
Title: Monotone Convergence Theorem (Proof and Application)

Series: Measure Theory

YouTubeTitle: Measure Theory 8  Monotone Convergence Theorem (Proof and Application)

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Subtitle in English
1 00:00:00,670 –> 00:00:02,529 Hello and welcome back to
2 00:00:02,539 –> 00:00:03,970 the next video in this
3 00:00:03,980 –> 00:00:04,769 series.
4 00:00:05,280 –> 00:00:07,269 I thank all the nice people
5 00:00:07,280 –> 00:00:08,689 that support this channel
6 00:00:08,699 –> 00:00:09,630 on Steady
7 00:00:10,510 –> 00:00:12,470 Well, this is measure theory
8 00:00:12,479 –> 00:00:13,779 part eight.
9 00:00:13,930 –> 00:00:15,359 And we are still talking
10 00:00:15,369 –> 00:00:17,049 about the monotone convergence
11 00:00:17,059 –> 00:00:17,670 theorem.
12 00:00:17,809 –> 00:00:19,590 But today I want to show
13 00:00:19,600 –> 00:00:21,329 you mainly the proof
14 00:00:21,340 –> 00:00:22,840 of this theorem
15 00:00:24,010 –> 00:00:25,600 for the start of this video.
16 00:00:25,709 –> 00:00:27,540 It might be helpful to
17 00:00:27,549 –> 00:00:29,350 give you a recap what the
18 00:00:29,360 –> 00:00:30,899 monotone convergence theorem
19 00:00:30,909 –> 00:00:32,729 states. For the
20 00:00:32,740 –> 00:00:34,479 ingredients, we need a measure
21 00:00:34,490 –> 00:00:35,939 space and
22 00:00:35,950 –> 00:00:37,380 measurable maps
23 00:00:38,139 –> 00:00:39,549 as always they are non
24 00:00:39,560 –> 00:00:40,200 negative.
25 00:00:40,389 –> 00:00:41,880 And we have a whole sequence
26 00:00:41,889 –> 00:00:43,520 of such measurable maps.
27 00:00:44,040 –> 00:00:45,720 And in addition, we also
28 00:00:45,729 –> 00:00:47,169 have the limit where we call
29 00:00:47,180 –> 00:00:48,720 this just F without an
30 00:00:48,729 –> 00:00:49,279 index.
31 00:00:50,049 –> 00:00:50,389 OK.
32 00:00:50,400 –> 00:00:52,119 Because the theorem is called
33 00:00:52,130 –> 00:00:53,869 monotone convergence theorem.
34 00:00:53,880 –> 00:00:55,669 We need two things first,
35 00:00:55,680 –> 00:00:57,349 a monotonic property and
36 00:00:57,360 –> 00:00:58,869 also a convergent property.
37 00:00:59,680 –> 00:01:01,340 The first thing is therefore
38 00:01:01,349 –> 00:01:02,509 the sequence is
39 00:01:02,520 –> 00:01:04,379 monotonically increasing
40 00:01:05,069 –> 00:01:07,049 but only almost everywhere
41 00:01:07,058 –> 00:01:08,579 with respect to our measure
42 00:01:08,588 –> 00:01:08,869 mu
43 00:01:09,558 –> 00:01:11,049 And the second property I
44 00:01:11,058 –> 00:01:12,888 already told you F should
45 00:01:12,899 –> 00:01:14,329 be the limit of
46 00:01:14,338 –> 00:01:16,308 FN and the limit should
47 00:01:16,319 –> 00:01:18,219 be in the pointwise sense which
48 00:01:18,228 –> 00:01:19,719 means I fix a point
49 00:01:19,728 –> 00:01:21,369 X and then I have a normal
50 00:01:21,378 –> 00:01:23,109 limit of real numbers.
51 00:01:23,359 –> 00:01:25,088 However, as before, this
52 00:01:25,098 –> 00:01:26,738 property only has to hold
53 00:01:26,749 –> 00:01:28,088 almost everywhere.
54 00:01:28,900 –> 00:01:30,639 So much about the premise.
55 00:01:30,650 –> 00:01:32,400 Now let’s talk about the
56 00:01:32,410 –> 00:01:33,169 conclusion.
