-
Title: Extending a Set to a Sigma-Algebra
-
Series: Exercises - Measure Theory
-
YouTube-Title: Exercises - Measure Theory 2 | Extending a Set to a Sigma-Algebra
-
Bright video: Watch on YouTube
-
Dark video: Watch on YouTube
-
Ad-free video: Watch Vimeo video
-
Original video for YT-Members (bright): Watch on YouTube
-
Original video for YT-Members (dark): Watch on YouTube
-
Forum: Ask a question in Mattermost
-
Dark-PDF: Download PDF version of the dark video
-
Print-PDF: Download printable PDF version
-
Thumbnail (bright): Download PNG
-
Thumbnail (dark): Download PNG
-
Subtitle on GitHub: ex-mt02_sub_eng.srt
-
Related videos:
-
Timestamps (n/a)
-
Subtitle in English
1 00:00:00.655 –> 00:00:03.045 Hello and welcome back to this video series
2 00:00:03.135 –> 00:00:06.485 where we talk about some exercises in measure theory.
3 00:00:07.415 –> 00:00:09.475 And now before we start with the problem solving,
4 00:00:09.875 –> 00:00:11.955 I really want to thank the members on Steady
5 00:00:12.015 –> 00:00:13.395 Patreon and YouTube.
6 00:00:14.315 –> 00:00:17.495 You make it possible that I can create such math videos.
7 00:00:18.515 –> 00:00:20.535 And the exercise of today is the following.
8 00:00:20.755 –> 00:00:24.575 We want to look at the Sigma Algebra A over the interval
9 00:00:24.605 –> 00:00:26.015 from zero to two.
10 00:00:26.975 –> 00:00:28.515 Of course there are different possibilities
11 00:00:28.975 –> 00:00:32.875 and here should check if the following collection is already
12 00:00:33.075 –> 00:00:34.955 a sigma algebra over X.
13 00:00:35.575 –> 00:00:38.235 And if that’s not the case, you should put some more sets
14 00:00:38.495 –> 00:00:41.435 to it until it is a Sigma algebra over X.
15 00:00:42.335 –> 00:00:44.555 So this is a standard exercise that checks,
16 00:00:44.775 –> 00:00:48.075 if you know the three properties of a Sigma algebra.
17 00:00:48.735 –> 00:00:51.995 Indeed the first property tells us that the empty set
18 00:00:52.255 –> 00:00:54.995 and the set X itself should be elements
19 00:00:55.015 –> 00:00:56.355 of the sigma algebra.
20 00:00:57.185 –> 00:01:00.795 This is what we call property (a) and it’s satisfied here.
21 00:01:01.585 –> 00:01:02.635 This is the first step.
22 00:01:02.655 –> 00:01:04.435 And now we can go to property (b),
23 00:01:04.645 –> 00:01:06.355 which is about the complements.
24 00:01:07.215 –> 00:01:09.075 The empty set and X are already done.
25 00:01:09.255 –> 00:01:11.355 And then the next step we can check if the set
26 00:01:11.355 –> 00:01:15.075 that only contains the element zero has its complement
27 00:01:15.295 –> 00:01:16.435 in A as well.
28 00:01:17.275 –> 00:01:20.935 In fact, since X is the whole interval here, we see
29 00:01:21.165 –> 00:01:24.975 that this one is exactly the complement of this set.
30 00:01:25.705 –> 00:01:26.725 So no problem there.
31 00:01:26.905 –> 00:01:29.925 And now we can go to the set that only has one in it.
32 00:01:30.475 –> 00:01:31.965 This means for the complement,
33 00:01:32.105 –> 00:01:35.325 if we consider the whole interval from zero to two,
34 00:01:35.595 –> 00:01:37.445 then we take out the middle point.
35 00:01:38.215 –> 00:01:41.445 Hence we actually now have two half open intervals
36 00:01:41.445 –> 00:01:42.885 here and there.
37 00:01:42.885 –> 00:01:44.485 You see this set as a union
38 00:01:44.545 –> 00:01:47.925 of two intervals is already included in A,
39 00:01:48.645 –> 00:01:51.305 so also there the complement makes no problems.
40 00:01:51.565 –> 00:01:53.825 And now we can go to the last element in A.
41 00:01:54.005 –> 00:01:56.545 Indeed it’s the last one
42 00:01:56.545 –> 00:01:57.665 because as we can see,
43 00:01:57.925 –> 00:02:01.185 the complement is the other one we have left in A.
44 00:02:01.935 –> 00:02:04.235 In other words, the complements are all in
45 00:02:04.655 –> 00:02:06.075 and (b) is satisfied.
46 00:02:06.925 –> 00:02:08.425 And then we can immediately go to (c),
47 00:02:08.555 –> 00:02:11.825 which in general is about the countable unions.
48 00:02:12.775 –> 00:02:15.265 However, in our case, this means that we only have
49 00:02:15.265 –> 00:02:17.185 to check the finite unions
50 00:02:17.255 –> 00:02:20.705 because we only have finitely many sets anyway.
51 00:02:21.475 –> 00:02:23.015 And there you might already recognize
52 00:02:23.015 –> 00:02:24.855 that this one is not satisfied
53 00:02:25.085 –> 00:02:28.415 because zero one is the union of zero
54 00:02:28.595 –> 00:02:30.615 and one is not in A.
55 00:02:31.515 –> 00:02:33.375 So with that we can directly conclude
56 00:02:33.375 –> 00:02:35.895 that we don’t have a sigma algebra given,
57 00:02:36.835 –> 00:02:39.735 but as you remember, the next part in the exercise already
58 00:02:39.735 –> 00:02:43.015 told us that in this case we should just extend it
59 00:02:43.015 –> 00:02:44.055 to a Sigma algebra.
