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Title: Proof of Lebesgue’s Dominated Convergence Theorem
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Series: Measure Theory
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YouTube-Title: Measure Theory 11 | Proof of Lebesgue’s Dominated Convergence Theorem
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Bright video: https://youtu.be/fCUj5WLiRCs
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Subtitle on GitHub: mt11_sub_eng.srt
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Subtitle in English
1 00:00:00,829 –> 00:00:02,490 Hello and welcome
2 00:00:02,500 –> 00:00:03,119 back.
3 00:00:03,130 –> 00:00:05,050 And as always, I thank
4 00:00:05,059 –> 00:00:06,690 all the nice people that
5 00:00:06,699 –> 00:00:08,310 support this channel on
6 00:00:08,319 –> 00:00:10,170 Steady today,
7 00:00:10,180 –> 00:00:11,649 we consider part
8 00:00:11,659 –> 00:00:13,390 11 in our measure
9 00:00:13,399 –> 00:00:14,649 theory series.
10 00:00:15,489 –> 00:00:17,149 And as promised, it will
11 00:00:17,159 –> 00:00:18,889 be about the proof
12 00:00:18,940 –> 00:00:20,750 of Lebesgue’s dominated
13 00:00:20,760 –> 00:00:22,110 convergence theorem.
14 00:00:23,159 –> 00:00:24,940 I already told you this is
15 00:00:24,950 –> 00:00:26,899 one of my favourite theorems.
16 00:00:26,909 –> 00:00:28,299 And you will see that the
17 00:00:28,309 –> 00:00:29,860 proof is indeed not so
18 00:00:29,870 –> 00:00:30,459 hard.
19 00:00:31,069 –> 00:00:31,329 OK.
20 00:00:31,340 –> 00:00:32,709 Then let’s recall the
21 00:00:32,720 –> 00:00:34,580 theorem from the last part
22 00:00:34,590 –> 00:00:35,580 of the series.
23 00:00:36,919 –> 00:00:38,279 Here you see all the
24 00:00:38,290 –> 00:00:39,919 assumptions we need for
25 00:00:39,979 –> 00:00:41,470 Lebesgue’s dominated convergence
26 00:00:41,479 –> 00:00:42,029 theorem.
27 00:00:42,509 –> 00:00:43,959 On the one hand, we have
28 00:00:43,970 –> 00:00:45,849 a sequence of measurable
29 00:00:45,860 –> 00:00:47,619 maps and
30 00:00:47,630 –> 00:00:49,150 also the pointwise limit
31 00:00:49,159 –> 00:00:50,639 function which we will call
32 00:00:50,650 –> 00:00:51,729 just F.
33 00:00:52,529 –> 00:00:53,950 And on the other hand, we
34 00:00:53,959 –> 00:00:55,909 have another function G
35 00:00:55,919 –> 00:00:57,669 which is integrable.
36 00:00:58,689 –> 00:01:00,430 And this function G is
37 00:01:00,439 –> 00:01:01,909 indeed the
38 00:01:01,919 –> 00:01:03,619 important ingredient in the
39 00:01:03,630 –> 00:01:04,510 whole theorem.
40 00:01:05,290 –> 00:01:06,309 It should lie
41 00:01:06,319 –> 00:01:07,900 above all the
42 00:01:07,910 –> 00:01:09,029 functions FN.
43 00:01:09,190 –> 00:01:10,589 And therefore we call it
44 00:01:10,680 –> 00:01:12,430 an integrable majorant.
45 00:01:13,300 –> 00:01:14,500 From these nice
46 00:01:14,510 –> 00:01:15,339 assumptions.
47 00:01:15,470 –> 00:01:17,160 We now can conclude
48 00:01:17,430 –> 00:01:19,080 that all the functions in
49 00:01:19,089 –> 00:01:20,800 the sequence are also
50 00:01:20,940 –> 00:01:21,750 integrable.
51 00:01:22,620 –> 00:01:24,580 And moreover, also the
52 00:01:24,589 –> 00:01:26,419 mu almost everywhere point-
53 00:01:26,430 –> 00:01:28,290 wise limit function F is
54 00:01:28,300 –> 00:01:29,510 also integrable.
55 00:01:30,449 –> 00:01:31,760 From this, we can conclude
56 00:01:31,769 –> 00:01:33,279 that all these integrals
57 00:01:33,290 –> 00:01:34,360 here make sense.
58 00:01:35,129 –> 00:01:37,069 And the equality tells us
59 00:01:37,080 –> 00:01:38,660 that we can pull in the
60 00:01:38,669 –> 00:01:40,389 limit into the integral.
61 00:01:41,190 –> 00:01:42,459 And that’s the reason we
62 00:01:42,470 –> 00:01:44,449 call it a convergence theorem.