57 00:01:33,830 –> 00:01:35,610 The theorem tells us that
58 00:01:35,620 –> 00:01:37,069 we can push the
59 00:01:37,080 –> 00:01:38,790 limit into the
60 00:01:38,800 –> 00:01:39,599 integral.
61 00:01:39,720 –> 00:01:41,410 This means that the sequence
62 00:01:41,419 –> 00:01:43,150 of integrals converges
63 00:01:43,160 –> 00:01:45,150 to an integral where in the
64 00:01:45,160 –> 00:01:46,779 integral there’s the
65 00:01:46,790 –> 00:01:48,650 limit of the sequence.
66 00:01:51,000 –> 00:01:52,989 And of course, this is just
67 00:01:53,000 –> 00:01:54,050 our F here.
68 00:01:55,910 –> 00:01:57,550 And now we have again our
69 00:01:57,559 –> 00:01:59,050 monotone convergence
70 00:01:59,059 –> 00:01:59,569 theorem.
71 00:02:00,260 –> 00:02:02,050 However, I should tell you
72 00:02:02,059 –> 00:02:04,050 there is an alternative formulation
73 00:02:04,059 –> 00:02:05,750 of this theorem which is
74 00:02:05,760 –> 00:02:07,250 a little bit more general
75 00:02:08,100 –> 00:02:09,679 what you do there is you
76 00:02:09,690 –> 00:02:10,750 include also
77 00:02:10,758 –> 00:02:12,619 functions that have the
78 00:02:12,630 –> 00:02:14,479 symbol infinity as an
79 00:02:14,490 –> 00:02:15,160 outcome.
80 00:02:15,919 –> 00:02:17,580 Of course, then you have
81 00:02:17,589 –> 00:02:19,270 to extend the Borel Sigma
82 00:02:19,279 –> 00:02:20,860 algebra to this
83 00:02:20,869 –> 00:02:22,380 new bigger set
84 00:02:22,710 –> 00:02:24,229 and also extend the notion
85 00:02:24,240 –> 00:02:25,199 of measurable.
86 00:02:25,750 –> 00:02:26,809 And indeed, that’s not a
87 00:02:26,820 –> 00:02:27,740 problem at all
88 00:02:29,070 –> 00:02:30,000 in the same way.
89 00:02:30,009 –> 00:02:31,509 Also the definition of the
90 00:02:31,520 –> 00:02:32,740 Integral does not change
91 00:02:32,750 –> 00:02:34,309 at all because in the worst
92 00:02:34,320 –> 00:02:35,919 case, as always, it could
93 00:02:35,929 –> 00:02:36,839 be infinity.
94 00:02:37,720 –> 00:02:39,279 The advantage of this formulation
95 00:02:39,289 –> 00:02:40,990 is now that you can get rid
96 00:02:41,000 –> 00:02:42,470 of the F completely.
97 00:02:43,250 –> 00:02:45,009 So we don’t need this
98 00:02:45,020 –> 00:02:46,660 and we don’t need
99 00:02:46,669 –> 00:02:47,470 this
100 00:02:48,479 –> 00:02:50,179 simply because everything
101 00:02:50,229 –> 00:02:52,009 follows from the monotonic
102 00:02:52,020 –> 00:02:53,800 property here, we have
103 00:02:53,809 –> 00:02:55,619 measurable functions and
104 00:02:55,630 –> 00:02:57,130 we know the supremum is also
105 00:02:57,139 –> 00:02:58,949 measurable and they are
106 00:02:58,960 –> 00:03:00,619 monotonically increasing.
107 00:03:00,669 –> 00:03:02,419 So we know the supremum is given
108 00:03:02,429 –> 00:03:03,789 almost everywhere as the
109 00:03:03,800 –> 00:03:04,259 limit.
110 00:03:05,350 –> 00:03:06,369 And there you see this is
111 00:03:06,380 –> 00:03:08,369 just a more compact formulation
112 00:03:08,380 –> 00:03:09,470 of the theorem.
