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Title: Monotone Convergence Theorem (and more)
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Series: Measure Theory
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YouTube-Title: Measure Theory 7 | Monotone Convergence Theorem (and more)
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Subtitle on GitHub: mt07_sub_eng.srt
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Subtitle in English
1 00:00:00,860 –> 00:00:02,789 Hello and welcome back.
2 00:00:03,200 –> 00:00:04,920 And I want to thank all the
3 00:00:04,929 –> 00:00:06,480 nice people that support
4 00:00:06,489 –> 00:00:08,159 this channel on Steady
5 00:00:09,180 –> 00:00:10,220 here we continue
6 00:00:10,229 –> 00:00:11,869 now our measure theory
7 00:00:11,880 –> 00:00:13,720 series and today
8 00:00:13,729 –> 00:00:15,600 we do part seven.
9 00:00:16,530 –> 00:00:17,889 The most important topic
10 00:00:17,899 –> 00:00:19,409 for today will be the
11 00:00:19,420 –> 00:00:20,930 monotone convergence
12 00:00:20,940 –> 00:00:21,469 theorem.
13 00:00:22,399 –> 00:00:24,319 Before explaining this very
14 00:00:24,329 –> 00:00:26,170 important convergence theorem,
15 00:00:26,379 –> 00:00:27,860 I first want to start
16 00:00:27,870 –> 00:00:29,670 showing you some essential
17 00:00:29,680 –> 00:00:30,870 properties of the Lebesgue
18 00:00:31,100 –> 00:00:31,719 integral.
19 00:00:32,529 –> 00:00:34,450 Recall, we have introduced
20 00:00:34,459 –> 00:00:36,349 the Lebesgue integral for
21 00:00:36,360 –> 00:00:36,700 non-negative
22 00:00:37,349 –> 00:00:38,669 measurable
23 00:00:38,680 –> 00:00:40,669 functions defined on
24 00:00:40,680 –> 00:00:42,610 some measure space X.
25 00:00:43,220 –> 00:00:44,389 For such functions
26 00:00:44,400 –> 00:00:45,880 We now know that the Lebesgue
27 00:00:45,990 –> 00:00:47,909 integral is well-defined.
28 00:00:48,580 –> 00:00:49,669 And the notation we have
29 00:00:49,680 –> 00:00:51,439 chosen is this integral
30 00:00:51,450 –> 00:00:52,990 symbol where we put the measure
31 00:00:53,000 –> 00:00:54,310 space X here.
32 00:00:54,319 –> 00:00:56,150 The function F here and the
33 00:00:56,159 –> 00:00:58,080 measure itself goes in with
34 00:00:58,090 –> 00:00:59,959 the mu where mu is the
35 00:00:59,970 –> 00:01:01,680 measure defined on X.
36 00:01:02,970 –> 00:01:04,540 The integral is well-defined
37 00:01:04,550 –> 00:01:06,510 Now means that this
38 00:01:06,519 –> 00:01:08,459 symbol is a number
39 00:01:08,519 –> 00:01:09,839 between zero and
40 00:01:09,849 –> 00:01:10,809 infinity.
41 00:01:10,910 –> 00:01:12,239 And in the worst case, it
42 00:01:12,250 –> 00:01:14,150 could be the symbol infinity.
43 00:01:15,330 –> 00:01:16,750 Also recall
44 00:01:17,010 –> 00:01:18,949 this symbol was defined
45 00:01:18,959 –> 00:01:20,940 by a supremum over all
46 00:01:20,949 –> 00:01:22,699 step functions that lie point-
47 00:01:22,709 –> 00:01:24,379 wisely below F
48 00:01:25,489 –> 00:01:26,150 OK.
49 00:01:26,160 –> 00:01:27,970 Now I want to collect some
50 00:01:27,980 –> 00:01:29,540 properties that follow
51 00:01:29,550 –> 00:01:31,430 immediately from this definition.
52 00:01:32,230 –> 00:01:34,110 So let’s choose two of
53 00:01:34,120 –> 00:01:35,790 these nice functions,
54 00:01:35,800 –> 00:01:37,500 which means they are non-
55 00:01:37,510 –> 00:01:39,309 negative and also
56 00:01:39,319 –> 00:01:40,309 measurable.
57 00:01:41,190 –> 00:01:42,930 Now we have the following.
58 00:01:42,940 –> 00:01:44,849 The first thing would be
59 00:01:44,860 –> 00:01:46,290 if both functions
60 00:01:46,300 –> 00:01:47,389 coincide,
61 00:01:48,190 –> 00:01:49,650 then also the
62 00:01:49,660 –> 00:01:51,290 integrals coincide.
63 00:01:52,360 –> 00:01:53,989 However, of course, this
64 00:01:54,000 –> 00:01:55,470 is trivially fulfilled.
65 00:01:56,400 –> 00:01:58,019 Therefore, I want to weaken
66 00:01:58,029 –> 00:01:59,129 the left hand side a little
67 00:01:59,139 –> 00:02:01,040 bit here, I want
68 00:02:01,050 –> 00:02:02,639 the functions to be equal
69 00:02:02,650 –> 00:02:03,849 mu-almost
70 00:02:03,860 –> 00:02:04,699 everywhere.
71 00:02:05,610 –> 00:02:07,169 The abbreviation one chooses
72 00:02:07,180 –> 00:02:08,949 here is always A
73 00:02:09,589 –> 00:02:10,029 E.
74 00:02:11,089 –> 00:02:13,009 So mu A E means
75 00:02:13,020 –> 00:02:14,910 mu-almost everywhere.
76 00:02:15,940 –> 00:02:17,350 This means now that the
77 00:02:17,360 –> 00:02:19,320 functions don’t have to be
78 00:02:19,330 –> 00:02:21,199 equal, but they should be
79 00:02:21,210 –> 00:02:23,119 equal almost everywhere
80 00:02:23,130 –> 00:02:24,839 with respect to the measure
81 00:02:24,850 –> 00:02:26,839 mu. more
82 00:02:26,850 –> 00:02:27,679 concretely
83 00:02:27,690 –> 00:02:29,259 This whole property here
84 00:02:29,270 –> 00:02:30,559 means that you
85 00:02:30,570 –> 00:02:32,160 look at the set
86 00:02:32,169 –> 00:02:33,399 of all
87 00:02:33,410 –> 00:02:35,210 lowercase X in
88 00:02:35,220 –> 00:02:37,139 our X for
89 00:02:37,149 –> 00:02:38,830 which this property
90 00:02:38,839 –> 00:02:40,119 does not hold.
