• Title: Monotone Convergence Theorem (and more)

  • Series: Measure Theory

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

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    1 00:00:00,860 –> 00:00:02,789 Hello and welcome back.

    2 00:00:03,200 –> 00:00:04,920 And I want to thank all the

    3 00:00:04,929 –> 00:00:06,480 nice people that support

    4 00:00:06,489 –> 00:00:08,159 this channel on Steady

    5 00:00:09,180 –> 00:00:10,220 here we continue

    6 00:00:10,229 –> 00:00:11,869 now our measure theory

    7 00:00:11,880 –> 00:00:13,720 series and today

    8 00:00:13,729 –> 00:00:15,600 we do part seven.

    9 00:00:16,530 –> 00:00:17,889 The most important topic

    10 00:00:17,899 –> 00:00:19,409 for today will be the

    11 00:00:19,420 –> 00:00:20,930 monotone convergence

    12 00:00:20,940 –> 00:00:21,469 theorem.

    13 00:00:22,399 –> 00:00:24,319 Before explaining this very

    14 00:00:24,329 –> 00:00:26,170 important convergence theorem,

    15 00:00:26,379 –> 00:00:27,860 I first want to start

    16 00:00:27,870 –> 00:00:29,670 showing you some essential

    17 00:00:29,680 –> 00:00:30,870 properties of the Lebesgue

    18 00:00:31,100 –> 00:00:31,719 integral.

    19 00:00:32,529 –> 00:00:34,450 Recall, we have introduced

    20 00:00:34,459 –> 00:00:36,349 the Lebesgue integral for

    21 00:00:36,360 –> 00:00:36,700 non-negative

    22 00:00:37,349 –> 00:00:38,669 measurable

    23 00:00:38,680 –> 00:00:40,669 functions defined on

    24 00:00:40,680 –> 00:00:42,610 some measure space X.

    25 00:00:43,220 –> 00:00:44,389 For such functions

    26 00:00:44,400 –> 00:00:45,880 We now know that the Lebesgue

    27 00:00:45,990 –> 00:00:47,909 integral is well-defined.

    28 00:00:48,580 –> 00:00:49,669 And the notation we have

    29 00:00:49,680 –> 00:00:51,439 chosen is this integral

    30 00:00:51,450 –> 00:00:52,990 symbol where we put the measure

    31 00:00:53,000 –> 00:00:54,310 space X here.

    32 00:00:54,319 –> 00:00:56,150 The function F here and the

    33 00:00:56,159 –> 00:00:58,080 measure itself goes in with

    34 00:00:58,090 –> 00:00:59,959 the mu where mu is the

    35 00:00:59,970 –> 00:01:01,680 measure defined on X.

    36 00:01:02,970 –> 00:01:04,540 The integral is well-defined

    37 00:01:04,550 –> 00:01:06,510 Now means that this

    38 00:01:06,519 –> 00:01:08,459 symbol is a number

    39 00:01:08,519 –> 00:01:09,839 between zero and

    40 00:01:09,849 –> 00:01:10,809 infinity.

    41 00:01:10,910 –> 00:01:12,239 And in the worst case, it

    42 00:01:12,250 –> 00:01:14,150 could be the symbol infinity.

    43 00:01:15,330 –> 00:01:16,750 Also recall

    44 00:01:17,010 –> 00:01:18,949 this symbol was defined

    45 00:01:18,959 –> 00:01:20,940 by a supremum over all

    46 00:01:20,949 –> 00:01:22,699 step functions that lie point-

    47 00:01:22,709 –> 00:01:24,379 wisely below F

    48 00:01:25,489 –> 00:01:26,150 OK.

    49 00:01:26,160 –> 00:01:27,970 Now I want to collect some

    50 00:01:27,980 –> 00:01:29,540 properties that follow

    51 00:01:29,550 –> 00:01:31,430 immediately from this definition.

    52 00:01:32,230 –> 00:01:34,110 So let’s choose two of

    53 00:01:34,120 –> 00:01:35,790 these nice functions,

    54 00:01:35,800 –> 00:01:37,500 which means they are non-

    55 00:01:37,510 –> 00:01:39,309 negative and also

    56 00:01:39,319 –> 00:01:40,309 measurable.

    57 00:01:41,190 –> 00:01:42,930 Now we have the following.

    58 00:01:42,940 –> 00:01:44,849 The first thing would be

    59 00:01:44,860 –> 00:01:46,290 if both functions

    60 00:01:46,300 –> 00:01:47,389 coincide,

    61 00:01:48,190 –> 00:01:49,650 then also the

    62 00:01:49,660 –> 00:01:51,290 integrals coincide.

    63 00:01:52,360 –> 00:01:53,989 However, of course, this

    64 00:01:54,000 –> 00:01:55,470 is trivially fulfilled.

    65 00:01:56,400 –> 00:01:58,019 Therefore, I want to weaken

    66 00:01:58,029 –> 00:01:59,129 the left hand side a little

    67 00:01:59,139 –> 00:02:01,040 bit here, I want

    68 00:02:01,050 –> 00:02:02,639 the functions to be equal

    69 00:02:02,650 –> 00:02:03,849 mu-almost

    70 00:02:03,860 –> 00:02:04,699 everywhere.

    71 00:02:05,610 –> 00:02:07,169 The abbreviation one chooses

    72 00:02:07,180 –> 00:02:08,949 here is always A

    73 00:02:09,589 –> 00:02:10,029 E.

    74 00:02:11,089 –> 00:02:13,009 So mu A E means

    75 00:02:13,020 –> 00:02:14,910 mu-almost everywhere.

    76 00:02:15,940 –> 00:02:17,350 This means now that the

    77 00:02:17,360 –> 00:02:19,320 functions don’t have to be

    78 00:02:19,330 –> 00:02:21,199 equal, but they should be

    79 00:02:21,210 –> 00:02:23,119 equal almost everywhere

    80 00:02:23,130 –> 00:02:24,839 with respect to the measure

    81 00:02:24,850 –> 00:02:26,839 mu. more

    82 00:02:26,850 –> 00:02:27,679 concretely

    83 00:02:27,690 –> 00:02:29,259 This whole property here

    84 00:02:29,270 –> 00:02:30,559 means that you

    85 00:02:30,570 –> 00:02:32,160 look at the set

    86 00:02:32,169 –> 00:02:33,399 of all

    87 00:02:33,410 –> 00:02:35,210 lowercase X in

    88 00:02:35,220 –> 00:02:37,139 our X for

    89 00:02:37,149 –> 00:02:38,830 which this property

    90 00:02:38,839 –> 00:02:40,119 does not hold.

