AI Assistant
Transcript
00:07:950Paolo Guiotto: I want to start going back a bit on the definition of integral, because there is something behind interesting to know, which is the following. So let…
00:22:910Paolo Guiotto: X… Sino… B… Measure… space.
00:33:500Paolo Guiotto: F… A measurable function defined on X or on set some subset E.
00:42:430Paolo Guiotto: of X, huh?
00:45:120Paolo Guiotto: positive.
00:46:710Paolo Guiotto: F greater than or equal than 0 on…
00:52:400Paolo Guiotto: Now, that construction suggests a way to do an approximation, a point-wise approximation, for a generic measurable, positive measurable function F,
01:05:420Paolo Guiotto: Through simple functions. So, functions having just a finite number of values, taking just a finite number of values.
01:14:510Paolo Guiotto: So the idea is the following. Let's say that the original axis stands for the SpaceX, and this is the real axis.
01:23:760Paolo Guiotto: We have our domain, which is a certain subset E of the space X, and our function that, for simplicity, we draw as if it is a continuous line, but maybe continuity is not even defined in this context.
01:40:670Paolo Guiotto: So now, the key idea behind the bank construction
01:45:660Paolo Guiotto: Is, to do a dual approach, to… to construct the… an approximation of the area.
01:54:490Paolo Guiotto: Delimited above by the graph of F, and below by the domain in.
02:02:250Paolo Guiotto: Now, the idea is that we divide the codomain in parts, in a finite number of parts, so fix NNN, the idea is that each of these parts will be, will have a size 1 over 2 to the n.
02:18:360Paolo Guiotto: So, these are points K over 2 to the N,
02:23:640Paolo Guiotto: K plus 1 over 2 to the N.
02:29:100Paolo Guiotto: And when k ranges from 0, we start from 0.0.
02:34:80Paolo Guiotto: When k is equal to the n, we get value 1.
02:37:990Paolo Guiotto: So since we can have a function that takes big values, possibly unbounded, we want to arrive up to the value that here we choose, 2 to the n.
02:49:930Paolo Guiotto: To arrive to this value, it means that K must range from 0, you are at quarter zero, k equals 1, 2, etc, and to arrive at that point, you need that K be equal 2 to the n squared, so 2 to 2N.
03:11:550Paolo Guiotto: Now, let's focus on this interval from value K over 2 to the n to value K plus 1 over 2 to the n.
03:23:390Paolo Guiotto: We divide the… Codomain in this way.
03:28:70Paolo Guiotto: And now, we build an approximation, so a function, that approximates from below our function f, which is the black line. The idea is to put on the set where the function f
03:42:550Paolo Guiotto: is between the two values. So, in this case, you see there are two pieces of this set.
03:50:00Paolo Guiotto: One is ER.
03:52:360Paolo Guiotto: The other one is here. These two are,
03:57:700Paolo Guiotto: make the set where F is above value K over 2 to the n, and let's say strictly below of k plus 1 over 2 to the n.
04:12:260Paolo Guiotto: Okay, so this is a subset of E, actually, that we call E, K, N. It depends on these two indexes.
04:22:510Paolo Guiotto: Well, on this, portion of E,
04:27:300Paolo Guiotto: We define the approximation as the minimal value, which is this one, k over 2 to the n. So we define a function like this, so define a function that will be called SN,
04:44:700Paolo Guiotto: Sn at point X is defined in this way.
04:50:650Paolo Guiotto: Well, it will be equal to value K over 2 to the n.
04:55:710Paolo Guiotto: when X belongs to the set of points.
05:00:310Paolo Guiotto: for which the value of F is between k over 2 to the n and k over the n, so on these two together basis. So we pick an X here, and the function is the green, the green value there. So when X belongs to EKN,
05:20:280Paolo Guiotto: Now, the function can get bigger than the maximum value to the nullode here, so let's, let me just,
05:28:870Paolo Guiotto: I guess that if I erase this.
05:32:520Paolo Guiotto: Let's do something like this, no? When the function goes above value 2 to the n.
