• Title: Not everything is Lebesgue measurable

  • Series: Measure Theory

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

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    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.

    131 00:03:46,669 –> 00:03:48,240 So I start with a, by

    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

    699 00:19:30,380 –> 00:19:30,969 additivity.

    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

    770 00:21:37,400 –> 00:21:39,109 about this in the future.

    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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