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Title: Riemann integral vs. Lebesgue integral
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YouTube-Title: Riemann integral vs. Lebesgue integral
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Subtitle on GitHub: rvsl01_sub_eng.srt
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Timestamps
0:00 Introduction
0:30 Riemann integral
2:00 Problems of Riemann integral
7:50 Riemann integral definition
9:13 Lebesgue integral - idea
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Subtitle in English
1 00:00:00,000 –> 00:00:04,500 Hello and welcome to Riemann integral vs. Lebesgue integral.
2 00:00:04,600 –> 00:00:09,842 Often students ask me why we have two integral notions at all.
3 00:00:09,942 –> 00:00:15,482 Or why we need the Lebesgue integral when we already have the Riemann integral.
4 00:00:15,582 –> 00:00:22,500 In this video, I explain why we go from the classical integral definition, which is the Riemann integral,
5 00:00:22,600 –> 00:00:26,100 to the more modern one, which is the Lebesgue integral.
6 00:00:26,200 –> 00:00:29,700 However, let us first start with the Riemann integral.
7 00:00:29,800 –> 00:00:36,300 Here, we want to integrate a function f that starts from R and goes into R.
8 00:00:36,400 –> 00:00:43,900 So a normal function defined on the real line and with values in the real line.
9 00:00:44,000 –> 00:00:50,500 And as you should know, this integral is related to the calculation of an area below a graph.
10 00:00:50,600 –> 00:00:56,100 So let me do a rough drawing of the graph of the function f here.
11 00:00:56,200 –> 00:01:03,700 And consider an interval on the real line. So maybe start with a and go to b.
12 00:01:03,800 –> 00:01:14,500 Then we are talking about this area below the graph. So exactly this one.
13 00:01:14,600 –> 00:01:25,100 In short, this means the Riemann integral of the function f, from the interval [a,b], is related to this area here.
14 00:01:25,200 –> 00:01:33,700 Therefore, we have the well-known idea to choose a partition of the x-axis and to approximate this area by choosing rectangles.
15 00:01:33,800 –> 00:01:39,300 And this approximation is known as the lower sum or the upper sum.
16 00:01:39,400 –> 00:01:44,900 So what I draw here is the well-known lower sum.
17 00:01:45,000 –> 00:01:48,500 And if we now choose finer and finer partitions,
18 00:01:48,600 –> 00:01:54,800 we call this limit of the lower or upper sums the Riemann integral if the limit is well-defined.
19 00:01:54,900 –> 00:01:58,400 Well, this was a short summary of the Riemann integral.
20 00:01:58,500 –> 00:02:02,000 And now we immediately find some problems here.
21 00:02:02,100 –> 00:02:07,600 These problems give us a reason to define another integral notion.
22 00:02:07,700 –> 00:02:15,200 Firstly, it is difficult to expand the Riemann integral to higher dimensions.
23 00:02:15,900 –> 00:02:23,400 Yes, you can expand to higher dimensions. So it is possible but very laborious.
24 00:02:23,500 –> 00:02:25,600 For example, if we would have a 2 here.
25 00:02:26,700 –> 00:02:34,200 This means that the x-axis here is now a two-dimensional space. So we can’t use rectangles for the partition.
26 00:02:35,300 –> 00:02:37,100 Of course, we now could use cuboids.
27 00:02:37,200 –> 00:02:56,200 So this is the two-dimensional space in the domain of definition R^2; then our partition is now just rectangles in the plane. And our integral approximation now is given by cuboids. So maybe in this way.
28 00:02:56,600 –> 00:03:07,200 As you can see, it gets very technical and it will be much more work with additional dimension. Just one more dimension but you have to much more work.
29 00:03:07,300 –> 00:03:11,500 For example, to build the limit process in the end.
30 00:03:11,600 –> 00:03:16,400 This is directly clear if we consider the domain of definition here.
31 00:03:16,500 –> 00:03:25,400 In the one-dimensional case, we only had an interval, which is an easy thing since it starts with a and ends with b. Just an interval.
32 00:03:25,500 –> 00:03:33,700 But even if we add even one dimension more. So here is the picture, x1, x2, f(x1,x2).
33 00:03:33,800 –> 00:03:38,300 So even if we even add just one dimension, there are a lot of more possibilities.
34 00:03:38,400 –> 00:03:44,000 For example, the domain of definition for the function could be a circle in this plane.
