• Title: Outer measures - Part 3: Proof

• Series: Measure Theory

• YouTube-Title: Measure Theory 22 | Outer measures - Part 3: Proof

• Bright video: https://youtu.be/iA6ATJFViUs

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• Timestamps (n/a)
• Subtitle in English

1 00:00:00,330 –> 00:00:02,250 Hello and welcome back to

2 00:00:02,259 –> 00:00:03,390 measure theory.

3 00:00:04,070 –> 00:00:05,500 And as always, first, I want

4 00:00:05,510 –> 00:00:06,949 to thank all the nice supporters

5 00:00:06,960 –> 00:00:08,189 on Steady or paypal.

6 00:00:09,020 –> 00:00:10,609 Now, in today’s part three

7 00:00:10,619 –> 00:00:12,199 about outer measures, we

8 00:00:12,210 –> 00:00:13,710 will fill in some details

9 00:00:13,720 –> 00:00:14,239 of a proof.

10 00:00:14,250 –> 00:00:15,590 From the first video

11 00:00:16,318 –> 00:00:17,959 there, please recall,

12 00:00:17,969 –> 00:00:19,459 we’ve discussed an important

13 00:00:19,469 –> 00:00:20,079 proposition.

14 00:00:20,969 –> 00:00:22,549 The proposition tells us

15 00:00:22,559 –> 00:00:24,069 that each outer measure

16 00:00:24,079 –> 00:00:25,989 phi can be transformed

17 00:00:26,000 –> 00:00:27,510 into an ordinary measure.

18 00:00:28,229 –> 00:00:29,510 More concretely, we have

19 00:00:29,520 –> 00:00:30,450 two parts.

20 00:00:30,479 –> 00:00:32,389 The first part A tells

21 00:00:32,400 –> 00:00:33,549 us what the corresponding

22 00:00:33,560 –> 00:00:35,349 Sigma algebra A Phi

23 00:00:35,360 –> 00:00:35,740 is.

24 00:00:36,319 –> 00:00:37,909 In fact, it turns out when

25 00:00:37,919 –> 00:00:39,650 we consider all subsets

26 00:00:39,659 –> 00:00:41,490 A in X that are phi

27 00:00:41,500 –> 00:00:43,380 measurable, we get a Sigma

28 00:00:43,389 –> 00:00:44,049 algebra.

29 00:00:44,520 –> 00:00:45,869 Of course, you know this

30 00:00:45,880 –> 00:00:47,209 is the first ingredient we

31 00:00:47,220 –> 00:00:49,150 need for an ordinary measure.

32 00:00:49,860 –> 00:00:51,840 And exactly this one we call

33 00:00:51,849 –> 00:00:53,319 mu in part B.

34 00:00:54,080 –> 00:00:55,799 Now the domain of mu is given

35 00:00:55,810 –> 00:00:57,459 by our new Sigma algebra.

36 00:00:57,569 –> 00:00:59,060 And the definition is given

37 00:00:59,069 –> 00:01:00,639 by the outer measure Phi.

38 00:01:01,360 –> 00:01:03,169 In other words, Mu of

39 00:01:03,180 –> 00:01:05,069 A is given by Phi of

40 00:01:05,080 –> 00:01:05,400 A.

41 00:01:06,099 –> 00:01:07,599 And then the claim is this

42 00:01:07,610 –> 00:01:09,389 is now a normal measure.

43 00:01:10,069 –> 00:01:10,540 OK.

44 00:01:10,550 –> 00:01:12,029 There you see this is a nice

45 00:01:12,040 –> 00:01:13,610 proposition we already

46 00:01:13,620 –> 00:01:15,199 introduced in part one.

47 00:01:15,800 –> 00:01:17,550 Therefore, now in this part

48 00:01:17,559 –> 00:01:19,120 I want to give you the ideas

49 00:01:19,129 –> 00:01:19,959 for the proof.

50 00:01:20,559 –> 00:01:21,930 This means that for the first

51 00:01:21,940 –> 00:01:23,750 part, we have to verify all

52 00:01:23,760 –> 00:01:25,129 the properties of a Sigma

53 00:01:25,139 –> 00:01:25,800 algebra.

54 00:01:26,349 –> 00:01:28,050 These are essentially three

55 00:01:28,059 –> 00:01:28,760 things.

56 00:01:28,779 –> 00:01:30,430 First, the empty

57 00:01:30,440 –> 00:01:32,279 set and X should be in

58 00:01:32,290 –> 00:01:33,110 a phi.

59 00:01:33,809 –> 00:01:35,230 Then the collection of sets

60 00:01:35,239 –> 00:01:36,699 should be closed under the

61 00:01:36,709 –> 00:01:38,500 complement and

62 00:01:38,510 –> 00:01:40,050 closed under countable

63 00:01:40,059 –> 00:01:40,900 unions.

64 00:01:41,610 –> 00:01:42,959 Hence, I would suggest we

65 00:01:42,970 –> 00:01:44,529 start with the simplest one

66 00:01:44,540 –> 00:01:46,160 checking if the empty set

67 00:01:46,169 –> 00:01:47,279 is in a Phi.

