• Title: Not everything is Lebesgue measurable

• Series: Measure Theory

• YouTube-Title: Measure Theory 4 | Not everything is Lebesgue measurable

• Bright video: https://youtu.be/Ur3ofJ61bpk

• Dark video: https://youtu.be/uq3sYQXzrOg

• Subtitle on GitHub: mt04_sub_eng.srt

• Other languages: German version

• Timestamps (n/a)
• Subtitle in English

1 00:00:00,490 –> 00:00:02,349 Hello and welcome back

2 00:00:02,359 –> 00:00:03,750 to measure theory.

3 00:00:04,239 –> 00:00:05,789 This is part four in our

4 00:00:05,800 –> 00:00:06,550 series.

5 00:00:06,559 –> 00:00:08,489 And today I will finally

6 00:00:08,500 –> 00:00:10,479 explain you why we can’t

7 00:00:10,489 –> 00:00:12,189 define the Lebesgue measure

8 00:00:12,199 –> 00:00:13,859 on the whole power set.

9 00:00:14,329 –> 00:00:15,460 And this should give you

10 00:00:15,470 –> 00:00:17,270 at least one motivation.

11 00:00:17,280 –> 00:00:19,260 Why it’s so important to

12 00:00:19,270 –> 00:00:20,770 study the measure theory

13 00:00:21,649 –> 00:00:22,680 for a start here.

14 00:00:22,690 –> 00:00:24,440 Let me state the measure

15 00:00:24,450 –> 00:00:25,280 problem again,

16 00:00:26,229 –> 00:00:27,940 we search and measure mu

17 00:00:28,409 –> 00:00:29,799 on the power set of

18 00:00:29,809 –> 00:00:31,530 R. So as you can see

19 00:00:31,540 –> 00:00:32,849 this is the one dimensional

20 00:00:32,860 –> 00:00:33,400 case.

21 00:00:33,950 –> 00:00:35,930 So we first deal with the

22 00:00:35,939 –> 00:00:36,919 real number line.

23 00:00:36,930 –> 00:00:38,889 Before we go to higher dimensions,

24 00:00:39,990 –> 00:00:41,130 we call this the measure

25 00:00:41,139 –> 00:00:43,049 problem because we want

26 00:00:43,060 –> 00:00:44,400 now two natural

27 00:00:44,409 –> 00:00:46,330 properties that are fulfilled

28 00:00:46,340 –> 00:00:47,580 by this measure mu

29 00:00:48,490 –> 00:00:50,360 first, the measure of a

30 00:00:50,369 –> 00:00:51,990 normal interval

31 00:00:52,020 –> 00:00:53,630 should be just the length

32 00:00:53,639 –> 00:00:54,669 of this interval.

33 00:00:55,029 –> 00:00:56,750 Hence just b

34 00:00:56,759 –> 00:00:58,590 minus a where I say

35 00:00:58,599 –> 00:01:00,290 that b is of course larger

36 00:01:00,299 –> 00:01:00,790 than a.

37 00:01:02,270 –> 00:01:03,610 Of course, this is what we

38 00:01:03,619 –> 00:01:03,900 want.

39 00:01:03,909 –> 00:01:05,790 We want to generalize

40 00:01:05,800 –> 00:01:07,720 the measuring of lengths.

41 00:01:08,459 –> 00:01:10,330 And now the second natural

42 00:01:10,339 –> 00:01:12,279 property is that we have

43 00:01:12,290 –> 00:01:14,220 some, we call translation

44 00:01:14,230 –> 00:01:15,190 invariance.

45 00:01:15,900 –> 00:01:17,680 This means that if we have

46 00:01:17,690 –> 00:01:19,559 a set A and shift that

47 00:01:19,569 –> 00:01:21,160 by a fixed vector X,

48 00:01:21,830 –> 00:01:23,370 we don’t change the length

49 00:01:23,430 –> 00:01:25,110 or the measure of the set.

50 00:01:26,120 –> 00:01:27,620 This means that we get out

51 00:01:27,629 –> 00:01:29,550 mu of a on

52 00:01:29,559 –> 00:01:30,580 the right hand side here

53 00:01:31,660 –> 00:01:32,559 and there you have it.

54 00:01:32,569 –> 00:01:34,180 That is what we call our

55 00:01:34,190 –> 00:01:35,319 measure problem.

56 00:01:35,870 –> 00:01:37,239 And such a measure mu

57 00:01:37,250 –> 00:01:38,699 would truly

58 00:01:38,709 –> 00:01:40,680 generalize the length

59 00:01:40,690 –> 00:01:42,239 measure of intervals.

60 00:01:42,980 –> 00:01:44,860 However, for more than 100

61 00:01:44,870 –> 00:01:46,720 years, it is known that

62 00:01:46,730 –> 00:01:48,339 this measure problem is not

63 00:01:48,349 –> 00:01:49,040 solvable.

64 00:01:49,160 –> 00:01:50,720 So we can’t define such a

65 00:01:50,730 –> 00:01:52,110 measure mu on the whole

66 00:01:52,120 –> 00:01:53,599 power set on R.

67 00:01:54,589 –> 00:01:56,059 Therefore, it makes sense

68 00:01:56,069 –> 00:01:57,389 to build the whole measure

69 00:01:57,400 –> 00:01:59,370 theory with Sigma algebras.

70 00:02:00,470 –> 00:02:02,199 In the end, we will see that

71 00:02:02,209 –> 00:02:03,680 we can choose a suitable

72 00:02:03,690 –> 00:02:05,480 Sigma algebra where we indeed

73 00:02:05,489 –> 00:02:07,239 can solve this measure problem.

74 00:02:08,309 –> 00:02:10,050 However, in this video, I

75 00:02:10,059 –> 00:02:11,809 start by proving

76 00:02:11,940 –> 00:02:12,990 that the measure problem

77 00:02:13,000 –> 00:02:14,779 on the power set does not

78 00:02:14,789 –> 00:02:15,570 have a solution.

79 00:02:16,259 –> 00:02:18,240 The claim I will prove now

80 00:02:18,250 –> 00:02:19,199 is the following.

81 00:02:20,009 –> 00:02:21,389 Let mu be a measure on the

82 00:02:21,399 –> 00:02:23,190 power set with the property

83 00:02:23,199 –> 00:02:25,149 one and two, but I will

84 00:02:25,179 –> 00:02:26,910 weaken the property

85 00:02:26,919 –> 00:02:27,960 one a little bit.

86 00:02:28,440 –> 00:02:30,039 I just want that the

87 00:02:30,050 –> 00:02:31,529 measure of the unit

88 00:02:31,539 –> 00:02:33,149 interval is finite.

89 00:02:33,229 –> 00:02:34,869 So not infinity.

90 00:02:36,589 –> 00:02:38,289 This is of course fulfilled

91 00:02:38,300 –> 00:02:39,839 when property one is

92 00:02:39,850 –> 00:02:40,570 fulfilled.