05:39:610Paolo Guiotto: So let's say that we take this, level.
05:43:800Paolo Guiotto: the level Y equal to 2DN, there will be a unique set where F is above that, so in this case, the set
05:55:280Paolo Guiotto: is down here.
05:57:440Paolo Guiotto: So this in, pink, is the set of points where…
06:02:940Paolo Guiotto: it has a different definition, slightly different from the former point of view. This guy here is the set of points where F is above 2 to the N.
06:19:50Paolo Guiotto: Also, this one is a subset of the domain, so it is contained in E, and I don't know what we…
06:29:590Paolo Guiotto: Well, let's call EN with just one index, N, okay? So we have two sets, two types of sets, sets EN, where F is above.
06:40:380Paolo Guiotto: 2 to the n, and the sets EKN, where f is between K over 2 to the n and k plus 1 over 2 to the n. Our simple function will be defined adjusted the value 2 to the n of that K. So, it is equal to k over 2 to the n when x is
06:59:240Paolo Guiotto: in the set EKN, and when X is in the set EN, it will be defined as 2 to the N.
07:06:780Paolo Guiotto: As you can see, since there are a finite number, and here is fixed, there is a finite number of values for K,
07:14:480Paolo Guiotto: This is a function that takes only a finite number of values, so it's a simple function. We can also write in the following way. We can write it is a sum for k going from value 0 to the last possible k, which is 2 to 2N minus 1.
07:33:510Paolo Guiotto: We have to subtract 1, because the K plus 1 would be above 2 to the n.
07:38:510Paolo Guiotto: He's young.
07:40:970Paolo Guiotto: That, that number here.
07:43:950Paolo Guiotto: The value is k over 2 to the n,
07:47:620Paolo Guiotto: indicator of EKN.
07:51:780Paolo Guiotto: plus value 2 to the n when we are in the set EN. This is the simple function SN of X that we define in this way.
08:05:540Paolo Guiotto: Okay. Now, what is interesting is the following proposition that we can prove.
08:14:720Paolo Guiotto: So, this function, so if, let, let's write the usual… Staff, access, new… B. Measure.
08:26:970Paolo Guiotto: Bates.
08:29:80Paolo Guiotto: F be a measurable function on E, positive, Way to equal than zero.
08:36:150Paolo Guiotto: done.
08:38:780Paolo Guiotto: We have that the sequence of functions as n, defined in this way, the sequence… the sequence.
08:50:170Paolo Guiotto: a function, SN.
08:53:60Paolo Guiotto: is made.
08:56:670Paolo Guiotto: off.
08:58:780Paolo Guiotto: simple, And measurable functions.
09:10:20Paolo Guiotto: So they are simple because they take only a finite number of values. They are measurable because, you remind the simple function is measurable
09:18:250Paolo Guiotto: if and only if these sets, the E, Kn, and DN, are measurable sets, which is the case here, because each of them is a set where F is in a sum interval, no? You see, this EKN is the set where F belongs to the interval from K over 2 to DN,
09:40:880Paolo Guiotto: And, K plus 1 over 2 period with the,
09:47:170Paolo Guiotto: right endpoint excluded. This is the set where F belongs to the half line 2 to the end, 2 plus infinity. So these are measurable sets, because F is measurable.
09:59:900Paolo Guiotto: So this is a simple, and these are simple and measurable functions, such that… and we have these two remarkable properties. Number one, well, first of all, they are positive.
10:13:180Paolo Guiotto: Yeah, because the values are positive, no? Either they are these numbers, k over 2 to the n or 2 to the n, so everything is positive, so this is not a particular…
10:27:290Paolo Guiotto: Relevant. But, so when you switch from n to n plus 1, what happens is that this function somehow moves up, no? So it's a sequence of functions that are moving upward, and this happens for every X in E.
10:46:10Paolo Guiotto: And number two, what they do when n is sent to infinity is that this sequence converges pointwise, so X by X, to the initial function f. So, the limit for n going to plus infinity
11:03:430Paolo Guiotto: of SNX is equal F of X for every X in E.