68 00:01:48,010 –> 00:01:49,629 So by definition, this means

69 00:01:49,639 –> 00:01:51,400 is the empty set phi

70 00:01:51,410 –> 00:01:52,300 measurable.

71 00:01:52,839 –> 00:01:54,190 So I think this is a good

72 00:01:54,199 –> 00:01:56,180 point, refreshing your memory.

73 00:01:56,199 –> 00:01:58,150 What phi measurable actually

74 00:01:58,160 –> 00:02:00,099 means a subset

75 00:02:00,110 –> 00:02:01,519 A is called phi

76 00:02:01,529 –> 00:02:02,370 measurable.

77 00:02:02,379 –> 00:02:03,839 If it fulfills the following

78 00:02:03,849 –> 00:02:05,610 relation, namely

79 00:02:05,620 –> 00:02:07,559 if a decomposes every

80 00:02:07,569 –> 00:02:09,300 other set Q into two

81 00:02:09,309 –> 00:02:11,130 parts such that we have this

82 00:02:11,139 –> 00:02:12,589 additive for Phi.

83 00:02:13,080 –> 00:02:14,729 Now, in the case that A is

84 00:02:14,740 –> 00:02:16,360 the empty set, we see this

85 00:02:16,369 –> 00:02:17,570 relation immediately

86 00:02:18,289 –> 00:02:19,759 because the first part is

87 00:02:19,770 –> 00:02:21,210 just the empty set again.

88 00:02:21,300 –> 00:02:23,009 And the second part is just

89 00:02:23,020 –> 00:02:23,619 Q.

90 00:02:24,190 –> 00:02:25,580 Of course, you see these

91 00:02:25,589 –> 00:02:27,050 intersections are not hard

92 00:02:27,059 –> 00:02:27,520 at all.

93 00:02:28,080 –> 00:02:29,190 And then we just need the

94 00:02:29,199 –> 00:02:30,869 fact that an outer measure

95 00:02:30,880 –> 00:02:32,830 sends the empty set to zero.

96 00:02:33,500 –> 00:02:35,210 Hence, the whole first part

97 00:02:35,220 –> 00:02:36,419 is just zero.

98 00:02:36,509 –> 00:02:38,000 And the relation holds

99 00:02:38,809 –> 00:02:40,460 and there you see one property

100 00:02:40,470 –> 00:02:41,850 for the Sigma algebra is

101 00:02:41,860 –> 00:02:43,190 already satisfied.

102 00:02:43,850 –> 00:02:45,229 Then the next question would

103 00:02:45,240 –> 00:02:46,970 be, can we do the same for

104 00:02:46,979 –> 00:02:48,350 the whole space X?

105 00:02:49,179 –> 00:02:50,839 So again, we have the question

106 00:02:50,850 –> 00:02:52,839 is the whole space X phi

107 00:02:52,850 –> 00:02:53,679 measurable.

108 00:02:54,300 –> 00:02:55,660 And to answer that we have

109 00:02:55,669 –> 00:02:57,080 to take the relation again

110 00:02:57,089 –> 00:02:58,520 for every set Q.

111 00:02:58,529 –> 00:03:00,360 But now we put in X

112 00:03:01,050 –> 00:03:02,429 and there you see this looks

113 00:03:02,440 –> 00:03:04,009 very similarly because the

114 00:03:04,020 –> 00:03:05,669 complement of the whole space

115 00:03:05,679 –> 00:03:07,270 is of course the empty set

116 00:03:07,279 –> 00:03:07,669 again.

117 00:03:08,570 –> 00:03:10,300 So you see this is exactly

118 00:03:10,309 –> 00:03:11,470 the same as before.

119 00:03:11,479 –> 00:03:12,619 Just the two terms in the

120 00:03:12,630 –> 00:03:14,429 sum have switched roles.

121 00:03:15,169 –> 00:03:16,910 Hence, we can also conclude

122 00:03:16,919 –> 00:03:18,860 X is phi measurable.

123 00:03:19,619 –> 00:03:19,880 OK.

124 00:03:19,889 –> 00:03:21,169 With this, let’s go to the

125 00:03:21,179 –> 00:03:22,770 next property of a Sigma

126 00:03:22,779 –> 00:03:24,729 algebra, namely the

127 00:03:24,740 –> 00:03:26,070 question is, is the whole

128 00:03:26,080 –> 00:03:27,940 collection of subsets closed

129 00:03:27,949 –> 00:03:29,000 under complements.

130 00:03:29,789 –> 00:03:31,169 Hence, we start with one

131 00:03:31,179 –> 00:03:33,009 subset A that is phi

132 00:03:33,020 –> 00:03:33,869 measurable.

133 00:03:34,880 –> 00:03:36,380 Then in the end of course,

134 00:03:36,389 –> 00:03:37,729 we want to conclude that

135 00:03:37,740 –> 00:03:39,490 the complement of A lies

136 00:03:39,500 –> 00:03:40,830 also in a phi

137 00:03:41,339 –> 00:03:43,289 indeed how this proof works.