63 00:01:45,680 –> 00:01:47,489 Well, then let’s start
64 00:01:47,500 –> 00:01:48,339 with the proof
65 00:01:49,419 –> 00:01:51,019 the ingredients we need for
66 00:01:51,029 –> 00:01:52,559 the proof are on the one
67 00:01:52,569 –> 00:01:54,519 side properties of the integral.
68 00:01:54,639 –> 00:01:56,519 And also on the other hand,
69 00:01:56,750 –> 00:01:57,279 Fatou’s lemma
70 00:01:58,660 –> 00:01:59,760 from the properties of the
71 00:01:59,769 –> 00:02:01,550 interval, we can immediately
72 00:02:01,559 –> 00:02:03,430 show the first property here.
73 00:02:04,360 –> 00:02:06,190 Please recall that lying
74 00:02:06,199 –> 00:02:08,020 in L1 means that the
75 00:02:08,029 –> 00:02:09,820 absolute value of
76 00:02:09,830 –> 00:02:11,559 FN has a finite
77 00:02:11,570 –> 00:02:12,300 integral.
78 00:02:13,220 –> 00:02:14,779 Please remember we have
79 00:02:14,789 –> 00:02:16,660 measurable maps and this
80 00:02:16,669 –> 00:02:18,369 is a non-negative function.
81 00:02:18,380 –> 00:02:20,160 Therefore, the integral always
82 00:02:20,169 –> 00:02:20,850 exists.
83 00:02:20,860 –> 00:02:22,139 But in the worst case, it
84 00:02:22,149 –> 00:02:23,399 could be infinity.
85 00:02:24,320 –> 00:02:26,020 Therefore lying in L1
86 00:02:26,029 –> 00:02:27,440 means this is not
87 00:02:27,449 –> 00:02:28,220 infinity.
88 00:02:28,229 –> 00:02:29,759 So smaller than
89 00:02:29,770 –> 00:02:30,419 infinity.
90 00:02:31,460 –> 00:02:31,759 OK.
91 00:02:31,770 –> 00:02:33,550 Now we can use our assumption
92 00:02:33,559 –> 00:02:35,240 we know that we have an integrable
93 00:02:35,990 –> 00:02:36,820 majorant
94 00:02:37,789 –> 00:02:38,779 called g.
95 00:02:40,320 –> 00:02:41,789 So we have an inequality
96 00:02:41,800 –> 00:02:42,309 here.
97 00:02:42,320 –> 00:02:44,210 And you remember we have
98 00:02:44,220 –> 00:02:46,050 a monotonicity property of
99 00:02:46,059 –> 00:02:46,820 the integral.
100 00:02:47,710 –> 00:02:49,550 This means we also have the
101 00:02:49,559 –> 00:02:51,240 inequality in the
102 00:02:51,250 –> 00:02:52,559 integral sense.
103 00:02:52,570 –> 00:02:53,910 So let’s put that to the
104 00:02:53,919 –> 00:02:54,440 left.
105 00:02:55,419 –> 00:02:57,190 And then I know this is less
106 00:02:57,199 –> 00:02:58,720 or equal than the
107 00:02:58,729 –> 00:03:00,350 integral of G.
108 00:03:02,419 –> 00:03:04,059 Now note that this right
109 00:03:04,070 –> 00:03:05,520 hand side is indeed our
110 00:03:05,529 –> 00:03:07,320 assumption that the integral
111 00:03:07,330 –> 00:03:08,759 of G is indeed
112 00:03:08,770 –> 00:03:09,600 finite.
113 00:03:10,729 –> 00:03:12,690 Our conclusion is now this
114 00:03:12,699 –> 00:03:14,199 integral is also
115 00:03:14,210 –> 00:03:14,990 finite.
116 00:03:15,869 –> 00:03:17,550 And this means all the
117 00:03:17,559 –> 00:03:19,330 FN lie in
118 00:03:19,339 –> 00:03:19,729 L one.
119 00:03:21,119 –> 00:03:22,589 Obviously we can do the
120 00:03:22,600 –> 00:03:24,389 same for F
121 00:03:24,399 –> 00:03:25,850 instead of FN
122 00:03:26,059 –> 00:03:27,380 because we know it’s a point
123 00:03:27,389 –> 00:03:28,979 where it’s limited almost
124 00:03:28,990 –> 00:03:29,740 everywhere.
125 00:03:30,550 –> 00:03:32,440 Therefore, this inequality
126 00:03:32,449 –> 00:03:33,850 holds almost
127 00:03:33,860 –> 00:03:34,720 everywhere.