35 00:03:44,100 –> 00:03:47,600 So let us focus on the domain of definition now.
36 00:03:47,700 –> 00:04:00,200 So I draw another picture, just the x1, x2 axis; and then we find the circle just there. Now, what we now want to do is to choose a partition of the circle and put the cuboids above.
37 00:04:00,300 –> 00:04:17,400 However, fo course, this is not exactly possible. So we can put rectangles inside the circle here. But as you can see: we even need a limit process to approximate the circle in the domain.
38 00:04:17,500 –> 00:04:23,000 Again I say it: it is feasible but it’s a lot more technical work.
39 00:04:23,100 –> 00:04:29,600 In summary, this is one problem of the Riemann integral, we directly understand.
40 00:04:29,700 –> 00:04:33,200 It is much more work to do this in higher dimensions.
41 00:04:33,300 –> 00:04:38,700 The second problem of the Riemann integral, and what you should know from this one dimensional case,
42 00:04:38,800 –> 00:04:44,350 It is that we need some continuity property for the functions we want to integrate
43 00:04:44,450 –> 00:04:45,950 in other words:
44 00:04:46,050 –> 00:04:49,050 we have some dependence on continuity.
45 00:04:49,150 –> 00:04:57,650 Recall that in the best case, the function we want to integrate should be continuous. Then we don’t have any problems.
46 00:04:57,750 –> 00:05:05,350 However, if have discontinuity points, then there should be only finitely many of them.
47 00:05:05,450 –> 00:05:10,050 But if we have infinitely many, it can destroy the integrability of the function.
48 00:05:10,150 –> 00:05:12,650 I hope you know the typical example.
49 00:05:12,750 –> 00:05:18,250 So an easy function that you can’t integrate using the Riemann integral.
50 00:05:18,350 –> 00:05:23,450 Hence, the dependence here is, indeed, a disadvantage of the Riemann integral.
51 00:05:23,550 –> 00:05:31,250 Now the most important disadvantage of the Riemann integral is the relationship to limit processes.
52 00:05:31,350 –> 00:05:44,850 The question here is in which situations is it allowed to interchange some limit processes? Or in other words, can I pull in the limit sign inside the integral sign.
53 00:05:44,950 –> 00:05:48,050 So let me show this in an example.
54 00:05:48,150 –> 00:05:50,150 So we have the limit n to infinity
55 00:05:50,250 –> 00:05:55,750 and we integrate a function from a to b (so this is sequence of functions f_n)
56 00:05:55,850 –> 00:06:00,050 and we integrate them with respect to x.
57 00:06:00,150 –> 00:06:08,850 The question is then: when is allowed to pull or push the limit inside the integral.
58 00:06:08,950 –> 00:06:12,850 When does this equality hold?
59 00:06:12,950 –> 00:06:17,450 For the Riemann integral, we don’t have many possibilities here.
60 00:06:17,550 –> 00:06:23,050 We also need the uniform convergence of this sequence of functions.
61 00:06:23,150 –> 00:06:28,650 only then we know that is allowed to pull the limit inside the integral.
62 00:06:28,750 –> 00:06:35,250 However, this uniform convergence is a very strong notion,
63 00:06:35,350 –> 00:06:36,850 and in some sense, it’s related to the continuity, again.
64 00:06:36,950 –> 00:06:40,450 And therefore, we want to weaken this notion.
65 00:06:40,550 –> 00:06:47,050 Indeed, we have a lot of examples where there is no uniform convergence of the sequences of functions at all,
66 00:06:47,150 –> 00:06:50,250 and we still have the same result on both sides.
67 00:06:50,350 –> 00:06:56,550 Hence we know it should be possible to generalise this convergence theorem.
68 00:06:57,250 –> 00:07:00,150 However, not by using the Riemann integral.
69 00:07:00,250 –> 00:07:06,750 The Riemann integral is just not flexible enough to prove such a convergence theorem.
70 00:07:06,850 –> 00:07:14,350 Well these are three important points that show the difficulties we may have when we use the Riemann integral.
71 00:07:14,450 –> 00:07:26,950 Most importantly, we have a wish to the generalisation to higher dimensions. So we want an integral notion that is not restricted to one dimension.
72 00:07:27,050 –> 00:07:34,550 So we don’t want to do all the work again if we just change the space of the domain of definition here.