93 00:02:41,009 –> 00:02:42,729 And why I excluded

94 00:02:42,740 –> 00:02:44,149 zero in this interval here

95 00:02:44,160 –> 00:02:45,089 you will see later.

96 00:02:46,399 –> 00:02:47,910 In addition, mu should now

97 00:02:47,919 –> 00:02:49,789 also fulfill property

98 00:02:49,800 –> 00:02:51,309 two, which means it is

99 00:02:51,320 –> 00:02:52,779 translation invariant.

100 00:02:53,649 –> 00:02:55,470 The result is now that there

101 00:02:55,479 –> 00:02:57,130 is only one measure that

102 00:02:57,139 –> 00:02:58,339 satisfies this

103 00:02:59,210 –> 00:03:00,830 and this is the

104 00:03:00,839 –> 00:03:01,990 zero measure.

105 00:03:03,389 –> 00:03:04,639 And of course, the trivial

106 00:03:04,649 –> 00:03:06,130 measure is not what we want.

107 00:03:06,139 –> 00:03:07,110 The trivial measure does

108 00:03:07,119 –> 00:03:08,899 not fulfill poverty one in

109 00:03:08,910 –> 00:03:09,929 our measure problem.

110 00:03:10,020 –> 00:03:11,690 So this explains this

111 00:03:11,699 –> 00:03:12,690 implication here.

112 00:03:14,380 –> 00:03:16,309 The whole video is now about

113 00:03:16,320 –> 00:03:17,720 proving this claim.

114 00:03:18,050 –> 00:03:19,839 And indeed, this is a typical

115 00:03:19,850 –> 00:03:21,720 exercise you find in

116 00:03:21,729 –> 00:03:23,250 measure theory lectures.

117 00:03:23,720 –> 00:03:25,399 Therefore, I strongly advise

118 00:03:25,410 –> 00:03:26,660 you to try for

119 00:03:26,669 –> 00:03:28,559 yourself and then maybe

120 00:03:28,570 –> 00:03:30,440 fill in the details by

121 00:03:30,449 –> 00:03:31,440 using this video.

122 00:03:32,169 –> 00:03:33,339 And of course, if you’re

123 00:03:33,350 –> 00:03:35,179 not interested in technical

124 00:03:35,190 –> 00:03:36,899 details, you can happily

125 00:03:36,910 –> 00:03:38,800 skip this video and go to

126 00:03:38,809 –> 00:03:40,539 the next one in the series

127 00:03:41,509 –> 00:03:42,380 for everyone else.

128 00:03:42,389 –> 00:03:44,179 Now comes the proof,

129 00:03:44,979 –> 00:03:46,309 I will structure it a little

130 00:03:46,320 –> 00:03:46,660 bit.

132 00:03:48,250 –> 00:03:49,759 giving some definitions,

133 00:03:50,570 –> 00:03:52,360 we will study the interval

134 00:03:52,369 –> 00:03:53,449 where we know that the measure

135 00:03:53,460 –> 00:03:54,110 is finite.

136 00:03:54,119 –> 00:03:55,779 So this is what I call

137 00:03:55,789 –> 00:03:57,070 capital I,

138 00:03:57,940 –> 00:03:59,490 which means the unit interval

139 00:03:59,500 –> 00:04:01,410 where I exclude the zero

140 00:04:01,419 –> 00:04:02,679 and include the one.

141 00:04:03,600 –> 00:04:05,300 And on I, I now

142 00:04:05,309 –> 00:04:07,240 define an equivalence

143 00:04:07,250 –> 00:04:07,820 relation.

144 00:04:08,500 –> 00:04:10,210 And this equivalent equivalence

145 00:04:10,220 –> 00:04:12,160 should ignore rational

146 00:04:12,169 –> 00:04:13,960 numbers inside the interval

147 00:04:13,970 –> 00:04:15,279 which means I

148 00:04:15,289 –> 00:04:16,940 define X and

149 00:04:16,950 –> 00:04:18,829 Y as equivalent

150 00:04:19,450 –> 00:04:21,178 if and only if the

151 00:04:21,190 –> 00:04:22,269 following holds

152 00:04:24,290 –> 00:04:25,859 the difference between both

153 00:04:25,869 –> 00:04:27,809 of them is only a rational

154 00:04:27,820 –> 00:04:28,269 number.

155 00:04:29,149 –> 00:04:31,109 So X minus

156 00:04:31,119 –> 00:04:32,880 Y is in

157 00:04:32,890 –> 00:04:33,399 Q,

158 00:04:35,410 –> 00:04:36,510 this means that we don’t

159 00:04:36,519 –> 00:04:38,190 want to distinguish

160 00:04:38,200 –> 00:04:39,940 numbers that only differ

161 00:04:39,950 –> 00:04:41,230 by a rational number.

162 00:04:41,519 –> 00:04:42,619 We want to deal with these

163 00:04:42,630 –> 00:04:44,220 numbers as the same.

164 00:04:44,700 –> 00:04:46,109 Therefore, the equivalence

165 00:04:46,119 –> 00:04:48,010 class of such an X can

166 00:04:48,019 –> 00:04:49,380 be written as X

167 00:04:49,390 –> 00:04:50,049 plus.

168 00:04:50,059 –> 00:04:51,820 All rational numbers

169 00:04:51,829 –> 00:04:52,480 are.

170 00:04:52,869 –> 00:04:54,619 So we put all these numbers

171 00:04:54,630 –> 00:04:56,570 in one box and call it

172 00:04:56,589 –> 00:04:58,109 the X with the brackets.

173 00:04:59,119 –> 00:05:00,480 However, keep in mind, we

174 00:05:00,489 –> 00:05:02,160 define the equivalence relation

175 00:05:02,170 –> 00:05:03,369 on our set.

176 00:05:03,380 –> 00:05:05,239 I, so now we just

177 00:05:05,250 –> 00:05:07,130 live on the unit interval.

178 00:05:07,260 –> 00:05:08,760 Therefore, these numbers

179 00:05:08,769 –> 00:05:10,160 should also live on the unit

180 00:05:10,170 –> 00:05:10,660 interval.

181 00:05:11,350 –> 00:05:12,679 Therefore, I have to add

182 00:05:12,690 –> 00:05:14,329 here that also X

183 00:05:14,339 –> 00:05:16,109 plus R lies

184 00:05:16,119 –> 00:05:18,089 still in the unit level.

185 00:05:18,100 –> 00:05:19,619 So I can’t add a

186 00:05:19,720 –> 00:05:21,260 big rational numbers because

187 00:05:21,269 –> 00:05:22,410 then I would leave the unit

188 00:05:22,510 –> 00:05:22,850 level.

189 00:05:23,440 –> 00:05:24,850 Otherwise now we hit

190 00:05:24,859 –> 00:05:26,630 all the numbers we want

191 00:05:27,670 –> 00:05:27,929 there.