91 00:02:40,130 –> 00:02:41,770 So you have F of
92 00:02:41,779 –> 00:02:43,679 X unequal to
93 00:02:43,690 –> 00:02:44,919 G of X,
94 00:02:45,600 –> 00:02:47,220 then this whole set
95 00:02:47,229 –> 00:02:48,919 should be of measure
96 00:02:48,929 –> 00:02:50,619 zero with respect to our
97 00:02:50,630 –> 00:02:51,460 measure mu.
98 00:02:52,490 –> 00:02:53,770 In other words, if you put
99 00:02:53,779 –> 00:02:55,369 the whole set into the measure,
100 00:02:55,699 –> 00:02:57,210 you get out zero.
101 00:02:58,809 –> 00:03:00,210 This means that the Lebesgue
102 00:03:00,220 –> 00:03:01,800 integral cannot see
103 00:03:01,809 –> 00:03:03,559 things that happen on
104 00:03:03,570 –> 00:03:04,869 zero measure sets.
105 00:03:05,960 –> 00:03:07,509 Maybe for visualization,
106 00:03:07,520 –> 00:03:09,110 it’s good to recall what
107 00:03:09,119 –> 00:03:11,020 we had for the Riemann integral.
108 00:03:12,339 –> 00:03:13,710 If you integrate a
109 00:03:13,720 –> 00:03:15,509 continuous function with
110 00:03:15,520 –> 00:03:17,339 our classical Riemann integral,
111 00:03:18,240 –> 00:03:19,460 you get out the
112 00:03:19,470 –> 00:03:21,139 area between
113 00:03:21,279 –> 00:03:23,130 the graph function and the
114 00:03:23,139 –> 00:03:23,839 X axis.
115 00:03:24,529 –> 00:03:26,100 And now if you change the
116 00:03:26,110 –> 00:03:27,789 function at one point,
117 00:03:27,979 –> 00:03:29,970 so the result is a non-
118 00:03:29,979 –> 00:03:31,710 continuous function now
119 00:03:32,490 –> 00:03:34,139 and then of course, you don’t
120 00:03:34,149 –> 00:03:36,029 change the area here at
121 00:03:36,039 –> 00:03:37,729 all, which then
122 00:03:37,740 –> 00:03:39,669 means the integral is
123 00:03:39,679 –> 00:03:41,410 the same for
124 00:03:41,419 –> 00:03:42,240 Lebesgue integral.
125 00:03:42,250 –> 00:03:43,649 Now, this works in an
126 00:03:43,660 –> 00:03:44,889 abstract sense
127 00:03:45,899 –> 00:03:47,009 which means that you can
128 00:03:47,020 –> 00:03:48,720 change the function as much
129 00:03:48,729 –> 00:03:49,559 as you want.
130 00:03:49,570 –> 00:03:51,460 As long as the set
131 00:03:51,639 –> 00:03:53,339 of all these changes
132 00:03:53,350 –> 00:03:54,679 has measure zero.
133 00:03:55,679 –> 00:03:57,039 I will talk about this later
134 00:03:57,050 –> 00:03:57,899 in another video.
135 00:03:57,919 –> 00:03:59,380 When I show you a lot of
136 00:03:59,389 –> 00:04:01,190 examples. Here, I
137 00:04:01,199 –> 00:04:02,500 want to continue with the
138 00:04:02,509 –> 00:04:04,389 second property which is
139 00:04:04,399 –> 00:04:05,539 the monotonicity.
140 00:04:06,080 –> 00:04:07,699 It is descriptively given
141 00:04:07,710 –> 00:04:09,699 by saying that if a function
142 00:04:09,710 –> 00:04:11,449 is bigger than another one,
143 00:04:11,460 –> 00:04:13,419 then also the area here
144 00:04:13,429 –> 00:04:14,320 should be bigger.
145 00:04:15,179 –> 00:04:17,010 However, as before we
146 00:04:17,019 –> 00:04:18,450 need the less of equal sign
147 00:04:18,459 –> 00:04:19,790 only almost
148 00:04:19,798 –> 00:04:20,488 everywhere.
149 00:04:21,450 –> 00:04:22,890 And then we can conclude
150 00:04:22,899 –> 00:04:24,049 that the integral of
151 00:04:24,059 –> 00:04:25,700 F is less or
152 00:04:25,709 –> 00:04:27,700 equal than the integral of
153 00:04:27,709 –> 00:04:28,130 G.
154 00:04:29,970 –> 00:04:31,790 And now to the last property
155 00:04:31,799 –> 00:04:33,600 which is related to the first
156 00:04:33,609 –> 00:04:35,350 one, I want my function
157 00:04:35,359 –> 00:04:36,950 F to be zero
158 00:04:37,119 –> 00:04:38,790 and also mu-
159 00:04:38,799 –> 00:04:40,230 almost everywhere.
160 00:04:41,070 –> 00:04:42,989 Now, by using a, I know
161 00:04:43,000 –> 00:04:44,859 that the integral of F
162 00:04:44,869 –> 00:04:46,500 is therefore also
163 00:04:46,510 –> 00:04:48,429 zero because the integral
164 00:04:48,440 –> 00:04:49,890 of the zero function is zero.
165 00:04:50,760 –> 00:04:52,649 However, here I want to emphasize
166 00:04:52,660 –> 00:04:53,829 another direction.
167 00:04:54,640 –> 00:04:56,100 If the function has an integral
168 00:04:56,109 –> 00:04:57,750 of zero, then we
169 00:04:57,760 –> 00:04:59,459 already know that the function
170 00:04:59,470 –> 00:05:01,109 has to be zero mu-
171 00:05:01,119 –> 00:05:02,179 almost everywhere.
172 00:05:02,959 –> 00:05:04,260 Please don’t forget that
173 00:05:04,269 –> 00:05:05,760 we only consider non-negative
174 00:05:06,130 –> 00:05:06,980 functions here.