11:11:390Paolo Guiotto: So we have a standard way to approximate it.
11:16:150Paolo Guiotto: a positive measurable function through a sequence of simple functions. Simple functions means… basically, all functions you can plot on a computer are of this type, because you don't plot a continuum of values, but just a discrete set of values.
11:33:900Paolo Guiotto: So these are the type of functions we usually use in numerical computation.
11:43:780Paolo Guiotto: And this is saying that we can always approximate point-wise, so it's a very, very weak way to
11:51:590Paolo Guiotto: do a limit in a sequence of functions, any positive measure function to this sequence of simple measurable functions. So let's, let's see the roof. It is basically
12:06:400Paolo Guiotto: An, an exerciser, okay?
12:10:550Paolo Guiotto: It's a little bit, the argument we have similar integral.
12:16:510Paolo Guiotto: Now,
12:30:820Paolo Guiotto: We have…
12:36:800Paolo Guiotto: with this property, that… Nope.
12:45:420Paolo Guiotto: Oh.
13:15:920Paolo Guiotto: Now, let's see what happens when we switch from SN to SN plus 1 of X, that…
13:25:720Paolo Guiotto: when I… I… we have seen last time. DSN plus 1, S1, which is exactly the double of the scale on the y-axis, let's say plus 1.
13:38:110Paolo Guiotto: plus 1, you see the value of S, R to this value, which is 2 to the n plus 1,
13:44:110Paolo Guiotto: This X, it means that…
13:47:80Paolo Guiotto: X belongs to the set where F is larger or equal to 2 to the n plus 1, and therefore, it would be also larger than 2 to the n. So it means that our X will be also in the set where F is above 2 to the n.
14:05:980Paolo Guiotto: And therefore, this means that the value of the function SN at point x, since for that X we are in the set where F is larger than 2 to the n, this is what we call the set EN. The value of Sn will be 2 to the n, which is less or equal than 2 to the n plus 1, which is the value of SN plus 1.
14:30:400Paolo Guiotto: All that. Okay.
14:33:990Paolo Guiotto: So, if the value of Sn plus 1 is this one, we know that this value will be greater than the value of Sn on the same point X.
14:45:990Paolo Guiotto: Now, if the value is less than 2 to the n plus 1, it will be one of these fractions, but with n replaced by n plus 1. So, let's see what happens in that case. If Sn plus 1
15:01:760Paolo Guiotto: Well, Vex.
15:03:780Paolo Guiotto: Let's say, in a… it is,
15:08:220Paolo Guiotto: If you do a figure, you can… you can do also a sort of graphical proof, because let's say that this is the function F. Let's see what happens between Sn and Sn plus 1.
15:22:350Paolo Guiotto: SN plus 1 is, we should divide, for, the,
15:28:750Paolo Guiotto: For the level n plus 1, for n equal n plus 1, we divide the y-axis in intervals of length 1 over 2 to the n plus 1.
15:41:340Paolo Guiotto: So they are exactly one half of the length of the previous ones. So let's say that if we are, let's say, here, between K over 2 to the n and k plus 1 over 2 to DN, this will be divided into sub parts, no? So you know that, let's use colors in this way. DSN
16:04:950Paolo Guiotto: does this. When you divide this strip on this set in the domain that you see down here, let the color in yellow, this is the set of points X square F,
16:18:810Paolo Guiotto: the value of F is between these two, k over 2 to the n and k plus 1. The value, the value of Sn will be this one, we say, then, huh? This is the value of SNX for X in that set.
16:36:730Paolo Guiotto: What happened for SN plus 1?
16:39:310Paolo Guiotto: Now, with different indexes, they are not exactly K and K plus 1. If we want, we can see what they are.
16:48:990Paolo Guiotto: This will be A, let's say, H over 2 to n plus 1, Then we will have here.