192 00:05:27,940 –> 00:05:29,230 We have the well-defined

193 00:05:29,239 –> 00:05:30,529 equivalence relation with

194 00:05:30,540 –> 00:05:32,040 the well-defined equivalence

195 00:05:32,049 –> 00:05:33,750 classes to give

196 00:05:33,760 –> 00:05:35,190 some visualization.

197 00:05:35,339 –> 00:05:37,209 I may draw these boxes

198 00:05:37,220 –> 00:05:37,950 for real.

199 00:05:38,390 –> 00:05:39,910 They are sets, sets with

200 00:05:39,920 –> 00:05:40,649 elements.

201 00:05:40,660 –> 00:05:42,549 So I could call this

202 00:05:43,579 –> 00:05:45,540 X one and I have

203 00:05:45,549 –> 00:05:46,250 one box here,

204 00:05:48,559 –> 00:05:50,059 then I have another one.

205 00:05:50,130 –> 00:05:52,010 So this would be X two,

206 00:05:54,369 –> 00:05:55,859 X three and X

207 00:05:55,869 –> 00:05:56,549 four.

208 00:05:56,679 –> 00:05:58,579 And of course, this picture

209 00:05:58,989 –> 00:06:00,410 goes on for

210 00:06:00,420 –> 00:06:01,149 forever.

211 00:06:01,619 –> 00:06:02,790 We don’t know if we have

212 00:06:02,799 –> 00:06:04,790 countable many boxes as the

213 00:06:04,799 –> 00:06:06,269 picture suggests here.

214 00:06:06,279 –> 00:06:07,670 So please be careful.

215 00:06:07,679 –> 00:06:09,000 But we know that we have

216 00:06:09,010 –> 00:06:10,880 a disjoint decomposition

217 00:06:10,890 –> 00:06:12,029 of the unit interval.

218 00:06:12,950 –> 00:06:14,929 This is what we always have

219 00:06:14,940 –> 00:06:16,579 if we define an

220 00:06:16,589 –> 00:06:17,670 equivalence relation.

221 00:06:19,420 –> 00:06:21,160 Now we already reached the

222 00:06:21,170 –> 00:06:22,519 essential part of the

223 00:06:22,529 –> 00:06:24,179 proof because we

224 00:06:24,190 –> 00:06:25,880 define a set

225 00:06:25,890 –> 00:06:27,440 A in

226 00:06:27,450 –> 00:06:29,209 I that describes

227 00:06:29,220 –> 00:06:30,619 all these boxes

228 00:06:31,359 –> 00:06:33,220 that means each element

229 00:06:33,230 –> 00:06:34,859 A describes or

230 00:06:34,869 –> 00:06:36,390 represents exactly

231 00:06:36,399 –> 00:06:38,059 one of these boxes.

232 00:06:38,859 –> 00:06:40,070 And this picture would look

233 00:06:40,079 –> 00:06:41,940 like OK, I choose

234 00:06:41,950 –> 00:06:43,200 one element here.

235 00:06:43,209 –> 00:06:45,149 Maybe I call this a one,

236 00:06:45,329 –> 00:06:47,049 then I go to the next box,

237 00:06:47,059 –> 00:06:49,040 I pick an element A two.

238 00:06:49,440 –> 00:06:51,239 Here I pick out an element

239 00:06:51,250 –> 00:06:53,000 A three here, a

240 00:06:53,010 –> 00:06:54,910 four and so on.

241 00:06:55,260 –> 00:06:57,059 And then I put all

242 00:06:57,070 –> 00:06:58,570 these elements in a

243 00:06:58,579 –> 00:07:00,299 set and this is

244 00:07:00,309 –> 00:07:02,089 what I call the set

245 00:07:02,100 –> 00:07:02,600 A,

246 00:07:03,779 –> 00:07:05,480 this looks nice now.

247 00:07:05,489 –> 00:07:07,029 But keep in mind if we have

248 00:07:07,040 –> 00:07:08,730 uncountable many boxes,

249 00:07:08,739 –> 00:07:10,480 this picture is not the

250 00:07:10,489 –> 00:07:12,239 correct way to represent

251 00:07:12,250 –> 00:07:12,720 this.

252 00:07:12,820 –> 00:07:13,970 And therefore we need another

253 00:07:13,980 –> 00:07:15,600 definition for set A.

254 00:07:16,290 –> 00:07:17,880 But of course, with exactly

255 00:07:17,890 –> 00:07:19,589 the same idea behind

256 00:07:20,579 –> 00:07:22,239 the first property is therefore

257 00:07:22,359 –> 00:07:23,690 for each box.

258 00:07:23,700 –> 00:07:25,109 So for each equivalence class

259 00:07:25,119 –> 00:07:26,890 X, I find such an

260 00:07:26,899 –> 00:07:28,359 a in A,

261 00:07:29,179 –> 00:07:30,760 this just means picking

262 00:07:30,769 –> 00:07:32,760 out an element A out

263 00:07:32,769 –> 00:07:33,559 of the box.

264 00:07:34,339 –> 00:07:35,769 And then the second property

265 00:07:35,779 –> 00:07:37,519 should tell us that this

266 00:07:37,529 –> 00:07:38,779 element is unique.

267 00:07:38,790 –> 00:07:40,529 So I only choose one

268 00:07:40,540 –> 00:07:42,279 element of each box.

269 00:07:43,079 –> 00:07:44,920 So for all A B and

270 00:07:44,929 –> 00:07:46,179 A, we have the

271 00:07:46,190 –> 00:07:47,920 property that A

272 00:07:47,929 –> 00:07:49,739 and B come out of the

273 00:07:49,750 –> 00:07:51,140 same box X,

274 00:07:52,769 –> 00:07:54,600 we can imply

275 00:07:55,049 –> 00:07:56,910 that A is equal

276 00:07:56,920 –> 00:07:57,510 to B.

277 00:07:58,380 –> 00:07:59,769 Hence the definition tells

278 00:07:59,779 –> 00:08:01,519 us that the set A

279 00:08:01,609 –> 00:08:03,279 has exactly one

280 00:08:03,290 –> 00:08:05,239 representative out of

281 00:08:05,250 –> 00:08:06,269 each box.

282 00:08:07,250 –> 00:08:08,799 The equivalence classes are

283 00:08:08,809 –> 00:08:10,109 therefore exactly

284 00:08:10,119 –> 00:08:11,869 represented by the set

285 00:08:11,880 –> 00:08:12,350 A.

286 00:08:12,869 –> 00:08:14,630 However, it’s not clear how

287 00:08:14,640 –> 00:08:16,470 to do this representation.

288 00:08:17,510 –> 00:08:18,980 You don’t find a

289 00:08:18,989 –> 00:08:20,809 construction how we can

290 00:08:20,820 –> 00:08:22,100 find the set A.

291 00:08:22,489 –> 00:08:23,630 It looks very nice in the

292 00:08:23,640 –> 00:08:24,250 picture.