    91 00:02:40,130 –> 00:02:41,770 So you have F of

    92 00:02:41,779 –> 00:02:43,679 X unequal to

    93 00:02:43,690 –> 00:02:44,919 G of X,

    94 00:02:45,600 –> 00:02:47,220 then this whole set

    95 00:02:47,229 –> 00:02:48,919 should be of measure

    96 00:02:48,929 –> 00:02:50,619 zero with respect to our

    97 00:02:50,630 –> 00:02:51,460 measure mu.

    98 00:02:52,490 –> 00:02:53,770 In other words, if you put

    99 00:02:53,779 –> 00:02:55,369 the whole set into the measure,

    100 00:02:55,699 –> 00:02:57,210 you get out zero.

    101 00:02:58,809 –> 00:03:00,210 This means that the Lebesgue

    102 00:03:00,220 –> 00:03:01,800 integral cannot see

    103 00:03:01,809 –> 00:03:03,559 things that happen on

    104 00:03:03,570 –> 00:03:04,869 zero measure sets.

    105 00:03:05,960 –> 00:03:07,509 Maybe for visualization,

    106 00:03:07,520 –> 00:03:09,110 it’s good to recall what

    107 00:03:09,119 –> 00:03:11,020 we had for the Riemann integral.

    108 00:03:12,339 –> 00:03:13,710 If you integrate a

    109 00:03:13,720 –> 00:03:15,509 continuous function with

    110 00:03:15,520 –> 00:03:17,339 our classical Riemann integral,

    111 00:03:18,240 –> 00:03:19,460 you get out the

    112 00:03:19,470 –> 00:03:21,139 area between

    113 00:03:21,279 –> 00:03:23,130 the graph function and the

    114 00:03:23,139 –> 00:03:23,839 X axis.

    115 00:03:24,529 –> 00:03:26,100 And now if you change the

    116 00:03:26,110 –> 00:03:27,789 function at one point,

    117 00:03:27,979 –> 00:03:29,970 so the result is a non-

    118 00:03:29,979 –> 00:03:31,710 continuous function now

    119 00:03:32,490 –> 00:03:34,139 and then of course, you don’t

    120 00:03:34,149 –> 00:03:36,029 change the area here at

    121 00:03:36,039 –> 00:03:37,729 all, which then

    122 00:03:37,740 –> 00:03:39,669 means the integral is

    123 00:03:39,679 –> 00:03:41,410 the same for

    124 00:03:41,419 –> 00:03:42,240 Lebesgue integral.

    125 00:03:42,250 –> 00:03:43,649 Now, this works in an

    126 00:03:43,660 –> 00:03:44,889 abstract sense

    127 00:03:45,899 –> 00:03:47,009 which means that you can

    128 00:03:47,020 –> 00:03:48,720 change the function as much

    129 00:03:48,729 –> 00:03:49,559 as you want.

    130 00:03:49,570 –> 00:03:51,460 As long as the set

    131 00:03:51,639 –> 00:03:53,339 of all these changes

    132 00:03:53,350 –> 00:03:54,679 has measure zero.

    133 00:03:55,679 –> 00:03:57,039 I will talk about this later

    134 00:03:57,050 –> 00:03:57,899 in another video.

    135 00:03:57,919 –> 00:03:59,380 When I show you a lot of

    136 00:03:59,389 –> 00:04:01,190 examples. Here, I

    137 00:04:01,199 –> 00:04:02,500 want to continue with the

    138 00:04:02,509 –> 00:04:04,389 second property which is

    139 00:04:04,399 –> 00:04:05,539 the monotonicity.

    140 00:04:06,080 –> 00:04:07,699 It is descriptively given

    141 00:04:07,710 –> 00:04:09,699 by saying that if a function

    142 00:04:09,710 –> 00:04:11,449 is bigger than another one,

    143 00:04:11,460 –> 00:04:13,419 then also the area here

    144 00:04:13,429 –> 00:04:14,320 should be bigger.

    145 00:04:15,179 –> 00:04:17,010 However, as before we

    146 00:04:17,019 –> 00:04:18,450 need the less of equal sign

    147 00:04:18,459 –> 00:04:19,790 only almost

    148 00:04:19,798 –> 00:04:20,488 everywhere.

    149 00:04:21,450 –> 00:04:22,890 And then we can conclude

    150 00:04:22,899 –> 00:04:24,049 that the integral of

    151 00:04:24,059 –> 00:04:25,700 F is less or

    152 00:04:25,709 –> 00:04:27,700 equal than the integral of

    153 00:04:27,709 –> 00:04:28,130 G.

    154 00:04:29,970 –> 00:04:31,790 And now to the last property

    155 00:04:31,799 –> 00:04:33,600 which is related to the first

    156 00:04:33,609 –> 00:04:35,350 one, I want my function

    157 00:04:35,359 –> 00:04:36,950 F to be zero

    158 00:04:37,119 –> 00:04:38,790 and also mu-

    159 00:04:38,799 –> 00:04:40,230 almost everywhere.

    160 00:04:41,070 –> 00:04:42,989 Now, by using a, I know

    161 00:04:43,000 –> 00:04:44,859 that the integral of F

    162 00:04:44,869 –> 00:04:46,500 is therefore also

    163 00:04:46,510 –> 00:04:48,429 zero because the integral

    164 00:04:48,440 –> 00:04:49,890 of the zero function is zero.

    165 00:04:50,760 –> 00:04:52,649 However, here I want to emphasize

    166 00:04:52,660 –> 00:04:53,829 another direction.

    167 00:04:54,640 –> 00:04:56,100 If the function has an integral

    168 00:04:56,109 –> 00:04:57,750 of zero, then we

    169 00:04:57,760 –> 00:04:59,459 already know that the function

    170 00:04:59,470 –> 00:05:01,109 has to be zero mu-

    171 00:05:01,119 –> 00:05:02,179 almost everywhere.

    172 00:05:02,959 –> 00:05:04,260 Please don’t forget that

    173 00:05:04,269 –> 00:05:05,760 we only consider non-negative

    174 00:05:06,130 –> 00:05:06,980 functions here.