175 00:05:07,769 –> 00:05:09,480 So all the areas we consider
176 00:05:09,489 –> 00:05:11,320 are above the X axis and
177 00:05:11,329 –> 00:05:12,230 therefore positive.
178 00:05:13,250 –> 00:05:15,019 Therefore, there is no cancellation
179 00:05:15,029 –> 00:05:16,700 whatsoever with positive
180 00:05:16,709 –> 00:05:17,429 and negative
181 00:05:17,440 –> 00:05:18,230 Areas here.
182 00:05:19,519 –> 00:05:21,230 Well, now I could say
183 00:05:21,239 –> 00:05:22,500 all the three properties
184 00:05:22,510 –> 00:05:23,829 are easy to prove.
185 00:05:23,839 –> 00:05:24,950 So you could do this for
186 00:05:24,959 –> 00:05:25,660 yourself.
187 00:05:26,130 –> 00:05:27,739 However, I also want you
188 00:05:27,750 –> 00:05:29,230 to learn some
189 00:05:29,239 –> 00:05:30,549 technical steps here.
190 00:05:30,559 –> 00:05:32,429 And therefore I will do the
191 00:05:32,440 –> 00:05:33,829 proof of B
192 00:05:34,790 –> 00:05:36,279 after seeing this proof,
193 00:05:36,290 –> 00:05:37,730 I think you will be
194 00:05:37,739 –> 00:05:39,369 capable of doing the proof
195 00:05:39,380 –> 00:05:41,170 A and C for yourself.
196 00:05:42,350 –> 00:05:44,190 If not, please ask in the
197 00:05:44,200 –> 00:05:46,040 comments for
198 00:05:46,049 –> 00:05:47,220 starting the proof.
199 00:05:47,230 –> 00:05:48,369 Let us choose a
200 00:05:48,380 –> 00:05:50,200 simple function h.
201 00:05:51,290 –> 00:05:53,130 This means that we have a
202 00:05:53,140 –> 00:05:54,549 representation with
203 00:05:54,559 –> 00:05:55,980 characteristic functions.
204 00:05:56,970 –> 00:05:58,489 So we know we can write it
205 00:05:58,579 –> 00:05:59,910 as a finite
206 00:05:59,920 –> 00:06:00,690 sum.
207 00:06:00,700 –> 00:06:02,510 Maybe we end with N
208 00:06:03,049 –> 00:06:04,500 and we have constant C
209 00:06:04,510 –> 00:06:06,309 I and also sets
210 00:06:06,320 –> 00:06:06,959 A I.
211 00:06:06,970 –> 00:06:08,750 When we put that into
212 00:06:08,760 –> 00:06:10,179 characteristic functions,
213 00:06:12,130 –> 00:06:13,579 the corresponding sets here
214 00:06:13,589 –> 00:06:15,029 should be measurable.
215 00:06:15,040 –> 00:06:16,450 And therefore, the simple
216 00:06:16,459 –> 00:06:17,690 function is always
217 00:06:17,700 –> 00:06:18,579 measurable.
218 00:06:19,410 –> 00:06:20,920 And the other property of
219 00:06:20,929 –> 00:06:22,470 a simple function is that
220 00:06:22,480 –> 00:06:23,750 it only has
221 00:06:23,760 –> 00:06:25,130 finitely many
222 00:06:25,140 –> 00:06:26,019 values.
223 00:06:26,929 –> 00:06:28,339 Therefore, there’s
224 00:06:28,350 –> 00:06:30,209 always a canonical way to
225 00:06:30,220 –> 00:06:32,100 write down a representation
226 00:06:32,109 –> 00:06:33,589 for a simple function.
227 00:06:34,010 –> 00:06:35,769 You’re forming the sum over
228 00:06:35,779 –> 00:06:36,959 all the values.
229 00:06:36,970 –> 00:06:38,309 So you have T
230 00:06:38,320 –> 00:06:40,130 in and now comes
231 00:06:40,140 –> 00:06:42,040 the image of the simple function.
232 00:06:42,049 –> 00:06:43,899 And I write that as h
233 00:06:43,910 –> 00:06:45,230 of X,
234 00:06:46,269 –> 00:06:47,890 then instead of C, we just
235 00:06:47,899 –> 00:06:49,790 have this T and
236 00:06:49,799 –> 00:06:51,769 then comes the characteristic
237 00:06:51,779 –> 00:06:53,269 function of a
238 00:06:53,279 –> 00:06:54,570 set I will now
239 00:06:54,579 –> 00:06:55,279 describe
240 00:06:56,390 –> 00:06:58,089 there we have all the X
241 00:06:58,100 –> 00:06:59,529 points that
242 00:06:59,540 –> 00:07:00,989 fulfill that.
243 00:07:01,000 –> 00:07:02,570 If I put in this
244 00:07:02,579 –> 00:07:04,380 point in my function, I
245 00:07:04,390 –> 00:07:06,209 get out this special
246 00:07:06,220 –> 00:07:06,690 T.
247 00:07:08,890 –> 00:07:10,019 Do you see this is a
248 00:07:10,029 –> 00:07:11,410 representation that is
249 00:07:11,420 –> 00:07:12,109 allowed.
250 00:07:12,750 –> 00:07:14,450 And often it is very helpful
251 00:07:14,459 –> 00:07:15,970 to choose this special one
252 00:07:16,200 –> 00:07:17,239 because you don’t have to
253 00:07:17,250 –> 00:07:19,170 define N or the CIS or
254 00:07:19,179 –> 00:07:20,970 the A is you just write down
255 00:07:20,980 –> 00:07:21,519 this representation.
256 00:07:23,600 –> 00:07:25,579 And also the integral
257 00:07:25,589 –> 00:07:27,339 can then easily be written
258 00:07:27,350 –> 00:07:29,309 down as
259 00:07:29,320 –> 00:07:30,799 always, it is defined by
260 00:07:30,809 –> 00:07:31,720 our sum.