138 00:03:43,300 –> 00:03:44,490 You might have already seen

139 00:03:44,500 –> 00:03:46,009 above when we considered the

140 00:03:46,020 –> 00:03:46,889 empty set.

141 00:03:46,910 –> 00:03:48,449 And the complement X

142 00:03:49,339 –> 00:03:50,699 of course, we need to do

143 00:03:50,710 –> 00:03:52,300 it in a similar way because

144 00:03:52,309 –> 00:03:53,919 we only have this relation

145 00:03:53,929 –> 00:03:55,639 to work with here.

146 00:03:55,649 –> 00:03:56,970 Please always keep in mind

147 00:03:56,979 –> 00:03:58,550 I don’t write it down this

148 00:03:58,559 –> 00:04:00,000 relation holds for every

149 00:04:00,009 –> 00:04:01,970 subset Q OK.

150 00:04:01,979 –> 00:04:02,940 Then let’s do what we have

151 00:04:02,949 –> 00:04:03,720 done above.

152 00:04:03,729 –> 00:04:05,389 Let’s exchange the two terms

153 00:04:05,399 –> 00:04:06,119 in the sum.

154 00:04:06,750 –> 00:04:08,639 In fact, this is almost what

155 00:04:08,649 –> 00:04:10,300 we want to have for the relation

156 00:04:10,309 –> 00:04:11,199 of AC.

157 00:04:11,830 –> 00:04:13,149 In the first part, we already

158 00:04:13,160 –> 00:04:14,130 have AC.

159 00:04:14,210 –> 00:04:15,479 And in the second part, we

160 00:04:15,490 –> 00:04:17,119 need the complement of AC.

161 00:04:17,700 –> 00:04:18,988 However, this is already

162 00:04:19,000 –> 00:04:20,559 given simply because the

163 00:04:20,570 –> 00:04:22,079 complement of the complement

164 00:04:22,089 –> 00:04:23,559 is the original set again.

165 00:04:24,230 –> 00:04:24,649 OK.

166 00:04:24,660 –> 00:04:25,739 Now here when you look at

167 00:04:25,750 –> 00:04:26,890 the left hand side and the

168 00:04:26,899 –> 00:04:28,609 right hand side, you see

169 00:04:28,619 –> 00:04:30,290 this is exactly the definition

170 00:04:30,299 –> 00:04:32,160 for ac being phi

171 00:04:32,170 –> 00:04:33,010 measurable.

172 00:04:33,519 –> 00:04:35,000 Hence, we see the second

173 00:04:35,010 –> 00:04:36,410 part for the Sigma algebra

174 00:04:36,420 –> 00:04:37,809 is also fulfilled.

175 00:04:38,320 –> 00:04:40,010 So only the hardest one

176 00:04:40,019 –> 00:04:41,149 remains to show.

177 00:04:41,809 –> 00:04:43,600 And here is as you can see

178 00:04:43,609 –> 00:04:45,119 the property that the Sigma

179 00:04:45,130 –> 00:04:46,750 algebra is close under

180 00:04:46,760 –> 00:04:48,260 countable unions.

181 00:04:48,880 –> 00:04:50,790 And we know this is related

182 00:04:50,799 –> 00:04:52,440 to the Sigma additive of

183 00:04:52,450 –> 00:04:53,529 the corresponding measure

184 00:04:53,540 –> 00:04:54,049 mu.

185 00:04:54,540 –> 00:04:55,899 Therefore, both things in

186 00:04:55,910 –> 00:04:57,299 the proof will go hand in

187 00:04:57,309 –> 00:04:57,829 hand.

188 00:04:58,299 –> 00:05:00,029 However, before we do this,

189 00:05:00,040 –> 00:05:01,480 we first should show that

190 00:05:01,489 –> 00:05:03,100 our A phi is

191 00:05:03,109 –> 00:05:04,540 indeed closed under a

192 00:05:04,549 –> 00:05:05,750 finite union,

193 00:05:06,429 –> 00:05:08,070 which means we have to consider

194 00:05:08,079 –> 00:05:08,970 two sets.

195 00:05:08,980 –> 00:05:10,809 So we choose a one and a

196 00:05:10,820 –> 00:05:12,440 two from our A Phi.

197 00:05:12,589 –> 00:05:14,079 And then we show that the

198 00:05:14,089 –> 00:05:15,929 union of both sets lies

199 00:05:15,940 –> 00:05:17,209 also in a phi

200 00:05:17,700 –> 00:05:18,529 as before.

201 00:05:18,540 –> 00:05:19,989 This simply means we have

202 00:05:20,000 –> 00:05:21,440 to show that this property

203 00:05:21,450 –> 00:05:23,230 here holds for every set

204 00:05:23,239 –> 00:05:23,769 Q.

205 00:05:24,619 –> 00:05:26,230 Therefore, we already know

206 00:05:26,239 –> 00:05:27,480 our goal here.