    175 00:05:07,769 –> 00:05:09,480 So all the areas we consider

    176 00:05:09,489 –> 00:05:11,320 are above the X axis and

    177 00:05:11,329 –> 00:05:12,230 therefore positive.

    178 00:05:13,250 –> 00:05:15,019 Therefore, there is no cancellation

    179 00:05:15,029 –> 00:05:16,700 whatsoever with positive

    180 00:05:16,709 –> 00:05:17,429 and negative

    181 00:05:17,440 –> 00:05:18,230 Areas here.

    182 00:05:19,519 –> 00:05:21,230 Well, now I could say

    183 00:05:21,239 –> 00:05:22,500 all the three properties

    184 00:05:22,510 –> 00:05:23,829 are easy to prove.

    185 00:05:23,839 –> 00:05:24,950 So you could do this for

    186 00:05:24,959 –> 00:05:25,660 yourself.

    187 00:05:26,130 –> 00:05:27,739 However, I also want you

    188 00:05:27,750 –> 00:05:29,230 to learn some

    189 00:05:29,239 –> 00:05:30,549 technical steps here.

    190 00:05:30,559 –> 00:05:32,429 And therefore I will do the

    191 00:05:32,440 –> 00:05:33,829 proof of B

    192 00:05:34,790 –> 00:05:36,279 after seeing this proof,

    193 00:05:36,290 –> 00:05:37,730 I think you will be

    194 00:05:37,739 –> 00:05:39,369 capable of doing the proof

    195 00:05:39,380 –> 00:05:41,170 A and C for yourself.

    196 00:05:42,350 –> 00:05:44,190 If not, please ask in the

    197 00:05:44,200 –> 00:05:46,040 comments for

    198 00:05:46,049 –> 00:05:47,220 starting the proof.

    199 00:05:47,230 –> 00:05:48,369 Let us choose a

    200 00:05:48,380 –> 00:05:50,200 simple function h.

    201 00:05:51,290 –> 00:05:53,130 This means that we have a

    202 00:05:53,140 –> 00:05:54,549 representation with

    203 00:05:54,559 –> 00:05:55,980 characteristic functions.

    204 00:05:56,970 –> 00:05:58,489 So we know we can write it

    205 00:05:58,579 –> 00:05:59,910 as a finite

    206 00:05:59,920 –> 00:06:00,690 sum.

    207 00:06:00,700 –> 00:06:02,510 Maybe we end with N

    208 00:06:03,049 –> 00:06:04,500 and we have constant C

    209 00:06:04,510 –> 00:06:06,309 I and also sets

    210 00:06:06,320 –> 00:06:06,959 A I.

    211 00:06:06,970 –> 00:06:08,750 When we put that into

    212 00:06:08,760 –> 00:06:10,179 characteristic functions,

    213 00:06:12,130 –> 00:06:13,579 the corresponding sets here

    214 00:06:13,589 –> 00:06:15,029 should be measurable.

    215 00:06:15,040 –> 00:06:16,450 And therefore, the simple

    216 00:06:16,459 –> 00:06:17,690 function is always

    217 00:06:17,700 –> 00:06:18,579 measurable.

    218 00:06:19,410 –> 00:06:20,920 And the other property of

    219 00:06:20,929 –> 00:06:22,470 a simple function is that

    220 00:06:22,480 –> 00:06:23,750 it only has

    221 00:06:23,760 –> 00:06:25,130 finitely many

    222 00:06:25,140 –> 00:06:26,019 values.

    223 00:06:26,929 –> 00:06:28,339 Therefore, there’s

    224 00:06:28,350 –> 00:06:30,209 always a canonical way to

    225 00:06:30,220 –> 00:06:32,100 write down a representation

    226 00:06:32,109 –> 00:06:33,589 for a simple function.

    227 00:06:34,010 –> 00:06:35,769 You’re forming the sum over

    228 00:06:35,779 –> 00:06:36,959 all the values.

    229 00:06:36,970 –> 00:06:38,309 So you have T

    230 00:06:38,320 –> 00:06:40,130 in and now comes

    231 00:06:40,140 –> 00:06:42,040 the image of the simple function.

    232 00:06:42,049 –> 00:06:43,899 And I write that as h

    233 00:06:43,910 –> 00:06:45,230 of X,

    234 00:06:46,269 –> 00:06:47,890 then instead of C, we just

    235 00:06:47,899 –> 00:06:49,790 have this T and

    236 00:06:49,799 –> 00:06:51,769 then comes the characteristic

    237 00:06:51,779 –> 00:06:53,269 function of a

    238 00:06:53,279 –> 00:06:54,570 set I will now

    239 00:06:54,579 –> 00:06:55,279 describe

    240 00:06:56,390 –> 00:06:58,089 there we have all the X

    241 00:06:58,100 –> 00:06:59,529 points that

    242 00:06:59,540 –> 00:07:00,989 fulfill that.

    243 00:07:01,000 –> 00:07:02,570 If I put in this

    244 00:07:02,579 –> 00:07:04,380 point in my function, I

    245 00:07:04,390 –> 00:07:06,209 get out this special

    246 00:07:06,220 –> 00:07:06,690 T.

    247 00:07:08,890 –> 00:07:10,019 Do you see this is a

    248 00:07:10,029 –> 00:07:11,410 representation that is

    249 00:07:11,420 –> 00:07:12,109 allowed.

    250 00:07:12,750 –> 00:07:14,450 And often it is very helpful

    251 00:07:14,459 –> 00:07:15,970 to choose this special one

    252 00:07:16,200 –> 00:07:17,239 because you don’t have to

    253 00:07:17,250 –> 00:07:19,170 define N or the CIS or

    254 00:07:19,179 –> 00:07:20,970 the A is you just write down

    255 00:07:20,980 –> 00:07:21,519 this representation.

    256 00:07:23,600 –> 00:07:25,579 And also the integral

    257 00:07:25,589 –> 00:07:27,339 can then easily be written

    258 00:07:27,350 –> 00:07:29,309 down as

    259 00:07:29,320 –> 00:07:30,799 always, it is defined by

    260 00:07:30,809 –> 00:07:31,720 our sum.

    261 00:07:31,730 –> 00:07:33,630 And our summation goes

    262 00:07:33,640 –> 00:07:35,429 over T in the

    263 00:07:35,440 –> 00:07:36,739 image of H

    264 00:07:37,579 –> 00:07:38,869 and then simply T

    265 00:07:38,880 –> 00:07:40,829 times the measure of

    266 00:07:40,839 –> 00:07:41,450 this set.