16:57:830Paolo Guiotto: H plus 1 over 2 to n plus 1, and this guy will be H plus 2 over 2 to the n plus 1.
17:07:829Paolo Guiotto: So we divide this strip into two substrips.
17:15:450Paolo Guiotto: And see, when F is between the value of F is between the smallest, this one and the intermediate, this one, it means we are in a substrip here, which is this part here, let's call her in…
17:34:830Paolo Guiotto: pink here.
17:36:410Paolo Guiotto: And the value of the SN plus 1 in that case, is the smallest value, so the SN plus 1 here will be the same, you see?
17:46:340Paolo Guiotto: But when I am in this other part, for the X for which F is between the intermediate and the upper value, here.
17:58:800Paolo Guiotto: So when I am… F is included between these two values, I am in this second part of the yellow set. On these points.
18:10:440Paolo Guiotto: on this point, the SN plus 1 will take value here.
18:16:700Paolo Guiotto: So, you see, it is above, it is always above, okay?
18:20:620Paolo Guiotto: So even if we don't want to do an analytic, discussion, that can be done,
18:30:150Paolo Guiotto: We did… and I went to, like, I said.
18:38:90Paolo Guiotto: In that time, we can see graphically that the function SN plus 1 is…
18:52:480Paolo Guiotto: Now, about the delimiter, The point is that, let's be connects, So, these, rules,
19:07:310Paolo Guiotto: that SN is less or equal than SN plus 1 for the X.
19:13:490Paolo Guiotto: Now, about the limiter…
19:15:880Paolo Guiotto: Of course, the second property is to prove that Sn of x goes to F of X.
19:24:150Paolo Guiotto: for every accident.
19:26:980Paolo Guiotto: that belongs to field.
19:30:480Paolo Guiotto: Now, what I know, I want to prove that this is true, this is equivalent to showing that the difference goes to zero.
19:38:610Paolo Guiotto: Now, since SN is below F by construction.
19:44:70Paolo Guiotto: Okay, what I assess is the difference f of x minus Sn , which is positive, below, because Sn is below F, and let's show that becomes smaller.
19:57:330Paolo Guiotto: Since we have to do this X by X, let's fix an X, okay? Take an X in E,
20:06:110Paolo Guiotto: Now, compute the value f of x, huh?
20:10:120Paolo Guiotto: That here, for us, is supposed to be a positive number, so a number in 0 plus infinity.
20:17:110Paolo Guiotto: It's not, infamous.
20:19:890Paolo Guiotto: So what happens is that
20:22:790Paolo Guiotto: For n big enough, the value 2 to the n will be greater than your F of X. So, let's try to do a figure like that. So this is the function.
20:33:290Paolo Guiotto: I'm picking a point X.
20:35:970Paolo Guiotto: anywhere in the domain. The value of the function is this value, f of x.
20:42:940Paolo Guiotto: I have this sequence of functions as standard.
20:46:840Paolo Guiotto: with the range which is limited above by the value 2 to the n.
20:53:50Paolo Guiotto: So, since 2 to the n goes to infinity, I can assume that for X fixed, my 2 to the n be big enough to be larger than value f of x. So, I can say this.
21:07:250Paolo Guiotto: let.
21:09:430Paolo Guiotto: N… big.
21:12:760Paolo Guiotto: Enough.
21:16:190Paolo Guiotto: such that 2 to the n be greater than F of X. So, as you can see, this means that
21:24:430Paolo Guiotto: the value of Sn is not 2 to the n, because the value of Sn is 2 to the n when x belongs to the set y n, which is the set of points where F is larger than 2 to the n.
21:37:560Paolo Guiotto: Okay, so the value of F at point SN will be what?
21:45:760Paolo Guiotto: So, now, let's say that this is the value 2 to the n. It is above.
21:52:80Paolo Guiotto: So, since I know that, I divide the range from 0 to 2 to the n in small intervals.
22:01:100Paolo Guiotto: of type k over 2 to the n, k plus 1, and since this is a partition of the range from 0 to 2 to the n, I can say that this f of x will belong to just one of these intervals.