113 00:03:09,770 –> 00:03:11,320 And as I told you before,
114 00:03:11,330 –> 00:03:12,380 in the worst case, you have
115 00:03:12,389 –> 00:03:13,979 infinity left and
116 00:03:13,990 –> 00:03:15,729 right in the conclusion here.
117 00:03:17,880 –> 00:03:19,419 Well, so this was the
118 00:03:19,429 –> 00:03:20,300 introduction.
119 00:03:20,309 –> 00:03:22,229 Now, let’s look at
120 00:03:22,240 –> 00:03:23,000 the proof,
121 00:03:25,929 –> 00:03:27,220 of course, we use what we
122 00:03:27,229 –> 00:03:28,199 know we have a
123 00:03:28,210 –> 00:03:29,389 monotonically
124 00:03:29,399 –> 00:03:31,309 increasing sequence
125 00:03:31,320 –> 00:03:32,389 of functions.
126 00:03:34,270 –> 00:03:36,009 This property holds almost
127 00:03:36,020 –> 00:03:36,759 everywhere.
128 00:03:37,050 –> 00:03:38,330 And if you remember what
129 00:03:38,339 –> 00:03:39,710 we did in the last video,
130 00:03:39,740 –> 00:03:41,039 we know that we have the
131 00:03:41,050 –> 00:03:42,880 monotonicity property of
132 00:03:42,889 –> 00:03:43,970 the Lebesgue integral.
133 00:03:44,470 –> 00:03:46,089 This means that the integral
134 00:03:46,100 –> 00:03:47,669 conserves
135 00:03:47,679 –> 00:03:49,429 these inequalities.
136 00:03:52,839 –> 00:03:54,320 Now we have a monotonically
137 00:03:54,330 –> 00:03:56,210 increasing sequence of real
138 00:03:56,220 –> 00:03:57,600 numbers where also
139 00:03:57,610 –> 00:03:59,059 infinity can occur
140 00:03:59,990 –> 00:04:01,440 in the same sense, we can
141 00:04:01,449 –> 00:04:03,250 write down the inequality
142 00:04:03,259 –> 00:04:04,270 for all N.
143 00:04:04,279 –> 00:04:05,929 So we know that the limit
144 00:04:06,179 –> 00:04:08,009 is always bigger or equal
145 00:04:08,020 –> 00:04:09,850 than FN for
146 00:04:09,860 –> 00:04:11,639 all N and almost
147 00:04:11,649 –> 00:04:12,429 everywhere,
148 00:04:13,619 –> 00:04:14,800 simply because the limit
149 00:04:14,809 –> 00:04:16,279 is almost everywhere, the
150 00:04:16,290 –> 00:04:18,200 supremum and because
151 00:04:18,209 –> 00:04:19,920 this holds almost everywhere,
152 00:04:19,928 –> 00:04:21,829 we know it holds for the
153 00:04:21,850 –> 00:04:23,309 integral as well.
154 00:04:24,269 –> 00:04:26,179 And there you see, we already
155 00:04:26,190 –> 00:04:27,690 have one part of the
156 00:04:27,700 –> 00:04:28,720 equality here.
157 00:04:28,730 –> 00:04:29,649 We want to show.
158 00:04:29,799 –> 00:04:31,489 So let’s call this star
159 00:04:31,500 –> 00:04:32,850 for a reference
160 00:04:33,989 –> 00:04:35,630 and let’s show the one
161 00:04:35,640 –> 00:04:36,380 inequality.
162 00:04:36,390 –> 00:04:37,970 So we formed the limit on
163 00:04:37,980 –> 00:04:39,309 both sides here.
164 00:04:39,929 –> 00:04:41,070 Of course, we don’t change
165 00:04:41,079 –> 00:04:42,160 the inequality here.
166 00:04:42,170 –> 00:04:43,609 And we have no limit on the
167 00:04:43,619 –> 00:04:44,809 right hand side because there’s
168 00:04:44,820 –> 00:04:46,429 no N involved.
169 00:04:46,440 –> 00:04:48,170 And therefore we have this
170 00:04:48,179 –> 00:04:49,329 as the inequality.
171 00:04:50,049 –> 00:04:51,510 And this is exactly one
172 00:04:51,519 –> 00:04:53,420 part of our
173 00:04:53,429 –> 00:04:54,739 equality star.