207 00:05:28,200 –> 00:05:29,640 So let’s put this to the

208 00:05:29,649 –> 00:05:30,019 bottom.

209 00:05:30,029 –> 00:05:31,880 Then here, please

210 00:05:31,890 –> 00:05:33,390 also recall it’s

211 00:05:33,399 –> 00:05:34,720 sufficient by the definition

212 00:05:34,730 –> 00:05:36,380 of an outer measure to show

213 00:05:36,390 –> 00:05:37,920 the one inequality here.

214 00:05:38,529 –> 00:05:38,920 OK.

215 00:05:38,929 –> 00:05:40,480 Then I would say let’s start

216 00:05:40,489 –> 00:05:41,679 bringing the two properties

217 00:05:41,690 –> 00:05:43,519 for a one and a two together.

218 00:05:43,690 –> 00:05:45,440 First, let’s apply it

219 00:05:45,450 –> 00:05:46,549 to a one.

220 00:05:47,070 –> 00:05:48,489 Then in order to bring a

221 00:05:48,500 –> 00:05:50,359 two in, let’s call the second

222 00:05:50,369 –> 00:05:52,279 part here Q tilde.

223 00:05:53,459 –> 00:05:55,179 Hence, there, we now can

224 00:05:55,190 –> 00:05:56,739 apply the property for A

225 00:05:56,750 –> 00:05:57,299 two.

226 00:05:57,440 –> 00:05:59,130 But now instead of Q for

227 00:05:59,140 –> 00:06:01,079 Q tilde, which means we

228 00:06:01,089 –> 00:06:02,519 don’t change the first part.

229 00:06:02,529 –> 00:06:03,760 But then we have two more

230 00:06:03,769 –> 00:06:05,420 terms which look

231 00:06:05,429 –> 00:06:07,200 like Phi of Q tilde

232 00:06:07,239 –> 00:06:08,820 intersected with a two

233 00:06:08,920 –> 00:06:10,720 plus the same with the

234 00:06:10,730 –> 00:06:11,399 complement.

235 00:06:12,049 –> 00:06:13,929 However, remember in

236 00:06:13,940 –> 00:06:15,660 the end, we only want two

237 00:06:15,670 –> 00:06:16,410 terms.

238 00:06:17,320 –> 00:06:18,920 Hence the overall idea now

239 00:06:18,929 –> 00:06:20,279 is that we put the first

240 00:06:20,290 –> 00:06:21,980 two terms here together.

241 00:06:22,779 –> 00:06:24,380 And in fact, we are allowed

242 00:06:24,390 –> 00:06:26,089 to do this using the Sigma

243 00:06:26,100 –> 00:06:27,940 sub additive of the

244 00:06:27,950 –> 00:06:29,000 outer measure Phi.

245 00:06:29,700 –> 00:06:31,399 However, this now means

246 00:06:31,410 –> 00:06:33,119 we get an inequality in

247 00:06:33,929 –> 00:06:35,649 this is simply the sub in

248 00:06:35,660 –> 00:06:36,899 subadditivity.

249 00:06:37,559 –> 00:06:38,809 And the rest of the property

250 00:06:38,820 –> 00:06:40,079 tells us that instead of

251 00:06:40,089 –> 00:06:41,739 the plus we have a union

252 00:06:41,750 –> 00:06:43,059 inside of Phi,

253 00:06:43,760 –> 00:06:44,260 OK.

254 00:06:44,269 –> 00:06:46,049 Here you can see our result,

255 00:06:46,059 –> 00:06:47,619 we have two terms for Phi.

256 00:06:47,630 –> 00:06:49,299 Now in the one, we

257 00:06:49,309 –> 00:06:51,040 find a one and a two

258 00:06:51,170 –> 00:06:52,529 and in the other one, we

259 00:06:52,540 –> 00:06:54,059 find the complement of both

260 00:06:54,070 –> 00:06:54,670 sets.

261 00:06:54,959 –> 00:06:56,660 So it seems like this is

262 00:06:56,670 –> 00:06:58,179 almost the thing we

263 00:06:58,190 –> 00:06:58,739 want.

264 00:06:59,130 –> 00:06:59,820 Indeed.

265 00:06:59,829 –> 00:07:01,269 The second part we immediately

266 00:07:01,279 –> 00:07:02,850 recognize as the correct

267 00:07:02,859 –> 00:07:04,760 one simply because Q

268 00:07:04,779 –> 00:07:06,320 tilde was defined as

269 00:07:06,329 –> 00:07:08,190 Q intersected with a

270 00:07:08,200 –> 00:07:09,579 one complement.

271 00:07:10,179 –> 00:07:11,619 In other words, we can write

272 00:07:11,630 –> 00:07:12,570 it like this.

273 00:07:13,239 –> 00:07:14,619 And then when we pull out

274 00:07:14,630 –> 00:07:16,260 the complement, we find the

275 00:07:16,269 –> 00:07:17,799 union, OK?

276 00:07:17,809 –> 00:07:18,559 Very good.

277 00:07:18,570 –> 00:07:20,179 This is exactly the term

278 00:07:20,190 –> 00:07:20,880 we wanted.