73 00:07:34,650 –> 00:07:39,150 We want an integral that works in every dimension the same.
74 00:07:39,250 –> 00:07:42,750 In order to understand this generalisation,
75 00:07:42,850 –> 00:07:45,350 we need more technical details.
76 00:07:45,450 –> 00:07:49,950 Hence, I would suggest, we start with the Riemann integral.
77 00:07:50,050 –> 00:07:58,550 Again, we do some sketch (so we have the x- and the y-axis). We also have a function, where I draw the graph in green.
78 00:07:58,650 –> 00:08:05,050 So we have a function with values in the real numbers.
79 00:08:05,150 –> 00:08:14,550 For the Riemann integral we now choose a partition of the x-axis as before; and now we can calculate the upper and lower sum.
80 00:08:14,650 –> 00:08:16,150 Here I have drawn the lower sum.
81 00:08:16,250 –> 00:08:23,450 This means that we take the infimum in each subinterval for the height of the rectangles.
82 00:08:23,550 –> 00:08:29,050 And for the upper sum, we will take the supremum.
83 00:08:29,150 –> 00:08:32,650 So here, for example, this rectangle. Hence we get the upper sum.
84 00:08:32,750 –> 00:08:42,050 If the difference between the upper and lower sum can be made arbitrarily small, for a finer partition,
85 00:08:42,150 –> 00:08:45,650 then we can define the Riemann integral.
86 00:08:45,750 –> 00:08:51,250 Then you can define it as the supremum of the lower sums or the infimum of the upper sums.
87 00:08:51,350 –> 00:08:58,550 This is then the same value and this value is then called the Riemann integral for the function f in the interval [a,b].
88 00:08:58,650 –> 00:09:07,150 In order to this, it was necessary that we have a function that goes from the real numbers into the real numbers.
89 00:09:07,250 –> 00:09:12,050 Otherwise, we can’t do the partition of the x-axis, and we don’t have a height.
90 00:09:12,150 –> 00:09:16,650 The idea of the Lebesgue integral is now different.
91 00:09:16,750 –> 00:09:18,250
92 00:09:18,350 –> 00:09:23,850 As you have seen before, it was very restrictive to find a partition of the domain of definition.
93 00:09:23,950 –> 00:09:27,050 (if it is not R itself)
94 00:09:27,150 –> 00:09:29,650 If, for example, we have a three-dimensional space,
95 00:09:29,750 –> 00:09:33,450 then we can’t find a partition easily.
96 00:09:33,550 –> 00:09:37,050 But we map still in the real numbers.
97 00:09:37,150 –> 00:09:41,150 So the right-hand side is still the same since we consider functions,
98 00:09:41,250 –> 00:09:43,750 so functions with values in the real numbers.
99 00:09:43,850 –> 00:09:47,850 So this is the same, but the left-hand side not the same.
100 00:09:47,950 –> 00:09:51,450 This means that the domain of definition, the left-hand side here,
101 00:09:51,550 –> 00:10:00,050 can be very high-, or even very abstract, and therefore, we don’t know what a meaningful partition of the left-hand side should be.
102 00:10:00,150 –> 00:10:06,650 Just because we don’t have these simple intervals in a higher dimensional or abstract space.
103 00:10:06,650 –> 00:10:13,250 However, on the right-hand side, we still find the real numbers, the one-dimensional space of the real numbers.
104 00:10:13,250 –> 00:10:21,350 So instead of finding for a partition of the x-axis; it would be easier to just find a partition of the y-axis.
105 00:10:21,450 –> 00:10:25,550 And indeed, this is what the Lebesgue integral does.
106 00:10:25,650 –> 00:10:36,880 Although, this is not the way, the Lebesgue measure is normally introduced. However it is still a very good idea to visualise the Lebesgue integral.
107 00:10:36,950 –> 00:10:40,450 So let us draw the graph again.
108 00:10:40,550 –> 00:10:52,050 And keep in mind, the x-axis may be high-dimensional or even abstract, but the y-axis is still the real line.
109 00:10:52,150 –> 00:10:59,650 And that is what we decompose now. Since this is only the real line, we can find the same interval partition as before.
110 00:10:59,750 –> 00:11:03,250 So we choose some values and define some intervals here.
