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Transcript
00:07:420Paolo Guiotto: Okay.
00:08:720Paolo Guiotto: Good morning.
00:13:890Paolo Guiotto: Last time, we,
00:19:290Paolo Guiotto: We discussed this important limit theorem, which is called the monotone convergence theorem.
00:28:510Paolo Guiotto: It says that we have a sequence of measurable functions on some measure space X,
00:34:840Paolo Guiotto: positive, and increasing with the index. This does not mean that they are increasing function, but it means that when we increase the index, certainly increase the function.
00:46:750Paolo Guiotto: Then, it holds that we can switch the limit with the integral, so the limit… we can pass your limit operation inside the integral, and we get this. The limit of the integral is the integral of the limit.
01:01:420Paolo Guiotto: Now, before we leave this, setup, a natural question would be, what if the sequence is not increasing, but decreasing within, no? So, if instead of rising up a day, goes down to something.
01:21:50Paolo Guiotto: Now, as you may expect, since this comes basically from the continuity from below, the other probability will be something like the continuity from above, and since that one is not always verified, we should expect that
01:35:490Paolo Guiotto: the monotone convergence with a decreasing sequence of function not always all. So, let's start with this remark.
01:50:170Paolo Guiotto: What?
01:53:150Paolo Guiotto: about… monotone… convergence, we've… decreasing… Sequences.
02:09:710Paolo Guiotto: So let's say that we have a sequence of functions FN, which is a sequence of measurable functions on some X, where X fu…
02:21:30Paolo Guiotto: Is a usual measure.
02:24:420Paolo Guiotto: space.
02:26:140Paolo Guiotto: Such that, now, we ask that FN plus 1 sorry, FN of X.
02:35:890Paolo Guiotto: is greater or equal than FN plus 1, and all of them are positive.
02:41:710Paolo Guiotto: almost every X in capital letters.
02:47:680Paolo Guiotto: So, the question is, can we still… can we say… Say… that… the, limit…
02:59:70Paolo Guiotto: of the integrals of the FN,
03:02:630Paolo Guiotto: The mu is still the integral of the limit.
03:07:50Paolo Guiotto: of the F and the muons.
03:12:790Paolo Guiotto: Also here, we may notice that since the sequence verified
03:17:00Paolo Guiotto: verifies this condition, the integral of N will be integrated than the integer of Fn plus 1. So the sequence of this number is an increasing sequence of numbers, so the limit will exist.
03:32:70Paolo Guiotto: And above the right-hand side, this is a pointwise need of measurable functions, so it is measurable and positive, because they are positive, so the interval makes sense. So the question is, are they the same? And of course, the answer is no.
03:51:00Paolo Guiotto: So… this.
03:55:30Paolo Guiotto: is… False.
04:01:650Paolo Guiotto: And actually, the example is more or less the same example we have seen for the fact that the measure is not continuous from above. I… instead of taking that example, I do another one, since now we have the Lebec measure. So, example.
04:24:850Paolo Guiotto: So we take as X, F,
04:28:600Paolo Guiotto: mu the space r with the… the big measure.
04:36:850Paolo Guiotto: We take as functions Fn, these functions… Yeah.
04:43:260Paolo Guiotto: of the af… And two, like…
04:51:620Paolo Guiotto: Well.
04:52:970Paolo Guiotto: familiar streets of naturalists, in that case, we had the counting male, and we took naturals greater than sum N.
05:01:350Paolo Guiotto: Now, in this case, we have this situation, this is,
05:05:980Paolo Guiotto: For the SpaceX, so the real line.
05:10:340Paolo Guiotto: These are our functions. It has zero.
05:15:10Paolo Guiotto: Until, some point, N.
05:18:310Paolo Guiotto: And then they are equal to 1.
05:21:940Paolo Guiotto: It doesn't matter useful.
05:24:180Paolo Guiotto: that point is included or less. So this is the plot of F and.
05:28:800Paolo Guiotto: As you can see, when you take FN plus 1,0 n plus 1,
05:37:950Paolo Guiotto: And then it is equal to 1.
05:44:740Paolo Guiotto: So, as you can see visually, the red line is always above the green line, or the green line is below, okay? Because at left of n, they are equal, at right of n plus 1, they are the same, and between n and n plus 1,
05:59:930Paolo Guiotto: the FN is 1, the FN plus 1 is 0, so it is always above. So, we have ex… the situation described. We have a…
06:09:180Paolo Guiotto: Of course, they are measurable functions.
06:13:120Paolo Guiotto: Because they are indicators of measurable sets.
06:17:00Paolo Guiotto: Fn is greater or equal than FN plus 1.
06:21:490Paolo Guiotto: greater or equal than zero for every ANA, and actually for every X, for almost every X.
06:28:700Paolo Guiotto: But… When we compute the integral on R, of FM…
06:37:690Paolo Guiotto: In the measure, which is the Lebec measure for the real line, this is the integral on R of the indicator of the offline from N to plus infinity.
06:48:340Paolo Guiotto: So we know that when we have an indicator, the integral is just the measure of the set.
06:53:910Paolo Guiotto: So the measure of the offline from n plus infinity, and for the LeBague measure, this is equal to plus infinity.
07:01:30Paolo Guiotto: So each of these numbers is equal to plus infinity. It's a constant sequence of plus infinities, they go to plus infinity.
07:11:240Paolo Guiotto: when we send the end to plus infinity.
07:15:70Paolo Guiotto: Why, Ella?
07:17:90Paolo Guiotto: If we do the calculation for the limit.
07:20:640Paolo Guiotto: integral on R of the limit of the FN.
07:26:300Paolo Guiotto: Now, what is the limit?
07:31:230Paolo Guiotto: What is this?
07:33:900Paolo Guiotto: So, the limit…
07:36:520Paolo Guiotto: Here we are doing the limit in N. N goes to plus infinity of Fn , 4x fixed. It's a pointwise limit, okay?
07:48:150Paolo Guiotto: Now, what happens when you fix an X?
07:51:30Paolo Guiotto: Now, the values of these FN can be only 0, 1, because these are indicators, right?
07:59:960Paolo Guiotto: So the question is, is it 0 or 1?
08:03:280Paolo Guiotto: Well, what can be said is that if this is the real line, and we fix an X, whatever this X is.
08:11:110Paolo Guiotto: As soon as n is bigger than this X, and this will happen because X is fixed and then it's moving to plus infinity, so there is a first natural n, which is at the right of point X. If you want, that capital N is the integer part of X,
08:30:10Paolo Guiotto: plus 1…
08:32:309Paolo Guiotto: Okay? Now, for this function, for this FN, Fn is made like that, this is 0, and then it is equal to 1.
08:42:620Paolo Guiotto: The value at the point x is 0, as you can see, no, for this FN Fn of X.
08:50:550Paolo Guiotto: Is equal to zero.
08:52:600Paolo Guiotto: And since the next ones are smaller L and positive, so this is greater or equal than FN plus 1,
09:01:980Paolo Guiotto: of X, which is positive.
09:05:470Paolo Guiotto: It cannot be… it cannot be anything else than equals zero.
09:09:510Paolo Guiotto: And the same happens with FN plus 2. If you want these functions, it's like they are moving to the right. You see this?
09:18:670Paolo Guiotto: You look at me as… you see, this sequence is like a sequence of graphs, they are moving, to the right to plus infinity. So, as soon as your n is larger than X, the sequence of values FN of X is 0, 0, forever.
09:37:740Paolo Guiotto: So the limit is… the limit of a sequence, which is a sequence that, from a certain point on, it is equal to 0, the limit will be 0. So here we are basically integrating on R, 0D lambda 1, which is 0. As you can see, these two values are different.
09:57:220Paolo Guiotto: So the limit of integral is different from the integral of the limit. So in general, this property cannot be used with just this,
10:10:780Paolo Guiotto: In the same way as we use the monotone convergence, so having the sequence of measurable functions, positive, that they decrease with them, okay? This is not a surprise, because we may remind
10:25:850Paolo Guiotto: That, the analogous property, which is the continuity from above of the measure, requires that something more than just having a sequence decreasing down to something.
10:39:250Paolo Guiotto: And it requires that the first set of the sequence should have a finite measure.