128 00:03:36,059 –> 00:03:37,720 And of course, also almost
129 00:03:37,729 –> 00:03:39,419 everywhere, the monotonicity
130 00:03:39,429 –> 00:03:41,320 property still holds, which
131 00:03:41,330 –> 00:03:42,889 means we get the same
132 00:03:42,899 –> 00:03:44,779 inequality here for the integral
133 00:03:44,789 –> 00:03:45,259 of F.
134 00:03:46,229 –> 00:03:48,000 In other words, we also
135 00:03:48,009 –> 00:03:49,669 have F in L
136 00:03:49,679 –> 00:03:50,149 one.
137 00:03:51,270 –> 00:03:51,619 OK.
138 00:03:51,630 –> 00:03:52,410 Very good.
139 00:03:52,419 –> 00:03:53,940 So what you have seen now
140 00:03:53,949 –> 00:03:55,610 is that the first part in
141 00:03:55,619 –> 00:03:57,449 the theorem was very easy to
142 00:03:57,460 –> 00:03:57,919 show.
143 00:03:58,880 –> 00:04:00,490 Hence, the crucial thing
144 00:04:00,500 –> 00:04:01,699 in the theorem is
145 00:04:01,710 –> 00:04:03,100 indeed this
146 00:04:03,110 –> 00:04:04,699 convergence statement here.
147 00:04:06,070 –> 00:04:06,490 OK.
148 00:04:06,500 –> 00:04:07,929 So this is what we do in
149 00:04:07,940 –> 00:04:09,449 the next five minutes.
150 00:04:09,460 –> 00:04:10,779 And indeed, I want to show
151 00:04:10,789 –> 00:04:12,690 something a little bit
152 00:04:12,699 –> 00:04:14,050 more stronger.
153 00:04:15,050 –> 00:04:16,250 We will show that the
154 00:04:16,260 –> 00:04:17,928 integral of the absolute
155 00:04:17,940 –> 00:04:19,230 value of
156 00:04:19,238 –> 00:04:21,010 FN minus the
157 00:04:21,019 –> 00:04:22,720 pointwise limit F
158 00:04:22,730 –> 00:04:24,299 goes to zero
159 00:04:24,320 –> 00:04:25,959 if N goes to
160 00:04:25,970 –> 00:04:26,750 infinity.
161 00:04:28,609 –> 00:04:30,350 And from this, we can immediately
162 00:04:30,359 –> 00:04:32,149 conclude the property we
163 00:04:32,160 –> 00:04:32,709 want here.
164 00:04:34,049 –> 00:04:35,679 However, let us start
165 00:04:35,690 –> 00:04:37,519 showing this property here.
166 00:04:38,989 –> 00:04:39,350 OK.
167 00:04:39,359 –> 00:04:40,760 In the integral, we have
168 00:04:40,769 –> 00:04:42,290 the function FN
169 00:04:42,299 –> 00:04:43,649 minus F.
170 00:04:43,660 –> 00:04:45,309 But in the absolute
171 00:04:45,320 –> 00:04:47,140 value now we
172 00:04:47,149 –> 00:04:49,040 know that for the absolute
173 00:04:49,049 –> 00:04:50,929 volume going a detour
174 00:04:50,950 –> 00:04:52,779 makes it greater or in the
175 00:04:52,790 –> 00:04:54,299 best case, it stays the same
176 00:04:55,250 –> 00:04:56,709 now going the detour to
177 00:04:56,720 –> 00:04:57,269 0.
178 00:04:57,279 –> 00:04:58,640 This means we have the absolute
179 00:04:58,649 –> 00:05:00,549 value of FN plus
180 00:05:00,559 –> 00:05:01,989 the absolute value of
181 00:05:02,000 –> 00:05:02,589 F.
182 00:05:03,549 –> 00:05:04,980 This is just the
183 00:05:04,989 –> 00:05:06,809 triangle inequality for the
184 00:05:06,820 –> 00:05:07,829 absolute value.
185 00:05:07,839 –> 00:05:09,279 When we read it point-
186 00:05:09,290 –> 00:05:11,130 wisely, this means
187 00:05:11,140 –> 00:05:12,829 we could put in x’s
188 00:05:12,839 –> 00:05:14,350 here for the functions,
189 00:05:15,190 –> 00:05:16,640 but it holds for all the
190 00:05:16,649 –> 00:05:17,029 x’s.
191 00:05:17,040 –> 00:05:18,489 Therefore, this is just the
192 00:05:18,500 –> 00:05:19,880 short notation we use
193 00:05:20,850 –> 00:05:22,220 also what we know is that
194 00:05:22,230 –> 00:05:23,869 we have our majorant
195 00:05:23,880 –> 00:05:25,660 G for FN.
196 00:05:25,670 –> 00:05:27,609 And also for F, as I told
197 00:05:27,619 –> 00:05:29,470 you, therefore,
198 00:05:29,480 –> 00:05:31,339 we have this as less
199 00:05:31,350 –> 00:05:33,019 or equal than
200 00:05:33,049 –> 00:05:33,950 two G,
201 00:05:34,709 –> 00:05:36,010 one G plus one G.