111 00:11:03,350 –> 00:11:09,550 So this is a partition of the real line as we did before on the Riemann integral on the x-axis.
112 00:11:09,750 –> 00:11:19,250 So maybe we call these partition elements C_i, while in the Riemann integral we usually call them x_i.
113 00:11:19,350 –> 00:11:27,350 Maybe as a reminder: In the Riemann integral we defined an upper sum of some partition of the x-axis.
114 00:11:27,450 –> 00:11:35,950 So you would write U(P) and this is the sum of the rectangles,
115 00:11:36,050 –> 00:11:50,050 which means we have the sum and then choose the supremum inside the interval times the length Delta x_i.
116 00:11:50,150 –> 00:12:02,450 And you can do the same for the lower sum L(P), given by infimum in the given interval
117 00:12:07,550 –> 00:12:15,050 times the length of the rectangles/intervals Delta x_i.
118 00:12:15,150 –> 00:12:17,650 In this way, we defined the Riemann integral.
119 00:12:17,750 –> 00:12:23,250 However, now we want to decompose the y-axis, for getting the Lebesgue integral.
120 00:12:23,350 –> 00:12:29,850 We have the following picture. For example, this line here …
121 00:12:29,950 –> 00:12:44,450 … this value gives you this line from here and we have this line here.
122 00:12:44,550 –> 00:12:50,550 Now, we want to find all the parts of the function that lie between these two lines.
123 00:12:50,650 –> 00:12:55,150 So meaning: inside this interval in the real line.
124 00:12:55,250 –> 00:13:04,750 This means that we don’t draw rectangles like before, but we look at which parts of the function fall between these two lines.
125 00:13:04,850 –> 00:13:17,350 So we find here that we have a part on the left-hand side, so we find all the arguments in x, that lie in this set,
126 00:13:17,450 –> 00:13:23,950 are sent to values inside this interval by the function f.
127 00:13:24,050 –> 00:13:25,550 And we also have here a part of the function.
128 00:13:25,650 –> 00:13:34,150 And we find all the arguments of x that are sent
129 00:13:34,250 –> 00:13:36,750 to the interval.
130 00:13:37,850 –> 00:13:44,350 Now these two parts are associated to the decomposition/partition of the y-axis, c_i, we chose.
131 00:13:45,450 –> 00:13:50,950 This is similar to the idea of the Riemann integral,
132 00:13:51,050 –> 00:13:53,150 but we immediately see
133 00:13:53,250 –> 00:13:59,550 that the decomposition of the x-axis below here is not connected.
134 00:13:59,650 –> 00:14:03,150 Then we have different sets that are not connected like the intervals that we saw before.
135 00:14:03,250 –> 00:14:08,750 And this is the whole idea. Instead of choosing a fixed partition of the x-axis,
136 00:14:08,850 –> 00:14:13,650 we now choose a partition of the y-axis, of the values of the function.
137 00:14:13,750 –> 00:14:17,050 And then we get a useful partition of the x-axis
138 00:14:17,150 –> 00:14:18,650 that we can work with.
139 00:14:19,750 –> 00:14:26,250 Now the next question could be how to measure these lengths on the x-axis below.
140 00:14:26,350 –> 00:14:36,850 Therefore on the domain of definition which is here R^3, (or it could be some space)
141 00:14:36,950 –> 00:14:42,450 We need a so-called measure. We want to measure volumes in the space.
142 00:14:42,550 –> 00:14:47,750 This means that we want to measure the size of sets inside this space.
143 00:14:47,850 –> 00:14:58,350 For example, if this space is R, we measure lengths. In R^2, we measure areas. In R^3, we measure volumes of sets.
144 00:14:58,450 –> 00:15:02,950 But, in an abstract sense, we will always speak of VOLUMES.
145 00:15:03,050 –> 00:15:13,550 Hence, on the x-axis here we could choose a general measure space. We just have to know how to measure the size of sets,
146 00:15:13,650 –> 00:15:15,150 the volumes of sets!
147 00:15:15,250 –> 00:15:18,750 Such a measure space is usually denoted by Ω.
148 00:15:18,850 –> 00:15:20,350
149 00:15:20,450 –> 00:15:27,950 This means that in the end we will have an integral that works on all functions that are defined on some measure space Ω.
150 00:15:28,350 –> 00:15:31,350 Indeed, this is all we need to define the integral.