60 00:02:44.755 –> 00:02:47.095 And now the thing is you can easily extend it
61 00:02:47.095 –> 00:02:51.015 to a Sigma algebra just by putting all other subsets to it,
62 00:02:51.145 –> 00:02:53.255 which means you would reach the power set.
63 00:02:53.945 –> 00:02:55.635 Obviously this is not so interesting.
64 00:02:55.775 –> 00:02:57.875 So let’s try to minimize the effort
65 00:02:58.135 –> 00:03:00.565 and just put sets to it that are needed
66 00:03:00.625 –> 00:03:02.605 to satisfy the condition in (c).
67 00:03:03.235 –> 00:03:07.415 So we already know that at least zero one as an element has
68 00:03:07.415 –> 00:03:08.415 to be added to A.
69 00:03:08.995 –> 00:03:12.135 And in the same way we also see that zero two
70 00:03:12.435 –> 00:03:14.535 and one two have to be added.
71 00:03:15.115 –> 00:03:17.255 So these are definitely unions we can form
72 00:03:17.475 –> 00:03:19.375 and they were not in A before.
73 00:03:20.225 –> 00:03:21.885 So we definitely solved this problem.
74 00:03:22.265 –> 00:03:24.285 But sadly we created a new one
75 00:03:24.285 –> 00:03:26.645 because we don’t know if the complements
76 00:03:26.665 –> 00:03:28.525 of the new sets are still in.
77 00:03:29.125 –> 00:03:31.375 Therefore, let’s simply also add the
78 00:03:31.375 –> 00:03:32.615 complements of these sets.
79 00:03:33.155 –> 00:03:34.215 So the complement
80 00:03:34.215 –> 00:03:37.655 of the first set here definitely misses the one element,
81 00:03:38.075 –> 00:03:39.575 but also the zero element.
82 00:03:40.275 –> 00:03:42.895 So it’s the open interval from zero to one
83 00:03:42.985 –> 00:03:44.455 where one is not included
84 00:03:44.835 –> 00:03:47.735 and then the open interval from one to two
85 00:03:47.945 –> 00:03:49.415 where two is included.
86 00:03:50.075 –> 00:03:51.975 And then for the next one we simply have
87 00:03:51.975 –> 00:03:53.695 to exclude zero and two.
88 00:03:54.035 –> 00:03:56.615 So we have the open interval zero to two.
89 00:03:57.275 –> 00:04:01.575 And then finally the last one is just missing one and two.
90 00:04:02.155 –> 00:04:04.615 So also here two has an open interval
91 00:04:04.615 –> 00:04:05.655 at the right hand side.
92 00:04:06.495 –> 00:04:08.955 So really nice, but maybe we are still not done
93 00:04:08.955 –> 00:04:11.715 because at the moment we only consider the union
94 00:04:11.815 –> 00:04:13.035 of just two sets.
95 00:04:13.845 –> 00:04:16.665 Indeed, the union with some other sets here does not make
96 00:04:16.665 –> 00:04:18.545 any problems because these are the complements.
97 00:04:18.885 –> 00:04:20.865 But now we can go to a higher union.
98 00:04:21.645 –> 00:04:23.065 In fact, what immediately should come
99 00:04:23.065 –> 00:04:26.225 to mind is the set zero, one and two.
100 00:04:26.995 –> 00:04:29.405 This one is the union of all the singletons.
101 00:04:29.625 –> 00:04:32.565 And this one was not included before, as you can see.
102 00:04:33.225 –> 00:04:36.805 And neither was the complement, which we also have to add.
103 00:04:36.865 –> 00:04:40.205 Now this one excludes zero one and two.
104 00:04:40.425 –> 00:04:43.525 So we have the union of these two open intervals.
105 00:04:44.225 –> 00:04:45.605 And with that we might have it
106 00:04:45.605 –> 00:04:49.005 because we immediately see if we add all these sets here,
107 00:04:49.315 –> 00:04:51.605 then (a) and (b) are definitely satisfied.
108 00:04:52.305 –> 00:04:55.405 And now the only question is can we form any new union,
109 00:04:55.615 –> 00:04:57.125 which was not there before
110 00:04:57.745 –> 00:05:00.245 But there it’s not hard to see that this is not possible.
111 00:05:00.785 –> 00:05:03.165 All the unions are covered with our sets.
112 00:05:03.955 –> 00:05:07.575 So the result is that we finally have a sigma algebra
113 00:05:07.725 –> 00:05:08.855 over the set X.
114 00:05:09.665 –> 00:05:11.645 So maybe let’s call it A tilde
115 00:05:11.865 –> 00:05:14.365 and let’s define it as we have discussed it.
116 00:05:14.955 –> 00:05:18.925 This means we take A and add all the sets from before.
117 00:05:19.745 –> 00:05:22.245 And with that, the exercise is solved.
118 00:05:23.015 –> 00:05:25.365 Hence this is the solution you can put at the
119 00:05:25.385 –> 00:05:26.565 end of the exercise.
120 00:05:27.425 –> 00:05:29.685 If you have more questions about this exercise,
121 00:05:29.865 –> 00:05:32.365 please use the comments or the community forum.
122 00:05:33.105 –> 00:05:35.365 And with that, I would say let’s meet in the next video
123 00:05:35.415 –> 00:05:39.405 again where we solve more exercises about measure theory.
124 00:05:40.345 –> 00:05:42.085 So have a nice day and bye-bye.
-
Quiz Content (n/a)
-
Date of video: 2025-11-07
-
Last update: 2025-11