22:17:380Paolo Guiotto: So there exist exactly for that and fixed, there exists exactly one single k for which my f of x is above or equal k over 2 to the n and below k plus 1, because they are a partition.
22:32:10Paolo Guiotto: So, there exists.
22:34:720Paolo Guiotto: K, such that F of X
22:38:600Paolo Guiotto: is greater or equal than K over 2 to the n.
22:43:00Paolo Guiotto: and less than K plus 1 over 2 to the n.
22:48:100Paolo Guiotto: But this means, what? That point X belongs to the set EKN.
22:54:500Paolo Guiotto: Therefore, the value of the function SN at point x
23:01:250Paolo Guiotto: As we defined the SN, when you are in the set EKN, the value is k over 2 to the n. It's written there.
23:10:380Paolo Guiotto: So, Sn of x is k over 2 to the n.
23:14:310Paolo Guiotto: And therefore, returning back to the difference F minus Sn, so F of X minus SN of X is positive. This is because Sn is below F.
23:31:380Paolo Guiotto: That also is equal to F of X minus k over 2 to the n.
23:38:100Paolo Guiotto: And what is this difference? Well.
23:40:920Paolo Guiotto: I have that my f of x belongs to this red interval, which is the interval from k over 2 to the n, and this one is k plus 1 over 2 to the n. I don't know where exactly is this f of x, but what I know for sure, that it will be in this interval. So the distance, the maximum distance between f of x and k over 2 to the n is at most the length of the interval.
24:05:390Paolo Guiotto: And that length is 1 over 2 to the n. So what I can say here is that this will be less or equal than 1 over 2 to the n.
24:14:850Paolo Guiotto: So what we proved is that, so…
24:18:770Paolo Guiotto: If X belongs to E, is… Fixed, huh?
24:28:870Paolo Guiotto: For every n larger than a certain initial N, such that 2 to the n
24:36:790Paolo Guiotto: is larger, than F of X, huh?
24:42:370Paolo Guiotto: So this happens, for every n larger than some initial n, because 2 to the n is going to infinity, f of x is a fixed number, okay? You see, the sequence 2 to the n goes to plus infinity, so sooner or later will be bigger than f of x. And once this is achieved, for every n larger than that initial N, it will stay bigger, because these powers are growing.
25:05:990Paolo Guiotto: Okay? They are doubling each time.
25:08:790Paolo Guiotto: And we have, in this case, that F of X minus Sn of x, which is always bounded by 0 below, above is bounded by 1 over 2 to the n, and this happens for every n greater than some initial capital N.
25:25:720Paolo Guiotto: Now you can send little n to plus infinity. So if you send little n to plus infinity, this guy, 1 over 2 to the n goes to 0, this is already 0, and by the squeeze theorem, this, which is in the middle, must go to 0.
25:41:960Paolo Guiotto: And there you get that… So F of X minus SN goes to 0 when
25:50:730Paolo Guiotto: And goes to plus infinity.
25:59:620Paolo Guiotto: And this means that exactly SN of X goes to F of X, for every… X in E.
26:12:140Paolo Guiotto: So we have the conclusion that this sequence approximates the function f in a pointwise wave.
26:20:270Paolo Guiotto: So now, we know that. This was a remarkable property that we have to keep always in mind for measurable functions. Whenever you do a pointwise limit of measurable functions.
26:34:170Paolo Guiotto: And this limit exists, so a pointwise convergent sequence of measurable functions, the limit is measurable. This was a property we just limit to the statement
26:47:680Paolo Guiotto: Perhaps a couple of… here we are, no? If you have a sequence of measurable functions, such that for every X, there exists the point-wise limit of the sequence, then the function, the limit function, is measurable.
27:01:430Paolo Guiotto: Now, we have a sort of vice versa, which says, if we have, at least for a positive measurable function.
27:09:650Paolo Guiotto: If we have a positive measurable function, we can always approximate this by a sequence of simple measurable functions that converges to F pointwise for every point of the domain.