202 00:05:37,200 –> 00:05:38,750 Now you could say this
203 00:05:38,760 –> 00:05:40,250 inequality only
204 00:05:40,260 –> 00:05:42,179 holds mu almost
205 00:05:42,190 –> 00:05:42,899 everywhere.
206 00:05:43,640 –> 00:05:45,049 However, it does not matter
207 00:05:45,059 –> 00:05:46,779 at all because the integral
208 00:05:46,790 –> 00:05:48,600 does not see changes that
209 00:05:48,609 –> 00:05:50,059 happen almost
210 00:05:50,359 –> 00:05:51,540 nowhere,
211 00:05:52,209 –> 00:05:53,820 which is the complement of
212 00:05:53,829 –> 00:05:54,940 almost everywhere.
213 00:05:55,660 –> 00:05:57,040 Therefore, we could change
214 00:05:57,049 –> 00:05:58,510 or choose another function
215 00:05:58,519 –> 00:05:59,820 G where this
216 00:05:59,829 –> 00:06:01,579 inequality here does
217 00:06:01,589 –> 00:06:03,269 indeed hold everywhere.
218 00:06:04,359 –> 00:06:05,660 Hence, we can assume that
219 00:06:05,670 –> 00:06:06,739 we do this here.
220 00:06:06,920 –> 00:06:08,720 And therefore, I can omit
221 00:06:08,730 –> 00:06:10,100 the mu almost everywhere
222 00:06:10,170 –> 00:06:11,369 and it makes the proof just
223 00:06:11,380 –> 00:06:11,899 shorter.
224 00:06:13,029 –> 00:06:13,500 OK.
225 00:06:13,510 –> 00:06:15,220 Now I can bring this on the
226 00:06:15,230 –> 00:06:16,769 other side and I get
227 00:06:16,779 –> 00:06:18,429 out a non negative
228 00:06:18,440 –> 00:06:19,929 function I want to call
229 00:06:20,019 –> 00:06:21,049 h_n.
230 00:06:21,089 –> 00:06:23,049 So this would be two G
231 00:06:23,760 –> 00:06:25,359 minus our absolute
232 00:06:25,369 –> 00:06:27,149 value FN minus
233 00:06:27,230 –> 00:06:28,000 F.
234 00:06:28,010 –> 00:06:29,489 And we know it’s non-
235 00:06:29,510 –> 00:06:30,269 negative.
236 00:06:31,670 –> 00:06:33,309 Obviously, this holds for
237 00:06:33,320 –> 00:06:35,209 all N and we know
238 00:06:35,220 –> 00:06:37,130 by the properties of
239 00:06:37,140 –> 00:06:39,019 measurable functions that
240 00:06:39,059 –> 00:06:40,809 h_n is
241 00:06:40,820 –> 00:06:42,320 also measurable.
242 00:06:43,720 –> 00:06:44,779 Now, I have written that
243 00:06:44,790 –> 00:06:46,739 down in such a way that you
244 00:06:46,750 –> 00:06:47,890 should recognize
245 00:06:47,899 –> 00:06:49,410 immediately Fatou’s
246 00:06:49,420 –> 00:06:51,089 Lemma simply
247 00:06:51,100 –> 00:06:52,420 because we have
248 00:06:52,429 –> 00:06:54,160 measurable functions
249 00:06:54,709 –> 00:06:56,250 and they are all non-
250 00:06:56,320 –> 00:06:57,079 negative.
251 00:06:58,299 –> 00:06:59,920 Therefore, now we can
252 00:06:59,929 –> 00:07:01,630 apply Fatou’s Lemma.
253 00:07:03,250 –> 00:07:05,200 Fatou’s Lemma tells us something
254 00:07:05,209 –> 00:07:06,799 about the limit
255 00:07:06,809 –> 00:07:07,779 inferior.
256 00:07:07,899 –> 00:07:08,910 So liminf.
257 00:07:11,440 –> 00:07:12,910 namely, it tells us that
258 00:07:12,920 –> 00:07:14,519 we can look at the integral
259 00:07:14,529 –> 00:07:16,269 of the liminf and we can
260 00:07:16,279 –> 00:07:18,209 pull it out with an
261 00:07:18,220 –> 00:07:19,589 inequality sign.
262 00:07:20,369 –> 00:07:22,220 In fact, it’s possible that
263 00:07:22,230 –> 00:07:24,179 it gets bigger if we
264 00:07:24,190 –> 00:07:25,500 pull the liminf
265 00:07:25,510 –> 00:07:26,420 outside.