261 00:07:31,730 –> 00:07:33,630 And our summation goes
262 00:07:33,640 –> 00:07:35,429 over T in the
263 00:07:35,440 –> 00:07:36,739 image of H
264 00:07:37,579 –> 00:07:38,869 and then simply T
265 00:07:38,880 –> 00:07:40,829 times the measure of
266 00:07:40,839 –> 00:07:41,450 this set.
267 00:07:41,459 –> 00:07:42,119 So we have
268 00:07:42,130 –> 00:07:43,929 mu of the
269 00:07:43,940 –> 00:07:45,769 whole set where I write all
270 00:07:45,779 –> 00:07:47,119 the X that
271 00:07:47,130 –> 00:07:48,880 fulfill HX
272 00:07:48,920 –> 00:07:50,820 equals to this T.
273 00:07:52,089 –> 00:07:53,250 And to make it a little bit
274 00:07:53,260 –> 00:07:54,649 easier, you can
275 00:07:54,660 –> 00:07:56,309 always ignore the zero.
276 00:07:56,320 –> 00:07:57,799 So you omit the zero as an
277 00:07:57,809 –> 00:07:59,350 element in the image.
278 00:07:59,359 –> 00:08:00,589 And because you don’t change
279 00:08:00,600 –> 00:08:02,100 anything for the integral
280 00:08:02,109 –> 00:08:03,519 here by multiplying with
281 00:08:03,529 –> 00:08:04,070 zero.
282 00:08:05,570 –> 00:08:07,029 The question for you is now
283 00:08:07,260 –> 00:08:08,910 what happens if we
284 00:08:08,920 –> 00:08:10,630 change the simple
285 00:08:10,640 –> 00:08:12,470 function on a
286 00:08:12,480 –> 00:08:14,220 set which has measure
287 00:08:14,230 –> 00:08:15,989 zero in
288 00:08:16,000 –> 00:08:17,309 order to investigate this,
289 00:08:17,320 –> 00:08:18,369 I need a little space here.
290 00:08:18,380 –> 00:08:20,309 So we push this one
291 00:08:20,320 –> 00:08:21,510 here to the bottom now.
292 00:08:22,540 –> 00:08:24,089 OK, maybe a quick sketch
293 00:08:24,100 –> 00:08:25,040 is helpful here.
294 00:08:25,170 –> 00:08:26,720 So this is the whole measure
295 00:08:26,730 –> 00:08:27,920 space X.
296 00:08:28,890 –> 00:08:30,350 And now let us split that
297 00:08:30,359 –> 00:08:31,670 up into two sets.
298 00:08:31,679 –> 00:08:33,669 So this one would be the
299 00:08:33,679 –> 00:08:35,390 big X tilde.
300 00:08:36,669 –> 00:08:38,609 And of course, the complement,
301 00:08:38,619 –> 00:08:40,249 the whole rest is just
302 00:08:40,258 –> 00:08:41,979 X tilde c.
303 00:08:41,989 –> 00:08:43,508 The compliment of X tilde,
304 00:08:44,280 –> 00:08:45,619 this means that our x is
305 00:08:45,630 –> 00:08:47,140 now divided into two
306 00:08:47,150 –> 00:08:47,820 sets.
307 00:08:49,320 –> 00:08:50,969 We want that
308 00:08:50,979 –> 00:08:52,820 the x tilde
309 00:08:52,830 –> 00:08:54,700 complement has
310 00:08:54,710 –> 00:08:55,820 measured zero.
311 00:08:58,349 –> 00:08:59,969 And with respect to this
312 00:08:59,979 –> 00:09:01,289 set, with measure zero, I
313 00:09:01,299 –> 00:09:02,770 want to change our simple
314 00:09:02,780 –> 00:09:03,210 function.
315 00:09:04,010 –> 00:09:05,179 And of course, I call it
316 00:09:05,229 –> 00:09:07,140 H tilde and define it
317 00:09:07,150 –> 00:09:08,419 by using two
318 00:09:08,429 –> 00:09:09,200 cases.
319 00:09:09,679 –> 00:09:10,940 The first case would be as
320 00:09:10,950 –> 00:09:11,530 before.
321 00:09:11,539 –> 00:09:13,469 So H tilde X
322 00:09:13,479 –> 00:09:15,250 is equal to HX for
323 00:09:15,260 –> 00:09:17,250 all X in
324 00:09:17,390 –> 00:09:18,340 X tilde.
325 00:09:19,239 –> 00:09:20,880 So nothing changes on the
326 00:09:20,890 –> 00:09:21,559 green set.
327 00:09:22,409 –> 00:09:23,929 But on the gray set, I will
328 00:09:23,940 –> 00:09:25,809 set it to a new value And
329 00:09:25,820 –> 00:09:27,190 I just choose an A.
330 00:09:27,320 –> 00:09:29,289 So this is for all X
331 00:09:29,299 –> 00:09:30,349 in X tilde
332 00:09:30,359 –> 00:09:32,190 compliment, and
333 00:09:32,200 –> 00:09:33,869 A is now just an arbitrary
334 00:09:33,880 –> 00:09:35,739 chosen positive number.
335 00:09:35,750 –> 00:09:37,429 So zero to infinity
336 00:09:39,000 –> 00:09:40,219 by definition, this is of
337 00:09:40,229 –> 00:09:41,770 course, again, a simple
338 00:09:41,780 –> 00:09:43,760 function because
339 00:09:43,770 –> 00:09:45,650 the set X tilde, I didn’t
340 00:09:45,659 –> 00:09:46,090 say it.
341 00:09:46,099 –> 00:09:47,169 But of course, it should
342 00:09:47,179 –> 00:09:48,739 be in the Sigma algebra.
343 00:09:48,750 –> 00:09:50,419 So it should be measurable.
344 00:09:51,130 –> 00:09:52,789 And then we can write down
345 00:09:52,799 –> 00:09:54,570 again a representation
346 00:09:54,580 –> 00:09:56,169 for this step or simple
347 00:09:56,179 –> 00:09:56,630 function.
348 00:09:57,270 –> 00:09:58,739 I can use the representation
349 00:09:58,750 –> 00:09:59,530 from before.
350 00:09:59,539 –> 00:10:01,409 So now let’s go over all
351 00:10:01,419 –> 00:10:02,929 the values of the
352 00:10:03,150 –> 00:10:04,849 original function h.