    267 00:07:41,459 –> 00:07:42,119 So we have

    268 00:07:42,130 –> 00:07:43,929 mu of the

    269 00:07:43,940 –> 00:07:45,769 whole set where I write all

    270 00:07:45,779 –> 00:07:47,119 the X that

    271 00:07:47,130 –> 00:07:48,880 fulfill HX

    272 00:07:48,920 –> 00:07:50,820 equals to this T.

    273 00:07:52,089 –> 00:07:53,250 And to make it a little bit

    274 00:07:53,260 –> 00:07:54,649 easier, you can

    275 00:07:54,660 –> 00:07:56,309 always ignore the zero.

    276 00:07:56,320 –> 00:07:57,799 So you omit the zero as an

    277 00:07:57,809 –> 00:07:59,350 element in the image.

    278 00:07:59,359 –> 00:08:00,589 And because you don’t change

    279 00:08:00,600 –> 00:08:02,100 anything for the integral

    280 00:08:02,109 –> 00:08:03,519 here by multiplying with

    281 00:08:03,529 –> 00:08:04,070 zero.

    282 00:08:05,570 –> 00:08:07,029 The question for you is now

    283 00:08:07,260 –> 00:08:08,910 what happens if we

    284 00:08:08,920 –> 00:08:10,630 change the simple

    285 00:08:10,640 –> 00:08:12,470 function on a

    286 00:08:12,480 –> 00:08:14,220 set which has measure

    287 00:08:14,230 –> 00:08:15,989 zero in

    288 00:08:16,000 –> 00:08:17,309 order to investigate this,

    289 00:08:17,320 –> 00:08:18,369 I need a little space here.

    290 00:08:18,380 –> 00:08:20,309 So we push this one

    291 00:08:20,320 –> 00:08:21,510 here to the bottom now.

    292 00:08:22,540 –> 00:08:24,089 OK, maybe a quick sketch

    293 00:08:24,100 –> 00:08:25,040 is helpful here.

    294 00:08:25,170 –> 00:08:26,720 So this is the whole measure

    295 00:08:26,730 –> 00:08:27,920 space X.

    296 00:08:28,890 –> 00:08:30,350 And now let us split that

    297 00:08:30,359 –> 00:08:31,670 up into two sets.

    298 00:08:31,679 –> 00:08:33,669 So this one would be the

    299 00:08:33,679 –> 00:08:35,390 big X tilde.

    300 00:08:36,669 –> 00:08:38,609 And of course, the complement,

    301 00:08:38,619 –> 00:08:40,249 the whole rest is just

    302 00:08:40,258 –> 00:08:41,979 X tilde c.

    303 00:08:41,989 –> 00:08:43,508 The compliment of X tilde,

    304 00:08:44,280 –> 00:08:45,619 this means that our x is

    305 00:08:45,630 –> 00:08:47,140 now divided into two

    306 00:08:47,150 –> 00:08:47,820 sets.

    307 00:08:49,320 –> 00:08:50,969 We want that

    308 00:08:50,979 –> 00:08:52,820 the x tilde

    309 00:08:52,830 –> 00:08:54,700 complement has

    310 00:08:54,710 –> 00:08:55,820 measured zero.

    311 00:08:58,349 –> 00:08:59,969 And with respect to this

    312 00:08:59,979 –> 00:09:01,289 set, with measure zero, I

    313 00:09:01,299 –> 00:09:02,770 want to change our simple

    314 00:09:02,780 –> 00:09:03,210 function.

    315 00:09:04,010 –> 00:09:05,179 And of course, I call it

    316 00:09:05,229 –> 00:09:07,140 H tilde and define it

    317 00:09:07,150 –> 00:09:08,419 by using two

    318 00:09:08,429 –> 00:09:09,200 cases.

    319 00:09:09,679 –> 00:09:10,940 The first case would be as

    320 00:09:10,950 –> 00:09:11,530 before.

    321 00:09:11,539 –> 00:09:13,469 So H tilde X

    322 00:09:13,479 –> 00:09:15,250 is equal to HX for

    323 00:09:15,260 –> 00:09:17,250 all X in

    324 00:09:17,390 –> 00:09:18,340 X tilde.

    325 00:09:19,239 –> 00:09:20,880 So nothing changes on the

    326 00:09:20,890 –> 00:09:21,559 green set.

    327 00:09:22,409 –> 00:09:23,929 But on the gray set, I will

    328 00:09:23,940 –> 00:09:25,809 set it to a new value And

    329 00:09:25,820 –> 00:09:27,190 I just choose an A.

    330 00:09:27,320 –> 00:09:29,289 So this is for all X

    331 00:09:29,299 –> 00:09:30,349 in X tilde

    332 00:09:30,359 –> 00:09:32,190 compliment, and

    333 00:09:32,200 –> 00:09:33,869 A is now just an arbitrary

    334 00:09:33,880 –> 00:09:35,739 chosen positive number.

    335 00:09:35,750 –> 00:09:37,429 So zero to infinity

    336 00:09:39,000 –> 00:09:40,219 by definition, this is of

    337 00:09:40,229 –> 00:09:41,770 course, again, a simple

    338 00:09:41,780 –> 00:09:43,760 function because

    339 00:09:43,770 –> 00:09:45,650 the set X tilde, I didn’t

    340 00:09:45,659 –> 00:09:46,090 say it.

    341 00:09:46,099 –> 00:09:47,169 But of course, it should

    342 00:09:47,179 –> 00:09:48,739 be in the Sigma algebra.

    343 00:09:48,750 –> 00:09:50,419 So it should be measurable.

    344 00:09:51,130 –> 00:09:52,789 And then we can write down

    345 00:09:52,799 –> 00:09:54,570 again a representation

    346 00:09:54,580 –> 00:09:56,169 for this step or simple

    347 00:09:56,179 –> 00:09:56,630 function.

    348 00:09:57,270 –> 00:09:58,739 I can use the representation

    349 00:09:58,750 –> 00:09:59,530 from before.

    350 00:09:59,539 –> 00:10:01,409 So now let’s go over all

    351 00:10:01,419 –> 00:10:02,929 the values of the

    352 00:10:03,150 –> 00:10:04,849 original function h.

    353 00:10:05,789 –> 00:10:07,090 And now I know it only

    354 00:10:07,099 –> 00:10:08,940 occurs for X in

    355 00:10:08,950 –> 00:10:09,539 X tilde.

