• Title: Monotone Convergence Theorem (and more)

• Series: Measure Theory

• YouTube-Title: Measure Theory 7 | Monotone Convergence Theorem (and more)

• Bright video: https://youtu.be/1tzaUiZJXm8

• Dark video: https://youtu.be/Dn5a741u_Pk

• Subtitle on GitHub: mt07_sub_eng.srt

• Other languages: German version

• Timestamps (n/a)
• 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.

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.

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.

• 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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