo Guiotto: So there is no contribution, and you have the integral of F in the mu. So this is just the first version of the inequality, the first line, no?
48:28:940Paolo Guiotto: So this says mu of F larger or equal than alpha is less or equal 1 over alpha integral on the set where f is larger than alpha over FB mu.
48:41:500Paolo Guiotto: Now, since this set is contained in 2E, if we integrate everything on the bigger domain, since F is supposed to be positive, the integral will increase, and therefore this is less or equal than 1 over alpha integral on e of f d mu. And that's the second inequality.
49:00:980Paolo Guiotto: There are some variations.
49:04:660Paolo Guiotto: Remark.
49:10:590Paolo Guiotto: variants.
49:15:30Paolo Guiotto: So, for example, you could do the same trick.
49:19:160Paolo Guiotto: But instead of saying, you want to assess the ma- how… the idea is that,
49:26:830Paolo Guiotto: If you look at this inequality, or whether we will use it several times soon, if you get this second form factor, this is the constant, it is the integral of F. Suppose that this value is 5, so the function is integral.
49:42:680Paolo Guiotto: So that's a constant divided by alpha. So it means better when you set alpha to infinity.
49:50:240Paolo Guiotto: As you may expect, it's not immediately validated, but it makes sense. The measure of the square F is that when alpha goes to zero.
50:01:420Paolo Guiotto: To why this should be true.
50:04:600Paolo Guiotto: So let's do… before we write the variance, so we say that mu of alpha is less or equal than 1 over alpha integral on e of f emu.
50:16:590Paolo Guiotto: So… If… F is integral on E,
50:25:150Paolo Guiotto: This means integral on e of f d mu is finite.
50:33:510Paolo Guiotto: We get that, mu.
50:36:650Paolo Guiotto: of F greater or equal than alpha.
50:40:340Paolo Guiotto: is less or equal than a constant divided alpha, so it goes to zero, and it goes to zero, it does not tell only that it goes to zero, because it goes to zero.
50:50:510Paolo Guiotto: Could be saved by some other argument.
50:53:980Paolo Guiotto: What's happening?
50:56:340Paolo Guiotto: I don't need the detection leak wanting to say that that thing goes to zero.
51:05:420Paolo Guiotto: At least, I'm saying, more so there is about Putin, you still need to do church in the world.
51:13:670Paolo Guiotto: Because what happens when alpha increases here?
51:18:830Paolo Guiotto: Look at these sets, the sets where F is greater than alpha. If I am greater than this, and now I increase, I pretend to be greater than this, the set of X that, for which F of X is greater than this is bigger or smaller of those.
51:37:150Paolo Guiotto: It's smaller, because you pretend to be bigger.
51:40:190Paolo Guiotto: So, when you increase alpha, this set is decreasing.
51:46:10Paolo Guiotto: No? You see?
51:48:400Paolo Guiotto: If you increase alpha, this goes down. So they are a decreasing sequence of sets.
51:56:440Paolo Guiotto: That's remind of the continuity from above.
52:00:10Paolo Guiotto: So, if you want to treat the problem of computing the limit when alpha goes to infinity, you may try to apply that result. But that result asks for something. You must have that the initial set must be fine.
52:15:960Paolo Guiotto: And that's why you need a bound that tells you that at least one of them has finite measure. This happens if the function f is integral, because the function integral is this one, which is fine, this is a constant, so this tells that these numbers are all fine.
52:30:870Paolo Guiotto: So you are authorized to apply the continuity from above, and you can prove that taking the limit, this quantity goes to zero.
52:42:560Paolo Guiotto: But now, they say something slightly more.
52:48:850Paolo Guiotto: That is more powerful than this?
52:55:510Paolo Guiotto: Let me do that.
52:59:240Paolo Guiotto: Sorry.
53:00:570Paolo Guiotto: The screen savable.
53:03:140Paolo Guiotto: So, this is saying that it goes to zero, but a little bit faster than 1 hour alpha, so it gives you the information about the speed.