    356 00:10:09,549 –> 00:10:11,289 So I write down T

    357 00:10:11,299 –> 00:10:12,729 times characteristic

    358 00:10:12,739 –> 00:10:13,359 function.

    359 00:10:13,510 –> 00:10:15,409 And now I put only the

    360 00:10:15,419 –> 00:10:17,320 XS from X Tilde

    361 00:10:17,570 –> 00:10:19,000 into this characteristic

    362 00:10:19,010 –> 00:10:19,450 function.

    363 00:10:20,080 –> 00:10:21,549 Not included in this sum

    364 00:10:21,559 –> 00:10:22,979 is what happens outside of

    365 00:10:22,989 –> 00:10:23,630 X tilde.

    366 00:10:23,640 –> 00:10:25,380 And therefore we add

    367 00:10:25,429 –> 00:10:26,820 this one as the value

    368 00:10:26,830 –> 00:10:28,179 a times

    369 00:10:28,239 –> 00:10:29,900 characteristic function.

    370 00:10:29,960 –> 00:10:31,770 And now we have all the

    371 00:10:31,780 –> 00:10:32,409 X

    372 00:10:33,539 –> 00:10:35,070 in X where

    373 00:10:35,080 –> 00:10:36,809 H tilde is equal to

    374 00:10:36,820 –> 00:10:38,599 a but this is simply

    375 00:10:38,609 –> 00:10:40,070 X tilde complement.

    376 00:10:40,359 –> 00:10:42,049 So we can write in

    377 00:10:42,059 –> 00:10:43,739 short just this

    378 00:10:43,750 –> 00:10:45,270 X tilde compliment here.

    379 00:10:46,390 –> 00:10:46,940 OK.

    380 00:10:47,039 –> 00:10:48,570 I have explicitly written

    381 00:10:48,580 –> 00:10:50,559 down the representation

    382 00:10:50,609 –> 00:10:52,210 because then we can write

    383 00:10:52,219 –> 00:10:54,010 down the integral as well.

    384 00:10:54,210 –> 00:10:55,919 Now I of

    385 00:10:55,929 –> 00:10:56,500 H tilde

    386 00:10:58,429 –> 00:10:59,979 equal to the

    387 00:10:59,989 –> 00:11:01,580 sum with T

    388 00:11:01,630 –> 00:11:02,280 over

    389 00:11:02,289 –> 00:11:03,940 HX

    390 00:11:04,760 –> 00:11:06,450 and you will have t times

    391 00:11:06,460 –> 00:11:08,150 the measure of this

    392 00:11:08,159 –> 00:11:08,679 set

    393 00:11:10,280 –> 00:11:12,250 plus a times

    394 00:11:12,260 –> 00:11:13,890 the measure of this set,

    395 00:11:15,140 –> 00:11:17,039 which is again, X tilde

    396 00:11:17,049 –> 00:11:18,390 the compliment.

    397 00:11:19,419 –> 00:11:21,359 However, you already know

    398 00:11:21,380 –> 00:11:23,229 the measure of X tilde compliment

    399 00:11:23,239 –> 00:11:23,799 is zero.

    400 00:11:23,809 –> 00:11:25,270 So this whole thing here

    401 00:11:25,280 –> 00:11:27,210 on the right, it’s

    402 00:11:27,219 –> 00:11:28,510 still zero.

    403 00:11:29,469 –> 00:11:30,929 That simply means we can

    404 00:11:30,940 –> 00:11:32,090 just omit it.

    405 00:11:32,780 –> 00:11:34,179 And now you should see the

    406 00:11:34,190 –> 00:11:36,140 differences above and below

    407 00:11:36,320 –> 00:11:37,719 are not so big.

    408 00:11:38,090 –> 00:11:39,419 The only real difference

    409 00:11:39,429 –> 00:11:40,549 is to tilde the here

    410 00:11:40,780 –> 00:11:42,340 because the zero does not

    411 00:11:42,349 –> 00:11:43,280 make any difference.

    412 00:11:43,289 –> 00:11:44,270 As I said before,

    413 00:11:45,270 –> 00:11:46,929 our task is now to

    414 00:11:46,940 –> 00:11:48,190 fill in the middle ground

    415 00:11:48,200 –> 00:11:48,549 here.

    416 00:11:49,909 –> 00:11:51,169 Now, it depends what you

    417 00:11:51,179 –> 00:11:52,049 find easier.

    418 00:11:52,059 –> 00:11:53,309 So maybe we come from the

    419 00:11:53,320 –> 00:11:55,080 bottom and just use

    420 00:11:55,789 –> 00:11:56,780 what we know there.

    421 00:11:56,789 –> 00:11:57,530 So we have

    422 00:11:58,429 –> 00:11:59,369 mu of

    423 00:12:00,590 –> 00:12:02,380 the whole X

    424 00:12:02,700 –> 00:12:03,580 and now we just split it

    425 00:12:03,590 –> 00:12:04,919 up into X tilde

    426 00:12:06,289 –> 00:12:08,000 where the condition is fulfilled

    427 00:12:09,400 –> 00:12:10,359 union.

    428 00:12:10,979 –> 00:12:12,820 And here we have

    429 00:12:12,830 –> 00:12:14,409 just x tilde

    430 00:12:14,419 –> 00:12:15,169 C.

    431 00:12:18,250 –> 00:12:19,960 Now, obviously this is the

    432 00:12:19,969 –> 00:12:21,409 same as the bottom part.

    433 00:12:21,609 –> 00:12:23,169 And we know it’s a disjoint

    434 00:12:23,179 –> 00:12:25,119 union which means we

    435 00:12:25,130 –> 00:12:27,030 can easily use the sigma

    436 00:12:27,039 –> 00:12:27,919 additivity here.

    437 00:12:29,200 –> 00:12:31,099 Therefore, parentheses here

    438 00:12:31,349 –> 00:12:32,539 and here I write

    439 00:12:32,549 –> 00:12:33,950 plus or

    440 00:12:33,960 –> 00:12:35,940 maybe first close

    441 00:12:35,950 –> 00:12:37,760 parenthesis here as well.

    442 00:12:38,659 –> 00:12:40,039 So this would be black

    443 00:12:40,989 –> 00:12:42,960 then plus

    444 00:12:43,140 –> 00:12:44,750 mu of

    445 00:12:45,179 –> 00:12:46,090 this set.