279 00:07:21,619 –> 00:07:22,850 Now looking at the first

280 00:07:22,859 –> 00:07:24,779 part, it looks more complicated.

281 00:07:24,850 –> 00:07:26,799 However, we can also transform

282 00:07:26,809 –> 00:07:27,779 it to this.

283 00:07:28,359 –> 00:07:30,049 And maybe we just verify

284 00:07:30,059 –> 00:07:31,959 this with a Venn diagram.

285 00:07:32,760 –> 00:07:33,980 This works because we have

286 00:07:33,989 –> 00:07:35,820 exactly three sets involved

287 00:07:35,829 –> 00:07:37,809 Q A one and A two.

288 00:07:38,450 –> 00:07:38,890 OK.

289 00:07:38,899 –> 00:07:40,420 Then let’s start with Q

290 00:07:40,429 –> 00:07:41,779 intersected A one.

291 00:07:42,579 –> 00:07:44,209 This intersection you see

292 00:07:44,220 –> 00:07:45,720 we simply find here

293 00:07:46,489 –> 00:07:48,390 and now the other intersection

294 00:07:48,399 –> 00:07:49,910 is an intersection with three

295 00:07:49,920 –> 00:07:50,519 sets.

296 00:07:50,720 –> 00:07:52,369 First we have Q

297 00:07:52,380 –> 00:07:54,149 then a one complement and

298 00:07:54,160 –> 00:07:55,950 then a two and

299 00:07:55,959 –> 00:07:57,679 then it’s not hard to see

300 00:07:57,690 –> 00:07:59,190 that this is exactly this

301 00:07:59,200 –> 00:08:00,350 small part here.

302 00:08:01,059 –> 00:08:01,549 OK.

303 00:08:01,559 –> 00:08:03,200 And then you see this Venn

304 00:08:03,209 –> 00:08:05,200 diagram explains exactly

305 00:08:05,209 –> 00:08:06,700 this intersection here.

306 00:08:07,410 –> 00:08:09,109 Hence the proof of the unit

307 00:08:09,119 –> 00:08:10,790 with two sets is indeed

308 00:08:10,799 –> 00:08:11,549 finished.

309 00:08:12,250 –> 00:08:13,829 The union A one with a

310 00:08:13,839 –> 00:08:15,790 two is also in a

311 00:08:15,799 –> 00:08:16,309 phi.

312 00:08:17,119 –> 00:08:17,609 OK.

313 00:08:17,619 –> 00:08:18,869 Then we can go to the next

314 00:08:18,880 –> 00:08:20,299 part where we have to consider

315 00:08:20,309 –> 00:08:22,049 countably many sets.

316 00:08:22,579 –> 00:08:23,899 Hence here we need the whole

317 00:08:23,910 –> 00:08:25,529 sequence of subsets A

318 00:08:25,540 –> 00:08:27,299 one A two and so on.

319 00:08:27,950 –> 00:08:29,149 They all should come from

320 00:08:29,160 –> 00:08:30,049 a Phi.

321 00:08:30,130 –> 00:08:31,720 And then the question is,

322 00:08:32,280 –> 00:08:34,200 is the union from J is

323 00:08:34,210 –> 00:08:35,369 equal to one to

324 00:08:35,380 –> 00:08:36,890 infinity or it’s an

325 00:08:36,900 –> 00:08:38,460 element in a phi

326 00:08:39,119 –> 00:08:40,630 for this, we want to find

327 00:08:40,640 –> 00:08:41,849 a positive answer.

328 00:08:42,409 –> 00:08:43,609 Hence, for the rest of the

329 00:08:43,619 –> 00:08:45,010 proof, it might be helpful

330 00:08:45,020 –> 00:08:46,390 to give this union a shorter

331 00:08:46,400 –> 00:08:46,900 name.

332 00:08:46,909 –> 00:08:48,549 So let’s simply call it a.

333 00:08:49,039 –> 00:08:50,669 Now, I can tell you instead

334 00:08:50,679 –> 00:08:52,369 of taking an arbitrary sequence

335 00:08:52,380 –> 00:08:54,090 of sets here, we can take

336 00:08:54,099 –> 00:08:55,489 a pairwise disjoint one.

337 00:08:56,090 –> 00:08:57,390 This is possible because

338 00:08:57,400 –> 00:08:58,570 we have already proven the

339 00:08:58,580 –> 00:09:00,030 closeness with respect to

340 00:09:00,039 –> 00:09:01,960 the normal union and also

341 00:09:01,969 –> 00:09:03,330 the closeness with respect

342 00:09:03,340 –> 00:09:04,299 to the complement.

343 00:09:04,869 –> 00:09:05,849 Therefore, we can simply

344 00:09:05,859 –> 00:09:06,640 do this.

345 00:09:06,650 –> 00:09:07,940 But it also has the big

346 00:09:07,950 –> 00:09:09,520 advantage that we immediately

347 00:09:09,530 –> 00:09:11,190 get this sigma additivity

348 00:09:11,200 –> 00:09:12,469 for the measure mu in the

349 00:09:12,479 –> 00:09:12,869 end.