266 00:07:27,890 –> 00:07:29,510 However, the inequality
267 00:07:29,519 –> 00:07:30,820 always holds and that is
268 00:07:30,829 –> 00:07:32,029 what we can need here.
269 00:07:33,720 –> 00:07:35,149 Of course, you should ask
270 00:07:35,160 –> 00:07:36,660 yourself, do we
271 00:07:36,670 –> 00:07:38,630 know the liminf here?
272 00:07:38,640 –> 00:07:39,329 And here,
273 00:07:40,269 –> 00:07:42,079 maybe we should look first
274 00:07:42,089 –> 00:07:43,380 on the left hand side
275 00:07:44,549 –> 00:07:46,179 inside the integral, it
276 00:07:46,190 –> 00:07:47,869 always means the point
277 00:07:47,880 –> 00:07:48,769 wise limit.
278 00:07:49,309 –> 00:07:51,119 Now it’s the pointwise Liminf
279 00:07:51,200 –> 00:07:51,450 .
280 00:07:51,890 –> 00:07:53,519 But we know that the point
281 00:07:53,529 –> 00:07:54,869 wise limit of
282 00:07:54,880 –> 00:07:56,329 HN indeed
283 00:07:56,339 –> 00:07:57,130 exists
284 00:07:58,320 –> 00:07:59,619 and therefore, it should
285 00:07:59,630 –> 00:08:01,130 be the same as the limits.
286 00:08:02,470 –> 00:08:02,839 OK.
287 00:08:02,850 –> 00:08:04,290 So let’s write it down.
288 00:08:04,700 –> 00:08:06,519 We know this is the integral
289 00:08:06,529 –> 00:08:08,250 of our point wise limit of
290 00:08:08,260 –> 00:08:10,200 HN two G
291 00:08:10,209 –> 00:08:11,029 is two G.
292 00:08:11,160 –> 00:08:12,899 But we know F is the point
293 00:08:13,049 –> 00:08:14,679 limit of FN and
294 00:08:14,690 –> 00:08:16,600 therefore this one is
295 00:08:16,609 –> 00:08:17,140 zero.
296 00:08:17,149 –> 00:08:18,679 So only two G
297 00:08:18,690 –> 00:08:19,600 remains here.
298 00:08:21,279 –> 00:08:21,619 OK.
299 00:08:21,630 –> 00:08:22,890 Then let’s look at the right
300 00:08:22,899 –> 00:08:24,179 hand side there.
301 00:08:24,190 –> 00:08:25,980 We have the integral of HN.
302 00:08:27,190 –> 00:08:28,630 However, the integral is
303 00:08:28,640 –> 00:08:29,279 linear.
304 00:08:29,309 –> 00:08:30,799 So we have indeed two
305 00:08:30,809 –> 00:08:32,609 integrals, one of two G
306 00:08:32,619 –> 00:08:33,979 and the other one of our
307 00:08:33,989 –> 00:08:35,390 F and minus F in the absolute
308 00:08:35,460 –> 00:08:35,890 value.
309 00:08:36,849 –> 00:08:38,400 For the first part, the liminf
310 00:08:38,409 –> 00:08:39,630 does not matter.
311 00:08:39,640 –> 00:08:40,840 So we can write down
312 00:08:40,849 –> 00:08:42,820 immediately we have
313 00:08:42,830 –> 00:08:44,159 two G here.
314 00:08:45,580 –> 00:08:47,210 Then for the second part,
315 00:08:47,349 –> 00:08:49,039 you have to be careful, we
316 00:08:49,049 –> 00:08:50,640 subtract something positive.
317 00:08:50,809 –> 00:08:51,840 And we look at the liminf
318 00:08:52,390 –> 00:08:54,080 which means to get
319 00:08:54,090 –> 00:08:56,000 out the liminf we have to
320 00:08:56,010 –> 00:08:57,909 subtract the limsup, the
321 00:08:57,919 –> 00:08:59,000 limit superior
322 00:09:00,010 –> 00:09:01,200 or maybe in other words,
323 00:09:01,500 –> 00:09:02,909 if you want to find the
324 00:09:02,919 –> 00:09:04,739 smallest outcome here
325 00:09:04,979 –> 00:09:06,539 as a non negative
326 00:09:06,549 –> 00:09:07,950 number, yeah, you have to
327 00:09:07,960 –> 00:09:08,989 subtract here.
328 00:09:09,159 –> 00:09:11,150 The biggest possible number.
329 00:09:12,200 –> 00:09:12,599 OK.
330 00:09:12,609 –> 00:09:13,919 So this explains why we have
331 00:09:13,929 –> 00:09:15,739 to limsup here, but we
332 00:09:15,750 –> 00:09:16,890 don’t change the integral
333 00:09:16,900 –> 00:09:17,289 at all.