151 00:15:31,450 –> 00:15:36,950 Now you see how we could define an “upper sum” in some sense.
152 00:15:37,050 –> 00:15:39,550 So we take these c_i here,
153 00:15:39,650 –> 00:15:56,150 and now I want the c_i to be this one. So the c_i is this stripe here. So this would be an upper sum of the integral if we know the measure of these sets together.
154 00:15:56,250 –> 00:16:04,750 Of course, these blue sets are the preimage of this interval under the function f.
155 00:16:04,850 –> 00:16:06,900 Maybe we just call it A.
156 00:16:07,000 –> 00:16:13,000 Now if I call the measure we have here in our measure space just by μ,
157 00:16:13,100 –> 00:16:17,600 then we calculate this generalised rectangle.
158 00:16:17,700 –> 00:16:35,200 It is c_i (the height) times the length on the bottom. This is the measure of our set A, μ(A). Now we have one part of the whole integral and now we can do it for the whole partition.
159 00:16:35,300 –> 00:16:45,300 So we have a finite sum over i. c_i times μ(Ai) Of course, I should also put an index i on the preimage. Then we have a sum.
160 00:16:45,400 –> 00:16:50,900 Since we chose a finite partition on the y-axis, we have finite sum here.
161 00:16:50,900 –> 00:16:56,000 and it will be the upper or lower sum depending which
162 00:16:56,100 –> 00:16:57,000 c_i we chose here.
163 00:16:57,300 –> 00:17:02,200 Then again, with some limit process with finer partitions
164 00:17:02,300 –> 00:17:04,800 we will get the Lebesgue integral.
165 00:17:04,900 –> 00:17:08,400 Usually denoted by integral f dμ.
166 00:17:08,500 –> 00:17:13,000 Since it is now depending on the measure we chose on the x-axis.
167 00:17:13,100 –> 00:17:21,600 This construction now works with arbitrary measure spaces, since the x-axis does not need a partition any more.
168 00:17:24,700 –> 00:17:31,200 We just need a partition of the axis that is R anyway. However, in this case, we really need the notion of a measure,
169 00:17:31,300 –> 00:17:33,800 even in one dimension.
170 00:17:33,900 –> 00:17:36,400 So here, in R, we know what the length of a subinterval is,
171 00:17:36,500 –> 00:17:40,000 it is simply the difference between the right side and the left side.
172 00:17:40,100 –> 00:17:42,200 But here we need more.
173 00:17:42,300 –> 00:17:51,800 Or in other words: In the Riemann integral we only measured the lengths of intervals but in the Lebesgue integral we can have any fragmentation
174 00:17:51,900 –> 00:17:54,900 and we have to know the measure of this fragmentation.
175 00:17:55,000 –> 00:17:59,500 But if we know how to measure any set in the real line,
176 00:17:59,600 –> 00:18:02,100 then we have a very robust integral definition.
177 00:18:02,200 –> 00:18:07,700 Well this generalizes the Riemann integral and eliminates
178 00:18:07,800 –> 00:18:11,000 all problems we had in higher dimensions.
179 00:18:11,100 –> 00:18:19,600 Since this measure theory works for any dimension you want. We just have to know how to measure volumes.
180 00:18:19,700 –> 00:18:30,200 We have also eliminated the problem of dependency on continuity since we can now measure the size of the set of points where there is discontinuity.
181 00:18:30,300 –> 00:18:40,800 For example, if we had a set with infinite points of discontinuity, but the measure of this whole set was 0,
182 00:18:40,900 –> 00:18:44,000 we don’t have a problem in the definition of the integral at all.
183 00:18:44,100 –> 00:18:49,100 And also for the last point, so these limit processes,
184 00:18:49,200 –> 00:18:51,100 we have good answers for.
185 00:18:51,200 –> 00:18:58,700 We will see that we have very good limit theorems that describe these limit processes for the Lebesgue integral.
186 00:18:58,800 –> 00:19:05,300 Even strong results for the one-dimensional case but also in the general case of measure spaces.
187 00:19:05,400 –> 00:19:12,900 Okay, to sum it up, the Riemann integral is the classical notion, it may be easier to understand at a beginner level,
188 00:19:13,000 –> 00:19:14,500 but in the end, everyone wants to work with the Lebesgue integral.
189 00:19:14,600 –> 00:19:16,100 Thank you very much and see you next time.
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