350 00:09:13,669 –> 00:09:14,820 Because in order to show

351 00:09:14,830 –> 00:09:16,580 this, you know, we need to

352 00:09:16,590 –> 00:09:17,979 consider pairwise disjoint

353 00:09:17,989 –> 00:09:18,409 sets.

354 00:09:19,159 –> 00:09:20,859 Hence, in summary, it’s indeed

355 00:09:20,869 –> 00:09:22,390 helpful to do the proof like

356 00:09:22,400 –> 00:09:22,840 this.

357 00:09:23,520 –> 00:09:24,960 Of course, here at the start

358 00:09:24,969 –> 00:09:26,299 of the proof, it might not

359 00:09:26,309 –> 00:09:27,950 surprise you that again,

360 00:09:27,960 –> 00:09:29,280 we use the property, the

361 00:09:29,289 –> 00:09:30,630 outer measure gives us.

362 00:09:31,140 –> 00:09:32,979 However, here I want to put

363 00:09:32,989 –> 00:09:34,599 in a special set

364 00:09:34,609 –> 00:09:36,500 Q first, we could

365 00:09:36,510 –> 00:09:37,320 have any set.

366 00:09:37,330 –> 00:09:39,099 So let’s simply call it Q

367 00:09:39,109 –> 00:09:39,690 hat.

368 00:09:40,320 –> 00:09:41,619 And then we put in the

369 00:09:41,630 –> 00:09:43,169 intersection of a

370 00:09:43,179 –> 00:09:44,830 one in the union with a

371 00:09:44,840 –> 00:09:45,260 two.

372 00:09:46,159 –> 00:09:47,719 And here please keep in mind

373 00:09:47,729 –> 00:09:49,489 a one and A two are

374 00:09:49,500 –> 00:09:50,280 disjoint.

375 00:09:50,700 –> 00:09:52,059 This fact will help us on

376 00:09:52,070 –> 00:09:53,380 the right hand side when

377 00:09:53,390 –> 00:09:54,940 we put in the same set for

378 00:09:54,950 –> 00:09:55,669 the set Q.

379 00:09:56,409 –> 00:09:57,599 For example, in the first

380 00:09:57,609 –> 00:09:59,190 part, you see the intersection

381 00:09:59,200 –> 00:10:01,049 with a one here will completely

382 00:10:01,059 –> 00:10:02,359 cancel out A two.

383 00:10:03,059 –> 00:10:04,460 Therefore, the first part

384 00:10:04,469 –> 00:10:06,359 is just Q head intersected

385 00:10:06,369 –> 00:10:07,239 with a one.

386 00:10:07,780 –> 00:10:09,090 Then in the second part,

387 00:10:09,099 –> 00:10:10,729 because we have the complement

388 00:10:10,739 –> 00:10:12,619 of a one, this will completely

389 00:10:12,630 –> 00:10:14,330 cancel out a one itself.

390 00:10:14,979 –> 00:10:16,869 So we simply have phi of

391 00:10:16,880 –> 00:10:18,590 Q head intersected with A

392 00:10:18,599 –> 00:10:19,010 two.

393 00:10:19,770 –> 00:10:20,190 OK.

394 00:10:20,200 –> 00:10:21,549 So you should see this is

395 00:10:21,559 –> 00:10:23,440 a nice trick to get the union

396 00:10:23,450 –> 00:10:24,770 on the left hand side.

397 00:10:24,890 –> 00:10:26,219 And the sum on the right

398 00:10:26,229 –> 00:10:28,070 hand side, hence, the

399 00:10:28,080 –> 00:10:29,520 next step would be to do

400 00:10:29,530 –> 00:10:30,830 an induction to get

401 00:10:30,840 –> 00:10:32,669 finitely many sets in.

402 00:10:33,359 –> 00:10:35,260 So the induction shows we

403 00:10:35,270 –> 00:10:36,969 have Phi of Q hat

404 00:10:37,530 –> 00:10:38,869 intersected with a

405 00:10:38,880 –> 00:10:40,299 finite union.

406 00:10:40,830 –> 00:10:42,320 So you see this whole thing

407 00:10:42,330 –> 00:10:44,020 here is simply the left hand

408 00:10:44,030 –> 00:10:45,799 side and then the

409 00:10:45,809 –> 00:10:47,330 right hand side is given

410 00:10:47,340 –> 00:10:48,270 by a sum.

411 00:10:48,700 –> 00:10:50,650 Of course, also from J is

412 00:10:50,659 –> 00:10:52,309 equal to one to N.

413 00:10:53,450 –> 00:10:55,239 And inside we have Phi of

414 00:10:55,250 –> 00:10:57,099 Q hat intersected with

415 00:10:57,109 –> 00:10:57,789 AJ.

416 00:10:58,340 –> 00:10:58,820 OK.

417 00:10:58,830 –> 00:11:00,380 So this is a nice result

418 00:11:00,390 –> 00:11:02,020 which holds for every N.