10:44:510Paolo Guiotto: Also here, if we add the condition.
10:48:150Paolo Guiotto: that is, the first, function of the sequence has finite integral, then the continuity from above hold also, okay? So,
11:02:550Paolo Guiotto: to… make… this… True.
11:12:80Paolo Guiotto: property…
11:18:570Paolo Guiotto: It doesn't… to… ask.
11:30:990Paolo Guiotto: And this is the, let's say… Corolla.
11:38:110Paolo Guiotto: of the monoton convergence. So we have a measure space, XF mu.
11:47:280Paolo Guiotto: We have a sequence FN.
11:50:380Paolo Guiotto: of measurable functions on X, such that
11:54:850Paolo Guiotto: Okay, we have the condition that they are decreasing, so FN of X is greater or equal than Fn plus 1 of X greater or equal than 0. This condition must be valid almost every x in capital X, and for every n natural.
12:13:380Paolo Guiotto: So, such that, number one, this holds, and we need a second condition, which is at least one of the functions has a finite integral.
12:23:570Paolo Guiotto: So let's say that this is the first one. If it is the function with index 1 million, we just discard the first 1 million of functions, and we start from that index. So, let's say that the integral on X of F1 dmu is finite.
12:41:860Paolo Guiotto: They are positive, so writing F1, or absolute value of F1, is the same thing, okay? So this is equal to the integral on X of modulus of F1,
12:53:880Paolo Guiotto: Which is, and so this means that we are just requiring that the first element of the list is an integral function, okay?
13:08:90Paolo Guiotto: Now, under this assumption, we have the conclusion. Then…
13:12:470Paolo Guiotto: the limit in N, there exists the limit in n of the integrals of the FN with respect to measure mu, and it is equal to the integral of the limit of the FN.
13:27:810Paolo Guiotto: So, in this extra condition, we can apply this.
13:33:810Paolo Guiotto: We see the little proof, because basically it's an application of the monotone convergence. So we will try to apply exactly as we apply the
13:44:850Paolo Guiotto: the parallelism with the properties of the measures is you have the continuity from below, which is the monotone convergence theorem. In fact, monotone convergence is an extension of that property.
13:57:630Paolo Guiotto: And then you have continuity from above.
14:00:620Paolo Guiotto: with the restriction, the first set must have finite measure. Here we have this monoton convergence
14:08:960Paolo Guiotto: for decreasing sequences, and actually use this to the other one. Let's see how.
14:15:130Paolo Guiotto: So…
14:18:930Paolo Guiotto: we need to build an increasing sequence of function to which we apply the monotone convergence theorem. So let's first define the limit here.
14:30:80Paolo Guiotto: Let F… of X be, by definition, the limit in N of F, n of X.
14:39:910Paolo Guiotto: Well, here, whenever we define something which is not just immediately trivial, and the limit, for example, is never trivial, as well as derivative and integral and whatever, so we should verify always that what is written is well-defined.
14:54:370Paolo Guiotto: Now, why decent… Fine. This is well defined.
14:58:130Paolo Guiotto: Because the sequence of Fn , when x is fixed, for almost every X is a decreasing sequence, property number 1.
15:06:540Paolo Guiotto: So the limit of that numerical sequence of these numbers, no, so you have to imagine that these numbers are moving on the real line, so we have F1 of X, then you have F2 of X, which is a little bit smaller, F3 of X, which is, again, a little bit smaller, and so on. They are all positive, so this sequence is moving now to the left.
15:27:610Paolo Guiotto: X by X. So freeze and X, you look at this sequence of values, this is a decreasing sequence, so this explains why there exists the limit of the sequence.
15:39:780Paolo Guiotto: Moreover, that limit is a pointwise limit, and we…
15:45:290Paolo Guiotto: We have seen the statement of an important result to say, whenever we do a point-wise limit of measurable functions, this is the measurable function. So this is a measurable function on X.
16:02:00Paolo Guiotto: So we have a positive, and also, since they are positive, this is a positive limit, no, by the permanence of sine, no? Whenever you have a limit of positive things, the limit must be greater or equal than zero.
16:17:260Paolo Guiotto: This F is positive, measurable, the integral of F makes sense.
16:21:700Paolo Guiotto: So… integral on X of F, d mu, is well defined.
16:30:790Paolo Guiotto: We reminded that, in principle, when we deal with a positive measurable function, this quantity could be even equal to plus infinity.
16:41:340Paolo Guiotto: However, here it is not the case, because, since,
16:47:250Paolo Guiotto: we say that this sequence is decreasing, so every value, every Fn , is bounded by this one, this guy, no? So, we can say that the limit
17:03:410Paolo Guiotto: in n of DFN of X,
17:07:10Paolo Guiotto: As you understand, it is less or equal than F1 of X.
17:11:290Paolo Guiotto: No? Because X by X, you just move to the left. The limit will be somewhere here. This is the limit in n of numbers f and of X. Of course, if you change X, it changes the sequence, but the story is always the same. And from this, it follows that our F, which is this one.
17:30:700Paolo Guiotto: is bounded by F1, so the integral of, of F
17:37:340Paolo Guiotto: measure mu will be less or equal than integral of F1 with respect to measure mu, which is supposed to be
17:44:560Paolo Guiotto: find it by assumption 2 we made in the statement, no? So, actually, that integral is not just
17:54:380Paolo Guiotto: well defined as positive as the integral of a positive function, but it is also finite. It is finite. It is, of course, greater than or equal than zero, because we are integrating a positive function. So, it is a finite number.
18:10:210Paolo Guiotto: Okay, now, what is the inc…
18:14:940Paolo Guiotto: Which we apply the monotone convergence.
18:22:700Paolo Guiotto: F1 minus Fn. Let's see if that… if this is the case. So, let now define Gn of X as… this is similar, if you… if you have an… if you remind of the proof.
18:37:980Paolo Guiotto: of the continuity. From, above, we did the finest, the initial set minus set of the sequence. It's similar, no? So…
18:47:170Paolo Guiotto: F1 of X minus FN of X.
18:51:910Paolo Guiotto: Let's see what happens to this GM. First, they are difference of measurable functions, so it is a measurable function, so this is a measurable function on X.
19:03:390Paolo Guiotto: Second, it is positive, because we know that each Fn is smaller than F1, so this quantity is positive, and here we have a positive measurable function.
19:15:720Paolo Guiotto: Third, what happens if we do GN plus 1?
19:19:530Paolo Guiotto: Gn plus 1 over X is F1 of X minus Fn over Fn plus 1 of X.
19:28:470Paolo Guiotto: Now, this FN plus 1 of X is, here is supposed, because of the, monotonic behavior, it is less or equal than Fn, and this means that the reverse will appear. This Gn plus 1 will be greater or equal…
19:46:280Paolo Guiotto: F1x minus FN.
19:51:180Paolo Guiotto: of X, so this is Gn of X.
19:56:750Paolo Guiotto: So, now we have an increasing sequence of positive, measurable functions, so we apply the monotone convergence to this sequence Gn.
20:07:260Paolo Guiotto: So, I… monotonement.
20:13:130Paolo Guiotto: convergence.
20:14:650Paolo Guiotto: We have that, the limit.
20:17:710Paolo Guiotto: in N of integrals of, GN.
20:23:380Paolo Guiotto: Dimension.
20:24:870Paolo Guiotto: is equal to the integrals of the limit in n of the GN. And now…
20:32:860Paolo Guiotto: what is the… what is written here, going back to the F. Now, this is the integral of F1 minus Fn.
20:46:70Paolo Guiotto: Okay, so here I would like to say that this is the integral of F1 d mu minus the integral of Fn d mu.
20:57:140Paolo Guiotto: Is that possible?
20:59:130Paolo Guiotto: Well, we know that the integral is linear.
21:02:850Paolo Guiotto: You understand this, even from this program? If the humanities are both members, so these individuals are the founder, if,
21:14:960Paolo Guiotto: the problem that this is a classic 50 minus relativity form, it means that that operation wouldn't be defined 5. But this is not the case, because all these points are 5. And why?
21:28:840Paolo Guiotto: And this is because we noticed that the integral of F1 dmu is finite, this is…
21:37:230Paolo Guiotto: Our assumption, number two, that we had to add here.