174 00:04:55,850 –> 00:04:57,269 Do you see this was not so
175 00:04:57,279 –> 00:04:57,790 hard.
176 00:04:57,799 –> 00:04:59,579 This part comes immediately
177 00:05:00,380 –> 00:05:01,959 but for the other part we
178 00:05:01,970 –> 00:05:03,410 need to use the definition
179 00:05:03,420 –> 00:05:04,640 of the Lebesgue integral.
180 00:05:05,179 –> 00:05:06,839 And this always means that
181 00:05:06,850 –> 00:05:08,790 we go back to step or simple
182 00:05:08,799 –> 00:05:09,709 functions.
183 00:05:09,829 –> 00:05:11,480 Therefore, let h be an
184 00:05:11,489 –> 00:05:13,390 arbitrary chosen simple function
185 00:05:13,529 –> 00:05:15,130 that lies below
186 00:05:15,140 –> 00:05:16,559 our function F
187 00:05:19,089 –> 00:05:20,239 and also choose a
188 00:05:20,250 –> 00:05:21,709 fixed error
189 00:05:21,720 –> 00:05:22,589 epsilon
190 00:05:23,809 –> 00:05:24,559 at this point.
191 00:05:24,570 –> 00:05:26,049 I think a short sketch is
192 00:05:26,059 –> 00:05:27,730 very helpful to get the idea.
193 00:05:28,880 –> 00:05:30,720 Here is our function F
194 00:05:31,140 –> 00:05:32,779 and this one should be the
195 00:05:32,790 –> 00:05:34,630 step function H so maybe
196 00:05:34,640 –> 00:05:35,559 it looks like this.
197 00:05:37,890 –> 00:05:38,299 OK.
198 00:05:38,309 –> 00:05:40,220 So this would be the usual
199 00:05:40,230 –> 00:05:40,799 idea.
200 00:05:41,720 –> 00:05:43,609 Now we know we have the pointwise
201 00:05:43,619 –> 00:05:45,160 convergence of this
202 00:05:45,170 –> 00:05:46,500 sequence of functions to
203 00:05:46,510 –> 00:05:47,160 F.
204 00:05:47,170 –> 00:05:48,850 So maybe we can choose some
205 00:05:48,859 –> 00:05:50,179 FN that lies
206 00:05:51,019 –> 00:05:52,600 very close to F.
207 00:05:52,709 –> 00:05:54,359 So maybe this is
208 00:05:54,369 –> 00:05:56,029 our chosen FN.
209 00:05:57,299 –> 00:05:59,040 The idea of the epsilon is
210 00:05:59,049 –> 00:06:00,929 now that we can always push
211 00:06:00,940 –> 00:06:02,709 down the step function by
212 00:06:02,720 –> 00:06:03,640 this epsilon.
213 00:06:03,869 –> 00:06:05,390 So this would be epsilon
214 00:06:05,399 –> 00:06:06,619 and then the new step function
215 00:06:06,630 –> 00:06:07,549 looks like this.
216 00:06:08,220 –> 00:06:09,859 And this should then always
217 00:06:09,869 –> 00:06:11,760 be a step function that lies
218 00:06:11,769 –> 00:06:13,000 below FN.
219 00:06:14,200 –> 00:06:15,859 Hence I define now
220 00:06:15,869 –> 00:06:17,739 sets XN
221 00:06:17,929 –> 00:06:19,660 where I put in all the
222 00:06:19,670 –> 00:06:21,109 points X
223 00:06:22,290 –> 00:06:24,089 where FN of X
224 00:06:24,200 –> 00:06:25,549 is bigger than the
225 00:06:25,559 –> 00:06:27,260 shifted step function.