419 00:11:02,030 –> 00:11:03,450 So we should give it a name,

420 00:11:03,460 –> 00:11:04,849 let’s call it star.

421 00:11:05,619 –> 00:11:07,510 Now let’s go one step back

422 00:11:07,520 –> 00:11:09,320 and recall what we already

423 00:11:09,330 –> 00:11:09,799 know.

424 00:11:10,179 –> 00:11:11,919 Namely, we know that this

425 00:11:11,929 –> 00:11:13,799 unit here is by our proof

426 00:11:13,809 –> 00:11:14,770 above all

427 00:11:15,150 –> 00:11:16,760 element in A five.

428 00:11:17,440 –> 00:11:18,750 So we don’t have the infinite

429 00:11:18,760 –> 00:11:20,619 union yet in a Phi but

430 00:11:20,630 –> 00:11:21,880 we are getting closer

431 00:11:22,369 –> 00:11:23,859 because when we know that

432 00:11:23,869 –> 00:11:25,770 this is an API we can use

433 00:11:25,780 –> 00:11:27,750 the property as always the

434 00:11:27,760 –> 00:11:29,299 proper for the outer measure.

435 00:11:30,059 –> 00:11:31,760 So this works as before.

436 00:11:31,770 –> 00:11:33,390 But now instead of Q, we

437 00:11:33,400 –> 00:11:35,090 use Q hat and instead of

438 00:11:35,099 –> 00:11:36,840 our A, we have here the union

439 00:11:36,849 –> 00:11:37,690 of the AJ.

440 00:11:38,609 –> 00:11:39,700 Now, for the first part,

441 00:11:39,710 –> 00:11:41,380 you should see we can use

442 00:11:41,390 –> 00:11:43,109 our star property to

443 00:11:43,119 –> 00:11:44,580 rewrite that as a sum.

444 00:11:45,250 –> 00:11:46,909 So you see with each step,

445 00:11:46,919 –> 00:11:48,000 we are getting closer and

446 00:11:48,010 –> 00:11:49,630 closer to the Sigma

447 00:11:49,640 –> 00:11:49,950 additivity.

448 00:11:50,469 –> 00:11:50,739 OK.

449 00:11:50,750 –> 00:11:52,039 Then for the second part

450 00:11:52,049 –> 00:11:53,669 here we use something we

451 00:11:53,679 –> 00:11:55,049 didn’t use before.

452 00:11:55,059 –> 00:11:56,650 But where we know it holds

453 00:11:56,659 –> 00:11:57,710 for an outer measure,

454 00:11:58,169 –> 00:11:59,669 indeed, it’s the second

455 00:11:59,679 –> 00:12:00,989 property and outer measure

456 00:12:01,000 –> 00:12:02,799 has the monotonicity.

457 00:12:03,229 –> 00:12:04,950 We can use that here because

458 00:12:04,960 –> 00:12:06,159 when we increase the index

459 00:12:06,169 –> 00:12:07,989 and here this whole set

460 00:12:08,000 –> 00:12:09,650 here would get smaller.

461 00:12:10,210 –> 00:12:11,750 Therefore, we can also do

462 00:12:11,760 –> 00:12:13,070 it for end to infinity,

463 00:12:13,640 –> 00:12:14,950 then we have the infinite

464 00:12:14,960 –> 00:12:16,669 union, which we simply called

465 00:12:16,679 –> 00:12:17,030 A.

466 00:12:17,770 –> 00:12:19,270 So the second part is greater

467 00:12:19,280 –> 00:12:20,960 or equal than Phi of

468 00:12:20,969 –> 00:12:22,770 Q hat intersected with

469 00:12:22,780 –> 00:12:24,349 AC OK.

470 00:12:24,359 –> 00:12:26,299 With this, we now have something

471 00:12:26,309 –> 00:12:27,799 where there is no index and

472 00:12:27,809 –> 00:12:29,369 in a union anymore.

473 00:12:30,010 –> 00:12:31,229 Indeed, we have here an

474 00:12:31,239 –> 00:12:33,109 inequality that holds for

475 00:12:33,119 –> 00:12:34,390 all natural numbers.

476 00:12:34,400 –> 00:12:36,200 And therefore this

477 00:12:36,210 –> 00:12:38,150 inequality also holds the

478 00:12:38,159 –> 00:12:39,710 limit and to infinity.

479 00:12:40,090 –> 00:12:41,739 Hence, Phi of Q head

480 00:12:41,750 –> 00:12:43,369 is create or equal.

481 00:12:43,719 –> 00:12:45,469 Then the infinite sum

482 00:12:45,479 –> 00:12:47,109 of Phi of Q hat

483 00:12:47,119 –> 00:12:48,669 intersected with AJ

484 00:12:49,200 –> 00:12:51,070 plus the second term Phi

485 00:12:51,080 –> 00:12:52,989 of Q hat intersected with

486 00:12:53,000 –> 00:12:53,580 AC.