334 00:09:17,299 –> 00:09:18,909 So this is FN
335 00:09:18,940 –> 00:09:20,729 minus F dmu
336 00:09:22,380 –> 00:09:22,719 OK?
337 00:09:22,729 –> 00:09:23,640 We have the left hand side
338 00:09:23,650 –> 00:09:24,880 here and the right hand side
339 00:09:24,890 –> 00:09:26,599 here, we have a very nice
340 00:09:26,609 –> 00:09:27,919 inequality
341 00:09:29,210 –> 00:09:31,159 if I then add what I
342 00:09:31,169 –> 00:09:32,349 missed before here.
343 00:09:32,359 –> 00:09:33,979 So the d mu and our
344 00:09:33,989 –> 00:09:35,969 X then you will recognize
345 00:09:35,979 –> 00:09:37,400 immediately that we have the
346 00:09:37,409 –> 00:09:38,549 same on the left and the
347 00:09:38,559 –> 00:09:39,270 right here.
348 00:09:42,919 –> 00:09:44,690 Of course, now you
349 00:09:44,700 –> 00:09:46,619 should subtract the same
350 00:09:46,630 –> 00:09:48,219 thing on both sides.
351 00:09:49,549 –> 00:09:51,530 If we do this, we find
352 00:09:51,539 –> 00:09:52,809 zero on the left
353 00:09:52,820 –> 00:09:53,640 obviously
354 00:09:54,460 –> 00:09:56,320 and only this part here
355 00:09:56,330 –> 00:09:57,469 on the right hand side.
356 00:09:57,479 –> 00:09:59,250 So minus our
357 00:09:59,260 –> 00:09:59,869 limsup
358 00:10:01,460 –> 00:10:01,849 OK.
359 00:10:01,859 –> 00:10:03,429 So the minus sign is not
360 00:10:03,440 –> 00:10:04,349 so beautiful.
361 00:10:04,359 –> 00:10:05,440 Therefore, I want to bring
362 00:10:05,450 –> 00:10:06,789 this on the other side,
363 00:10:07,260 –> 00:10:09,130 which means we now
364 00:10:09,140 –> 00:10:11,070 have the inequality here
365 00:10:11,159 –> 00:10:12,770 on the right, which means
366 00:10:12,780 –> 00:10:14,200 this would be without the
367 00:10:14,210 –> 00:10:16,080 minus sign less or
368 00:10:16,090 –> 00:10:17,169 equal than zero.
369 00:10:18,719 –> 00:10:19,150 OK.
370 00:10:19,159 –> 00:10:21,140 So now please note this is
371 00:10:21,150 –> 00:10:22,409 very interesting.
372 00:10:22,750 –> 00:10:24,390 The limsup of non-
373 00:10:24,419 –> 00:10:25,940 negative numbers
374 00:10:25,979 –> 00:10:27,250 should be non-
375 00:10:27,260 –> 00:10:28,109 positives.
376 00:10:29,219 –> 00:10:30,840 Hence, from this, we
377 00:10:30,849 –> 00:10:32,739 can conclude that the
378 00:10:32,750 –> 00:10:34,080 limit exists.
379 00:10:35,239 –> 00:10:36,599 We do this in the following
380 00:10:36,609 –> 00:10:38,520 way, we say OK, I
381 00:10:38,530 –> 00:10:38,880 have to limsup
382 00:10:38,929 –> 00:10:40,580 here, so this is always
383 00:10:40,590 –> 00:10:42,520 greater or equal than
384 00:10:42,530 –> 00:10:43,260 the liminf
385 00:10:44,140 –> 00:10:46,099 Of course, this holds for
386 00:10:46,109 –> 00:10:47,659 all sequence of
387 00:10:47,669 –> 00:10:48,460 numbers.
388 00:10:49,140 –> 00:10:51,059 The liminf is always less
389 00:10:51,070 –> 00:10:52,719 or equal than the limsup.
390 00:10:53,760 –> 00:10:55,530 But still we have non-
391 00:10:55,539 –> 00:10:57,039 negative numbers here.
392 00:10:57,049 –> 00:10:58,760 Therefore, the liminf should
393 00:10:58,770 –> 00:11:00,739 be also nonnegative.
394 00:11:00,909 –> 00:11:02,770 So we have this inequality
395 00:11:02,780 –> 00:11:03,179 here.
396 00:11:03,989 –> 00:11:05,559 This one here is now always
397 00:11:05,570 –> 00:11:06,200 nice.
398 00:11:06,210 –> 00:11:08,169 We have inequalities but
399 00:11:08,179 –> 00:11:09,869 on the left is the
400 00:11:09,880 –> 00:11:11,690 same as on the right.