293 00:08:24,420 –> 00:08:25,690 And there you might see we

294 00:08:25,700 –> 00:08:27,540 have a lot of possibilities

295 00:08:27,549 –> 00:08:29,329 to define such a set A

296 00:08:29,670 –> 00:08:31,480 But keep in mind if we are

297 00:08:31,489 –> 00:08:33,250 in the uncountable picture,

298 00:08:33,530 –> 00:08:35,169 we might not know if such

299 00:08:35,179 –> 00:08:36,849 a set A could

300 00:08:36,859 –> 00:08:37,650 exist.

301 00:08:38,580 –> 00:08:40,479 And indeed, this justification

302 00:08:40,489 –> 00:08:41,880 is very strong here.

303 00:08:42,808 –> 00:08:44,460 What we need here is indeed

304 00:08:44,520 –> 00:08:46,309 the axiom of choice

305 00:08:46,450 –> 00:08:48,039 that is given in the set

306 00:08:48,049 –> 00:08:48,549 theory.

307 00:08:49,299 –> 00:08:50,960 Therefore, it’s an axiom

308 00:08:50,969 –> 00:08:52,440 that guarantees the

309 00:08:52,450 –> 00:08:54,320 existence of such a set A

310 00:08:54,330 –> 00:08:55,640 with exactly these two

311 00:08:55,650 –> 00:08:56,400 properties.

312 00:08:57,359 –> 00:08:57,799 OK.

313 00:08:57,809 –> 00:08:58,859 That was a lot.

314 00:08:58,869 –> 00:09:00,849 But we are still not finished

315 00:09:00,859 –> 00:09:02,229 with all the definitions

316 00:09:02,239 –> 00:09:03,210 I want to give.

317 00:09:03,549 –> 00:09:04,809 Now we have fixed the set

318 00:09:04,820 –> 00:09:06,530 A and now I want to shift

319 00:09:06,539 –> 00:09:07,489 it a little bit.

320 00:09:07,500 –> 00:09:09,190 So I translated by a

321 00:09:09,200 –> 00:09:10,010 rational number.

322 00:09:10,020 –> 00:09:11,340 So this is what I define

323 00:09:11,349 –> 00:09:12,690 as A N.

324 00:09:12,700 –> 00:09:13,739 So this would be

325 00:09:14,010 –> 00:09:15,719 RN plus the

326 00:09:15,729 –> 00:09:17,690 set A and

327 00:09:17,700 –> 00:09:19,369 RN is a rational number.

328 00:09:19,380 –> 00:09:21,200 More concretely, I want a

329 00:09:21,210 –> 00:09:22,690 sequence RN.

330 00:09:22,969 –> 00:09:24,479 So this goes over the natural

331 00:09:24,489 –> 00:09:25,979 numbers that

332 00:09:26,080 –> 00:09:28,039 enumerate the whole rational

333 00:09:28,049 –> 00:09:28,650 numbers.

334 00:09:29,460 –> 00:09:31,010 That’s not completely correct.

335 00:09:31,020 –> 00:09:32,239 I just want an

336 00:09:32,250 –> 00:09:33,760 enumeration of rational

337 00:09:33,770 –> 00:09:35,320 numbers intersected

338 00:09:35,330 –> 00:09:37,010 with the real

339 00:09:37,020 –> 00:09:38,929 interval minus one

340 00:09:39,000 –> 00:09:39,960 till one.

341 00:09:41,380 –> 00:09:42,710 And of course, why I want

342 00:09:42,719 –> 00:09:44,159 that you will see later.

343 00:09:45,080 –> 00:09:46,590 However, you should see that

344 00:09:46,599 –> 00:09:48,299 we can use that the rational

345 00:09:48,309 –> 00:09:50,239 numbers accountable when

346 00:09:50,250 –> 00:09:51,780 we want to apply the sigma

347 00:09:51,789 –> 00:09:52,849 additivity later.

348 00:09:54,340 –> 00:09:56,080 And now we can finally go

349 00:09:56,090 –> 00:09:57,859 over to part B of

350 00:09:57,869 –> 00:09:58,520 my proof.

351 00:09:59,250 –> 00:10:00,750 First, we show here that

352 00:10:00,760 –> 00:10:02,460 the sets are here defined

353 00:10:02,469 –> 00:10:04,349 are indeed disjoint.

354 00:10:04,409 –> 00:10:06,099 So we have a

355 00:10:06,299 –> 00:10:07,669 intersected with

356 00:10:07,679 –> 00:10:09,549 AM and this gives

357 00:10:09,559 –> 00:10:11,190 you the empty set

358 00:10:11,260 –> 00:10:13,250 if N is

359 00:10:13,260 –> 00:10:14,739 not equal to M,

360 00:10:15,859 –> 00:10:17,640 the proof works easily

361 00:10:17,650 –> 00:10:19,049 by contraposition.

362 00:10:19,409 –> 00:10:20,469 This means you could read

363 00:10:20,479 –> 00:10:21,710 this one here as an

364 00:10:21,719 –> 00:10:22,630 implication.

365 00:10:22,659 –> 00:10:24,049 So if N is

366 00:10:24,059 –> 00:10:26,039 unequal M, this

367 00:10:26,049 –> 00:10:27,960 implies that

368 00:10:27,969 –> 00:10:29,799 this sets are disjoint.

369 00:10:30,260 –> 00:10:31,479 Now contraposition now

370 00:10:31,489 –> 00:10:33,140 means OK, they are not disjoint

371 00:10:33,150 –> 00:10:34,799 and this implies

372 00:10:34,809 –> 00:10:36,130 N equals M.

373 00:10:36,239 –> 00:10:37,849 So this is logically

374 00:10:37,859 –> 00:10:38,539 equivalent,

375 00:10:39,419 –> 00:10:41,130 however, not being disjoint

376 00:10:41,140 –> 00:10:42,489 means there is an element

377 00:10:42,500 –> 00:10:44,409 we can choose out of this

378 00:10:44,419 –> 00:10:45,130 intersection.

379 00:10:46,280 –> 00:10:47,239 So that’s what we do.

380 00:10:48,049 –> 00:10:49,349 And this implies

381 00:10:49,359 –> 00:10:51,349 immediately to properties

382 00:10:51,700 –> 00:10:53,669 being in A N means

383 00:10:53,679 –> 00:10:55,099 I can write X

384 00:10:55,109 –> 00:10:56,710 as rn

385 00:10:56,840 –> 00:10:58,830 plus some, well, the

386 00:10:59,440 –> 00:11:01,419 lower case a in A so

387 00:11:01,429 –> 00:11:03,219 I have an A here or

388 00:11:03,229 –> 00:11:05,150 being in A_m means I

389 00:11:05,159 –> 00:11:06,359 can write it as

390 00:11:06,369 –> 00:11:07,450 r_m

391 00:11:07,530 –> 00:11:09,130 plus some a.

392 00:11:09,729 –> 00:11:11,130 But of course, these could

393 00:11:11,140 –> 00:11:12,830 be two different A’s.

394 00:11:12,840 –> 00:11:14,349 So I also give you an

395 00:11:14,359 –> 00:11:15,919 index to say

396 00:11:15,929 –> 00:11:17,890 this, OK.