27:27:660Paolo Guiotto: So, at the end, the idea of the integral is that, so…
27:33:760Paolo Guiotto: If SN of X is this thing, sum for k going from 0 to 2 to 2N minus 1, K of a 2 to the n, indicator of EKN,
27:49:800Paolo Guiotto: plus a 2KBN indicator of the set EN.
27:56:30Paolo Guiotto: Now, it is a natural idea to define the integral of this thing, and this is to the definition we have, you know, seen last time, no?
28:06:310Paolo Guiotto: If you have a function like this, you have on your space X a subset E. Of course, here in this figure, you see intervals, but X is not necessarily R, RD in the Euclidean space can be whatever.
28:24:610Paolo Guiotto: And you have a constant function, a function constant equal to C, Well,
28:31:550Paolo Guiotto: What could be a reasonable definition of the, let's say, the area
28:37:670Paolo Guiotto: Of course, it's not an area, because it depends on what is X. But let's say an area of this, no? What should be the reasonable definition of integral on X for a function which is a constant equal to 1 on the set E with respect to the measure mu?
28:58:400Paolo Guiotto: Well, you would expect that this will follow the rule based on hate.
29:03:00Paolo Guiotto: Now, the 8 is C, so I would say it is C times the measure of the set E, so mu of E.
29:12:830Paolo Guiotto: So this is the natural position we should set, okay? So… We expect…
29:26:600Paolo Guiotto: D.
29:28:300Paolo Guiotto: Integr Thus.
29:32:120Paolo Guiotto: These.
29:34:790Paolo Guiotto: So this explains why, so if we define IN
29:40:360Paolo Guiotto: the integral of Sn with respect to measure mu on X as
29:46:70Paolo Guiotto: well, then we expect other things. The integral of a sum be the sum of the integrals, right? Because this is what happens normally. So, if we do the sum here, you see that we are summing
29:57:530Paolo Guiotto: K from 0 to 2 to 2N minus 1. What? Constant times indicator. C times 1E. And the integral of this thing should be the constant, so k over 2 to DN times now the measure of the set EKN,
30:17:780Paolo Guiotto: plus another constant times indicator. So we expect that this will have integral, the constant, 2 to the n times the measure of the set.
30:28:190Paolo Guiotto: And this leads to the definition we have given last time, no? That, in fact, should be the integral of the function SN. Last time, we proved that… we proved, we tried to prove this theorem that… so, let…
30:46:460Paolo Guiotto: X F, mu.
30:49:940Paolo Guiotto: be a measure space.
30:54:920Paolo Guiotto: Then, F be a measurable function on E and positive.
31:03:210Paolo Guiotto: Then, we proved that there exists the limit… actually, the sequence CN is an increasing sequence of numbers.
31:13:160Paolo Guiotto: And this is reasonable, because if you increase n, the idea is that you raise this function here, so this becomes bigger, and the area below will become bigger. So that's… of course, this is not really evident from that definition.
31:30:440Paolo Guiotto: biased, and it's a proof that intuition should be working. So, there existed a limit
31:41:270Paolo Guiotto: When n goes to plus infinity of this IN, and we call this, by definition, the integral on E of, or integral on X,
31:51:640Paolo Guiotto: Actually, it's on E, sorry.
31:54:440Paolo Guiotto: This domain was E.
31:57:100Paolo Guiotto: The integral on e of the function
32:25:920Paolo Guiotto: Now, in this way, we are…
32:37:810Paolo Guiotto: She…
32:43:350Paolo Guiotto: Definitely.
32:53:470Paolo Guiotto: too many…
33:03:570Paolo Guiotto: Okay, so…
33:33:870Paolo Guiotto: So… It can be proved that for example.
34:06:720Paolo Guiotto: Property.
34:19:20Paolo Guiotto: It can be improved by using this liquidation.
34:29:980Paolo Guiotto: It's not the boss.
34:41:820Paolo Guiotto: So…