226 00:06:27,269 –> 00:06:28,609 And the shifted step function
227 00:06:28,619 –> 00:06:30,190 is just one minus
228 00:06:30,200 –> 00:06:32,029 epsilon h
229 00:06:34,200 –> 00:06:35,299 because we have our
230 00:06:35,309 –> 00:06:36,700 convergence of the
231 00:06:36,709 –> 00:06:38,670 sequence FN to F
232 00:06:38,679 –> 00:06:39,899 almost everywhere
233 00:06:40,420 –> 00:06:41,859 we know that
234 00:06:42,089 –> 00:06:43,899 almost every X
235 00:06:43,980 –> 00:06:45,730 in X lies
236 00:06:45,739 –> 00:06:47,410 in at least one of these
237 00:06:47,420 –> 00:06:47,929 XN.
238 00:06:48,929 –> 00:06:50,480 In other words, we can look
239 00:06:50,489 –> 00:06:52,399 at the union of all these
240 00:06:52,410 –> 00:06:54,369 XN and
241 00:06:54,380 –> 00:06:56,369 let’s call it X tilde,
242 00:06:56,850 –> 00:06:58,450 then we know that the complement
243 00:06:58,459 –> 00:07:00,450 of X tilde is a set with
244 00:07:00,459 –> 00:07:01,489 measure zero.
245 00:07:04,390 –> 00:07:05,950 Knowing this, we can now
246 00:07:05,959 –> 00:07:07,429 look at the integral again.
247 00:07:08,549 –> 00:07:10,010 Now we start here with the
248 00:07:10,019 –> 00:07:11,089 left hand side.
249 00:07:11,100 –> 00:07:12,670 So I want to look at the
250 00:07:12,679 –> 00:07:14,529 integral of FN.
251 00:07:16,309 –> 00:07:17,890 This integral gets not
252 00:07:17,899 –> 00:07:19,850 bigger if I change the set
253 00:07:19,859 –> 00:07:20,850 to a smaller set.
254 00:07:20,859 –> 00:07:22,160 So here I want to choose
255 00:07:22,170 –> 00:07:23,010 XN
256 00:07:24,260 –> 00:07:25,700 Fn dmu.
257 00:07:26,350 –> 00:07:28,329 Now inside the set XN,
258 00:07:28,339 –> 00:07:30,239 we have this inequality
259 00:07:30,299 –> 00:07:31,570 and can use the normal
260 00:07:31,579 –> 00:07:33,390 monotonicity of the integral.
261 00:07:34,260 –> 00:07:35,250 Therefore, I have the same
262 00:07:35,260 –> 00:07:36,500 inequality here.
263 00:07:36,510 –> 00:07:38,109 And a new function inside
264 00:07:38,119 –> 00:07:39,480 the integral which is now
265 00:07:39,489 –> 00:07:40,959 one minus epsilon
266 00:07:40,970 –> 00:07:42,480 h dmu.
267 00:07:43,750 –> 00:07:45,070 And of course, we want to
268 00:07:45,079 –> 00:07:46,209 look at the limit.
269 00:07:46,369 –> 00:07:48,089 So we found the limit here
270 00:07:48,100 –> 00:07:49,250 on both sides,
271 00:07:49,880 –> 00:07:50,859 then we get
272 00:07:51,519 –> 00:07:52,420 limit
273 00:07:53,399 –> 00:07:54,829 of the integrals.
274 00:07:55,709 –> 00:07:57,250 And on the right hand side,
275 00:07:57,290 –> 00:07:59,089 we have the limit of
276 00:07:59,100 –> 00:08:00,329 these integrals.
277 00:08:01,149 –> 00:08:01,769 OK.
278 00:08:01,970 –> 00:08:03,720 At this point, I want to
279 00:08:03,730 –> 00:08:05,529 omit some details because
280 00:08:05,540 –> 00:08:06,869 otherwise we will lose our
281 00:08:06,880 –> 00:08:07,989 focus completely.
282 00:08:08,380 –> 00:08:10,010 What one can show now is
283 00:08:10,019 –> 00:08:11,890 that this is indeed
284 00:08:12,250 –> 00:08:13,869 the integral over
285 00:08:13,950 –> 00:08:14,720 X tilde.
286 00:08:15,260 –> 00:08:17,170 And we don’t change the
287 00:08:17,179 –> 00:08:18,320 function at all.
288 00:08:19,459 –> 00:08:20,690 And what we need in the details
289 00:08:20,700 –> 00:08:22,239 is of course that this is
290 00:08:22,250 –> 00:08:23,109 a step function.