487 00:12:54,270 –> 00:12:55,650 So here you see we have an

488 00:12:55,659 –> 00:12:57,510 infinite sum and remember

489 00:12:57,520 –> 00:12:59,169 we want to show the Sigma

490 00:12:59,179 –> 00:13:00,880 additivity but

491 00:13:00,890 –> 00:13:02,520 also don’t forget for

492 00:13:02,530 –> 00:13:04,429 Phi, we already have the

493 00:13:04,440 –> 00:13:05,820 Sigma subadditivity.

494 00:13:06,809 –> 00:13:08,469 And indeed, this is what

495 00:13:08,479 –> 00:13:10,390 we should use here, which

496 00:13:10,400 –> 00:13:12,190 means we get an inequality

497 00:13:12,200 –> 00:13:13,770 when we substitute the infinite

498 00:13:13,780 –> 00:13:15,250 sum with an infinite union

499 00:13:15,260 –> 00:13:16,669 inside Phi.

500 00:13:17,130 –> 00:13:18,859 However, this infinite union

501 00:13:18,869 –> 00:13:20,109 already has a name.

502 00:13:20,119 –> 00:13:21,820 We called it A at the beginning.

503 00:13:22,460 –> 00:13:24,039 And in fact, with this last

504 00:13:24,049 –> 00:13:25,799 step, the inequality now

505 00:13:25,809 –> 00:13:27,719 looks very short and

506 00:13:27,729 –> 00:13:29,280 we can make it even shorter

507 00:13:29,289 –> 00:13:30,880 by using the Sigma sub edit

508 00:13:30,890 –> 00:13:32,809 tity again, which

509 00:13:32,820 –> 00:13:34,799 means the sum here now

510 00:13:34,809 –> 00:13:36,239 goes to a union inside

511 00:13:36,250 –> 00:13:36,840 Phi.

512 00:13:37,239 –> 00:13:38,799 However, because we have

513 00:13:38,809 –> 00:13:40,320 a and the complement of a,

514 00:13:40,469 –> 00:13:42,280 just Q hat remains.

515 00:13:42,770 –> 00:13:44,679 And now, maybe surprisingly,

516 00:13:44,690 –> 00:13:46,190 on the right hand side, we

517 00:13:46,200 –> 00:13:47,500 have the same thing as on

518 00:13:47,510 –> 00:13:49,359 the left hand side, which

519 00:13:49,369 –> 00:13:51,190 means for all the inequalities

520 00:13:51,200 –> 00:13:53,159 here that we actually have

521 00:13:53,169 –> 00:13:55,090 equalities in

522 00:13:55,099 –> 00:13:56,760 particular we get because

523 00:13:56,770 –> 00:13:58,599 this whole equality holds

524 00:13:58,609 –> 00:14:00,030 no matter which subset Q

525 00:14:00,039 –> 00:14:02,000 had we choose the defining

526 00:14:02,010 –> 00:14:04,000 property for a being an

527 00:14:04,010 –> 00:14:04,739 A Phi.

528 00:14:05,099 –> 00:14:07,000 Please recall this was our

529 00:14:07,010 –> 00:14:08,890 goal all along because

530 00:14:08,900 –> 00:14:10,849 it shows now that a Phi

531 00:14:10,859 –> 00:14:12,760 is indeed a Sigma algebra.

532 00:14:13,520 –> 00:14:15,150 In fact, we have also proven

533 00:14:15,159 –> 00:14:16,840 part B in the proposition

534 00:14:16,989 –> 00:14:18,719 because this equality

535 00:14:18,760 –> 00:14:20,520 gives us the sigma additivity

536 00:14:20,530 –> 00:14:21,359 for our measure.

537 00:14:22,039 –> 00:14:23,719 You see this because Q hat

538 00:14:23,729 –> 00:14:25,440 can be any set in

539 00:14:25,450 –> 00:14:27,070 particular, it could be a

540 00:14:27,080 –> 00:14:28,849 itself and then we

541 00:14:28,859 –> 00:14:30,820 have measure of the infinite

542 00:14:30,830 –> 00:14:32,650 union is equal to the

543 00:14:32,659 –> 00:14:33,559 infinite sum.

544 00:14:34,179 –> 00:14:34,599 OK.

545 00:14:34,609 –> 00:14:36,520 So here you see we have everything

546 00:14:36,530 –> 00:14:37,950 this closes the proof.

547 00:14:38,890 –> 00:14:40,599 Now with this nice proven

548 00:14:40,609 –> 00:14:42,549 result, we are way closer

549 00:14:42,559 –> 00:14:44,020 to be able to prove the

550 00:14:44,030 –> 00:14:45,340 famous Carathéodory’s

551 00:14:45,590 –> 00:14:47,150 extension theorem.

552 00:14:47,659 –> 00:14:49,539 And maybe we do this

553 00:14:49,549 –> 00:14:50,780 in the next video.

554 00:14:51,489 –> 00:14:52,780 Therefore, I hope I see you

555 00:14:52,789 –> 00:14:54,260 there and have a nice day.

556 00:14:54,309 –> 00:14:55,070 Bye.

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