353 00:10:05,789 –> 00:10:07,090 And now I know it only
354 00:10:07,099 –> 00:10:08,940 occurs for X in
355 00:10:08,950 –> 00:10:09,539 X tilde.
356 00:10:09,549 –> 00:10:11,289 So I write down T
357 00:10:11,299 –> 00:10:12,729 times characteristic
358 00:10:12,739 –> 00:10:13,359 function.
359 00:10:13,510 –> 00:10:15,409 And now I put only the
360 00:10:15,419 –> 00:10:17,320 XS from X Tilde
361 00:10:17,570 –> 00:10:19,000 into this characteristic
362 00:10:19,010 –> 00:10:19,450 function.
363 00:10:20,080 –> 00:10:21,549 Not included in this sum
364 00:10:21,559 –> 00:10:22,979 is what happens outside of
365 00:10:22,989 –> 00:10:23,630 X tilde.
366 00:10:23,640 –> 00:10:25,380 And therefore we add
367 00:10:25,429 –> 00:10:26,820 this one as the value
368 00:10:26,830 –> 00:10:28,179 a times
369 00:10:28,239 –> 00:10:29,900 characteristic function.
370 00:10:29,960 –> 00:10:31,770 And now we have all the
371 00:10:31,780 –> 00:10:32,409 X
372 00:10:33,539 –> 00:10:35,070 in X where
373 00:10:35,080 –> 00:10:36,809 H tilde is equal to
374 00:10:36,820 –> 00:10:38,599 a but this is simply
375 00:10:38,609 –> 00:10:40,070 X tilde complement.
376 00:10:40,359 –> 00:10:42,049 So we can write in
377 00:10:42,059 –> 00:10:43,739 short just this
378 00:10:43,750 –> 00:10:45,270 X tilde compliment here.
379 00:10:46,390 –> 00:10:46,940 OK.
380 00:10:47,039 –> 00:10:48,570 I have explicitly written
381 00:10:48,580 –> 00:10:50,559 down the representation
382 00:10:50,609 –> 00:10:52,210 because then we can write
383 00:10:52,219 –> 00:10:54,010 down the integral as well.
384 00:10:54,210 –> 00:10:55,919 Now I of
385 00:10:55,929 –> 00:10:56,500 H tilde
386 00:10:58,429 –> 00:10:59,979 equal to the
387 00:10:59,989 –> 00:11:01,580 sum with T
388 00:11:01,630 –> 00:11:02,280 over
389 00:11:02,289 –> 00:11:03,940 HX
390 00:11:04,760 –> 00:11:06,450 and you will have t times
391 00:11:06,460 –> 00:11:08,150 the measure of this
392 00:11:08,159 –> 00:11:08,679 set
393 00:11:10,280 –> 00:11:12,250 plus a times
394 00:11:12,260 –> 00:11:13,890 the measure of this set,
395 00:11:15,140 –> 00:11:17,039 which is again, X tilde
396 00:11:17,049 –> 00:11:18,390 the compliment.
397 00:11:19,419 –> 00:11:21,359 However, you already know
398 00:11:21,380 –> 00:11:23,229 the measure of X tilde compliment
399 00:11:23,239 –> 00:11:23,799 is zero.
400 00:11:23,809 –> 00:11:25,270 So this whole thing here
401 00:11:25,280 –> 00:11:27,210 on the right, it’s
402 00:11:27,219 –> 00:11:28,510 still zero.
403 00:11:29,469 –> 00:11:30,929 That simply means we can
404 00:11:30,940 –> 00:11:32,090 just omit it.
405 00:11:32,780 –> 00:11:34,179 And now you should see the
406 00:11:34,190 –> 00:11:36,140 differences above and below
407 00:11:36,320 –> 00:11:37,719 are not so big.
408 00:11:38,090 –> 00:11:39,419 The only real difference
409 00:11:39,429 –> 00:11:40,549 is to tilde the here
410 00:11:40,780 –> 00:11:42,340 because the zero does not
411 00:11:42,349 –> 00:11:43,280 make any difference.
412 00:11:43,289 –> 00:11:44,270 As I said before,
413 00:11:45,270 –> 00:11:46,929 our task is now to
414 00:11:46,940 –> 00:11:48,190 fill in the middle ground
415 00:11:48,200 –> 00:11:48,549 here.
416 00:11:49,909 –> 00:11:51,169 Now, it depends what you
417 00:11:51,179 –> 00:11:52,049 find easier.
418 00:11:52,059 –> 00:11:53,309 So maybe we come from the
419 00:11:53,320 –> 00:11:55,080 bottom and just use
420 00:11:55,789 –> 00:11:56,780 what we know there.
421 00:11:56,789 –> 00:11:57,530 So we have
422 00:11:58,429 –> 00:11:59,369 mu of
423 00:12:00,590 –> 00:12:02,380 the whole X
424 00:12:02,700 –> 00:12:03,580 and now we just split it
425 00:12:03,590 –> 00:12:04,919 up into X tilde
426 00:12:06,289 –> 00:12:08,000 where the condition is fulfilled
427 00:12:09,400 –> 00:12:10,359 union.
428 00:12:10,979 –> 00:12:12,820 And here we have
429 00:12:12,830 –> 00:12:14,409 just x tilde
430 00:12:14,419 –> 00:12:15,169 C.
431 00:12:18,250 –> 00:12:19,960 Now, obviously this is the
432 00:12:19,969 –> 00:12:21,409 same as the bottom part.
433 00:12:21,609 –> 00:12:23,169 And we know it’s a disjoint
434 00:12:23,179 –> 00:12:25,119 union which means we
435 00:12:25,130 –> 00:12:27,030 can easily use the sigma
436 00:12:27,039 –> 00:12:27,919 additivity here.
437 00:12:29,200 –> 00:12:31,099 Therefore, parentheses here
438 00:12:31,349 –> 00:12:32,539 and here I write
439 00:12:32,549 –> 00:12:33,950 plus or
440 00:12:33,960 –> 00:12:35,940 maybe first close
441 00:12:35,950 –> 00:12:37,760 parenthesis here as well.