    446 00:12:47,570 –> 00:12:49,229 However, we already know

    447 00:12:49,239 –> 00:12:51,150 that X tilde complement

    448 00:12:51,219 –> 00:12:53,030 is a set with measure

    449 00:12:53,039 –> 00:12:54,989 zero with or without this

    450 00:12:55,000 –> 00:12:56,679 condition, it only gets

    451 00:12:56,690 –> 00:12:58,070 smaller with the condition,

    452 00:12:58,080 –> 00:12:59,989 which means this is still

    453 00:13:00,000 –> 00:13:01,429 a set with measure

    454 00:13:01,440 –> 00:13:02,030 zero.

    455 00:13:03,460 –> 00:13:05,159 And now we have all the

    456 00:13:05,169 –> 00:13:06,239 equalities here.

    457 00:13:06,580 –> 00:13:08,059 We just add a zero here.

    458 00:13:08,070 –> 00:13:09,179 So this is equal and this

    459 00:13:09,190 –> 00:13:11,109 is equal, everything is equal,

    460 00:13:11,119 –> 00:13:12,820 which means the integral

    461 00:13:12,830 –> 00:13:14,710 of H tilde is equal to

    462 00:13:14,719 –> 00:13:15,880 the integral of H.

    463 00:13:16,369 –> 00:13:17,919 This means now that we can

    464 00:13:17,929 –> 00:13:19,510 change the simple function

    465 00:13:19,520 –> 00:13:21,219 on a set with measure

    466 00:13:21,229 –> 00:13:22,919 zero as much as we want.

    467 00:13:24,080 –> 00:13:25,780 Indeed, this is a very important

    468 00:13:25,789 –> 00:13:26,469 result.

    469 00:13:26,479 –> 00:13:27,900 And you can use that

    470 00:13:28,090 –> 00:13:29,940 for proving some of these

    471 00:13:29,950 –> 00:13:30,760 parts here.

    472 00:13:30,789 –> 00:13:32,549 But we wanted to prove part

    473 00:13:32,559 –> 00:13:33,010 B

    474 00:13:34,950 –> 00:13:36,650 there, we have two measurable

    475 00:13:36,659 –> 00:13:38,169 functions whether one

    476 00:13:38,179 –> 00:13:39,820 is bigger than the other

    477 00:13:39,830 –> 00:13:41,570 one almost everywhere.

    478 00:13:42,400 –> 00:13:44,059 Therefore, now we know how

    479 00:13:44,070 –> 00:13:45,690 to choose our x tilde.

    480 00:13:48,510 –> 00:13:50,229 It simply should be the

    481 00:13:50,239 –> 00:13:51,400 set of all x

    482 00:13:53,599 –> 00:13:55,109 where F of

    483 00:13:55,119 –> 00:13:56,700 X is less or

    484 00:13:56,710 –> 00:13:58,469 equal than G of

    485 00:13:58,479 –> 00:13:59,239 X.

    486 00:14:00,340 –> 00:14:02,169 Then by assumption we

    487 00:14:02,179 –> 00:14:04,020 also know that X tilde

    488 00:14:04,059 –> 00:14:05,640 complement has

    489 00:14:05,650 –> 00:14:06,859 measure zero.

    490 00:14:07,359 –> 00:14:09,140 Well, then let’s look at

    491 00:14:09,150 –> 00:14:10,960 the integral of F

    492 00:14:12,049 –> 00:14:13,419 by definition, it’s the

    493 00:14:13,429 –> 00:14:14,849 supremum of

    494 00:14:14,859 –> 00:14:16,659 all integrals of step

    495 00:14:16,669 –> 00:14:17,510 functions.

    496 00:14:18,260 –> 00:14:20,200 And we denoted them by S

    497 00:14:20,210 –> 00:14:22,159 plus where the

    498 00:14:22,169 –> 00:14:23,500 step function or the simple

    499 00:14:23,510 –> 00:14:25,219 function is less or

    500 00:14:25,229 –> 00:14:26,909 equal than F.

    501 00:14:28,380 –> 00:14:29,869 Now by the results from

    502 00:14:29,880 –> 00:14:31,650 before we know that we

    503 00:14:31,659 –> 00:14:33,340 don’t change the integral

    504 00:14:33,349 –> 00:14:35,000 value when we change the

    505 00:14:35,010 –> 00:14:36,739 step function on a

    506 00:14:36,750 –> 00:14:38,260 set with measure zero.

    507 00:14:38,909 –> 00:14:40,179 Therefore, the supremum is

    508 00:14:40,190 –> 00:14:41,719 still the same when I put

    509 00:14:41,729 –> 00:14:43,299 in our step functions

    510 00:14:43,549 –> 00:14:45,080 where I have H

    511 00:14:45,119 –> 00:14:46,760 tilde less than

    512 00:14:46,770 –> 00:14:48,159 F but

    513 00:14:48,169 –> 00:14:49,929 only on

    514 00:14:49,940 –> 00:14:50,880 X tilde.

    515 00:14:53,169 –> 00:14:54,429 That is the whole point.

    516 00:14:54,520 –> 00:14:56,289 If we change something outside

    517 00:14:56,299 –> 00:14:57,989 of X tilde, we

    518 00:14:58,000 –> 00:14:59,789 can’t change I of

    519 00:14:59,799 –> 00:15:00,690 H tilde.

    520 00:15:00,849 –> 00:15:02,820 So the whole supremum is the

    521 00:15:02,830 –> 00:15:04,059 same as the supremum here

    522 00:15:04,070 –> 00:15:04,710 on the left.

    523 00:15:05,830 –> 00:15:06,609 Very good.

    524 00:15:06,619 –> 00:15:08,200 And now we can use what we

    525 00:15:08,210 –> 00:15:09,859 know from F and

    526 00:15:09,869 –> 00:15:11,549 G on

    527 00:15:11,559 –> 00:15:13,500 X tilde G

    528 00:15:13,510 –> 00:15:14,969 is always bigger than F.

    529 00:15:14,979 –> 00:15:16,419 So we have here

    530 00:15:16,429 –> 00:15:17,719 always this

    531 00:15:17,729 –> 00:15:18,500 inequality.