401 00:11:12,609 –> 00:11:14,309 This simply means that
402 00:11:14,320 –> 00:11:16,119 all the inequalities here
403 00:11:16,130 –> 00:11:17,219 are in fact
404 00:11:17,229 –> 00:11:18,140 equalities.
405 00:11:19,010 –> 00:11:20,169 There is simply no other
406 00:11:20,179 –> 00:11:21,799 way, which means the liminf
407 00:11:22,119 –> 00:11:23,710 is equal to the limsup,
408 00:11:23,719 –> 00:11:25,250 which means the limit
409 00:11:25,260 –> 00:11:27,229 exists and is equal to
410 00:11:27,239 –> 00:11:29,190 the limsup and liminf which
411 00:11:29,200 –> 00:11:29,909 is zero.
412 00:11:31,000 –> 00:11:32,659 So let’s write it down limit
413 00:11:32,669 –> 00:11:34,549 exists and the limit
414 00:11:34,760 –> 00:11:36,659 of this integral
415 00:11:38,020 –> 00:11:39,520 is equal to
416 00:11:39,530 –> 00:11:40,200 zero.
417 00:11:42,419 –> 00:11:44,359 Well, this is what we wanted
418 00:11:44,369 –> 00:11:46,210 to show and I explained
419 00:11:46,219 –> 00:11:47,880 before this is a stronger
420 00:11:47,890 –> 00:11:49,409 result than that what we
421 00:11:49,419 –> 00:11:51,080 have in Lebesgue’s
422 00:11:51,090 –> 00:11:51,679 theorem.
423 00:11:52,159 –> 00:11:54,020 But I will now show you how
424 00:11:54,030 –> 00:11:55,869 we get to the result
425 00:11:55,880 –> 00:11:57,820 in Lebesgue’s dominated convergence
426 00:11:57,830 –> 00:11:58,239 theorem.
427 00:11:58,979 –> 00:11:59,289 OK.
428 00:11:59,299 –> 00:12:00,630 Let’s do that now.
429 00:12:01,830 –> 00:12:03,099 So we want to
430 00:12:03,109 –> 00:12:04,940 show that
431 00:12:05,640 –> 00:12:07,049 the limit of the
432 00:12:07,059 –> 00:12:07,979 functions
433 00:12:08,929 –> 00:12:10,760 and the integral is
434 00:12:10,770 –> 00:12:12,690 equal to the
435 00:12:12,700 –> 00:12:14,080 integral of our
436 00:12:14,090 –> 00:12:14,700 function.
437 00:12:14,710 –> 00:12:15,349 F
438 00:12:16,539 –> 00:12:17,710 therefore, we can look at
439 00:12:17,719 –> 00:12:19,200 the difference and in the
440 00:12:19,210 –> 00:12:21,080 absolute value and show
441 00:12:21,210 –> 00:12:22,640 that this goes to
442 00:12:22,650 –> 00:12:24,140 zero for N to
443 00:12:24,150 –> 00:12:24,880 infinity.
444 00:12:25,900 –> 00:12:26,200 OK.
445 00:12:26,210 –> 00:12:27,369 The first part we can note
446 00:12:27,380 –> 00:12:29,020 is this is also
447 00:12:29,030 –> 00:12:30,719 nonnegative because it’s
448 00:12:30,729 –> 00:12:31,719 an absolute value.
449 00:12:32,909 –> 00:12:34,059 And then the next steps we
450 00:12:34,070 –> 00:12:35,690 can use the properties we
451 00:12:35,700 –> 00:12:37,450 already know from the integral,
452 00:12:37,539 –> 00:12:38,979 for example, the
453 00:12:38,989 –> 00:12:39,890 linearity.
454 00:12:39,900 –> 00:12:41,130 So we know this is indeed
455 00:12:41,140 –> 00:12:43,080 just one integral where we
456 00:12:43,090 –> 00:12:44,530 have F and
457 00:12:44,539 –> 00:12:46,349 minus F in
458 00:12:46,359 –> 00:12:47,619 the integral
459 00:12:47,900 –> 00:12:48,840 itself.
460 00:12:48,919 –> 00:12:50,570 And the absolute value around.
461 00:12:51,919 –> 00:12:53,690 Now, in the next part, we
462 00:12:53,700 –> 00:12:55,109 use something that is also
463 00:12:55,119 –> 00:12:56,270 called triangle
464 00:12:56,280 –> 00:12:57,409 inequality.
465 00:12:57,919 –> 00:12:59,580 But here now for
466 00:12:59,919 –> 00:13:01,719 integrals, which means now
467 00:13:02,020 –> 00:13:03,909 put the absolute value inside
468 00:13:03,919 –> 00:13:05,340 and then we can get bigger
469 00:13:05,349 –> 00:13:06,340 or stay the same.