442 00:12:38,659 –> 00:12:40,039 So this would be black
443 00:12:40,989 –> 00:12:42,960 then plus
444 00:12:43,140 –> 00:12:44,750 mu of
445 00:12:45,179 –> 00:12:46,090 this set.
446 00:12:47,570 –> 00:12:49,229 However, we already know
447 00:12:49,239 –> 00:12:51,150 that X tilde complement
448 00:12:51,219 –> 00:12:53,030 is a set with measure
449 00:12:53,039 –> 00:12:54,989 zero with or without this
450 00:12:55,000 –> 00:12:56,679 condition, it only gets
451 00:12:56,690 –> 00:12:58,070 smaller with the condition,
452 00:12:58,080 –> 00:12:59,989 which means this is still
453 00:13:00,000 –> 00:13:01,429 a set with measure
454 00:13:01,440 –> 00:13:02,030 zero.
455 00:13:03,460 –> 00:13:05,159 And now we have all the
456 00:13:05,169 –> 00:13:06,239 equalities here.
457 00:13:06,580 –> 00:13:08,059 We just add a zero here.
458 00:13:08,070 –> 00:13:09,179 So this is equal and this
459 00:13:09,190 –> 00:13:11,109 is equal, everything is equal,
460 00:13:11,119 –> 00:13:12,820 which means the integral
461 00:13:12,830 –> 00:13:14,710 of H tilde is equal to
462 00:13:14,719 –> 00:13:15,880 the integral of H.
463 00:13:16,369 –> 00:13:17,919 This means now that we can
464 00:13:17,929 –> 00:13:19,510 change the simple function
465 00:13:19,520 –> 00:13:21,219 on a set with measure
466 00:13:21,229 –> 00:13:22,919 zero as much as we want.
467 00:13:24,080 –> 00:13:25,780 Indeed, this is a very important
468 00:13:25,789 –> 00:13:26,469 result.
469 00:13:26,479 –> 00:13:27,900 And you can use that
470 00:13:28,090 –> 00:13:29,940 for proving some of these
471 00:13:29,950 –> 00:13:30,760 parts here.
472 00:13:30,789 –> 00:13:32,549 But we wanted to prove part
473 00:13:32,559 –> 00:13:33,010 B
474 00:13:34,950 –> 00:13:36,650 there, we have two measurable
475 00:13:36,659 –> 00:13:38,169 functions whether one
476 00:13:38,179 –> 00:13:39,820 is bigger than the other
477 00:13:39,830 –> 00:13:41,570 one almost everywhere.
478 00:13:42,400 –> 00:13:44,059 Therefore, now we know how
479 00:13:44,070 –> 00:13:45,690 to choose our x tilde.
480 00:13:48,510 –> 00:13:50,229 It simply should be the
481 00:13:50,239 –> 00:13:51,400 set of all x
482 00:13:53,599 –> 00:13:55,109 where F of
483 00:13:55,119 –> 00:13:56,700 X is less or
484 00:13:56,710 –> 00:13:58,469 equal than G of
485 00:13:58,479 –> 00:13:59,239 X.
486 00:14:00,340 –> 00:14:02,169 Then by assumption we
487 00:14:02,179 –> 00:14:04,020 also know that X tilde
488 00:14:04,059 –> 00:14:05,640 complement has
489 00:14:05,650 –> 00:14:06,859 measure zero.
490 00:14:07,359 –> 00:14:09,140 Well, then let’s look at
491 00:14:09,150 –> 00:14:10,960 the integral of F
492 00:14:12,049 –> 00:14:13,419 by definition, it’s the
493 00:14:13,429 –> 00:14:14,849 supremum of
494 00:14:14,859 –> 00:14:16,659 all integrals of step
495 00:14:16,669 –> 00:14:17,510 functions.
496 00:14:18,260 –> 00:14:20,200 And we denoted them by S
497 00:14:20,210 –> 00:14:22,159 plus where the
498 00:14:22,169 –> 00:14:23,500 step function or the simple
499 00:14:23,510 –> 00:14:25,219 function is less or
500 00:14:25,229 –> 00:14:26,909 equal than F.
501 00:14:28,380 –> 00:14:29,869 Now by the results from
502 00:14:29,880 –> 00:14:31,650 before we know that we
503 00:14:31,659 –> 00:14:33,340 don’t change the integral
504 00:14:33,349 –> 00:14:35,000 value when we change the
505 00:14:35,010 –> 00:14:36,739 step function on a
506 00:14:36,750 –> 00:14:38,260 set with measure zero.
507 00:14:38,909 –> 00:14:40,179 Therefore, the supremum is
508 00:14:40,190 –> 00:14:41,719 still the same when I put
509 00:14:41,729 –> 00:14:43,299 in our step functions
510 00:14:43,549 –> 00:14:45,080 where I have H
511 00:14:45,119 –> 00:14:46,760 tilde less than
512 00:14:46,770 –> 00:14:48,159 F but
513 00:14:48,169 –> 00:14:49,929 only on
514 00:14:49,940 –> 00:14:50,880 X tilde.
515 00:14:53,169 –> 00:14:54,429 That is the whole point.
516 00:14:54,520 –> 00:14:56,289 If we change something outside
517 00:14:56,299 –> 00:14:57,989 of X tilde, we
518 00:14:58,000 –> 00:14:59,789 can’t change I of
519 00:14:59,799 –> 00:15:00,690 H tilde.
520 00:15:00,849 –> 00:15:02,820 So the whole supremum is the
521 00:15:02,830 –> 00:15:04,059 same as the supremum here
522 00:15:04,070 –> 00:15:04,710 on the left.
523 00:15:05,830 –> 00:15:06,609 Very good.
524 00:15:06,619 –> 00:15:08,200 And now we can use what we
525 00:15:08,210 –> 00:15:09,859 know from F and
526 00:15:09,869 –> 00:15:11,549 G on
527 00:15:11,559 –> 00:15:13,500 X tilde G
528 00:15:13,510 –> 00:15:14,969 is always bigger than F.
529 00:15:14,979 –> 00:15:16,419 So we have here
530 00:15:16,429 –> 00:15:17,719 always this
531 00:15:17,729 –> 00:15:18,500 inequality.