21:42:530Paolo Guiotto: And what about the DFN? That the FN are smaller than DF1, no, because this…
21:49:900Paolo Guiotto: is pointing to the left. So, you see, FN is always less or equal than F1, X by X.
21:59:610Paolo Guiotto: So the integral of Fn will be smaller by monotonicity.
22:04:40Paolo Guiotto: Of the integral of Fn.
22:07:360Paolo Guiotto: And it will be positive. So, automatically, you get that will be fined. So, the integral of FN d mu
22:16:700Paolo Guiotto: is fine. So, in practice, with the assumption 2, not only, of course, F1 is integral, but all the Fn are integral, okay? So, it means that this can be done because…
22:31:530Paolo Guiotto: Both F1 and FN are actually integral functions, and so the integral is a linear alteration.
22:41:290Paolo Guiotto: You remind that, I don't know if you remind these kind of details, but…
22:46:00Paolo Guiotto: We have… we said that the integral is linear for positive function when we have a positive linear combination.
22:52:810Paolo Guiotto: coefficients of the combination are positive numbers. In general, so we can do… we can say the integral of the… is the sum of the integrals, so positive functions, but not the difference, because the difference might be plus infinity plus infinity.
23:08:330Paolo Guiotto: Therefore, integral with a general sign, whenever you have integral functions, so the integrals are finite, you can do any linear combination. Integral linear combination is the linear combination of the integrals.
23:22:700Paolo Guiotto: Okay, so, at left, we have, at the end, So let's continue here.
23:29:760Paolo Guiotto: the limit in n of integral on X of F1 d mu minus integral on X of Fn d mu. This is the left
23:44:530Paolo Guiotto: side of, let's give a name star, to the identity we have here.
23:52:190Paolo Guiotto: Okay.
23:54:280Paolo Guiotto: So that, that limit of, sorry, let's do this.
23:59:890Paolo Guiotto: The limit in n of integrals of Gn is…
24:08:160Paolo Guiotto: is this one.
24:10:260Paolo Guiotto: Yeah, it's the limit of these parameters.
24:15:210Paolo Guiotto: Now, integral of the fun is a constant.
24:18:370Paolo Guiotto: So this becomes mu minus the limit In a nutshell.
24:27:920Paolo Guiotto: of the F and the mu. So this is the left-hand side of the T-star we deduced from the monotone convergence. Let's look now at the right-hand side. At the right-hand side, we have integral of the limit of GN.
24:44:920Paolo Guiotto: So, the right-hand side…
24:47:210Paolo Guiotto: is a limit, integral on X of limit in n of gn d mu. Let's write this. This is integral on X of limit in n of gn. Gn was, by definition F1 minus Fn.
25:03:430Paolo Guiotto: So, again, F1 does not depend on N, so this becomes F1 minus the limit of F. N.
25:15:80Paolo Guiotto: Inanna?
25:17:740Paolo Guiotto: So this is the integral on X of this thing.
25:21:220Paolo Guiotto: Now, the limit has been called F, okay?
25:26:90Paolo Guiotto: beginning of proof, we define the F as the limit, the point-wise limit of the FN.
25:35:320Paolo Guiotto: And remind that it turns out that except…
25:40:210Paolo Guiotto: fine integral. It's positive, has fine integrals, so it is an L1 function. So, here, I can write that this is the integral on X of F1 minus F, d mu.
25:52:520Paolo Guiotto: And since both F1 and F are in L1, this sees, by linearity, the integral of F1 minus the integral
26:02:50Paolo Guiotto: Minus?
26:04:140Paolo Guiotto: F in Vimeo.
26:08:650Paolo Guiotto: Okay, so this is the right-hand side.
26:11:810Paolo Guiotto: And now let's, let's draw the conclusion, quoting these two. So…
26:18:800Paolo Guiotto: At the right side, we have an integral of F1.
26:23:910Paolo Guiotto: In the mu minus the limit.
26:27:640Paolo Guiotto: of the integrals of DFN d mu, this is the left-hand side, this is equal to the integral on X of F1 d mu, minus the integral on X of F,
26:40:190Paolo Guiotto: Which is the limit.
26:42:570Paolo Guiotto: Since, as you can see, this thing is the same here and there, these numbers are all finite, so we can cancel.
26:49:970Paolo Guiotto: And then we simplify also the minus, yeah.
26:54:120Paolo Guiotto: And we got the conclusion, because you see, limit of the integrals is the integral of F, which is the limit of the effect. So, limit of integrals equals integral of the limit.
27:06:380Paolo Guiotto: And this is the convolution.
27:12:170Paolo Guiotto: Okay, so this is about the monotonic, monotone convergence.
27:17:980Paolo Guiotto: from above. This is not important. The monotone convergence is not frequently used, but it is important to prove the next important results that will be the true convergence theorem, which is called the dominated convergence.
27:34:580Paolo Guiotto: But before we leave the monotone convergence, let's show another, let's say, interesting application of monotone convergence.
27:44:120Paolo Guiotto: So,
27:50:850Paolo Guiotto: Which is this proposition.
27:55:970Paolo Guiotto: So, let… F?
27:59:00Paolo Guiotto: be a measurable function on X, positive.
28:04:680Paolo Guiotto: Now, this is basically a way to define a large class of measures from a given measure.
28:13:810Paolo Guiotto: Define, huh?
28:16:740Paolo Guiotto: mu F, defined on the same family of measurable sets of the space.
28:24:60Paolo Guiotto: Positive valued, possibly equal to plus infinity.
28:28:820Paolo Guiotto: as… mu F of E
28:33:930Paolo Guiotto: is, by definition, the integral on e of the function f.
28:38:370Paolo Guiotto: in measure mu. Okay, so you have a fixed function after.
28:43:130Paolo Guiotto: And you, consider your assignment to set E, this number, which is the integral of the function f on the set E.
28:50:990Paolo Guiotto: This functionality is required to be just positive.
28:54:780Paolo Guiotto: measurable, so this quantum is always well defined, and overall, it can be also plus infinity, because this definition does not imply that necessarily this quantity is well. Now, it turns out that this function is a measure
29:12:160Paolo Guiotto: mu F, Jeez.
29:15:590Paolo Guiotto: measure.
29:19:90Paolo Guiotto: on S.
29:21:230Paolo Guiotto: So, in particular, since,
29:23:870Paolo Guiotto: basically, we have, normally, let's say, if you think to the case, X equals RD.
29:30:180Paolo Guiotto: mu de la Bag measure, S, any positive measurable function, there are a lot.
29:37:890Paolo Guiotto: We can define a lot of measures on RB.
29:42:190Paolo Guiotto: with this, with this mechanism, okay?
29:46:740Paolo Guiotto: Well, this is basically a sort of exercise of… so we do the proof.
29:55:480Paolo Guiotto: So what do we have to check? First of all, there is no problem about the definition, the well-posedness, mu F of E.
30:06:670Paolo Guiotto: is well.
30:09:90Paolo Guiotto: defined the… For every measurable sector.
30:17:800Paolo Guiotto: Because, the integral on E of F
30:20:980Paolo Guiotto: The mu can be seen also as the integral on X of F times indicator of
30:27:860Paolo Guiotto: set E. So, this product is zero outside of the set E. So, if you split the integral into integral on E of F times indicator, plus integral on e complementary of F times indicator.
30:47:510Paolo Guiotto: As you can see, the second integral is 0, because in the first case, the indicator of it on the seti is constantly equal to 1, so you get that this is just F.
30:59:430Paolo Guiotto: The second one, when I evaluate the indicator of E on points of E complementary, I get always 0. So this thing is constantly equal to 0, and therefore you are integrating 0, you get 0. So this integral is a fake integral equal to zero. And the other one is the integral on E of F, linear.
31:20:10Paolo Guiotto: So, this means that this is the integral of this function, which is the product of F times an indicator. So, this is a measurable function. I write improperly like that, it's not a set, but we understand it's a measurable function, if you want, is an L, more precise notation.
31:39:770Paolo Guiotto: is an LX function. Also, this one is an LX function when E is in the family F, and the product of measurable functions is measurable.
31:50:350Paolo Guiotto: It is positive because both are positive, so this is a positive, measurable function you can integrate, you get the value. So, the quantity muf is well defined.