397 00:11:18,090 –> 00:11:19,700 But of course, the X is the

398 00:11:19,710 –> 00:11:20,619 same on the left.

399 00:11:20,630 –> 00:11:21,969 So I can equal them.

400 00:11:22,669 –> 00:11:24,150 This means that I have

401 00:11:24,159 –> 00:11:25,710 RN plus A

402 00:11:25,719 –> 00:11:27,690 N equals two,

403 00:11:28,520 –> 00:11:29,950 two RM

404 00:11:29,960 –> 00:11:31,380 plus AM.

405 00:11:32,010 –> 00:11:32,890 And there you see what I

406 00:11:32,900 –> 00:11:34,859 can do is put all

407 00:11:34,869 –> 00:11:36,789 the A’s on the one side and

408 00:11:36,799 –> 00:11:38,109 all the RS on the other side.

409 00:11:39,049 –> 00:11:40,349 And now on the right, we

410 00:11:40,359 –> 00:11:42,109 subtract two rational numbers.

411 00:11:42,119 –> 00:11:44,109 So you know what comes out

412 00:11:44,119 –> 00:11:45,950 is also a rational number.

413 00:11:46,729 –> 00:11:48,309 Now remind yourself that

414 00:11:48,320 –> 00:11:50,059 we defined an equivalence

415 00:11:50,070 –> 00:11:51,780 relation exactly

416 00:11:51,789 –> 00:11:53,150 when the difference of two

417 00:11:53,159 –> 00:11:54,650 numbers is a rational number.

418 00:11:54,950 –> 00:11:56,859 So in other words, a_n and

419 00:11:56,869 –> 00:11:58,109 a_m are now

420 00:11:58,119 –> 00:12:00,020 equivalent or

421 00:12:00,030 –> 00:12:00,969 to put this.

422 00:12:00,979 –> 00:12:02,469 In other words, we could

423 00:12:02,479 –> 00:12:04,219 also say our A

424 00:12:04,229 –> 00:12:05,900 N is in the

425 00:12:05,909 –> 00:12:07,250 equivalence class

426 00:12:07,369 –> 00:12:08,919 represented by

427 00:12:08,929 –> 00:12:09,590 AM.

428 00:12:10,700 –> 00:12:11,900 If you want, you could also

429 00:12:11,909 –> 00:12:13,479 now add AM

430 00:12:13,609 –> 00:12:14,520 here as well.

431 00:12:14,530 –> 00:12:16,219 So of course, am is also

432 00:12:16,229 –> 00:12:17,369 in the equivalence class

433 00:12:17,380 –> 00:12:17,719 here.

434 00:12:18,140 –> 00:12:19,849 And then you see we have

435 00:12:19,859 –> 00:12:21,719 property two for

436 00:12:21,729 –> 00:12:23,590 set A and it tells

437 00:12:23,599 –> 00:12:25,429 you if two elements come

438 00:12:25,440 –> 00:12:27,070 out of the same box or the

439 00:12:27,080 –> 00:12:28,510 same equivalence class, they

440 00:12:28,520 –> 00:12:29,429 have to be the same.

441 00:12:30,130 –> 00:12:31,789 So the conclusion here is

442 00:12:31,799 –> 00:12:33,679 a_n is equal

443 00:12:33,690 –> 00:12:34,750 to a_m.

444 00:12:35,820 –> 00:12:36,359 OK.

445 00:12:36,369 –> 00:12:38,159 So this tells us the left

446 00:12:38,169 –> 00:12:39,599 hand side here is zero,

447 00:12:40,000 –> 00:12:41,299 but then also the right hand

448 00:12:41,309 –> 00:12:42,159 side is zero.

449 00:12:42,479 –> 00:12:44,440 And therefore, we can again

450 00:12:44,450 –> 00:12:46,000 imply that

451 00:12:46,010 –> 00:12:47,320 also the rational numbers

452 00:12:47,330 –> 00:12:48,289 here are the same.

453 00:12:48,299 –> 00:12:50,039 So RN is equal

454 00:12:50,049 –> 00:12:51,940 to RM but

455 00:12:51,950 –> 00:12:53,520 the R MS were chosen

456 00:12:53,530 –> 00:12:55,250 as an enumeration of the

457 00:12:55,260 –> 00:12:56,280 rational numbers.

458 00:12:56,289 –> 00:12:58,039 And therefore here also

459 00:12:58,090 –> 00:12:59,440 the indices have to

460 00:12:59,450 –> 00:13:00,260 coincide

461 00:13:01,330 –> 00:13:02,989 and this proves now the claim

462 00:13:03,010 –> 00:13:04,190 by contraposition

463 00:13:05,049 –> 00:13:05,789 very good.

464 00:13:05,799 –> 00:13:07,359 Now I want to go over to

465 00:13:07,369 –> 00:13:08,229 the next part.

466 00:13:09,289 –> 00:13:11,200 In part C I want to look

467 00:13:11,210 –> 00:13:12,570 at the union of

468 00:13:12,580 –> 00:13:14,460 these disjoint

469 00:13:14,469 –> 00:13:15,049 sets.

470 00:13:15,960 –> 00:13:16,530 So I have

471 00:13:17,559 –> 00:13:19,080 union n

472 00:13:19,270 –> 00:13:20,190 over N,

473 00:13:21,289 –> 00:13:21,630 OK.

474 00:13:21,640 –> 00:13:22,789 Now keep in mind what the

475 00:13:22,799 –> 00:13:24,719 definition of A was that

476 00:13:24,729 –> 00:13:26,309 was defined by our set

477 00:13:26,320 –> 00:13:28,080 A that lives in the unit

478 00:13:28,090 –> 00:13:29,859 interval, shifted by

479 00:13:29,869 –> 00:13:31,830 rational numbers that live

480 00:13:31,840 –> 00:13:33,630 in minus 1 to 1.

481 00:13:34,369 –> 00:13:35,789 This means by using the

482 00:13:35,799 –> 00:13:37,669 union, I still should not

483 00:13:37,679 –> 00:13:39,289 be able to leave the

484 00:13:39,299 –> 00:13:40,109 interval.

485 00:13:40,119 –> 00:13:42,090 That is given by minus

486 00:13:42,099 –> 00:13:43,869 one and now I

487 00:13:43,880 –> 00:13:45,390 shift one with

488 00:13:45,400 –> 00:13:46,690 maximum one.

489 00:13:46,700 –> 00:13:48,229 So here I have two.

490 00:13:49,750 –> 00:13:50,919 And on the other side, you

491 00:13:50,929 –> 00:13:52,369 can use how we

492 00:13:52,380 –> 00:13:53,549 defined a.

493 00:13:53,559 –> 00:13:54,500 So this was

494 00:13:54,840 –> 00:13:56,580 representation of all the

495 00:13:56,590 –> 00:13:58,299 equivalence classes and where

496 00:13:58,309 –> 00:13:59,549 the equivalence classes were

497 00:13:59,559 –> 00:14:00,929 defined by the

498 00:14:00,940 –> 00:14:02,330 differences with the rational

499 00:14:02,369 –> 00:14:03,000 numbers.