291 00:08:23,119 –> 00:08:24,820 So the integral is very easy.
292 00:08:25,140 –> 00:08:26,510 Essentially, it’s just a
293 00:08:26,519 –> 00:08:27,880 sum of our measures.
294 00:08:28,619 –> 00:08:30,420 And we know that the
295 00:08:30,429 –> 00:08:31,799 sets here are
296 00:08:31,809 –> 00:08:33,760 increasing, you
297 00:08:33,770 –> 00:08:35,619 have to find the sets appropriately.
298 00:08:35,630 –> 00:08:36,830 But then you get out that
299 00:08:36,840 –> 00:08:38,260 you have a monotonically
300 00:08:38,270 –> 00:08:40,229 increasing sequence
301 00:08:40,239 –> 00:08:41,020 of sets.
302 00:08:41,789 –> 00:08:42,450 OK.
303 00:08:42,640 –> 00:08:43,700 In summary, I’ll leave the
304 00:08:43,710 –> 00:08:44,969 details here for you.
305 00:08:44,979 –> 00:08:46,950 But remember we can do
306 00:08:46,960 –> 00:08:48,580 this because this is just
307 00:08:48,590 –> 00:08:50,010 a step function and integral.
308 00:08:50,020 –> 00:08:51,010 It’s very easy there.
309 00:08:51,840 –> 00:08:52,200 OK.
310 00:08:52,210 –> 00:08:54,099 But after this, you see it
311 00:08:54,109 –> 00:08:55,219 doesn’t make any difference
312 00:08:55,229 –> 00:08:57,059 to choose X tilde here or X
313 00:08:58,280 –> 00:08:59,359 because for the integral,
314 00:08:59,369 –> 00:09:01,320 it does not matter what happens
315 00:09:01,409 –> 00:09:03,140 on a set with measure
316 00:09:03,150 –> 00:09:03,729 zero.
317 00:09:04,280 –> 00:09:05,299 And in the last video, I
318 00:09:05,309 –> 00:09:06,799 have shown it explicitly
319 00:09:06,809 –> 00:09:07,969 for the step functions.
320 00:09:09,229 –> 00:09:09,929 OK.
321 00:09:09,940 –> 00:09:10,849 Now we have an
322 00:09:10,859 –> 00:09:12,729 inequality that holds
323 00:09:12,739 –> 00:09:14,530 for all epsilon here on the
324 00:09:14,539 –> 00:09:16,409 right hand side, this
325 00:09:16,419 –> 00:09:18,219 means now that we can choose
326 00:09:18,229 –> 00:09:19,630 the epsilon as
327 00:09:19,640 –> 00:09:21,340 small as we want.
328 00:09:22,229 –> 00:09:23,900 And therefore we have a
329 00:09:23,909 –> 00:09:25,719 limit process for the epsilon
330 00:09:25,729 –> 00:09:26,989 and still have the
331 00:09:27,000 –> 00:09:28,650 inequality if we
332 00:09:28,659 –> 00:09:30,530 would set epsilon to
333 00:09:30,539 –> 00:09:31,289 zero.
334 00:09:31,299 –> 00:09:33,159 So this means here we
335 00:09:33,169 –> 00:09:34,750 have the integral greater
336 00:09:34,760 –> 00:09:36,390 or equal than the
337 00:09:36,400 –> 00:09:37,229 integral.
338 00:09:37,309 –> 00:09:38,469 And here just h
339 00:09:38,909 –> 00:09:39,909 remains.
340 00:09:41,609 –> 00:09:43,270 Now in the same sense,
341 00:09:43,280 –> 00:09:44,909 this h was arbitrarily
342 00:09:44,919 –> 00:09:45,479 chosen.
343 00:09:45,989 –> 00:09:47,669 It was just a step function
344 00:09:47,679 –> 00:09:49,229 that lies below our
345 00:09:49,239 –> 00:09:50,299 function F.
346 00:09:51,440 –> 00:09:52,940 This means now that we can
347 00:09:52,950 –> 00:09:53,909 go to the supremum
348 00:09:54,219 –> 00:09:56,049 over all the step
349 00:09:56,059 –> 00:09:56,880 functions.