32:03:100Paolo Guiotto: Now, we have to check…
32:10:170Paolo Guiotto: the two characteristic, properties of any measure, that mu F, number 1, mu F of MP is 0, and number two, that mu F is countably
32:26:310Paolo Guiotto: additive.
32:27:880Paolo Guiotto: So the measure of this joint union is the sum of the measures.
32:32:260Paolo Guiotto: Now, mu f of empty equals 0 is trivial, because muf
32:37:400Paolo Guiotto: of empty means that you are integrating on the empty set the function F.
32:42:990Paolo Guiotto: We know that when we integrate on a measure zero set, one of the properties of the integral says, as natural, that the integral is zero, so there is nothing, basically, to prove. Number two, what if I have a disjoint, countable disjoint union of measurable sets?
33:03:450Paolo Guiotto: Now, according to the definition, this is the integral on this set, the disjoint union of the EN of the function F.
33:14:240Paolo Guiotto: So what I can do is just to do this trick I've done here, transform the integral on domain as the integral on domain, the full space times the indicator of the set E.
33:26:410Paolo Guiotto: So I do this. This becomes integral on X of F times the indicator of the disjoint union of the y.
33:37:70Paolo Guiotto: In video.
33:38:470Paolo Guiotto: Now, let's look at this indicator. This is the indicator of the disjoint unit. It means it is 1 if and only if you are in the disjoint union, and it means if and only if you are in exactly one of the sets, okay?
33:53:70Paolo Guiotto: So, you understand that this is the same of some of the indicators of the yield.
33:59:30Paolo Guiotto: Because even if this sum is an infinite sum.
34:03:160Paolo Guiotto: It's in fact, somewhere. All terms are zero, except at most one of them, no?
34:08:750Paolo Guiotto: Because if this… if you are the norm, he says, yeah.
34:14:659Paolo Guiotto: All these things are zero, so it means that if you are not in the union, That's some useful.
34:23:80Paolo Guiotto: If you are in the union, you will be only disjoint in one of them. So all these terms will be a zero except one of them. So this term will give you value one. So they are the same.
34:36:570Paolo Guiotto: written in this format.
34:39:350Paolo Guiotto: I have that this can be seen as I carry F, which is a constant factor inside, as integral of sum from 0 to infinity of F times indicator of EN d mu.
34:53:510Paolo Guiotto: Now, these are… Measurable functions on Excel.
35:00:810Paolo Guiotto: So, let's call them FN, measurable function on X, and positive.
35:07:770Paolo Guiotto: And you remind that one of the consequences of the monotone convergence is that we can switch infinity sum with the integral when we add the positive functions, positive measurable functions. So, this is still the
35:22:680Paolo Guiotto: We still call monotone convergence.
35:26:20Paolo Guiotto: It is the version 4 series.
35:28:550Paolo Guiotto: And it says that we can carry out this… Because terms are positive.
35:34:510Paolo Guiotto: So the fact is that we apply the monotone convergence to the partial sums of the series. That's the increasing series sequence of function. So this becomes an integral of F times indicator of EN, but going back.
35:52:380Paolo Guiotto: This is the integral on EN of function f mu.
35:58:300Paolo Guiotto: And these are, by definition, the measures of the set CNN. So we have sum for n going from 0 to infinity of muf
36:06:790Paolo Guiotto: Yeah, and that's.
36:08:760Paolo Guiotto: They end up the profile.
36:12:330Paolo Guiotto: So now we have a criteria.
36:16:80Paolo Guiotto: To have a lot of measures.
36:19:840Paolo Guiotto: On a space where there is a radiomesh.
36:23:370Paolo Guiotto: We just take our measure, and we multiply by a positive measure of both functions, and we get a new measure, okay?
36:32:460Paolo Guiotto: So, we can construct lots of measures here.
36:37:160Paolo Guiotto: Now, as I told, actually, the, the monotone convergence, is,
36:46:970Paolo Guiotto: A very limited tool, because of its main restrictions.
36:53:540Paolo Guiotto: Which are, first of all, it applies only to positive things. Well, if they are negative, you can change the sign, and you understand that there will be a version for negative things, no?
37:04:370Paolo Guiotto: multiply by minus this equality, looking at FN is less than 0, okay? This will become FN greater than Fn plus 1, so it works for decreasing sequences of negative functions.
37:19:520Paolo Guiotto: No? But see that there is a constant sign.
37:23:470Paolo Guiotto: Okay, involved. So, that's a strong limitation. The second strong implication is, of course, that it's a very rigid condition, this one, huh? That you must have that functional direct one above the other.
37:37:590Paolo Guiotto: So, that's why this is not the best tool we can use, but fortunately, from this tool.
37:45:550Paolo Guiotto: we can deduce another very important, that's the most important tool, we could say, of the entire theorem. So let's write in red this theorem.
37:56:480Paolo Guiotto: This is called the LeBague.
38:04:540Paolo Guiotto: dominated…
38:12:70Paolo Guiotto: convergence.
38:18:160Paolo Guiotto: It requires something a bit stronger on the functions FN, so the problem is always the same. You want to do limit of integrals and say that this is the integral of the limit, okay?
38:30:650Paolo Guiotto: Now, here, we assume that there is a usual measure of space, XF mu.
38:37:500Paolo Guiotto: the… We measure space.
38:43:640Paolo Guiotto: sequence FN, that now this is a little bit stronger requirement, but if you think a second, it seems to be a…
38:52:960Paolo Guiotto: require that these are integral functions on X.
38:58:610Paolo Guiotto: Because we mostly are interested to integral function when these values are finite, okay?
39:05:20Paolo Guiotto: And the interesting point is that here there is no restriction about the values of the FN. They can even be complex-valued, okay?
39:13:220Paolo Guiotto: F… N. Can. B.
39:17:880Paolo Guiotto: Both.
39:19:550Paolo Guiotto: real.
39:20:880Paolo Guiotto: valued them.
39:24:530Paolo Guiotto: Or… he valued.
39:31:410Paolo Guiotto: It works in any case.
39:34:930Paolo Guiotto: Suppose that… well, let's say such that… B.
39:42:640Paolo Guiotto: such.
39:44:540Paolo Guiotto: that. Now, there are the requirements on this sequence. The first one is a very mild assumption.
39:52:360Paolo Guiotto: It says that this sequence, you know, for the monotone convergence, we do not have this, because once we know that the sequence is ordered, is an increase with N, automatically, we have that the limit of the FN is
40:09:840Paolo Guiotto: Because X by X, you have a sequence of numbers, which is an increasing sequence, okay? So the limit X by X is well defined.
40:20:430Paolo Guiotto: Here, if I… if I have just the sequence of function, the sequence FN of X can be whatever, can be going around without any limit. So we require that there is a limit, but the requirement on the limit is very weak, so the weakest, point-wise.
40:38:10Paolo Guiotto: We assume that there exists the limit.
40:41:320Paolo Guiotto: Inanna?
40:44:20Paolo Guiotto: of Fn of X.
40:47:380Paolo Guiotto: almost every X in capital X. So that's a very, very, very weak requirement. You cannot ask anything less than this, basically. And number two.
41:03:790Paolo Guiotto: And that's where this world comes dominated. We require that these functions are controlled
41:12:760Paolo Guiotto: independently of the index n, that's important, by an integral function. So this means that there exists a function, say G, integral, so it must be an L1 function.
41:25:320Paolo Guiotto: on X, such that this function dominates the FN. Now.
41:31:510Paolo Guiotto: What is the sense of this? The FN can be a real value, so the bound is not FN less or equal G, because these are low values for F to be negative, whatever, and they can be even complex, so maybe there is not the order. So what I take is the absolute value.
41:48:500Paolo Guiotto: of Fn, must be controlled above by G . Now, this control must be almost everywhere with respect to X.
41:59:40Paolo Guiotto: And it must be for every na with respect to the index n.
42:05:250Paolo Guiotto: So, it's a sort of… form of control, but not uniform in the value of X.
42:11:610Paolo Guiotto: That's a pointwise bound in the value, because you say X by X bar. It's uniform in the index n.