500 00:14:03,700 –> 00:14:05,229 And now I add

501 00:14:05,450 –> 00:14:07,159 back all the rational numbers.

502 00:14:07,169 –> 00:14:08,960 So I should

503 00:14:08,969 –> 00:14:10,880 get at least the unit interval

504 00:14:10,890 –> 00:14:11,880 out again.

505 00:14:13,429 –> 00:14:13,780 OK.

506 00:14:13,789 –> 00:14:15,590 So this one is to claim

507 00:14:15,830 –> 00:14:17,450 we should prove here in part

508 00:14:17,460 –> 00:14:18,119 C.

509 00:14:18,460 –> 00:14:20,250 However, I already

510 00:14:20,260 –> 00:14:21,780 told you most of the things

511 00:14:21,789 –> 00:14:22,250 you need.

512 00:14:23,140 –> 00:14:24,729 Therefore, I think it’s not

513 00:14:24,739 –> 00:14:26,090 hard for you to do the proof

514 00:14:26,099 –> 00:14:26,859 for yourself.

515 00:14:26,919 –> 00:14:28,609 So the proof is an

516 00:14:28,619 –> 00:14:30,460 exercise for you just

517 00:14:30,469 –> 00:14:32,219 put all the ideas I gave

518 00:14:32,229 –> 00:14:34,000 you now into formulas,

519 00:14:34,900 –> 00:14:35,510 OK?

520 00:14:35,530 –> 00:14:37,450 With all these parts in mind,

521 00:14:37,539 –> 00:14:39,489 I can now go to the core

522 00:14:39,500 –> 00:14:41,369 of the proof, we

523 00:14:41,380 –> 00:14:42,419 now assume

524 00:14:43,309 –> 00:14:44,570 that we have a measure on

525 00:14:44,580 –> 00:14:45,969 the whole power set of the

526 00:14:45,979 –> 00:14:46,690 real line.

527 00:14:47,070 –> 00:14:48,570 And it also should fulfill

528 00:14:48,580 –> 00:14:50,210 the two properties we have

529 00:14:50,219 –> 00:14:51,609 given in the claim.

530 00:14:52,549 –> 00:14:54,440 And now we can use all the

531 00:14:54,450 –> 00:14:56,369 things above, for example,

532 00:14:56,590 –> 00:14:58,390 by our translation

533 00:14:58,400 –> 00:15:00,349 invariance (2), we

534 00:15:00,359 –> 00:15:02,210 can write that

535 00:15:02,219 –> 00:15:03,119 the measure

536 00:15:04,210 –> 00:15:05,179 of a

537 00:15:05,289 –> 00:15:07,169 ARN plus

538 00:15:07,179 –> 00:15:09,010 A is the same as the

539 00:15:09,020 –> 00:15:10,809 measure of

540 00:15:10,820 –> 00:15:11,289 A.

541 00:15:11,650 –> 00:15:13,450 And this holds for all

542 00:15:13,460 –> 00:15:15,010 natural numbers N.

543 00:15:16,340 –> 00:15:17,950 And maybe let us, you see

544 00:15:17,960 –> 00:15:19,799 here immediately, we know

545 00:15:19,890 –> 00:15:21,799 that the measure is always

546 00:15:21,809 –> 00:15:23,489 monotonic, which

547 00:15:23,500 –> 00:15:25,140 means the measure of

548 00:15:25,150 –> 00:15:27,090 this set is less or

549 00:15:27,099 –> 00:15:28,359 equal than the measure of

550 00:15:28,369 –> 00:15:29,159 this set.

551 00:15:29,169 –> 00:15:30,359 And this is less or equal

552 00:15:30,369 –> 00:15:31,130 than the measure of this

553 00:15:31,140 –> 00:15:31,549 set.

554 00:15:32,229 –> 00:15:33,859 So we have exactly this

555 00:15:33,869 –> 00:15:34,989 inequality here.

556 00:15:36,030 –> 00:15:37,229 We need it later again.

557 00:15:37,239 –> 00:15:38,869 So I call the inequality

558 00:15:38,880 –> 00:15:40,409 here by star.

559 00:15:41,530 –> 00:15:43,369 Before we go further, let

560 00:15:43,380 –> 00:15:45,239 us use now our second

561 00:15:45,250 –> 00:15:47,169 condition here that the

562 00:15:47,179 –> 00:15:48,950 measure of the unit interval

563 00:15:48,960 –> 00:15:50,609 is at least finite.

564 00:15:51,020 –> 00:15:52,609 Let us give this number a

565 00:15:52,619 –> 00:15:53,169 name.

566 00:15:53,179 –> 00:15:54,940 So I write the measure of

567 00:15:54,950 –> 00:15:56,090 01

568 00:15:56,099 –> 00:15:57,950 equals to

569 00:15:57,960 –> 00:15:59,530 number and I call it just

570 00:15:59,539 –> 00:16:00,570 capital C.

571 00:16:01,239 –> 00:16:02,590 With this, we can indeed

572 00:16:02,599 –> 00:16:04,539 calculate the measure here.

573 00:16:05,260 –> 00:16:06,479 We can use the properties

574 00:16:06,489 –> 00:16:08,150 of a measure namely the

575 00:16:08,159 –> 00:16:09,359 Sigma Additivity.

576 00:16:09,580 –> 00:16:11,289 So I split the set

577 00:16:11,299 –> 00:16:12,659 into unit

578 00:16:12,669 –> 00:16:14,609 intervals or shifted unit

579 00:16:14,619 –> 00:16:15,390 intervals.

580 00:16:15,989 –> 00:16:17,849 Here I go to zero and include

581 00:16:17,859 –> 00:16:18,179 it.

582 00:16:18,400 –> 00:16:19,849 And then I have a disjoint

583 00:16:19,859 –> 00:16:21,669 unit when I exclude

584 00:16:21,679 –> 00:16:23,419 the zero here and go

585 00:16:23,429 –> 00:16:24,679 to one

586 00:16:24,919 –> 00:16:26,119 included here

587 00:16:26,440 –> 00:16:27,599 again, union

588 00:16:28,239 –> 00:16:29,869 one and he had

589 00:16:29,880 –> 00:16:30,340 two.

590 00:16:32,390 –> 00:16:32,809 OK.

591 00:16:32,820 –> 00:16:34,270 And here we can now use the

592 00:16:34,280 –> 00:16:35,489 Sigma additivity.

593 00:16:35,559 –> 00:16:36,969 Now you can write this as

594 00:16:36,979 –> 00:16:38,260 measure of this set plus

595 00:16:38,270 –> 00:16:39,380 measure of this set plus

596 00:16:39,390 –> 00:16:40,580 measure of this set.