    532 00:15:19,359 –> 00:15:21,090 Hence, if we write down the

    533 00:15:21,099 –> 00:15:22,549 same thing again,

    534 00:15:23,070 –> 00:15:24,659 but now with G

    535 00:15:24,669 –> 00:15:26,020 instead of F,

    536 00:15:26,200 –> 00:15:27,989 then this set gets

    537 00:15:28,000 –> 00:15:29,830 bigger than this set because

    538 00:15:29,840 –> 00:15:31,640 there are more step functions

    539 00:15:31,650 –> 00:15:33,640 maybe inside this set.

    540 00:15:33,979 –> 00:15:35,200 Therefore, we have an

    541 00:15:35,210 –> 00:15:37,010 inequality at

    542 00:15:37,020 –> 00:15:37,869 this point.

    543 00:15:40,039 –> 00:15:41,500 Now with the same reasoning

    544 00:15:41,510 –> 00:15:43,309 as before the supremum

    545 00:15:43,580 –> 00:15:45,059 is equal to the

    546 00:15:45,070 –> 00:15:46,739 supremum where I

    547 00:15:46,750 –> 00:15:48,739 ignore x tilde and

    548 00:15:48,750 –> 00:15:50,049 use step functions

    549 00:15:50,690 –> 00:15:52,289 on the whole set X.

    550 00:15:53,530 –> 00:15:54,940 And this one is

    551 00:15:54,950 –> 00:15:56,549 exactly the definition of

    552 00:15:56,559 –> 00:15:58,440 the integral of

    553 00:15:58,450 –> 00:15:59,859 our function G.

    554 00:16:01,340 –> 00:16:02,539 And if we put everything

    555 00:16:02,549 –> 00:16:04,380 together, so this one, the

    556 00:16:04,390 –> 00:16:06,070 inequality and this

    557 00:16:06,080 –> 00:16:08,000 one, we have proven our

    558 00:16:08,010 –> 00:16:09,950 claim and this

    559 00:16:09,960 –> 00:16:11,789 is the monotonicity property

    560 00:16:11,799 –> 00:16:13,429 of the Lebesgue integral where

    561 00:16:13,440 –> 00:16:14,960 we only need this

    562 00:16:14,969 –> 00:16:16,109 inequality

    563 00:16:16,489 –> 00:16:17,900 almost everywhere.

    564 00:16:18,340 –> 00:16:18,760 OK.

    565 00:16:18,770 –> 00:16:20,549 So this one was a long

    566 00:16:20,559 –> 00:16:22,229 proof and I showed you all

    567 00:16:22,239 –> 00:16:23,549 the technical details

    568 00:16:24,210 –> 00:16:25,869 because we need them again.

    569 00:16:26,090 –> 00:16:27,909 When we prove the monotone

    570 00:16:27,919 –> 00:16:29,150 convergence theorem.

    571 00:16:29,909 –> 00:16:31,349 However, maybe let’s

    572 00:16:31,359 –> 00:16:32,669 first state the

    573 00:16:32,679 –> 00:16:34,299 monotone convergence

    574 00:16:34,309 –> 00:16:34,770 theorem.

    575 00:16:35,700 –> 00:16:37,559 Remember this was our

    576 00:16:37,570 –> 00:16:39,369 goal from the beginning of

    577 00:16:39,380 –> 00:16:40,000 the video.

    578 00:16:42,590 –> 00:16:44,250 The first condition is that

    579 00:16:44,260 –> 00:16:46,099 we have a measure space.

    580 00:16:46,109 –> 00:16:47,820 So set X Sigma algebra

    581 00:16:48,289 –> 00:16:49,460 and a measure mu.

    582 00:16:50,140 –> 00:16:51,880 And we also have some

    583 00:16:51,890 –> 00:16:52,409 nonne

    584 00:16:53,000 –> 00:16:54,119 measurable

    585 00:16:54,130 –> 00:16:55,000 functions

    586 00:16:55,619 –> 00:16:57,299 FN and

    587 00:16:57,309 –> 00:16:57,979 F

    588 00:16:59,929 –> 00:17:01,840 from X to zero

    589 00:17:01,929 –> 00:17:02,919 to infinity.

    590 00:17:03,419 –> 00:17:04,680 And as I said, they are

    591 00:17:04,689 –> 00:17:06,368 measurable for

    592 00:17:06,380 –> 00:17:08,300 all N in

    593 00:17:08,310 –> 00:17:08,719 N.

    594 00:17:10,858 –> 00:17:12,160 And in addition, they

    595 00:17:12,170 –> 00:17:13,650 satisfy two

    596 00:17:13,660 –> 00:17:14,719 properties.

    597 00:17:14,969 –> 00:17:16,170 First, they are

    598 00:17:16,180 –> 00:17:17,420 monotonically

    599 00:17:17,430 –> 00:17:18,348 increasing.

    600 00:17:19,598 –> 00:17:21,388 So we have F one less or

    601 00:17:21,398 –> 00:17:23,348 equal than F two, less or

    602 00:17:23,358 –> 00:17:24,858 equal than F three

    603 00:17:25,078 –> 00:17:26,188 and so on.

    604 00:17:26,198 –> 00:17:27,338 And this holds

    605 00:17:28,079 –> 00:17:29,430 almost everywhere.

    606 00:17:29,439 –> 00:17:30,349 So mu

    607 00:17:30,719 –> 00:17:32,349 almost everywhere.

    608 00:17:32,770 –> 00:17:34,650 This always means that the

    609 00:17:34,660 –> 00:17:36,560 points X where this condition

    610 00:17:36,569 –> 00:17:38,150 is not satisfied

    611 00:17:38,319 –> 00:17:40,069 form a set with

    612 00:17:40,079 –> 00:17:41,189 measure zero.

    613 00:17:42,140 –> 00:17:43,329 And the second condition

    614 00:17:43,339 –> 00:17:45,109 is that the pointwise

    615 00:17:45,119 –> 00:17:46,510 limit of the

    616 00:17:46,520 –> 00:17:48,189 sequence of functions

    617 00:17:49,239 –> 00:17:51,060 is equal to

    618 00:17:51,069 –> 00:17:52,819 F of X

    619 00:17:53,130 –> 00:17:54,900 also mu

    620 00:17:55,000 –> 00:17:56,739 almost everywhere

    621 00:17:57,239 –> 00:17:58,920 for X in X.

    622 00:18:00,619 –> 00:18:00,910 OK.