532 00:15:19,359 –> 00:15:21,090 Hence, if we write down the
533 00:15:21,099 –> 00:15:22,549 same thing again,
534 00:15:23,070 –> 00:15:24,659 but now with G
535 00:15:24,669 –> 00:15:26,020 instead of F,
536 00:15:26,200 –> 00:15:27,989 then this set gets
537 00:15:28,000 –> 00:15:29,830 bigger than this set because
538 00:15:29,840 –> 00:15:31,640 there are more step functions
539 00:15:31,650 –> 00:15:33,640 maybe inside this set.
540 00:15:33,979 –> 00:15:35,200 Therefore, we have an
541 00:15:35,210 –> 00:15:37,010 inequality at
542 00:15:37,020 –> 00:15:37,869 this point.
543 00:15:40,039 –> 00:15:41,500 Now with the same reasoning
544 00:15:41,510 –> 00:15:43,309 as before the supremum
545 00:15:43,580 –> 00:15:45,059 is equal to the
546 00:15:45,070 –> 00:15:46,739 supremum where I
547 00:15:46,750 –> 00:15:48,739 ignore x tilde and
548 00:15:48,750 –> 00:15:50,049 use step functions
549 00:15:50,690 –> 00:15:52,289 on the whole set X.
550 00:15:53,530 –> 00:15:54,940 And this one is
551 00:15:54,950 –> 00:15:56,549 exactly the definition of
552 00:15:56,559 –> 00:15:58,440 the integral of
553 00:15:58,450 –> 00:15:59,859 our function G.
554 00:16:01,340 –> 00:16:02,539 And if we put everything
555 00:16:02,549 –> 00:16:04,380 together, so this one, the
556 00:16:04,390 –> 00:16:06,070 inequality and this
557 00:16:06,080 –> 00:16:08,000 one, we have proven our
558 00:16:08,010 –> 00:16:09,950 claim and this
559 00:16:09,960 –> 00:16:11,789 is the monotonicity property
560 00:16:11,799 –> 00:16:13,429 of the Lebesgue integral where
561 00:16:13,440 –> 00:16:14,960 we only need this
562 00:16:14,969 –> 00:16:16,109 inequality
563 00:16:16,489 –> 00:16:17,900 almost everywhere.
564 00:16:18,340 –> 00:16:18,760 OK.
565 00:16:18,770 –> 00:16:20,549 So this one was a long
566 00:16:20,559 –> 00:16:22,229 proof and I showed you all
567 00:16:22,239 –> 00:16:23,549 the technical details
568 00:16:24,210 –> 00:16:25,869 because we need them again.
569 00:16:26,090 –> 00:16:27,909 When we prove the monotone
570 00:16:27,919 –> 00:16:29,150 convergence theorem.
571 00:16:29,909 –> 00:16:31,349 However, maybe let’s
572 00:16:31,359 –> 00:16:32,669 first state the
573 00:16:32,679 –> 00:16:34,299 monotone convergence
574 00:16:34,309 –> 00:16:34,770 theorem.
575 00:16:35,700 –> 00:16:37,559 Remember this was our
576 00:16:37,570 –> 00:16:39,369 goal from the beginning of
577 00:16:39,380 –> 00:16:40,000 the video.
578 00:16:42,590 –> 00:16:44,250 The first condition is that
579 00:16:44,260 –> 00:16:46,099 we have a measure space.
580 00:16:46,109 –> 00:16:47,820 So set X Sigma algebra
581 00:16:48,289 –> 00:16:49,460 and a measure mu.
582 00:16:50,140 –> 00:16:51,880 And we also have some
583 00:16:51,890 –> 00:16:52,409 nonne
584 00:16:53,000 –> 00:16:54,119 measurable
585 00:16:54,130 –> 00:16:55,000 functions
586 00:16:55,619 –> 00:16:57,299 FN and
587 00:16:57,309 –> 00:16:57,979 F
588 00:16:59,929 –> 00:17:01,840 from X to zero
589 00:17:01,929 –> 00:17:02,919 to infinity.
590 00:17:03,419 –> 00:17:04,680 And as I said, they are
591 00:17:04,689 –> 00:17:06,368 measurable for
592 00:17:06,380 –> 00:17:08,300 all N in
593 00:17:08,310 –> 00:17:08,719 N.
594 00:17:10,858 –> 00:17:12,160 And in addition, they
595 00:17:12,170 –> 00:17:13,650 satisfy two
596 00:17:13,660 –> 00:17:14,719 properties.
597 00:17:14,969 –> 00:17:16,170 First, they are
598 00:17:16,180 –> 00:17:17,420 monotonically
599 00:17:17,430 –> 00:17:18,348 increasing.
600 00:17:19,598 –> 00:17:21,388 So we have F one less or
601 00:17:21,398 –> 00:17:23,348 equal than F two, less or
602 00:17:23,358 –> 00:17:24,858 equal than F three
603 00:17:25,078 –> 00:17:26,188 and so on.
604 00:17:26,198 –> 00:17:27,338 And this holds
605 00:17:28,079 –> 00:17:29,430 almost everywhere.
606 00:17:29,439 –> 00:17:30,349 So mu
607 00:17:30,719 –> 00:17:32,349 almost everywhere.
608 00:17:32,770 –> 00:17:34,650 This always means that the
609 00:17:34,660 –> 00:17:36,560 points X where this condition
610 00:17:36,569 –> 00:17:38,150 is not satisfied
611 00:17:38,319 –> 00:17:40,069 form a set with
612 00:17:40,079 –> 00:17:41,189 measure zero.
613 00:17:42,140 –> 00:17:43,329 And the second condition
614 00:17:43,339 –> 00:17:45,109 is that the pointwise
615 00:17:45,119 –> 00:17:46,510 limit of the
616 00:17:46,520 –> 00:17:48,189 sequence of functions
617 00:17:49,239 –> 00:17:51,060 is equal to
618 00:17:51,069 –> 00:17:52,819 F of X
619 00:17:53,130 –> 00:17:54,900 also mu
620 00:17:55,000 –> 00:17:56,739 almost everywhere
621 00:17:57,239 –> 00:17:58,920 for X in X.
622 00:18:00,619 –> 00:18:00,910 OK.