42:19:200Paolo Guiotto: Okay? Well, what is important of this theory is also to understand this, sort of, setup, no? Because you will, find again in the consequences of this, like theorem where we completed the
42:35:360Paolo Guiotto: It's like this, well, we need some control which is uniform, but in something.
42:42:840Paolo Guiotto: Can you think it makes sense.
42:44:690Paolo Guiotto: Amen.
42:47:80Paolo Guiotto: Actually, you may notice that…
42:50:450Paolo Guiotto: At least formally, there is a bit of relevance in these assumptions, because once you have this bound.
43:00:130Paolo Guiotto: The integral of Fn is bounded by the integral of G, because of the monotonicity, and if G is Z1, automatically has a finite integral absolute value, so it's L1. So what is relevant about this condition?
43:16:850Paolo Guiotto: So even if you write a measurable here, automatically here you get that they are integral. So that's why I emphasize there, to remind that at the end, these can be on integral functions.
43:29:930Paolo Guiotto: Okay, under these assumptions, then… We can do the desired operation that exists.
43:38:380Paolo Guiotto: Well, here, it's non-trivial that there exists the limit of integrals, because we have not anything like the ordering of the integrals, so there exists the limit in n.
43:50:610Paolo Guiotto: of integrals of DFN and d mu, and this turns out to be the integral of the limit in n of DFN d mu.
44:00:940Paolo Guiotto: Well, actually, If I have to write this statement a bit more precisely than this.
44:08:940Paolo Guiotto: The statement says, more precisely.
44:17:600Paolo Guiotto: I just write in this form because, to remind you that at the end of the day, what we want to do is to limit… to carry the limited inside. That's what we need to be able to do. But, under this thing.
44:33:600Paolo Guiotto: there are some, hidden, details. For example, this leaf here.
44:40:920Paolo Guiotto: Is number one, is it still the point of wisening effect?
44:45:450Paolo Guiotto: The hypothesis 1 says that this limit exists, so there is no problem with the definition of this thing.
44:52:760Paolo Guiotto: Second, it is a limit, pointwise limit of measurable functions, so it will be measured, there is no problem even with this. But third, as you see here that is the integral, and this thing can be a real value of C-valid integer.
45:06:890Paolo Guiotto: Whenever you write this integral, you must specify that this thing is integral, otherwise it used to be something not defined. So the more precisely means, I should say, the more precisely, the limit
45:21:260Paolo Guiotto: The pointwise limit, so the function f of x, defined pointwise as the limit of f, function which is well-defined because of the assumption 1,
45:37:80Paolo Guiotto: F is in L1X, so it's an integral function, this would be a consequence of this statement.
45:47:990Paolo Guiotto: Ender?
45:49:750Paolo Guiotto: Now, I can say that this conclusion is true.
45:53:430Paolo Guiotto: Okay?
45:56:650Paolo Guiotto: Okay, so this is the statement. Now, do you want to take a break, or we continue?
46:04:860Paolo Guiotto: If just one of you wants a break, I have to give it a…
46:09:390Paolo Guiotto: Want the break?
46:11:620Paolo Guiotto: Yeah, okay. Is there anyone who wants a break?
46:16:890Paolo Guiotto: Okay, we continue. Okay, let's see… mmm…
46:21:820Paolo Guiotto: This is actually a concept, a corollary of the monotone convergence, so I won't do the proof in the full details, I will give you an idea for where it comes from, okay? So, let's say it's catch…
46:39:520Paolo Guiotto: off, it should be.
46:41:710Paolo Guiotto: proof. It's not long, there is a certain moment, a little detail that you will accept. Let's say, you will accept because it sounds, to be, not particularly complicated to be proved.
46:56:50Paolo Guiotto: Okay, let's start,
47:00:250Paolo Guiotto: with this part, with this final part. So, define the limit, let f of x
47:09:100Paolo Guiotto: be, by definition, the limit in n of DFN of X.
47:14:60Paolo Guiotto: Well, it must be clear that this is a pointwise mean, so the next
47:20:530Paolo Guiotto: you could meet the limit of Fn of X, you get a number. When you change accent, this number will change, because you have limited signals. These signals depend on X. Now, remind the example…
47:34:970Paolo Guiotto: I do not, probably, here.
47:39:350Paolo Guiotto: Here… We have seen for the moment just one example of this case, a limit Where is it?
47:51:20Paolo Guiotto: A limit of, A pointwise limit of a sequence of measurable functions.
48:03:990Paolo Guiotto: Yes, it is this one, no? We will see it after the next future, because we will often be the convergence of the sequences of functions, but we need the first to figure out the integration here, otherwise we cannot do the kind of problems.
48:20:420Paolo Guiotto: But we are almost at the end of this battle.
48:23:850Paolo Guiotto: So, see, you see the sequence is this one.
48:27:200Paolo Guiotto: So X by X, you have the sequence X to the n. If X is 0, this is 0 to the N, constantly equal to 0.
48:36:230Paolo Guiotto: If X is 1, you have 1 to begin, constantly goes to 1. You see? You change X, you have different sequences.
48:44:50Paolo Guiotto: And X equals 1 half is 1 over 2 to the i. It's another sequence. So for every X, you have a different sequence.
48:51:40Paolo Guiotto: When you freeze action, and you take the lead in action, what you're doing is what is the point-wise limit.
48:58:870Paolo Guiotto: Fix this X, and contribute sending N to infinity. So I understand that here we are two variables, n and X, but only one is going to be the limit is N. The other one is just a parameter, huh?
49:12:900Paolo Guiotto: So here… We say, fix X, compute the limit FN of X.
49:20:450Paolo Guiotto: by 2, by 1, this mini is supposed to exist almost every X in the space. So I can say that this f of x is defined for almost every X in capital X.
49:35:950Paolo Guiotto: And moreover, since DFN are integral, and in particular they are measurable functions.
49:44:690Paolo Guiotto: This, is a point-wise limit of measurable functions if it is a measurable function.
49:53:510Paolo Guiotto: So now we know that it is measurable. I think that it is an L1. L1 is a bit more, because it means that the integral of the absolute value of F is finite.
50:05:460Paolo Guiotto: So, actually, F is L1X. This can be proved because
50:15:750Paolo Guiotto: Okay? In fact, if you take 2… Indeed.
50:24:320Paolo Guiotto: by…
50:25:480Paolo Guiotto: 2, assumption 2, we have modulus Fn of X is less or equal than G of X. This holds almost every x in capital X, and for every n.
50:38:800Paolo Guiotto: So you freeze now an X, an X for which this thing is valid.
50:42:940Paolo Guiotto: Okay?
50:43:980Paolo Guiotto: And this will be almost every X, okay? Now here, there are little details, because the almost every should change with N, but you can… we have already seen a kind of problem like this. You can put all the bed sets together, and this will be a measure zero set. However, let's assume that the almost every means for every X, for almost every X, and for almost every N.
51:07:780Paolo Guiotto: Okay? Then you pass to the limit here, this will go to F of X by definition of F, and therefore its absolute value will go to the absolute value of F.
51:20:70Paolo Guiotto: And since all of them, they are less or equal than G , also the limit will be less or equal. That's the permanence of sine, no? Cannot be greater. So, to be less or equal than G . And this will be valid almost every X in capital X. There is no more N, because n has gone to plus infinity.
51:40:660Paolo Guiotto: But from this, now I integrate, and I deduce that integral on X of modulus of F in D mu.
51:47:630Paolo Guiotto: It is less or equal than the integral of G in the mu.
51:54:50Paolo Guiotto: G is supposed to be what? It's supposed to be L1. It's written here, you see?
52:00:750Paolo Guiotto: So, what does it mean? That the integral of the absolute value of G is fine, but this coincides with the integral of G. Why? Because when you have this bound here, automatically, implicitly, you are saying that G is positive.
52:16:460Paolo Guiotto: Because G is greater than an absolute value, G is a positive function, okay? It cannot be negative. So, it means that this is the same of absolute value of G,
52:27:40Paolo Guiotto: And therefore, this is finite because the 2 says that G is in L1, is supposed to be in L1.
52:38:110Paolo Guiotto: So, by this, we have, efforts.
52:41:760Paolo Guiotto: the point-wise limit of our sequence is in L1, as all the FN, okay?