470 00:13:08,440 –> 00:13:10,059 Here, we now reach something
471 00:13:10,080 –> 00:13:11,750 that we already know at
472 00:13:11,760 –> 00:13:12,780 least in a limit.
473 00:13:13,770 –> 00:13:14,590 So it goes to
474 00:13:14,599 –> 00:13:16,549 for N
475 00:13:16,659 –> 00:13:17,789 to infinity.
476 00:13:19,520 –> 00:13:20,859 Hence the last step we need
477 00:13:20,869 –> 00:13:22,419 here now is just a
478 00:13:22,429 –> 00:13:24,289 small sandwich theorem.
479 00:13:25,030 –> 00:13:26,400 Now, on the limit we have
480 00:13:26,409 –> 00:13:27,950 the zero left and
481 00:13:27,960 –> 00:13:28,440 right.
482 00:13:28,570 –> 00:13:30,090 And therefore we know this
483 00:13:30,099 –> 00:13:31,969 limit also exists
484 00:13:32,179 –> 00:13:33,609 and is equal to zero.
485 00:13:34,710 –> 00:13:36,359 And therefore the whole thing
486 00:13:36,369 –> 00:13:37,840 in the absolute value has
487 00:13:37,849 –> 00:13:39,409 a limit and is equal to
488 00:13:39,419 –> 00:13:39,940 zero.
489 00:13:40,250 –> 00:13:41,409 So putting that on the other
490 00:13:41,419 –> 00:13:43,369 side, we now can conclude
491 00:13:44,229 –> 00:13:46,059 limit of the integrals of
492 00:13:46,070 –> 00:13:47,289 FN is
493 00:13:47,299 –> 00:13:48,650 equal to the
494 00:13:48,659 –> 00:13:50,559 integral of F.
495 00:13:52,710 –> 00:13:54,330 And here you see this is
496 00:13:54,340 –> 00:13:55,770 the convergence statement
497 00:13:55,780 –> 00:13:57,520 that we wanted to prove in
498 00:13:57,530 –> 00:13:58,280 the beginning.
499 00:14:00,390 –> 00:14:02,270 And that was the proof of
500 00:14:02,280 –> 00:14:03,630 Lebesgue’s dominated
501 00:14:03,640 –> 00:14:04,849 convergence theorem.
502 00:14:05,940 –> 00:14:07,460 I already told you
503 00:14:07,539 –> 00:14:09,440 this part dominated
504 00:14:09,450 –> 00:14:11,340 is the important ingredient
505 00:14:11,349 –> 00:14:13,239 for the theorem because
506 00:14:13,250 –> 00:14:15,159 we need such an
507 00:14:15,359 –> 00:14:17,239 integrable majorant
508 00:14:18,330 –> 00:14:20,030 we call this majorant just
509 00:14:20,039 –> 00:14:21,539 G but you saw
510 00:14:21,880 –> 00:14:23,549 we need this and
511 00:14:23,559 –> 00:14:25,510 then we can apply
512 00:14:25,700 –> 00:14:27,630 Lebesgue’s dominated convergence
513 00:14:27,640 –> 00:14:28,099 theorem.
514 00:14:29,479 –> 00:14:31,109 If you want to see some
515 00:14:31,119 –> 00:14:32,739 applications of
516 00:14:32,750 –> 00:14:34,619 this theorem, please let
517 00:14:34,630 –> 00:14:36,590 me know because this
518 00:14:36,599 –> 00:14:37,950 could be a very good idea
519 00:14:38,020 –> 00:14:39,450 for the next part in the
520 00:14:39,460 –> 00:14:40,150 series.
521 00:14:40,940 –> 00:14:42,479 So thank you very much and
522 00:14:42,489 –> 00:14:43,719 see you next time.
523 00:14:44,080 –> 00:14:44,679 Bye.
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Quiz Content
Q1: Consider a measure space $(X, \mathcal{A}, \mu)$. Let $f: X \rightarrow \mathbb{R}$ be measurable and $g \in \mathcal{L}^1(\mu)$ with $|f| \leq g$. What is the correct conclusion?
A1: $\int |f| , d\mu < \infty$
A2: $f \notin \mathcal{L}^1(\mu)$
A3: $\int |f| , d\mu > \int g , d\mu = \infty$
Q2: Consider a sequence of positive real numbers $(a_n)$. Which claim is always correct?
A1: $\limsup\limits_{n \rightarrow \infty} (-a_n) = - \liminf\limits_{n \rightarrow \infty} a_n$
A2: $\liminf\limits_{n \rightarrow \infty} a_n$ is zero.
A3: $\limsup\limits_{n \rightarrow \infty} a_n$ is finite.
A4: $\limsup\limits_{n \rightarrow \infty} a_n = \liminf\limits_{n \rightarrow \infty} a_n$
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Last update: 2024-10