597 00:16:40,900 –> 00:16:42,530 And by using the

598 00:16:42,539 –> 00:16:43,969 translation invariance in

599 00:16:43,979 –> 00:16:45,770 two, you know all this

600 00:16:45,780 –> 00:16:47,530 measure have the same value

601 00:16:47,539 –> 00:16:48,830 namely C

602 00:16:49,219 –> 00:16:50,780 hence we have C plus C

603 00:16:50,789 –> 00:16:51,609 plus C.

604 00:16:51,619 –> 00:16:53,059 So we

605 00:16:53,070 –> 00:16:53,729 c

606 00:16:55,289 –> 00:16:55,750 OK.

607 00:16:55,760 –> 00:16:56,580 Very good.

608 00:16:57,030 –> 00:16:58,869 Now, I want to use this in

609 00:16:58,880 –> 00:17:00,799 the quality I called star

610 00:17:00,809 –> 00:17:01,570 before.

611 00:17:04,020 –> 00:17:05,958 So this one, what

612 00:17:05,968 –> 00:17:07,019 you should see immediately

613 00:17:07,029 –> 00:17:08,558 now is that on the left we

614 00:17:08,568 –> 00:17:10,368 have C itself but on the

615 00:17:10,378 –> 00:17:11,999 right we now calculate three

616 00:17:12,009 –> 00:17:12,388 C.

617 00:17:12,989 –> 00:17:14,310 So let us write it down.

618 00:17:14,319 –> 00:17:16,040 So I have C less so

619 00:17:16,050 –> 00:17:17,050 equal then.

620 00:17:17,449 –> 00:17:19,050 And now I can use the Sigma

621 00:17:19,060 –> 00:17:21,050 additivity as always

622 00:17:21,060 –> 00:17:22,510 because this one is

623 00:17:22,520 –> 00:17:23,329 disjoint.

624 00:17:23,670 –> 00:17:25,410 Now this was part B

625 00:17:25,420 –> 00:17:26,400 from before.

626 00:17:27,060 –> 00:17:28,510 So I have now here the

627 00:17:28,520 –> 00:17:30,459 sum or the series

628 00:17:30,469 –> 00:17:31,890 from one to

629 00:17:31,900 –> 00:17:32,729 infinity

630 00:17:34,410 –> 00:17:36,390 of mu of

631 00:17:36,400 –> 00:17:38,109 A N and this is

632 00:17:38,119 –> 00:17:39,969 less for equal than three

633 00:17:39,979 –> 00:17:40,790 times C.

634 00:17:42,349 –> 00:17:43,729 And now we also know that

635 00:17:43,739 –> 00:17:45,530 we can get rid of the

636 00:17:45,540 –> 00:17:47,530 N here because

637 00:17:47,540 –> 00:17:49,410 here you see it, this is

638 00:17:49,420 –> 00:17:49,910 a N.

639 00:17:50,729 –> 00:17:52,349 So the translated

640 00:17:52,359 –> 00:17:54,270 version of A but by

641 00:17:54,280 –> 00:17:55,500 translation invariance, we

642 00:17:55,510 –> 00:17:56,650 know the measure is the same.

643 00:17:56,660 –> 00:17:58,180 So we can write mu

644 00:17:58,189 –> 00:17:59,869 of A on this case.

645 00:18:00,670 –> 00:18:01,869 And I want to write that

646 00:18:01,880 –> 00:18:03,500 down as the important

647 00:18:03,510 –> 00:18:04,459 inequality here.

648 00:18:04,469 –> 00:18:06,229 So C less or

649 00:18:06,239 –> 00:18:08,089 equal the series of

650 00:18:08,099 –> 00:18:09,790 MU A and less or equal than

651 00:18:09,800 –> 00:18:10,569 three C.

652 00:18:11,359 –> 00:18:12,609 So please look

653 00:18:12,619 –> 00:18:14,260 closely at this

654 00:18:14,270 –> 00:18:15,390 inequality.

655 00:18:15,560 –> 00:18:17,260 You see a fixed number

656 00:18:17,270 –> 00:18:19,219 mu of A in the middle and

657 00:18:19,229 –> 00:18:20,180 then the series

658 00:18:21,020 –> 00:18:22,219 then you know the series

659 00:18:22,229 –> 00:18:23,680 will explode, it gives you

660 00:18:23,689 –> 00:18:25,250 infinity if

661 00:18:25,260 –> 00:18:26,939 mu of A is greater than

662 00:18:26,949 –> 00:18:27,479 zero.

663 00:18:27,729 –> 00:18:29,510 So the only possible

664 00:18:29,530 –> 00:18:31,199 case when this is finite

665 00:18:31,209 –> 00:18:32,890 needs mu of A

666 00:18:32,900 –> 00:18:34,140 equals to zero.

667 00:18:34,420 –> 00:18:35,800 And this is the case because

668 00:18:35,810 –> 00:18:37,540 we know that C is

669 00:18:37,550 –> 00:18:39,160 less than infinity.

670 00:18:39,609 –> 00:18:41,599 So the series has to be

671 00:18:41,609 –> 00:18:43,550 convergent by

672 00:18:43,560 –> 00:18:44,930 this calculation, we now

673 00:18:44,939 –> 00:18:46,930 can conclude that the measure

674 00:18:46,939 –> 00:18:48,920 of our set A has to be

675 00:18:48,930 –> 00:18:49,489 zero.

676 00:18:51,680 –> 00:18:53,530 However, this means now the

677 00:18:53,540 –> 00:18:55,449 value of the series is

678 00:18:55,459 –> 00:18:56,000 zero.

679 00:18:56,030 –> 00:18:57,760 So we have zero in the middle

680 00:18:57,770 –> 00:18:59,319 and left and right, we have

681 00:18:59,329 –> 00:19:00,400 C and three C.

682 00:19:00,449 –> 00:19:02,209 So there’s also no other

683 00:19:02,219 –> 00:19:04,099 possible way other

684 00:19:04,109 –> 00:19:05,660 than C to be

685 00:19:05,670 –> 00:19:06,180 zero.

686 00:19:07,319 –> 00:19:08,979 However, remind yourself

687 00:19:08,989 –> 00:19:10,780 that C was just a short

688 00:19:10,790 –> 00:19:12,180 notation for the measure

689 00:19:12,189 –> 00:19:13,180 of the unit interval.

690 00:19:14,150 –> 00:19:15,770 Hence, this measure

691 00:19:15,910 –> 00:19:17,689 is also just

692 00:19:17,699 –> 00:19:18,390 zero.

693 00:19:20,420 –> 00:19:22,239 And indeed, this helps us

694 00:19:22,250 –> 00:19:24,079 now calculating the

695 00:19:24,089 –> 00:19:25,489 measure of the whole real

696 00:19:25,500 –> 00:19:27,160 line just

697 00:19:27,170 –> 00:19:28,760 because I have translation

698 00:19:28,770 –> 00:19:30,369 invariance and sigma

700 00:19:31,949 –> 00:19:33,290 So we split the real

701 00:19:33,300 –> 00:19:35,109 line into unit

702 00:19:35,119 –> 00:19:36,469 intervals and shift them.