53:12:930Paolo Guiotto: Of going to zero. Now, you see?
53:18:190Paolo Guiotto: We can have another type of bounds, so you can say other… Bouncer.
53:30:760Paolo Guiotto: If you take an exponent P, which is greater or equal than 1, so contains also the case P equal 1,
53:38:140Paolo Guiotto: You can say that the mu of F greater or equal than alpha, you do the same argument. Now, this is the integral of the indicator where f is larger or equal than alpha on the set E.
53:53:200Paolo Guiotto: Okay, now put a 1 here, and notice that when this thing is 1, so when F is larger than alpha, you can write 1 less or equal than F divided by alpha.
54:06:350Paolo Guiotto: Now, raise everything to power P. If P is larger than 1, the order of power is respected.
54:16:550Paolo Guiotto: If they are larger than 1, okay? Otherwise, it is invertible.
54:22:150Paolo Guiotto: So, I can say that 1 to P is less or equal than F to alpha to P, so F to P divided by alpha to P.
54:31:750Paolo Guiotto: And this is, 1 to P, of course, is 1, is that if I put here this ratio, F to power P divided by alpha to power P,
54:42:480Paolo Guiotto: I still increase my function, because 1 is less than that. So I have less or equal than integral on e of f to power P divided alpha to power P,
54:55:570Paolo Guiotto: indicator, F greater than or equal alpha, the immune.
55:00:670Paolo Guiotto: Now, carry outside the constant.
55:03:800Paolo Guiotto: 1 over alpha to PE,
55:06:90Paolo Guiotto: You can throw away this one integrate on the full domain, integral on e of f to power P d mu.
55:14:880Paolo Guiotto: So you get this bound… sorry, this is a P. The measure of set where n ph greater or equal than alpha is controlled by this quantity.
55:27:350Paolo Guiotto: Now, if this quantity for some P is 5,
55:31:630Paolo Guiotto: This bound says that this measurement is fine, but also, it decays to 0 adds 1 to answer to P. So, for example, P equals 2.
55:42:610Paolo Guiotto: So… P equal 2.
55:48:680Paolo Guiotto: If integral on e of f squared d mu is finite.
55:55:300Paolo Guiotto: So not only the integral of F, but the integral of F squared, then the measure where f is larger or equal than alpha decays
56:06:520Paolo Guiotto: Because here you see the exponent is 2, decays as 1 over alpha squared.
56:13:560Paolo Guiotto: times this constant, integral of F squared d mu.
56:17:640Paolo Guiotto: So, you see, if I assume that this is established by bound, if F is in L1,
56:26:650Paolo Guiotto: So the integral of F in spine inter this measure decays as at least faster than one over at alpha.
56:37:120Paolo Guiotto: Which post was in?
56:39:80Paolo Guiotto: Now, if I require a little bit more, like this integral of F squared, find it. Now, this decays as 1 over alpha squared, so in the case be faster.
56:53:190Paolo Guiotto: Okay? So this is interesting, we will see, because we have to assess this, we wanted this decay to zero. It's fast enough, we maybe have to require some…
57:05:950Paolo Guiotto: integral of this type to be found, and use this one to get the correct duplicate speed we need, okay? However, we wish you all this.
57:16:590Paolo Guiotto: Okay, that's all for today.
57:20:210Paolo Guiotto: Please… So today, I will publish a solution of part of the exercises.
57:28:750Paolo Guiotto: Of, previous classes.
57:33:80Paolo Guiotto: Do… Exercises, the… 4… 3… 2…
57:46:600Paolo Guiotto: No, I think it's true.
57:53:670Paolo Guiotto: No, 434… 435… For 36, it's what we have done now.
58:08:90Paolo Guiotto: 437…
58:15:140Paolo Guiotto: Well, the remaining, maybe we have… we see tomorrow some application of which we should build quality. Okay.
58:22:240Paolo Guiotto: That's it.