623 00:18:00,920 –> 00:18:02,660 Here is now an X and we
624 00:18:02,670 –> 00:18:04,079 say it holds almost
625 00:18:04,089 –> 00:18:06,050 everywhere which just means
626 00:18:06,060 –> 00:18:07,709 OK, there is a set X tilde
627 00:18:07,719 –> 00:18:09,459 where this one holds and
628 00:18:09,469 –> 00:18:11,000 a compliment of X tilde
629 00:18:11,010 –> 00:18:12,510 has measured zero.
630 00:18:13,079 –> 00:18:14,099 And this one is just the
631 00:18:14,109 –> 00:18:16,040 common abbreviation of this.
632 00:18:16,359 –> 00:18:18,329 So we just say X and X mu-
633 00:18:18,339 –> 00:18:19,680 almost everywhere and then
634 00:18:19,689 –> 00:18:21,380 everyone knows what we mean.
635 00:18:22,390 –> 00:18:23,900 And now the monotone
636 00:18:23,910 –> 00:18:25,900 convergence theorem states
637 00:18:26,880 –> 00:18:28,359 that we can push the
638 00:18:28,369 –> 00:18:30,180 limit inside
639 00:18:30,339 –> 00:18:31,319 the integral.
640 00:18:31,819 –> 00:18:33,660 Hence the limit of
641 00:18:33,670 –> 00:18:34,939 the integrals
642 00:18:34,949 –> 00:18:36,469 f n d mu
643 00:18:37,449 –> 00:18:39,349 is equal to OK.
644 00:18:39,359 –> 00:18:41,040 Limit inside means
645 00:18:41,079 –> 00:18:42,449 integral over
646 00:18:42,459 –> 00:18:43,329 X.
647 00:18:43,349 –> 00:18:45,020 And the limit inside is
648 00:18:45,030 –> 00:18:46,989 just the function F
649 00:18:47,060 –> 00:18:48,359 mu-almost everywhere.
650 00:18:48,369 –> 00:18:49,550 So we can put
651 00:18:49,880 –> 00:18:51,209 f dmu here.
652 00:18:52,250 –> 00:18:53,540 And that is the convergence
653 00:18:53,550 –> 00:18:54,030 theorem.
654 00:18:54,040 –> 00:18:55,930 Now, you know, when you
655 00:18:55,939 –> 00:18:57,469 can push the limit
656 00:18:57,479 –> 00:18:59,219 inside the, the call, when
657 00:18:59,229 –> 00:19:01,189 you have a monotonic sequence
658 00:19:01,199 –> 00:19:01,910 of functions.
659 00:19:02,890 –> 00:19:04,489 Indeed, such convergence
660 00:19:04,500 –> 00:19:05,579 theorems are
661 00:19:05,589 –> 00:19:07,300 mostly the best
662 00:19:07,310 –> 00:19:08,989 advantages the
663 00:19:09,060 –> 00:19:10,430 Lebesgue integral has over
664 00:19:10,439 –> 00:19:11,510 the Riemann integral.
665 00:19:12,430 –> 00:19:14,199 And therefore, I really want
666 00:19:14,209 –> 00:19:15,869 to show you the proof of
667 00:19:15,880 –> 00:19:17,130 this monotone
668 00:19:17,140 –> 00:19:18,359 convergence theorem.
669 00:19:19,219 –> 00:19:20,959 However, this is a thing
670 00:19:20,969 –> 00:19:22,550 we will do in the next video
671 00:19:22,560 –> 00:19:24,300 because this video is
672 00:19:24,489 –> 00:19:26,239 already very long
673 00:19:26,969 –> 00:19:28,859 and it is good to do a short
674 00:19:28,869 –> 00:19:29,339 break.
675 00:19:30,349 –> 00:19:31,770 Maybe now recall
676 00:19:31,780 –> 00:19:33,630 everything we did here in
677 00:19:33,640 –> 00:19:34,390 this video.
678 00:19:34,619 –> 00:19:36,260 And then maybe you can
679 00:19:36,270 –> 00:19:37,930 come to the next video where
680 00:19:37,939 –> 00:19:39,329 we do the proof of the
681 00:19:39,339 –> 00:19:40,979 monotone convergence
682 00:19:40,989 –> 00:19:41,489 theorem.
683 00:19:42,209 –> 00:19:43,469 I really hope you will be
684 00:19:43,479 –> 00:19:43,849 there.
685 00:19:43,859 –> 00:19:45,739 So then see you next
686 00:19:45,750 –> 00:19:46,030 time.
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Quiz Content
Q1: Let $(X, \mathcal{A}, \mu)$ be a measure space. What does $f = g ~~\mu\text{-a.e.}$ mean?
A1: $\mu({ x \in X \mid f(x) = g(x) }) = 1$.
A2: $\mu({ x \in X \mid f(x) = g(x) }) = 0$.
A3: $\mu({ x \in X \mid f(x) \neq g(x) }) = 1$.
A4: $\mu({ x \in X \mid f(x) \neq g(x) }) = 0$.
Q2: Let $(X, \mathcal{A}, \mu)$ be a measure space. If $f = g ~~\mu\text{-a.e}$, then:
A1: $\int_X f , d\mu = 0$.
A2: $\int_X f , d\mu = \int_X g , d\mu$.
A3: $\int_X f , d\mu = 1$.
A4: $\int_X f , d\mu \neq \int_X g , d\mu$.
Q3: Let $(X, \mathcal{A}, \mu)$ be a measure space. If $f \leq g ~~\mu\text{-a.e}$, then:
A1: $\int_X f , d\mu \leq 0$.
A2: $\int_X f , d\mu \leq \int_X g , d\mu$.
A3: $\int_X f , d\mu \leq 1$.
A4: $\int_X f , d\mu \neq \int_X g , d\mu$.
Q4: 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: $f_n: X \rightarrow [0,\infty)$ is measurable.
A2: $f_1 \leq f_2 \leq f_3 \leq \cdots$ holds $\mu\text{-a.e}$.
A3: $\lim_{n \rightarrow \infty} f_n(x) = f(x)$ holds $\mu\text{-a.e}$.
A4: $\int_X f_n , d\mu < \infty$.
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Last update: 2024-10