350 00:09:57,020 –> 00:09:58,229 And then we get on the right
351 00:09:58,239 –> 00:09:59,940 hand side, the integral of
352 00:09:59,950 –> 00:10:01,710 F and there you have
353 00:10:01,719 –> 00:10:01,960 it.
354 00:10:01,969 –> 00:10:03,630 This is the second part of
355 00:10:03,640 –> 00:10:05,260 our equality star.
356 00:10:05,710 –> 00:10:07,070 Both things together
357 00:10:07,210 –> 00:10:08,960 gives us the proof of the
358 00:10:08,969 –> 00:10:10,789 monotone convergence theorem.
359 00:10:11,940 –> 00:10:12,330 OK.
360 00:10:12,340 –> 00:10:13,719 For the end of the video,
361 00:10:13,849 –> 00:10:15,450 I want to show you an
362 00:10:15,460 –> 00:10:17,090 application of this very
363 00:10:17,099 –> 00:10:18,869 important monotone convergence
364 00:10:18,880 –> 00:10:19,320 theorem.
365 00:10:20,510 –> 00:10:22,409 One can always use this
366 00:10:22,419 –> 00:10:24,409 for series if
367 00:10:24,419 –> 00:10:26,309 you have any sequence of
368 00:10:26,320 –> 00:10:27,960 functions, then I use now the
369 00:10:27,979 –> 00:10:29,799 letter G that are
370 00:10:29,809 –> 00:10:31,479 nonnegative and also
371 00:10:31,489 –> 00:10:32,369 measurable.
372 00:10:33,440 –> 00:10:35,119 Then you can apply the
373 00:10:35,130 –> 00:10:36,900 monotone convergence theorem.
374 00:10:37,530 –> 00:10:39,219 So here I use also to assume
375 00:10:39,229 –> 00:10:41,030 the infinity for the functions
376 00:10:41,289 –> 00:10:43,179 and measurable.
377 00:10:44,030 –> 00:10:45,659 So we don’t need a monotonic
378 00:10:45,669 –> 00:10:46,580 behavior here.
379 00:10:47,890 –> 00:10:49,770 Then we can say that the
380 00:10:49,780 –> 00:10:50,710 series
381 00:10:51,320 –> 00:10:53,239 which now goes from one
382 00:10:53,250 –> 00:10:54,140 to infinity.
383 00:10:54,820 –> 00:10:56,309 This is also a well defined
384 00:10:56,320 –> 00:10:58,010 function and also measurable.
385 00:10:59,099 –> 00:11:00,659 The trick is now of course,
386 00:11:00,669 –> 00:11:02,239 that the partial sums
387 00:11:02,250 –> 00:11:03,700 form a monotonically
388 00:11:03,710 –> 00:11:05,179 increasing sequence of
389 00:11:05,190 –> 00:11:05,840 functions.
390 00:11:07,119 –> 00:11:08,840 Therefore, we can write down
391 00:11:08,849 –> 00:11:10,539 the integral of the
392 00:11:10,549 –> 00:11:11,419 series.
393 00:11:11,510 –> 00:11:13,179 So it goes to infinity.
394 00:11:13,559 –> 00:11:15,539 g n de mu
395 00:11:16,320 –> 00:11:18,179 is equal, putting
396 00:11:18,190 –> 00:11:20,080 the series outside the
397 00:11:20,090 –> 00:11:20,599 integrals
398 00:11:21,349 –> 00:11:23,340 So here’s the series of
399 00:11:23,349 –> 00:11:24,429 the integrals
400 00:11:24,710 –> 00:11:25,869 So gn
401 00:11:26,059 –> 00:11:26,770 d mu
402 00:11:27,780 –> 00:11:29,729 this means that the monotone
403 00:11:29,739 –> 00:11:31,690 convergence theorem allows
404 00:11:31,700 –> 00:11:33,130 us to switch
405 00:11:33,140 –> 00:11:34,599 integral and series.