    623 00:18:00,920 –> 00:18:02,660 Here is now an X and we

    624 00:18:02,670 –> 00:18:04,079 say it holds almost

    625 00:18:04,089 –> 00:18:06,050 everywhere which just means

    626 00:18:06,060 –> 00:18:07,709 OK, there is a set X tilde

    627 00:18:07,719 –> 00:18:09,459 where this one holds and

    628 00:18:09,469 –> 00:18:11,000 a compliment of X tilde

    629 00:18:11,010 –> 00:18:12,510 has measured zero.

    630 00:18:13,079 –> 00:18:14,099 And this one is just the

    631 00:18:14,109 –> 00:18:16,040 common abbreviation of this.

    632 00:18:16,359 –> 00:18:18,329 So we just say X and X mu-

    633 00:18:18,339 –> 00:18:19,680 almost everywhere and then

    634 00:18:19,689 –> 00:18:21,380 everyone knows what we mean.

    635 00:18:22,390 –> 00:18:23,900 And now the monotone

    636 00:18:23,910 –> 00:18:25,900 convergence theorem states

    637 00:18:26,880 –> 00:18:28,359 that we can push the

    638 00:18:28,369 –> 00:18:30,180 limit inside

    639 00:18:30,339 –> 00:18:31,319 the integral.

    640 00:18:31,819 –> 00:18:33,660 Hence the limit of

    641 00:18:33,670 –> 00:18:34,939 the integrals

    642 00:18:34,949 –> 00:18:36,469 f n d mu

    643 00:18:37,449 –> 00:18:39,349 is equal to OK.

    644 00:18:39,359 –> 00:18:41,040 Limit inside means

    645 00:18:41,079 –> 00:18:42,449 integral over

    646 00:18:42,459 –> 00:18:43,329 X.

    647 00:18:43,349 –> 00:18:45,020 And the limit inside is

    648 00:18:45,030 –> 00:18:46,989 just the function F

    649 00:18:47,060 –> 00:18:48,359 mu-almost everywhere.

    650 00:18:48,369 –> 00:18:49,550 So we can put

    651 00:18:49,880 –> 00:18:51,209 f dmu here.

    652 00:18:52,250 –> 00:18:53,540 And that is the convergence

    653 00:18:53,550 –> 00:18:54,030 theorem.

    654 00:18:54,040 –> 00:18:55,930 Now, you know, when you

    655 00:18:55,939 –> 00:18:57,469 can push the limit

    656 00:18:57,479 –> 00:18:59,219 inside the, the call, when

    657 00:18:59,229 –> 00:19:01,189 you have a monotonic sequence

    658 00:19:01,199 –> 00:19:01,910 of functions.

    659 00:19:02,890 –> 00:19:04,489 Indeed, such convergence

    660 00:19:04,500 –> 00:19:05,579 theorems are

    661 00:19:05,589 –> 00:19:07,300 mostly the best

    662 00:19:07,310 –> 00:19:08,989 advantages the

    663 00:19:09,060 –> 00:19:10,430 Lebesgue integral has over

    664 00:19:10,439 –> 00:19:11,510 the Riemann integral.

    665 00:19:12,430 –> 00:19:14,199 And therefore, I really want

    666 00:19:14,209 –> 00:19:15,869 to show you the proof of

    667 00:19:15,880 –> 00:19:17,130 this monotone

    668 00:19:17,140 –> 00:19:18,359 convergence theorem.

    669 00:19:19,219 –> 00:19:20,959 However, this is a thing

    670 00:19:20,969 –> 00:19:22,550 we will do in the next video

    671 00:19:22,560 –> 00:19:24,300 because this video is

    672 00:19:24,489 –> 00:19:26,239 already very long

    673 00:19:26,969 –> 00:19:28,859 and it is good to do a short

    674 00:19:28,869 –> 00:19:29,339 break.

    675 00:19:30,349 –> 00:19:31,770 Maybe now recall

    676 00:19:31,780 –> 00:19:33,630 everything we did here in

    677 00:19:33,640 –> 00:19:34,390 this video.

    678 00:19:34,619 –> 00:19:36,260 And then maybe you can

    679 00:19:36,270 –> 00:19:37,930 come to the next video where

    680 00:19:37,939 –> 00:19:39,329 we do the proof of the

    681 00:19:39,339 –> 00:19:40,979 monotone convergence

    682 00:19:40,989 –> 00:19:41,489 theorem.

    683 00:19:42,209 –> 00:19:43,469 I really hope you will be

    684 00:19:43,479 –> 00:19:43,849 there.

    685 00:19:43,859 –> 00:19:45,739 So then see you next

    686 00:19:45,750 –> 00:19:46,030 time.

  • Quiz Content

    Q1: Let $(X, \mathcal{A}, \mu)$ be a measure space. What does $f = g ~~\mu\text{-a.e.}$ mean?

    A1: $\mu({ x \in X \mid f(x) = g(x) }) = 1$.

    A2: $\mu({ x \in X \mid f(x) = g(x) }) = 0$.

    A3: $\mu({ x \in X \mid f(x) \neq g(x) }) = 1$.

    A4: $\mu({ x \in X \mid f(x) \neq g(x) }) = 0$.

    Q2: Let $(X, \mathcal{A}, \mu)$ be a measure space. If $f = g ~~\mu\text{-a.e}$, then:

    A1: $\int_X f , d\mu = 0$.

    A2: $\int_X f , d\mu = \int_X g , d\mu$.

    A3: $\int_X f , d\mu = 1$.

    A4: $\int_X f , d\mu \neq \int_X g , d\mu$.

    Q3: Let $(X, \mathcal{A}, \mu)$ be a measure space. If $f \leq g ~~\mu\text{-a.e}$, then:

    A1: $\int_X f , d\mu \leq 0$.

    A2: $\int_X f , d\mu \leq \int_X g , d\mu$.

    A3: $\int_X f , d\mu \leq 1$.

    A4: $\int_X f , d\mu \neq \int_X g , d\mu$.

    Q4: Let $(X, \mathcal{A}, \mu)$ be a measure space. The monotone convergence theorem tells us something about a sequence of functions. What is not a premise of this theorem?

    A1: $f_n: X \rightarrow [0,\infty)$ is measurable.

    A2: $f_1 \leq f_2 \leq f_3 \leq \cdots$ holds $\mu\text{-a.e}$.

    A3: $\lim_{n \rightarrow \infty} f_n(x) = f(x)$ holds $\mu\text{-a.e}$.

    A4: $\int_X f_n , d\mu < \infty$.

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