52:50:80Paolo Guiotto: Now, the big part of the proof is to prove that the limit of the integrals exists, and it is equal to the integral of the
52:59:200Paolo Guiotto: You got it.
53:01:960Paolo Guiotto: These… Peace… Which is something we proved, basically, in detail.
53:11:770Paolo Guiotto: Step 2 is to prove that
53:15:670Paolo Guiotto: The integral of FN d mu
53:20:320Paolo Guiotto: the limit in n of these, which are finite values. N are integral, so these quantities are well-defined. What is not to be at is if the limit exists.
53:32:970Paolo Guiotto: Because that's different from the monotone convergence. The monotone convergence, you have that the sequence of DF increases, so also the sequence of the interpass is an increasing sequence.
53:45:960Paolo Guiotto: So you know that that sequence is an immediate sequence of numbers, the sequence of integrals.
53:51:300Paolo Guiotto: And therefore, we have a region. We don't have to spend time to build better than just the regions. But now, here, we have…
53:59:630Paolo Guiotto: A sequence of integrals of generic function that can be…
54:04:220Paolo Guiotto: Complex standards. So these are complex numbers, maybe, and we'll also be tailored.
54:10:920Paolo Guiotto: You see, that's the difference, so we actually have to prove that this limit exists, so let's put in parentheses, because this is also part of the proof, and it is equal to the integral of the limit that we called n.
54:26:670Paolo Guiotto: Now, here there is a literal, which has nothing to do with regards. You can say that this is equivalent
54:36:290Paolo Guiotto: Since everything here is finite, these quantities are finite, are real or complex, if you are in the complex case, this is real, or complex if you are in the complex case. Numbers. So, they are finite values.
54:51:700Paolo Guiotto: Now, you can carry, for example, this one, the limit, at left.
54:57:140Paolo Guiotto: It is a constant, and you can carry inside the limit, so you can say that this is equivalent of saying that the limit in n of integral of Fn d mu minus
55:09:420Paolo Guiotto: L of F d mu, this thing has limit 0.
55:14:650Paolo Guiotto: No, I'm just saying that… Fall.
55:20:320Paolo Guiotto: N minus L…
55:32:40Paolo Guiotto: number integrals, we can use a bit of integral properties, so this is the difference between two integrals, we can…
55:38:960Paolo Guiotto: say that it is the integral of the difference. So this is once again equivalent to limiting n of a unique integral of Fn minus F, d mu is 0.
55:52:80Paolo Guiotto: And let's say that this is what we are going to prove. Actually, we will prove a bit more than this.
55:59:100Paolo Guiotto: In fact, the statement could be written with a little bit stronger conclusion than this one, okay?
56:06:370Paolo Guiotto: Now, however, how can we prove that this,
56:11:490Paolo Guiotto: limit of these quantities is zero. Here, there is a standard, but simple idea. Very simple. When you want to prove that the sequence of numbers goes to zero, you take the absolute value and you try to prove that it goes to zero.
56:28:390Paolo Guiotto: So, this is equivalent of saying that the lead… Inan, of… Use of these numbers.
56:43:350Paolo Guiotto: is zero.
56:45:370Paolo Guiotto: Now, take these numbers, the absolute values of the integrals n minus F.
56:56:780Paolo Guiotto: And whenever you have absolute value of an integral, the temptation is to apply the triangular inequality and say that this is, for sure, less or equal than the integral on x of fn minus F. You carry side the absolute value. That's by the triangular
57:16:310Paolo Guiotto: inequality.
57:18:440Paolo Guiotto: So, since I want to prove that these guys… yeah, I wasn't CEO.
57:22:960Paolo Guiotto: I will actually prove that this guy here also, which is a bit more than this, okay? So, we actually prove…
57:32:770Paolo Guiotto: Soon.
57:35:290Paolo Guiotto: If we… Prove.
57:39:780Paolo Guiotto: that.
57:42:40Paolo Guiotto: the integral on X of modules FN minus F in mu. This goes to zero, We are done.
57:54:200Paolo Guiotto: And this is actually more, no? Because the inequality can be a strict inequality, a triangular inequality.
58:01:800Paolo Guiotto: Now, why this is the key point where, let's say, the idea of using the unique limit theorem we have here is the monotone convergence at this stage?
58:12:860Paolo Guiotto: Because these functions are important.
58:16:320Paolo Guiotto: Yeah, we have a problem, but we are.
58:18:920Paolo Guiotto: requirements of certain functions here, which are positive, measurable, no, is going to zero. And look what happens inside. If you look at the point-wise field of this.
58:31:380Paolo Guiotto: Since F has been defined as the pointwise limit of EFN, X is different.
58:39:160Paolo Guiotto: Absolute value goes to zero, so the quantity inside the input is going to zero expat.
58:45:430Paolo Guiotto: Pointwise, the quantity inside goes to zero.
58:50:160Paolo Guiotto: Got it.
58:51:440Paolo Guiotto: Now, if it goes to zero in some monotonic way, we could hope that maybe the monoton voltages apply. Now…
59:03:180Paolo Guiotto: What would be a funny play?
59:06:100Paolo Guiotto: But when a quantity goes to zero, it's not true that it goes to zero.
59:12:30Paolo Guiotto: If you expect this force is going to zero, the unipmonotolic wave could be a decreasing wave, not increasing. Now, if you start from 1 and go to zero, you must go down, you know, so you must expect that this goes down to zero, but actually, you could go to zero. Now, imagine this is my initial dependency is the ground. I do this, then I jump up, then I go down like this, no.
59:36:470Paolo Guiotto: up and down, and descending at 0. So, it's not true that this will go to zero in the decreasing rate in N. X by X.
59:44:860Paolo Guiotto: But here is where the trick and the non-trivial trick comes in. Okay, this does not go to zero, but I define this quantity
59:53:90Paolo Guiotto: DeFi, you know?
59:55:70Paolo Guiotto: Let's call, in China.
00:02:50Paolo Guiotto: You do not take the absolute value of Fn minus F, which is going to zero,
00:07:560Paolo Guiotto: Notice that, let's write this, remark…
00:12:00Paolo Guiotto: modules FN of X minus F of X, goes to zero by… definition of of effort.
00:25:210Paolo Guiotto: And so, by assumption one. For the moment, we have not yet used the assumption number 2, no?
00:31:540Paolo Guiotto: more or less, yes, we used to show that F was integral, but it does not be… it does not yet really used.
00:41:310Paolo Guiotto: Now, what you take is the following. Take the sequence of the distances between Fn and F, starting from index n.
00:50:530Paolo Guiotto: So you take the distance between F and X and death.
00:56:760Paolo Guiotto: Then, the distance between Fn plus 1 X, and death.
01:03:550Paolo Guiotto: This infinite list of numbers, no?
01:08:350Paolo Guiotto: So, in a… in a few words, this is the distances FKX to F.
01:17:110Paolo Guiotto: for all index, K starting from N infinity, so k larger or equal than n.
01:25:160Paolo Guiotto: So this is a set of numbers.
01:29:290Paolo Guiotto: And, take the biggest possible of them, so what is the supreme option.
01:36:390Paolo Guiotto: So, the more and maximum.
01:39:160Paolo Guiotto: You have now defined a function hn.
01:43:170Paolo Guiotto: Okay? Let's say that,
01:45:810Paolo Guiotto: More or less, this is the worst of the distances between Fn and the f of x, the worst, starting from the index n.
01:56:630Paolo Guiotto: Okay?
01:58:560Paolo Guiotto: Now, what you understand, elite… so they are definitely positive.
02:07:80Paolo Guiotto: And this is clear, you are doing the maximum of positive quantities, there is no question here. But what happens when we take HN plus 1?
02:16:00Paolo Guiotto: Look, this is the same operation, you do the supremum of these distances, but starting from index n plus 1, so you will have distance between Fn plus 1X and FX.
02:29:390Paolo Guiotto: Then you will have distance between Fn plus 2, X, and f of x, and so on, forever.
02:37:320Paolo Guiotto: So, you see that in the second set, you're doing the maximum, I call the maximum property, I just say the minimum, but it's more or less the same. I'm doing the maximum of these quantities. Here, I'm doing the maximum of the same quantities plus this one.
02:54:310Paolo Guiotto: So, do you see any relation between these two maximum?