406 00:11:35,479 –> 00:11:36,520 The only thing we have to
407 00:11:36,530 –> 00:11:38,299 put in is that we have
408 00:11:38,309 –> 00:11:40,039 nonnegative measurable
409 00:11:40,049 –> 00:11:41,880 functions and we
410 00:11:41,890 –> 00:11:43,469 need no other assumptions
411 00:11:43,479 –> 00:11:43,960 at all.
412 00:11:44,789 –> 00:11:46,049 And there you see the power
413 00:11:46,059 –> 00:11:47,710 of this theorem, it gives
414 00:11:47,719 –> 00:11:49,179 you immediately such a nice
415 00:11:49,190 –> 00:11:50,250 property as this
416 00:11:51,450 –> 00:11:52,309 very well.
417 00:11:52,500 –> 00:11:53,919 That’s all I wanted to tell
418 00:11:53,929 –> 00:11:55,830 you about the monotone convergence
419 00:11:55,840 –> 00:11:56,369 theorem.
420 00:11:56,429 –> 00:11:58,150 And I hope you learned
421 00:11:58,159 –> 00:11:59,460 how one can prove it.
422 00:11:59,549 –> 00:12:01,179 And more importantly, you
423 00:12:01,190 –> 00:12:02,969 now learned when you can
424 00:12:02,979 –> 00:12:04,700 apply it, you
425 00:12:04,710 –> 00:12:06,169 need a monotonically
426 00:12:06,179 –> 00:12:07,559 increasing sequence of
427 00:12:07,570 –> 00:12:09,109 functions that are non
428 00:12:09,119 –> 00:12:09,900 negative.
429 00:12:10,679 –> 00:12:12,150 Later, we will see how we
430 00:12:12,159 –> 00:12:13,909 can weaken a little bit this
431 00:12:13,919 –> 00:12:14,590 assumption.
432 00:12:15,659 –> 00:12:17,369 However, for the moment this
433 00:12:17,380 –> 00:12:18,969 version here is powerful
434 00:12:18,979 –> 00:12:19,409 enough.
435 00:12:21,299 –> 00:12:22,869 In the following next videos
436 00:12:22,880 –> 00:12:24,489 we will talk about other
437 00:12:24,500 –> 00:12:26,059 convergence theorem that are
438 00:12:26,070 –> 00:12:27,349 also very important.
439 00:12:28,390 –> 00:12:29,890 I hope I see you there and
440 00:12:29,900 –> 00:12:31,500 of course, I wish you a nice
441 00:12:31,510 –> 00:12:31,929 day.
442 00:12:32,229 –> 00:12:32,849 Bye.

Quiz Content
Q1: Let $(X, \mathcal{A}, \mu)$ be a measure space. The monotone convergence theorem tells us something about a sequence of functions. What is not a premise of this theorem?
A1: $f_1 \leq f_2 \leq f_3 \leq \cdots$ holds $\mu\text{a.e}$.
A2: $f_n: X \rightarrow [0,\infty)$ is measurable.
A3: $\int_X f_n , d\mu < \infty$.
A4: $\lim_{n \rightarrow \infty} f_n(x) = f(x)$ holds $\mu\text{a.e}$.
Q2: Let $(X, \mathcal{A}, \mu)$ be a measure space. Is the following implication correct? $$ f_1 \leq f_2 \leq f_3 \leq \cdots ~ \mu\text{a.e}$$ $$~\Rightarrow~$$ $$\int_X f_1 \leq \int_X f_2 \leq \int_X f_3 \leq \cdots $$
A1: Yes
A2: No
Q3: Let $(X, \mathcal{A}, \mu)$ be a measure space. If for all simple functions $h$ with $h \leq f$ we have $$ 5 \geq \int_X h , d \mu , ,$$ then we also have $$ 5 \geq \int_X f , d \mu , .$$
A1: Yes, the statement is correct.
A2: No, this statement is not correct.
Q4: Let $(X, \mathcal{A}, \mu)$ be a measure space. If you have positive functions $g_n$ that are measurable. Is the following implication correct? $$ \int_X \sum_{n=1}^\infty g_n , d \mu ~\Rightarrow~ \sum_{n=1}^\infty \int_X g_n , d \mu$$
A1: Yes, the statement is correct.
A2: No, this statement is not correct.

Last update: 202410