703 00:19:36,479 –> 00:19:38,160 So what we can do is

704 00:19:38,170 –> 00:19:39,530 just use an interval

705 00:19:39,540 –> 00:19:40,949 starting with an integer

706 00:19:40,959 –> 00:19:42,939 M and then go to M

707 00:19:42,949 –> 00:19:43,550 plus one.

708 00:19:44,420 –> 00:19:46,359 And if I use the union here,

709 00:19:46,369 –> 00:19:47,790 which is then the disjoint

710 00:19:47,800 –> 00:19:49,680 union where M goes over

711 00:19:49,689 –> 00:19:51,420 all integers, I

712 00:19:51,430 –> 00:19:52,359 have what I want.

713 00:19:52,369 –> 00:19:54,239 Now I use Sigma additivity

714 00:19:54,339 –> 00:19:56,020 and then translation variances

715 00:19:56,180 –> 00:19:58,030 and then I get also

716 00:19:58,040 –> 00:19:59,869 out adding up zero

717 00:19:59,880 –> 00:20:01,119 stays at zero.

718 00:20:02,209 –> 00:20:03,560 This now means that the

719 00:20:03,569 –> 00:20:05,300 volume or length of the

720 00:20:05,310 –> 00:20:07,040 whole line measured in mu

721 00:20:07,079 –> 00:20:08,550 is just zero.

722 00:20:09,089 –> 00:20:10,670 So we are dealing with the

723 00:20:10,979 –> 00:20:12,780 real measure this zero measure.

724 00:20:14,890 –> 00:20:16,579 And in fact, this is what

725 00:20:16,589 –> 00:20:17,709 we wanted to prove.

726 00:20:20,069 –> 00:20:21,890 And there we have it the

727 00:20:21,900 –> 00:20:23,670 full proof how to

728 00:20:23,680 –> 00:20:25,089 see that the measure

729 00:20:25,099 –> 00:20:26,550 problem is not

730 00:20:26,560 –> 00:20:27,250 solvable.

731 00:20:28,099 –> 00:20:29,359 I know it was a long

732 00:20:29,650 –> 00:20:31,400 proof with a lot of technical

733 00:20:31,410 –> 00:20:33,400 details, but I hope you learn

734 00:20:33,410 –> 00:20:34,160 something here

735 00:20:35,150 –> 00:20:37,050 and maybe I should close

736 00:20:37,270 –> 00:20:38,920 this proof with some

737 00:20:38,930 –> 00:20:40,209 interpretation of the whole

738 00:20:40,219 –> 00:20:42,189 thing you saw

739 00:20:42,199 –> 00:20:43,900 that it is possible to

740 00:20:43,910 –> 00:20:45,680 construct such a set

741 00:20:45,689 –> 00:20:47,390 a where we

742 00:20:47,400 –> 00:20:49,089 can’t have a reasonable

743 00:20:49,099 –> 00:20:50,469 length or measure,

744 00:20:51,219 –> 00:20:52,979 we would get contradictions

745 00:20:53,020 –> 00:20:54,569 if we’re not dealing with

746 00:20:54,579 –> 00:20:56,530 the trivial zero measure.

747 00:20:57,319 –> 00:20:59,189 So the only possibility

748 00:20:59,199 –> 00:21:00,579 to deal with such

749 00:21:00,589 –> 00:21:02,119 sets that behave so

750 00:21:02,130 –> 00:21:04,060 strangely that we can’t measure

751 00:21:04,069 –> 00:21:06,060 them is to exclude them

752 00:21:06,069 –> 00:21:07,020 from the beginning.

753 00:21:07,349 –> 00:21:09,089 We won’t measure all

754 00:21:09,099 –> 00:21:10,530 possible subsets.

755 00:21:11,359 –> 00:21:12,810 We just deal with sets, we

756 00:21:12,819 –> 00:21:14,359 then call the measurable

757 00:21:14,369 –> 00:21:14,890 sets.

758 00:21:15,569 –> 00:21:17,349 Now these are the sets that

759 00:21:17,359 –> 00:21:18,900 behave nicely

760 00:21:18,910 –> 00:21:20,630 enough such that we

761 00:21:20,640 –> 00:21:22,189 can solve the measure

762 00:21:22,199 –> 00:21:24,060 problem in this so

763 00:21:24,069 –> 00:21:25,369 called Sigma algebra then .

764 00:21:26,459 –> 00:21:27,920 In fact, the Borel Sigma

765 00:21:27,930 –> 00:21:29,819 algebra you learned in

766 00:21:29,829 –> 00:21:31,510 part two of the series about

767 00:21:31,750 –> 00:21:33,449 is a correct choice to

768 00:21:33,459 –> 00:21:34,869 solve the measure problem.

769 00:21:35,689 –> 00:21:37,390 We will go into details

771 00:21:39,119 –> 00:21:40,920 But first in the next videos,

772 00:21:40,930 –> 00:21:42,589 I want to talk about

773 00:21:42,599 –> 00:21:44,310 maps that preserve

774 00:21:44,319 –> 00:21:45,709 our measurable structure

775 00:21:45,719 –> 00:21:46,109 here.

776 00:21:46,229 –> 00:21:47,890 So these are so called

777 00:21:47,900 –> 00:21:49,069 measurable maps.

778 00:21:49,670 –> 00:21:50,109 OK.

779 00:21:50,119 –> 00:21:51,540 Then see you in the next

780 00:21:51,550 –> 00:21:51,959 video.

781 00:21:52,199 –> 00:21:52,739 Bye.

• Quiz Content

Q1: Is there a measure $\mu: \mathcal{P}(\mathbb{R}) \rightarrow \mathbb{R}$ with the two properties: $$\mu([a,b]) = b-a$$ for all intervals $[a,b]$ with $a < b$ and $$\mu(x+A) = \mu(A)$$ for all $A \in \mathcal{P}(\mathbb{R})$ and $x \in \mathbb{R}$?

A1: Yes, there is exactly one.

A2: Yes, there are a lot of such measures.

A3: No, there is no such measure.

Q2: Is there a measure $\mu: \mathcal{P}(\mathbb{R}) \rightarrow \mathbb{R}$ with the two properties: $$\mu((0,1]) < \infty$$ and $$\mu(x+A) = \mu(A)$$ for all $A \in \mathcal{P}(\mathbb{R})$ and $x \in \mathbb{R}$?

A1: Yes, there is exactly one.

A2: Yes, there are a lot of such measures.

A3: No, there is no such measure.

Q3: By using the axiom of choice, one can construct sets $A_n \in\mathcal{P}(\mathbb{R})$ such that $$C \leq \sum_{n=1}^\infty \mu(A_n) \leq 3 C$$ where $\mu$ is translation-invariant with $C := \mu( (0,1] )$.

A1: No, that is not correct because $C = \mu(\mathbb{R})$.

A2: No, that is not correct because $\mu$ does not have to be translation-invariant.

A3: Yes, that is correct.

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