02:58:990Paolo Guiotto: Exactly, because here, you factor the same numbers I have there, that's… One more.
03:10:110Paolo Guiotto: So the maximum can be bigger, but not… cannot be smaller. So this will be greater or equal.
03:18:540Paolo Guiotto: attendees.
03:19:690Paolo Guiotto: So, in other words, this sequence HN is a decreasing sequence of positive functions.
03:26:750Paolo Guiotto: And that is the sequence to which we will apply the monotone convergence. So here we have that 0 is less or equal than HN plus 1 of X, which is less or equal than HN of X. This happens for every X.
03:47:20Paolo Guiotto: for every X. Well, actually, we should say for almost every X, but…
03:51:920Paolo Guiotto: These are details, and for every N.
03:56:330Paolo Guiotto: Now, to apply the monotone convergence, since this is the decreasing monotone convergence.
04:03:220Paolo Guiotto: We need to know. They are measurable functions, and this is a technical part that could be done, forget it, we accept. Now, these…
04:15:110Paolo Guiotto: We have that H and are measurable functions on X.
04:20:210Paolo Guiotto: We accept them.
04:24:410Paolo Guiotto: And, second, second point… Is that DJ Anna?
04:30:750Paolo Guiotto: What can we expect, when M goes to pass infinity?
04:36:220Paolo Guiotto: Well, let's say that H n for X pixels is the maximum of the distances between Fn n and f of x, which is the root of the F of N.
04:46:940Paolo Guiotto: So you expect that since 1N goes to infinity F of Y goes back to the X, that maximum will go to 0. Now, this can be proved, it requires a little bit of work, but…
04:59:320Paolo Guiotto: it's not impossible. So, but you see, those are the properties I told you we can accept, because that seems to be reasonable. So, HN of X goes to 0.
05:10:90Paolo Guiotto: And the third factor…
05:12:820Paolo Guiotto: Since we are going to use the monotone convergence from, for the decreasing Ks, we need to show that the first, H is an L1 function. So this happens for every X, and third, H1 is in L1.
05:30:800Paolo Guiotto: Well… This follows from the… Because, if you see the function… the distance…
05:40:300Paolo Guiotto: And it is FN of X minus F of X,
05:44:550Paolo Guiotto: By the triangular inequality of the absolute value, this is mod less than or equal than Fn of x plus modus f of x.
05:52:990Paolo Guiotto: Now, by 2, this is bounded by G of X.
05:57:550Paolo Guiotto: This is the assumption, too.
05:59:690Paolo Guiotto: And if you remind, above, we proved somewhere that also modulus of F is X area.
06:08:880Paolo Guiotto: Also, modulus of F is controlled by that G.
06:13:540Paolo Guiotto: Okay? That's the crucial point where the G comes, because it helps to prove that these functions are N1, which is what we need to apply the
06:24:350Paolo Guiotto: monotone convergence. So, also, 1 is bounded by G of X,
06:29:400Paolo Guiotto: And therefore, this is bounded by 2 times G of X.
06:34:840Paolo Guiotto: Porvriana?
06:37:230Paolo Guiotto: Because this is not depending on N. And therefore, when you take the maximum of these increments, these distances, so HN,
06:47:920Paolo Guiotto: Which is the supremum.
06:50:570Paolo Guiotto: of these distances, F, k, x minus F, X for K greater or equal than n. This is bounded by… each of them is bounded by 2G of X, the maximum is bounded by 2G of X as well.
07:07:40Paolo Guiotto: So you see that this is the bound that says HN is controlled above by 2G. If G is L1, 2G will be L1. So this is an L1 function, still by 2, assumption 2.
07:21:600Paolo Guiotto: So now we have all the ingredients to conclude, because we have a sequence A chain.
07:27:330Paolo Guiotto: Which is decreasing.
07:29:710Paolo Guiotto: sequence in N is going to 0, and the first element of the sequence is in L1, so the limit
07:37:830Paolo Guiotto: of the integrals of the HN,
07:41:510Paolo Guiotto: The mu will be the integral, this is because of the monotone.
07:45:990Paolo Guiotto: convergence.
07:47:510Paolo Guiotto: Will be the integral of the limit.
07:50:190Paolo Guiotto: of DHNA.
07:53:540Paolo Guiotto: In the mule?
07:55:120Paolo Guiotto: But…
07:56:800Paolo Guiotto: they go to zero, so that limit is… the limit is zero. So you are integrating zero, and this is zero. So the limit of the HN is 0.
08:08:240Paolo Guiotto: We… we add the… to show that this quantity is going to zero. But… And since…
08:20:920Paolo Guiotto: When you take the integral on X of modules F and X, minus F of X, GMU,
08:29:450Paolo Guiotto: Now, DHN is defined as the maximum of the quantities, it's better if we see here… is defined as the maximum of the quantities FK minus F when K is larger than N.
08:46:140Paolo Guiotto: So, in particular, this is less or equal than HN, because that distance is one of those that we find inside this supremum, no? Among all these quantities, there is also Fn of X minus F of X.
09:02:540Paolo Guiotto: So HN, which is the maximum of all these, will be bigger than that, so this is less or equal than the integral of HN.
09:11:950Paolo Guiotto: Which is going to zero, because we proved here.
09:16:340Paolo Guiotto: And therefore, we get the conclusion. This one goes to zero as well. It is positive, so it goes to zero.
09:24:260Paolo Guiotto: This finishes the proof of dominated convergence here.
09:29:950Paolo Guiotto: Okay, now, this is one of the most important facts that comes with the bacteria, and for this course, it will have a lot of applications. So, initially, we start… we start with some
09:49:490Paolo Guiotto: let's say, concrete direct application of this theorem. So, I will start with an example here, which is the example 622.
10:00:860Paolo Guiotto: The problem is compute…
10:05:590Paolo Guiotto: the limit in n of integral 0 to plus infinity of… we have this, n square 1 minus cos x divided n times e to minus n over n plus 1x dx.
10:25:900Paolo Guiotto: Now, do not… do not be worried about what is the interest in doing this. Now, the interest is just to learning, how to use the theorem, no? How to verify the hypothesis of the theorem, okay? That's the point, yeah.
10:45:390Paolo Guiotto: Okay, as you can see, this is the classical limit problem. We have a limit of integrals of certain functions Fn , which are these ones.
10:56:270Paolo Guiotto: Okay? Now, we see, of course, we…
11:01:920Paolo Guiotto: In principle, let's look, since now we have basically two different setups, the monotone convergence and the dominated convergence.
11:12:220Paolo Guiotto: Let's see if we can apply, both, just to understand, what are the limitations.
11:22:400Paolo Guiotto: Especially for the first one.
11:25:310Paolo Guiotto: So, we have… to… Computer…
11:34:550Paolo Guiotto: the limit in n of the integrals of this FN of X. And of course, we would like to have inside the limit. Of course, an idea could be, why don't we compute the integral? No, it's an integral in one variable.
11:51:580Paolo Guiotto: Actually, if we can carry the limiting side.
11:55:640Paolo Guiotto: Let's see. Now, the problem will be, can we… can we do that? Can we say that this is inequality? Well, let's see what is first this one, to see if we gain something by doing this.
12:06:630Paolo Guiotto: Because if the calculation of that right-hand side is too difficult, impossible, we don't waste time in trying to justify this identity, okay?
12:17:130Paolo Guiotto: If true, Sing, sir.
12:24:940Paolo Guiotto: the limit…
12:26:460Paolo Guiotto: in n of FNX would be the limit. So, as you can see here, we have an example, another example of pointwise limit. N squared 1 minus cos x divided N t minus n over n plus 1X.
12:44:430Paolo Guiotto: Now, in this limit, it is end variable. X is fixed.
12:49:970Paolo Guiotto: Okay, perhaps we will have to distinguish certain exceptional cases, we will see, but let's say that, in general, this is a limiting N.
13:00:120Paolo Guiotto: Now, what happens in N? If you look in n, this is going to plus infinity.
13:06:370Paolo Guiotto: What about the parentheses? X over N, X is fixed, x over n goes to 0, so cosine goes to 1, so parentheses goes to 0, so we have a first problem, infinity times 0. About the exponential, the exponent, the fraction n over n plus 1 goes to 1, so the exponential will go to E minus X.
13:25:770Paolo Guiotto: So this is a finite value.
13:27:960Paolo Guiotto: Now, the question is, of course, what happens here?
13:31:980Paolo Guiotto: Well, here there are… there are several, several…
13:38:950Paolo Guiotto: methods that can be used. For example, I may notice that this is cosine when the argument is small, cosine at 0. So it sounds like 1 minus cos T when t is going to 0.
13:53:620Paolo Guiotto: Okay, so here, what do you know about this? Well, for example, you may remind that there is a fundamental limit for cosine, that when you do limit for t going to zero of 1 minus T, this goes like T squared, no? And this limit is equal to… you remind?
14:11:320Paolo Guiotto: One map.
14:12:750Paolo Guiotto: on.
14:14:890Paolo Guiotto: you may remind, and that's more interesting, you may remind of the Taylor-Mechlorian formulas.
14:23:710Paolo Guiotto: You may remind that cosine of t, still when T is going to 0,
14:29:60Paolo Guiotto: becomes a polynomial, 1 minus T squared over 2, plus a correction, which is a little O of T squared. This is the first approximation. If this is sufficient, we just stop here, we don't need to write more terms.
14:43:170Paolo Guiotto: So let's say that we use this second one. The first one is similar. So, in this case, I would say that… so when I have 1 minus cosine of x divided N,
14:54:660Paolo Guiotto: This would be 1 minus… the cosine is 1 minus… I have to replace X over N in t in place of T, divided 2 plus little O of X over N squared.
15:07:960Paolo Guiotto: Now, be careful, because here you have to remind, the calculus with little o, for example. So this one simplified, 1 goes away with 1, so we have plus X squared divided 2N squared.
15:24:50Paolo Guiotto: plus a little O of, I should write X squared divided N squared.
15:28:360Paolo Guiotto: Here, be careful, because it is N, the variable for the limit, okay? So, X is a constant. You know that the constants are normally eliminated in little O, because this, you say, it is a quantity that goes to zero faster than.
15:45:810Paolo Guiotto: Okay? So a factor, a multiplying factor, does not change the speed of going to zero. So this could be, safely written as 1 over n squared, not to change.
15:56:970Paolo Guiotto: Okay?
15:57:940Paolo Guiotto: So, this says that 1 minus cosine is this, so when I multiply by n squared, it means that n squared times 1 minus cos x divided n, this is n squared times… I replaced the polynomial approximation.
16:20:750Paolo Guiotto: Now, when I do the algebraic calculations, you see that DN square simplifies this denominator, so it remains X squared over 2 plus n squared times O of N1 over n squared.
16:34:650Paolo Guiotto: Another feature of the little O's that you may remind is that they work like powers. So we can do the algebraic multiplication as if they were powers. So, roughly, we could carry this factor inside.
16:51:120Paolo Guiotto: if it is the variable, respect, we are completing the limit, so we have X squared over 2 plus, we get…
16:57:900Paolo Guiotto: O of N squared times 1 over n squared, which is little O , which means just something that goes to 0, okay? There's a quantity here, even if you don't see any hole in N, the n is still there, because you are computing a little bit.
17:13:190Paolo Guiotto: Okay, so this would be a middle order of 1 is a quantity, which is middle order of 1, it's just a quantity that equals to 0. So, this will go to X squared
17:25:210Paolo Guiotto: of, 2.
17:27:170Paolo Guiotto: And this happens for every X fixed, right? For every X real. So, in particular, since our integration domain is 0 plus infinity for every X positive, okay?
17:41:990Paolo Guiotto: So this holds for every X real.
17:45:750Paolo Guiotto: So, for every X in the integration numer, which is 0 plus infinity. Okay, so this, returning back to the limit of the FN,
17:57:340Paolo Guiotto: We can say that, so, the limit in n of the functions FN of X
18:04:680Paolo Guiotto: is X by X, huh?
18:07:400Paolo Guiotto: So we say that this part here, we now know that this part here The indeterminate form goes to
18:17:330Paolo Guiotto: X squared over 2.
18:19:530Paolo Guiotto: The other one goes to e to minus X, so there is no problem. The limit is their product, so it is X squared over 2E2 minus X.
18:29:920Paolo Guiotto: You see? And this also for every X in the integration interval, 0 plus infinity. So, it's not only for almost every, it's for every, this thing.
18:41:250Paolo Guiotto: Okay, that's the pointwise limit, the function we may call F of X.
18:46:880Paolo Guiotto: So, this means that… Going back.
18:50:120Paolo Guiotto: If the, swap between limit and integral, is, applies.
18:57:360Paolo Guiotto: We can, at the end, compute this quantity here, which will be the integral of this thing, X squared over 2 times C to minus X.
19:07:100Paolo Guiotto: That looks to be easier than computing directly the integrals of these FN, and then doing the limit of the values of the integrals, which is
19:17:500Paolo Guiotto: let's say, a direct way, no? I compute the integrals.
19:21:670Paolo Guiotto: it will come something, depending on n, I do the limit. Now, that's the… but here we are… we are doing this calculation by carrying inside the limit, and perhaps we get something easier to be computed. But of course, here, there is a bigger point. We need to justify this step.
19:40:400Paolo Guiotto: Okay? So now we have to decide what of the two results we could apply to carry the limit inside.
19:51:60Paolo Guiotto: Now, let's give a look if we can, let's say that. So… In conclusion…
20:02:650Paolo Guiotto: If, These.
20:06:980Paolo Guiotto: the… the… Applies.
20:13:810Paolo Guiotto: We… We are.
20:18:160Paolo Guiotto: reduced the… to compute…
20:24:930Paolo Guiotto: the integral on 0 plus infinity of this thing, x squared over 2E2 minus X, which is something that can be done with an integration by parts easily. We will do maybe later.
20:37:350Paolo Guiotto: Now, the question is, This passage, no?
20:43:560Paolo Guiotto: Can we carry the limit inside?
21:02:250Paolo Guiotto: Now, since time is over, well, I will do the fast way, which is we apply determinated convergence. Well, let's just give a look if we can try to… if we see that we can use
21:18:550Paolo Guiotto: monoconstant versions, okay?
21:21:540Paolo Guiotto: Now, if I have it… if I want to apply the machinery of one of the verses, I mean, positive functions, and they are off, because this is positive, 1 minus 12 is positive.
21:36:130Paolo Guiotto: exponential is positive, so they are positive. They are positive, measurable, so that's okay. Then I need the very strong assumption, result of monotonic behavior in the head, like increasing or decreasing.
21:50:960Paolo Guiotto: And here, things would be different, because this is increasingly there.
21:56:210Paolo Guiotto: What about these paranglades?
21:58:360Paolo Guiotto: When n increases, so 1 over X, or X over N equals X is positive, decreases.
22:06:350Paolo Guiotto: What happens to cosine when the argument decreases?
22:10:670Paolo Guiotto: Appreciate it.
22:11:770Paolo Guiotto: in the bags.
22:14:60Paolo Guiotto: Because society, no? If X is, one…
22:19:950Paolo Guiotto: And now you take X over 1, X over 2, so 1 million drop, who knows what happens to these values. So, I have here…
22:30:600Paolo Guiotto: So when X is more, I definitely do this for X fits the, and then bigger, this is going down to zero around 0. So cosine will be increasing, and this will be the…
22:44:610Paolo Guiotto: And so, the gas decreasing, and this is another problem, because increasing that, decreasing the…
22:50:950Paolo Guiotto: It's not clear who we…
22:53:270Paolo Guiotto: So, even if it is possible to say that X has a 1 point to the X, I have a problem then determining if it is increasing by liquidity. Moreover, it is not a total monotonic, because when X is bigger, this has made X over X will be big, so cosine goes up and down, and that's why the noise is parenthesis has one point in the other.
23:13:960Paolo Guiotto: So I hope this guy… and let's see what happens if we… Attack this with,
23:27:530Paolo Guiotto: with the dominated convergence, okay? So, we apply…
23:37:100Paolo Guiotto: dominated.
23:38:830Paolo Guiotto: conversions.
23:40:320Paolo Guiotto: So… To do that, we have to verify… the two conditions.
23:46:490Paolo Guiotto: Number one.